Solutions to Royden & Fitzpatrick’s Real Analysis

Throughout we use only the Field Axioms.

Step 0: \(c \cdot 0 = 0\) for every real number \(c\). Since \(0\) is the additive identity, \(0 + 0 = 0\), and hence by the Distributive Property

\begin{equation*} c \cdot 0 = c(0 + 0) = c \cdot 0 + c \cdot 0 . \end{equation*}

Let \(d\) be an additive inverse of \(c \cdot 0\). Adding \(d\) to both sides and using associativity of addition,

\begin{equation*} 0 = c\cdot 0 + d = (c \cdot 0 + c \cdot 0) + d = c \cdot 0 + (c \cdot 0 + d) = c \cdot 0 + 0 = c \cdot 0 . \end{equation*}

Step 1: \(ab \neq 0\), so that \((ab)^{-1}\) is defined. Suppose to the contrary that \(ab = 0\). Since \(a \neq 0\), the multiplicative inverse \(a^{-1}\) exists, and by associativity of multiplication together with Step 0,

\begin{equation*} b = 1 \cdot b = (a^{-1} a) b = a^{-1}(ab) = a^{-1} \cdot 0 = 0 , \end{equation*}

contradicting \(b \neq 0\). Hence \(ab \neq 0\), and the Multiplicative Inverse axiom provides \((ab)^{-1}\).

Step 2: Multiplicative inverses are unique. Suppose \(x \neq 0\) and \(xy = 1 = xz\). Then, using commutativity and associativity of multiplication,

\begin{equation*} y = 1 \cdot y = (zx) y = z(xy) = z \cdot 1 = z . \end{equation*}

Thus a nonzero real number has exactly one real number whose product with it is \(1\); this is the uniqueness noted in the text just after the Field Axioms.

Step 3: The computation. Using commutativity and associativity of multiplication repeatedly,

\begin{equation*} (ab)(a^{-1} b^{-1}) = (a b) (b^{-1} a^{-1}) = a \big( b (b^{-1} a^{-1}) \big) = a \big( (b b^{-1}) a^{-1} \big) = a (1 \cdot a^{-1}) = a a^{-1} = 1 . \end{equation*}

So \(a^{-1} b^{-1}\) is a real number whose product with \(ab\) equals \(1\). Since \(ab \neq 0\) by Step 1, the uniqueness established in Step 2 forces

\begin{equation*} (ab)^{-1} = a^{-1} b^{-1} , \end{equation*}

as claimed.

Recall the definitions in force: \(\mathcal{P}\) denotes the set of positive numbers, subject to P1 (if \(a, b \in \mathcal{P}\) then \(ab \in \mathcal{P}\) and \(a + b \in \mathcal{P}\)) and P2 (for each real \(a\) exactly one of \(a \in \mathcal{P}\), \(-a \in \mathcal{P}\), \(a = 0\) holds); and \(a > b\) means \(a - b \in \mathcal{P}\). In particular \(a > 0\) means \(a - 0 = a \in \mathcal{P}\), i.e. \(a\) is positive.

We first record two consequences of the Field Axioms. As in Problem 1, \(c \cdot 0 = 0\) for every \(c\). Next, for all \(a\) and \(b\),

\begin{equation*} ab + (-a)b = \big(a + (-a)\big) b = 0 \cdot b = 0 , \end{equation*}

so \((-a)b\) is the additive inverse of \(ab\); since additive inverses are unique, \((-a)b = -(ab)\). Applying this twice,

\begin{equation*} (-a)(-b) = -\big(a(-b)\big) = -\big(-(ab)\big) = ab , \end{equation*}

the last equality because \(ab\) is an additive inverse of \(-(ab)\) and additive inverses are unique.

(i) Let \(a \neq 0\). By P2, either \(a \in \mathcal{P}\) or \(-a \in \mathcal{P}\). In the first case \(a^2 = a \cdot a \in \mathcal{P}\) by P1. In the second case, by the identity above and P1,

\begin{equation*} a^2 = a \cdot a = (-a)(-a) \in \mathcal{P} . \end{equation*}

In either case \(a^2 \in \mathcal{P}\), i.e. \(a^2 > 0\). By the Nontriviality Assumption \(1 \neq 0\), and \(1 = 1 \cdot 1 = 1^2\) by the Multiplicative Identity axiom; hence \(1 > 0\).

(ii) Let \(a \in \mathcal{P}\). Then \(a \neq 0\) by P2, so \(a^{-1}\) exists. Also \(a^{-1} \neq 0\): otherwise \(1 = a a^{-1} = a \cdot 0 = 0\), contradicting Nontriviality. By P2, either \(a^{-1} \in \mathcal{P}\) or \(-a^{-1} \in \mathcal{P}\). Suppose the latter. Then, by P1, \(a \cdot (-a^{-1}) \in \mathcal{P}\). But

\begin{equation*} a \cdot (-a^{-1}) = -\big(a a^{-1}\big) = -1 , \end{equation*}

so \(-1 \in \mathcal{P}\). This contradicts P2, since by part (i) we already have \(1 \in \mathcal{P}\), and exactly one of \(1 \in \mathcal{P}\), \(-1 \in \mathcal{P}\), \(1 = 0\) may hold. Therefore \(a^{-1} \in \mathcal{P}\), i.e. \(a^{-1} > 0\).

(iii) Assume \(a > b\), that is, \(a - b \in \mathcal{P}\). By the Distributive Property and the identity \((-b)c = -(bc)\),

\begin{equation*} (a - b) c = \big(a + (-b)\big) c = ac + (-b)c = ac + \big(-(bc)\big) = ac - bc . \end{equation*}

Suppose \(c > 0\), so \(c \in \mathcal{P}\). Then P1 gives \((a-b)c \in \mathcal{P}\), i.e. \(ac - bc \in \mathcal{P}\), which by definition says \(ac > bc\).

Suppose instead \(c < 0\), i.e. \(0 > c\), which means \(0 - c = -c \in \mathcal{P}\). Then P1 gives \((a - b)(-c) \in \mathcal{P}\). Now

\begin{equation*} (a-b)(-c) = -\big((a-b)c\big) = -(ac - bc) = bc - ac , \end{equation*}

where the last equality holds because \(bc + (-(ac))\) added to \(ac + (-(bc))\) gives \(0\), and additive inverses are unique. Hence \(bc - ac \in \mathcal{P}\), i.e. \(bc > ac\), which is exactly \(ac < bc\).

Throughout, \(\sup E\) and \(\inf E\) are understood in the extended real numbers, as in the text: \(\sup E = \infty\) if \(E\) is not bounded above and \(\inf E = -\infty\) if \(E\) is not bounded below, so that every nonempty set of real numbers has both a supremum and an infimum.

First a remark on finiteness. Since \(E \neq \varnothing\), choose \(x_0 \in E\). Then \(\inf E \leq x_0 \leq \sup E\), because an infimum is a lower bound and a supremum is an upper bound. Hence \(\inf E < \infty\) and \(\sup E > -\infty\). Consequently, if \(\inf E = \sup E\), this common value \(c\) satisfies \(c < \infty\) and \(c > -\infty\), so \(c\) is a real number; in particular \(E\) is then bounded above and below, and the Completeness Axiom (together with Problem 6 for the infimum) guarantees that \(\sup E\) and \(\inf E\) are the least upper bound and the greatest lower bound of \(E\).

Suppose \(\inf E = \sup E\), and call the common (real) value \(c\). Let \(x \in E\) be arbitrary. Since \(\sup E\) is an upper bound for \(E\),

\begin{equation*} x \leq \sup E = c , \end{equation*}

and since \(\inf E\) is a lower bound for \(E\),

\begin{equation*} x \geq \inf E = c . \end{equation*}

Thus \(x \leq c\) and \(c \leq x\). If \(x \neq c\), then by P2 exactly one of \(x - c\) and \(-(x-c) = c - x\) is positive; but \(x \leq c\) gives \(c - x \in \mathcal{P}\) or \(x = c\), and \(c \leq x\) gives \(x - c \in \mathcal{P}\) or \(x = c\), so \(x \neq c\) would force both \(c - x\) and \(x - c\) to be positive, contradicting P2. Hence \(x = c\). As \(x \in E\) was arbitrary and \(E \neq \varnothing\), we conclude \(E = \{c\}\), a single point.

Conversely, suppose \(E = \{x_0\}\) for some real number \(x_0\). Then \(x_0\) is an upper bound for \(E\) (indeed \(x_0 \leq x_0\)), and if \(b\) is any upper bound for \(E\) then \(b \geq x_0\) since \(x_0 \in E\). So \(x_0\) is the least upper bound: \(\sup E = x_0\). Symmetrically, \(x_0\) is a lower bound for \(E\) and every lower bound \(a\) satisfies \(a \leq x_0\), so \(x_0\) is the greatest lower bound: \(\inf E = x_0\). Therefore

\begin{equation*} \inf E = x_0 = \sup E . \end{equation*}

(i) Suppose \(ab = 0\) and \(a \neq 0\). Then \(a^{-1}\) exists, and since \(a^{-1} \cdot 0 = 0\) (Step 0 of Problem 1),

\begin{equation*} b = 1 \cdot b = (a^{-1} a) b = a^{-1}(ab) = a^{-1} \cdot 0 = 0 . \end{equation*}

So either \(a = 0\) or \(b = 0\).

(ii) By the Distributive Property and commutativity of multiplication, and using \((-b)x = -(bx)\) from Problem 2,

\begin{equation*} (a-b)(a+b) = a(a+b) + (-b)(a+b) = \big(a^2 + ab\big) + \big(-(ba) - b^2\big) = a^2 + (ab - ab) - b^2 = a^2 - b^2 . \end{equation*}

Now assume \(a^2 = b^2\). Then \(a^2 - b^2 = 0\), so \((a-b)(a+b) = 0\), and part (i) gives \(a - b = 0\) or \(a + b = 0\). In the first case \(a = b\); in the second, \(a\) is the additive inverse of \(b\), i.e. \(a = -b\).

(iii) Let \(c > 0\) and \(E = \{x \in \mathbb{R} \mid x^2 < c\}\).

\(E\) is nonempty: \(0^2 = 0 < c\), so \(0 \in E\).

\(E\) is bounded above by \(1 + c\): note \(1 > 0\) by Problem 2(i), and \((1+c) - 1 = c\) is positive, so \(1 + c > 1 > 0\). Let \(x \in E\) and suppose \(x > 1 + c\). Then \(x > 1 > 0\), so \(x\) is positive. Multiplying the inequality \(x > 1 + c\) by the positive number \(x\) gives \(x^2 > (1+c)x\), and multiplying the inequality \(x > 1\) by the positive number \(1 + c\) gives \((1+c)x > (1+c) \cdot 1\); both multiplications are legitimate by Problem 2(iii). Combining,

\begin{equation*} x^2 > (1+c)x > (1+c) \cdot 1 = 1 + c > c , \end{equation*}

contradicting \(x^2 < c\). So no \(x \in E\) satisfies \(x > 1+c\), and hence, by P2 applied to \(x - (1+c)\), every \(x \in E\) satisfies \(x \leq 1 + c\).

By the Completeness Axiom, \(x_0 = \sup E\) exists. We record two facts.

First, \(x_0 > 0\). Put \(t = c(1+c)^{-1}\). Since \(c > 0\) and \(1 + c > 0\), Problem 2(ii) and P1 give \(t > 0\). From \(c < 1 + c\), multiplying by the positive number \((1+c)^{-1}\) gives \(t < 1\); from \(1 \leq 1 + c\), multiplying by the positive number \(c(1+c)^{-1}\) gives \(t \leq c\). Since \(0 < t < 1\), multiplying \(t < 1\) by \(t > 0\) gives \(t^2 < t \leq c\), so \(t \in E\) and therefore \(x_0 \geq t > 0\).

Second, a monotonicity lemma: if \(0 < u < v\) then \(u^2 < v^2\). Indeed, multiplying \(u < v\) by \(u > 0\) gives \(u^2 < uv\), and multiplying \(u<v\) by \(v>0\) gives \(uv < v^2\); combine.

We now show \(x_0^2 = c\) by excluding the other two alternatives of P2 applied to \(x_0^2 - c\).

Suppose \(x_0^2 < c\) and set \(\epsilon = c - x_0^2 > 0\). Since \(2x_0 + 1 > 0\), the number

\begin{equation*} h = \min \left\{ \tfrac{1}{2}, \; \epsilon \big( 2(2x_0+1) \big)^{-1} \right\} \end{equation*}

is positive, and satisfies \(h < 1\) and \(h (2x_0 + 1) < \epsilon\). Because \(0 < h < 1\) we have \(h^2 < h\), hence

\begin{equation*} (x_0 + h)^2 = x_0^2 + 2 x_0 h + h^2 < x_0^2 + 2x_0 h + h = x_0^2 + h(2x_0+1) < x_0^2 + \epsilon = c . \end{equation*}

So \(x_0 + h \in E\) while \(x_0 + h > x_0 = \sup E\), contradicting that \(x_0\) is an upper bound for \(E\).

Suppose instead \(x_0^2 > c\) and set \(\epsilon = x_0^2 - c > 0\). Since \(2 x_0 > 0\), the number

\begin{equation*} k = \min \left\{ \tfrac{1}{2} x_0 , \; \epsilon \big( 2 \cdot 2 x_0 \big)^{-1} \right\} \end{equation*}

is positive and satisfies \(k < x_0\) and \(2 x_0 k < \epsilon\). Then \(x_0 - k > 0\) and, since \(k^2 > 0\),

\begin{equation*} (x_0 - k)^2 = x_0^2 - 2x_0 k + k^2 > x_0^2 - 2 x_0 k > x_0^2 - \epsilon = c . \end{equation*}

We claim \(x_0 - k\) is an upper bound for \(E\). Let \(x \in E\). If \(x \leq 0\) then \(x < x_0 - k\) since \(x_0 - k > 0\). If \(x > 0\) and \(x \geq x_0 - k\), then either \(x = x_0 - k\), whence \(x^2 = (x_0-k)^2 > c\), or \(x > x_0 - k > 0\), whence the monotonicity lemma gives \(x^2 > (x_0-k)^2 > c\); both contradict \(x^2 < c\). So \(x < x_0 - k\) in every case, and \(x_0 - k\) is an upper bound for \(E\) strictly smaller than \(x_0\), contradicting that \(x_0\) is the least upper bound.

By P2 the only remaining alternative is \(x_0^2 - c = 0\), that is, \(x_0^2 = c\), and \(x_0 > 0\).

Finally, uniqueness. Suppose \(x > 0\) and \(x^2 = c = x_0^2\). By part (ii), \(x = x_0\) or \(x = -x_0\). Since \(x_0 > 0\), P2 gives \(-x_0 < 0\), so \(x = -x_0\) is incompatible with \(x > 0\). Hence \(x = x_0\). Thus there is exactly one positive real number whose square is \(c\); it is denoted \(\sqrt{c}\).

Write \(2 = 1 + 1\) and \(4 = 2 \cdot 2\). By Problem 2(i), \(1 > 0\), so P1 gives \(2 > 0\) and \(4 > 0\); in particular \(2 \neq 0\) and \(4 \neq 0\) by P2. Since also \(a \neq 0\), Problem 4(i) shows \(2a \neq 0\) and \(4a \neq 0\), so \((2a)^{-1}\) and \((4a)^{-1}\) exist.

Completing the square. For every real \(x\), the Field Axioms give

\begin{equation*} (2ax + b)^2 = 4a^2 x^2 + 4abx + b^2 = 4a\big(a x^2 + b x + c\big) + \big(b^2 - 4ac\big) . \end{equation*}

Hence

\begin{equation*} 4a \big( a x^2 + bx + c \big) = (2ax+b)^2 - \big(b^2 - 4ac\big) . \end{equation*}

Since \(4a \neq 0\), we have \(4a(ax^2+bx+c) = 0\) if and only if \(ax^2+bx+c = 0\): one direction is \(4a \cdot 0 = 0\), and the other is Problem 4(i). Therefore, for every real \(x\),

\begin{equation*} a x^2 + bx + c = 0 \iff (2ax + b)^2 = b^2 - 4ac . \end{equation*}

Call this equivalence \((\ast)\).

(i) Suppose \(d := b^2 - 4ac > 0\). By Problem 4(iii) there is a unique real number \(\sqrt{d} > 0\) with \((\sqrt d)^2 = d\). By \((\ast)\), \(x\) solves the quadratic equation if and only if \((2ax+b)^2 = (\sqrt d)^2\), which by Problem 4(ii) happens if and only if

\begin{equation*} 2ax + b = \sqrt{d} \qquad \text{or} \qquad 2ax + b = -\sqrt{d} . \end{equation*}

Since \(2a \neq 0\), the first alternative is equivalent to \(2ax = -b + \sqrt d\), i.e.

\begin{equation*} x = (2a)^{-1}\big(-b + \sqrt{d}\,\big) = \frac{-b + \sqrt{b^2 - 4ac}}{2a} , \end{equation*}

and the second is equivalent, in the same way, to

\begin{equation*} x = (2a)^{-1}\big(-b - \sqrt{d}\,\big) = \frac{-b - \sqrt{b^2-4ac}}{2a} . \end{equation*}

Every step above is an equivalence, so these two numbers are solutions and there are no others. They are distinct: if they were equal, multiplying by \(2a\) would give \(-b + \sqrt d = -b - \sqrt d\), hence \(2\sqrt d = 0\), and since \(2 \neq 0\) Problem 4(i) would force \(\sqrt d = 0\), contradicting \(\sqrt d > 0\). Thus the equation has exactly two solutions, as stated.

(ii) Suppose \(b^2 - 4ac < 0\) and suppose, for contradiction, that some real \(x\) solves the equation. Put \(y = 2ax + b\). By \((\ast)\), \(y^2 = b^2 - 4ac < 0\). But every square is nonnegative: if \(y \neq 0\) then \(y^2 > 0\) by Problem 2(i), and if \(y = 0\) then \(y^2 = 0\). In either case \(y^2 \geq 0 > b^2 - 4ac\), and P2 forbids \(y^2\) from being both equal to and greater than \(b^2 - 4ac\). This contradiction shows the quadratic equation has no solution when \(b^2 - 4ac < 0\).

We first record the reflection lemma: for real numbers \(u\) and \(v\),

\begin{equation*} u \leq v \iff -v \leq -u . \end{equation*}

Indeed, since \(-(-v) = v\) (additive inverses are unique), we have \((-u) - (-v) = (-u) + v = v - u\). So \(v - u\) is positive or zero exactly when \((-u) - (-v)\) is positive or zero, which is the assertion.

Let \(E\) be a nonempty set of real numbers that is bounded below, and set

\begin{equation*} -E = \{ -x \mid x \in E \} . \end{equation*}

Since \(E \neq \varnothing\), also \(-E \neq \varnothing\). Let \(m\) be a lower bound for \(E\), so \(m \leq x\) for all \(x \in E\). By the reflection lemma, \(-x \leq -m\) for all \(x \in E\), i.e. \(-m\) is an upper bound for \(-E\). Thus \(-E\) is a nonempty set of real numbers that is bounded above, and the Completeness Axiom provides its least upper bound

\begin{equation*} s = \sup \{-x \mid x \in E\} . \end{equation*}

We claim \(-s\) is the greatest lower bound of \(E\).

\(-s\) is a lower bound for \(E\): let \(x \in E\). Then \(-x \in -E\), so \(-x \leq s\), and the reflection lemma gives \(-s \leq -(-x) = x\).

\(-s\) is the greatest lower bound: let \(\ell\) be any lower bound for \(E\). Then \(\ell \leq x\) for all \(x \in E\), so by the reflection lemma \(-x \leq -\ell\) for all \(x \in E\), i.e. \(-\ell\) is an upper bound for \(-E\). Since \(s\) is the least upper bound of \(-E\), we get \(s \leq -\ell\), and the reflection lemma once more gives \(\ell = -(-\ell) \leq -s\).

Hence among the lower bounds of \(E\) there is a largest, namely \(-s\); that is, \(E\) has an infimum and

\begin{equation*} \inf E = -s = -\sup \{-x \mid x \in E\} . \end{equation*}

Recall the definition: \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\). Two immediate consequences will be used repeatedly.

Fact A: \(|x| \geq 0\), and \(|x|\) equals \(x\) or \(-x\). If \(x \geq 0\) this is clear; if \(x < 0\) then \(-x > 0\) by P2 and \(|x| = -x > 0\).

Fact B: \(|-x| = |x|\). If \(x > 0\) then \(-x < 0\), so \(|-x| = -(-x) = x = |x|\); if \(x < 0\) then \(-x > 0\), so \(|-x| = -x = |x|\); and \(|-0| = |0| = 0\).

Fact C: \(-|x| \leq x \leq |x|\). If \(x \geq 0\) then \(|x| = x\), so \(x \leq |x|\), while \(-|x| = -x \leq 0 \leq x\). If \(x < 0\) then \(|x| = -x > 0 > x\), so \(x < |x|\), while \(-|x| = x\).

Fact D: inequalities add. If \(u \leq v\) and \(u^{\prime} \leq v^{\prime}\), then

\begin{equation*} (v + v^{\prime}) - (u + u^{\prime}) = (v-u) + (v^{\prime}-u^{\prime}) , \end{equation*}

and each of \(v - u\), \(v^{\prime} - u^{\prime}\) is positive or zero. If both are positive their sum is positive by P1; if exactly one is zero the sum equals the other, which is positive; if both are zero the sum is zero. In every case \((v+v^{\prime}) - (u+u^{\prime})\) is positive or zero, i.e. \(u + u^{\prime} \leq v + v^{\prime}\).

(i) We use \((-x)y = -(xy)\) and \((-x)(-y) = xy\), established in Problem 2.

If \(a = 0\) or \(b = 0\), then \(ab = 0\) and both sides equal \(0\), since \(|0| = 0\).

If \(a > 0\) and \(b > 0\), then \(ab > 0\) by P1, so \(|ab| = ab = |a|\,|b|\).

If \(a > 0\) and \(b < 0\), then \(-b > 0\), so \(a(-b) = -(ab) > 0\) by P1, hence \(ab < 0\) by P2, and

\begin{equation*} |ab| = -(ab) = a(-b) = |a| \, |b| . \end{equation*}

The case \(a < 0\), \(b > 0\) is identical with the roles of \(a\) and \(b\) exchanged.

If \(a < 0\) and \(b < 0\), then \(-a > 0\) and \(-b > 0\), so \(ab = (-a)(-b) > 0\) by P1, and

\begin{equation*} |ab| = ab = (-a)(-b) = |a| \, |b| . \end{equation*}

All cases are covered by P2, so \(|ab| = |a|\,|b|\) always.

(ii) By Fact C, \(a \leq |a|\) and \(b \leq |b|\), so Fact D gives

\begin{equation*} a + b \leq |a| + |b| . \end{equation*}

Again by Fact C, \(-|a| \leq a\) and \(-|b| \leq b\), so Fact D gives \((-|a|) + (-|b|) \leq a + b\). Now \((-|a|) + (-|b|) = -\big(|a| + |b|\big)\), since adding \(|a| + |b|\) to \((-|a|) + (-|b|)\) yields \(0\) and additive inverses are unique. Hence \(-\big(|a| + |b|\big) \leq a + b\), and applying the reflection lemma of Problem 6,

\begin{equation*} -(a+b) \leq |a| + |b| . \end{equation*}

By Fact A, \(|a+b|\) is either \(a+b\) or \(-(a+b)\), and both of these are at most \(|a| + |b|\). Hence

\begin{equation*} |a+b| \leq |a| + |b| . \end{equation*}

(iii) Let \(\epsilon > 0\) and let \(x\) and \(a\) be real numbers.

Suppose \(|x - a| < \epsilon\). By Fact C, \(x - a \leq |x-a| < \epsilon\), so \(x - a < \epsilon\) and therefore \(x < a + \epsilon\). Also, by Fact C applied to \(-(x-a)\) together with Fact B,

\begin{equation*} -(x-a) \leq |-(x-a)| = |x-a| < \epsilon , \end{equation*}

so \(a - x < \epsilon\), which gives \(a - \epsilon < x\). Hence \(a - \epsilon < x < a + \epsilon\).

Conversely, suppose \(a - \epsilon < x < a + \epsilon\). From \(x < a + \epsilon\) we get \(x - a < \epsilon\), and from \(a - \epsilon < x\) we get \(-(x-a) = a - x < \epsilon\). By Fact A, \(|x-a|\) equals either \(x - a\) or \(-(x-a)\), and each of these is less than \(\epsilon\); hence \(|x-a| < \epsilon\).

Recall that a set \(E\) of real numbers is inductive provided \(1 \in E\) and \(x \in E\) implies \(x + 1 \in E\), and that \(\mathbb{N}\) is defined to be the intersection of all inductive subsets of \(\mathbb{R}\). Consequently, to prove that every natural number belongs to a set \(E\) it suffices to prove that \(E\) is inductive; this is exactly the content of the Principle of Mathematical Induction.

We first record two lemmas.

Lemma 1. Every natural number \(n\) satisfies \(n \geq 1\).

Indeed, the set \(\{x \in \mathbb{R} \mid x \geq 1\}\) is inductive: it contains \(1\), and if \(x \geq 1\) then \(x + 1 \geq 1 + 1 > 1\), since \(1 > 0\). Hence \(\mathbb{N}\) is contained in this set.

Lemma 2. If \(n\) is a natural number and \(n \neq 1\), then \(n - 1\) is a natural number.

Define

\begin{equation*} F = \{x \in \mathbb{R} \mid x = 1 \text{ or } x - 1 \in \mathbb{N}\}. \end{equation*}

Then \(1 \in F\). Suppose \(x \in F\). If \(x = 1\), then \((x + 1) - 1 = 1 \in \mathbb{N}\), so \(x + 1 \in F\). If \(x - 1 \in \mathbb{N}\), then, since \(\mathbb{N}\) is inductive, \(x = (x - 1) + 1 \in \mathbb{N}\), and therefore \((x + 1) - 1 = x \in \mathbb{N}\), so again \(x + 1 \in F\). Thus \(F\) is inductive and hence \(\mathbb{N} \subseteq F\), which is the assertion of the lemma. (This is the first assertion of Problem 9; its proof uses only the definition of \(\mathbb{N}\) as an intersection of inductive sets, so there is no circularity.)

We now prove, by induction on \(n\), the assertion

\begin{equation*} S(n): \quad (n, n + 1) \cap \mathbb{N} = \emptyset . \end{equation*}

The base case \(S(1)\). Define

\begin{equation*} E = \{1\} \cup \{x \in \mathbb{R} \mid x \geq 2\}. \end{equation*}

Then \(1 \in E\). If \(x \in E\) and \(x = 1\), then \(x + 1 = 2 \geq 2\), so \(x + 1 \in E\); if \(x \in E\) and \(x \geq 2\), then \(x + 1 \geq 3 > 2\), so \(x + 1 \in E\). Hence \(E\) is inductive and therefore \(\mathbb{N} \subseteq E\). Since no point of \(E\) lies strictly between \(1\) and \(2\), no natural number lies in \((1, 2)\). Thus \(S(1)\) holds.

The induction step. Suppose \(S(k)\) holds for the natural number \(k\), that is, no natural number lies in \((k, k + 1)\). We claim \(S(k + 1)\) holds. Suppose, to the contrary, that \(n\) is a natural number with

\begin{equation*} k + 1 < n < k + 2 . \end{equation*}

By Lemma 1, \(k \geq 1\), so \(n > k + 1 \geq 2 > 1\) and in particular \(n \neq 1\). By Lemma 2, \(n - 1\) is a natural number. Subtracting \(1\) from each term of the displayed inequality gives

\begin{equation*} k < n - 1 < k + 1 , \end{equation*}

so \(n - 1\) is a natural number lying in \((k, k + 1)\). This contradicts \(S(k)\). Hence no natural number lies in \((k + 1, k + 2)\), that is, \(S(k + 1)\) holds.

By the Principle of Mathematical Induction, \(S(n)\) holds for every natural number \(n\): for each natural number \(n\) the interval \((n, n + 1)\) contains no natural number.

Throughout we use the definition of \(\mathbb{N}\) as the intersection of all inductive subsets of \(\mathbb{R}\); equivalently, we use the Principle of Mathematical Induction, which asserts that a set of real numbers which is inductive contains \(\mathbb{N}\).

First assertion. Define

\begin{equation*} F = \{x \in \mathbb{R} \mid x = 1 \text{ or } x - 1 \in \mathbb{N}\}. \end{equation*}

We claim \(F\) is inductive. Certainly \(1 \in F\). Let \(x \in F\). There are two cases. If \(x = 1\), then \((x + 1) - 1 = 1\), and \(1 \in \mathbb{N}\), so \(x + 1 \in F\). If \(x \neq 1\), then \(x - 1 \in \mathbb{N}\); since \(\mathbb{N}\) is inductive, \(x = (x - 1) + 1\) belongs to \(\mathbb{N}\), and therefore \((x + 1) - 1 = x \in \mathbb{N}\), so \(x + 1 \in F\). In either case \(x + 1 \in F\), so \(F\) is inductive.

Since \(F\) is inductive, \(\mathbb{N} \subseteq F\). Now let \(n\) be a natural number with \(n > 1\). Then \(n \in F\) and \(n \neq 1\), so by the definition of \(F\) we conclude that \(n - 1 \in \mathbb{N}\). This proves the first assertion.

Second assertion. For a natural number \(m\), let \(S(m)\) be the assertion

\begin{equation*} S(m): \quad \text{for every natural number } n \text{ with } n > m, \text{ the number } n - m \text{ is a natural number.} \end{equation*}

The base case \(S(1)\) is precisely the first assertion just proved.

The induction step. Suppose \(S(m)\) holds for the natural number \(m\). Let \(n\) be a natural number with \(n > m + 1\). Since \(1 > 0\) we have \(m + 1 > m\), and hence \(n > m\). By the induction hypothesis \(S(m)\), the number \(n - m\) is a natural number. Moreover, subtracting \(m\) from both sides of \(n > m + 1\) gives

\begin{equation*} n - m > 1 . \end{equation*}

Therefore \(n - m\) is a natural number strictly greater than \(1\), so by the first assertion \((n - m) - 1\) is a natural number. But

\begin{equation*} (n - m) - 1 = n - (m + 1), \end{equation*}

so \(n - (m + 1)\) is a natural number. As \(n\) was an arbitrary natural number exceeding \(m + 1\), the assertion \(S(m + 1)\) holds.

By the Principle of Mathematical Induction, \(S(m)\) holds for every natural number \(m\). That is, if \(m\) and \(n\) are natural numbers with \(n > m\), then \(n - m\) is a natural number.

Recall that the set \(\mathbb{Z}\) of integers consists of the natural numbers, their negatives, and the number \(0\); thus

\begin{equation*} \mathbb{Z} = \mathbb{N} \cup \{0\} \cup (-\mathbb{N}), \qquad -\mathbb{N} = \{-p \mid p \in \mathbb{N}\}. \end{equation*}

As observed in the solution of Problem 8, every natural number is greater than or equal to \(1\), since \(\{x \in \mathbb{R} \mid x \geq 1\}\) is inductive. Hence every member of \(-\mathbb{N}\) is less than or equal to \(-1\), and in particular is negative. This yields the positivity observation: if \(k\) is an integer and \(k > 0\), then \(k\) belongs to neither \(\{0\}\) nor \(-\mathbb{N}\), so \(k\) is a natural number and therefore \(k \geq 1\).

We also need the following closure property.

Lemma A. \(\mathbb{N}\) is closed under addition: if \(p\) and \(m\) are natural numbers, then \(p + m\) is a natural number.

Fix a natural number \(m\) and set \(G = \{x \in \mathbb{R} \mid x + m \in \mathbb{N}\}\). Since \(\mathbb{N}\) is inductive and \(m \in \mathbb{N}\), we have \(1 + m = m + 1 \in \mathbb{N}\), so \(1 \in G\). If \(x \in G\), then \(x + m \in \mathbb{N}\), so \((x + 1) + m = (x + m) + 1 \in \mathbb{N}\) because \(\mathbb{N}\) is inductive; hence \(x + 1 \in G\). Thus \(G\) is inductive and therefore \(\mathbb{N} \subseteq G\), which proves the lemma.

Lemma B (the difference lemma). If \(m\) and \(n\) are integers with \(n > m\), then \(n - m\) is a natural number.

We consider cases according to the trichotomy for \(m\).

Case \(m \in \mathbb{N}\). If \(n \in \mathbb{N}\), then \(n - m \in \mathbb{N}\) by Problem 9. If \(n = 0\) or \(n \in -\mathbb{N}\), then \(n \leq 0 < 1 \leq m\), contradicting \(n > m\); so these subcases do not occur.

Case \(m = 0\). Then \(n > 0\), so by the positivity observation \(n \in \mathbb{N}\), and \(n - m = n \in \mathbb{N}\).

Case \(m = -q\) with \(q \in \mathbb{N}\). If \(n \in \mathbb{N}\), then \(n - m = n + q \in \mathbb{N}\) by Lemma A. If \(n = 0\), then \(n - m = q \in \mathbb{N}\). If \(n = -p\) with \(p \in \mathbb{N}\), then \(-p > -q\) gives \(q > p\), and \(n - m = q - p \in \mathbb{N}\) by Problem 9.

The three cases are exhaustive, since \(\mathbb{Z} = \mathbb{N} \cup \{0\} \cup (-\mathbb{N})\). In every case \(n - m\) is a natural number, and Lemma B is proved. (It is used again in Problems 11 and 12.)

Now let \(n\) be an integer. Since \(n \leq n\) and \(n < n + 1\), the integer \(n\) itself lies in \([n, n + 1)\), so the interval contains at least one integer.

For uniqueness, suppose \(m\) is an integer with \(m \in [n, n + 1)\) and \(m \neq n\). Since \(m \geq n\) and \(m \neq n\), we have \(m > n\). By the lemma, \(m - n\) is a natural number, hence \(m - n \geq 1\), that is, \(m \geq n + 1\). This contradicts \(m < n + 1\). Therefore \(n\) is the only integer in \([n, n + 1)\).

Let \(E\) be a nonempty set of integers that is bounded above. Since \(E\) is a nonempty set of real numbers that is bounded above, the Completeness Axiom provides a supremum; define

\begin{equation*} c = \sup E . \end{equation*}

Since \(c - 1 < c\) and \(c\) is the least upper bound for \(E\), the number \(c - 1\) is not an upper bound for \(E\). Hence there is a member \(m\) of \(E\) with

\begin{equation*} m > c - 1 . \end{equation*}

We claim \(m\) is the largest member of \(E\). Suppose otherwise. Then there is a member \(n\) of \(E\) with \(n > m\). Both \(m\) and \(n\) are integers, so by the difference lemma (Lemma B in the solution of Problem 10), \(n - m\) is a natural number, and therefore \(n - m \geq 1\), since every natural number is at least \(1\). Thus

\begin{equation*} n \geq m + 1 > (c - 1) + 1 = c , \end{equation*}

so \(n > c\). This contradicts the fact that \(c\) is an upper bound for \(E\) and \(n \in E\).

Therefore no member of \(E\) exceeds \(m\); that is, \(m\) is a member of \(E\) with \(n \leq m\) for all \(n \in E\). Hence \(E\) has a largest member, namely \(m\). (In fact \(m = c = \sup E\), so the supremum of such a set is attained.)

Recall the definition: a set \(E\) of real numbers is dense in \(\mathbb{R}\) provided that between any two real numbers there lies a member of \(E\). So we must show that if \(a\) and \(b\) are real numbers with \(a < b\), then there is an irrational number \(x\) with \(a < x < b\).

We first record that \(\mathbb{Q}\) is closed under subtraction; for this we need that \(\mathbb{Z}\) is closed under subtraction and under multiplication.

Recall from the solution of Problem 10 that \(\mathbb{N}\) is closed under addition (Lemma A there). It is also closed under multiplication: for a fixed natural number \(m\) the set \(\{x \in \mathbb{R} \mid xm \in \mathbb{N}\}\) contains \(1\), since \(1 \cdot m = m \in \mathbb{N}\), and if \(xm \in \mathbb{N}\) then \((x + 1)m = xm + m \in \mathbb{N}\) by additive closure; hence this set is inductive and contains \(\mathbb{N}\).

Now let \(m\) and \(n\) be integers, so that each lies in \(\mathbb{N} \cup \{0\} \cup (-\mathbb{N})\). For the product: if either factor is \(0\), then \(mn = 0 \in \mathbb{Z}\); otherwise write \(m = \epsilon p\) and \(n = \delta q\) with \(p, q \in \mathbb{N}\) and \(\epsilon, \delta \in \{1, -1\}\), so that \(mn = (\epsilon \delta)(pq)\), and \(pq \in \mathbb{N}\) by multiplicative closure, whence \(mn \in \mathbb{N} \cup (-\mathbb{N}) \subseteq \mathbb{Z}\). For the difference: if \(n > m\), then \(n - m \in \mathbb{N} \subseteq \mathbb{Z}\) by the difference lemma (Lemma B in the solution of Problem 10); if \(n = m\), then \(n - m = 0 \in \mathbb{Z}\); and if \(n < m\), then \(m - n \in \mathbb{N}\) by that same lemma, so \(n - m = -(m - n) \in -\mathbb{N} \subseteq \mathbb{Z}\). Thus \(\mathbb{Z}\) is closed under multiplication and subtraction.

Consequently \(\mathbb{Q}\) is closed under subtraction. Indeed, let \(s\) and \(t\) be rational, say \(s = a_1 / a_2\) and \(t = b_1 / b_2\) with \(a_1, a_2, b_1, b_2\) integers and \(a_2 \neq 0\), \(b_2 \neq 0\). Then

\begin{equation*} s - t = \frac{a_1 b_2 - b_1 a_2}{a_2 b_2}. \end{equation*}

By the closure properties just established, the numerator and the denominator are integers, and \(a_2 b_2 \neq 0\) by Problem 4(i). Hence \(s - t\) is a quotient of integers with nonzero denominator, that is, \(s - t\) is rational.

By Problem 4(iii) there is a unique positive real number whose square is \(2\); it is denoted by \(\sqrt{2}\). As shown in the text of Section 1.2, using the prime factorization theorem, \(\sqrt{2}\) is irrational.

Now let \(a\) and \(b\) be real numbers with \(a < b\). Then

\begin{equation*} a - \sqrt{2} < b - \sqrt{2}, \end{equation*}

so by Theorem 2, which asserts that the rational numbers are dense in \(\mathbb{R}\), there is a rational number \(r\) with

\begin{equation*} a - \sqrt{2} < r < b - \sqrt{2} . \end{equation*}

Set \(x = r + \sqrt{2}\). Adding \(\sqrt{2}\) to each term of the last display gives

\begin{equation*} a < x < b . \end{equation*}

Finally, \(x\) is irrational. For if \(x\) were rational, then, since \(r\) is rational and the rationals are closed under subtraction, the number

\begin{equation*} \sqrt{2} = x - r \end{equation*}

would be rational, contradicting the irrationality of \(\sqrt{2}\).

Thus between any two real numbers there lies an irrational number, that is, the irrational numbers are dense in \(\mathbb{R}\).

Let \(x\) be a real number.

The rational case. Define

\begin{equation*} E = \{r \in \mathbb{Q} \mid r < x\}, \end{equation*}

a set of rational numbers. We check that \(E\) is nonempty and bounded above, and that \(\sup E = x\).

\(E\) is nonempty: since \(x - 1 < x\), Theorem 2 (the rational numbers are dense in \(\mathbb{R}\)) provides a rational number \(r\) with \(x - 1 < r < x\); this \(r\) belongs to \(E\).

\(E\) is bounded above by \(x\), directly from its definition. Hence by the Completeness Axiom \(c = \sup E\) exists, and since \(x\) is an upper bound for \(E\) while \(c\) is the least upper bound,

\begin{equation*} c \leq x . \end{equation*}

Suppose \(c < x\). By Theorem 2 there is a rational number \(r\) with \(c < r < x\). Since \(r\) is rational and \(r < x\), we have \(r \in E\); but \(r > c = \sup E\), contradicting the fact that \(c\) is an upper bound for \(E\). Therefore \(c = x\), that is,

\begin{equation*} x = \sup \{r \in \mathbb{Q} \mid r < x\}, \end{equation*}

and \(x\) is the supremum of a set of rational numbers.

The irrational case. The argument is identical, with Problem 12 in place of Theorem 2. Define

\begin{equation*} E^{\prime} = \{t \in \mathbb{R} \mid t \text{ is irrational and } t < x\}, \end{equation*}

a set of irrational numbers.

\(E^{\prime}\) is nonempty: by Problem 12 the irrational numbers are dense in \(\mathbb{R}\), so, as \(x - 1 < x\), there is an irrational \(t\) with \(x - 1 < t < x\), and this \(t\) belongs to \(E^{\prime}\).

\(E^{\prime}\) is bounded above by \(x\), so by the Completeness Axiom \(c^{\prime} = \sup E^{\prime}\) exists and \(c^{\prime} \leq x\).

If \(c^{\prime} < x\), then by Problem 12 there is an irrational number \(t\) with \(c^{\prime} < t < x\); then \(t \in E^{\prime}\) while \(t > c^{\prime} = \sup E^{\prime}\), a contradiction. Hence \(c^{\prime} = x\), that is,

\begin{equation*} x = \sup \{t \mid t \text{ irrational}, \ t < x\}, \end{equation*}

and \(x\) is also the supremum of a set of irrational numbers.

Fix a real number \(r > 0\). For a real number \(x\) the natural powers are defined recursively by \(x^1 = x\) and \(x^{k+1} = x^k \cdot x\). For a natural number \(n\) let \(S(n)\) be the assertion

\begin{equation*} S(n): \quad (1 + r)^n \geq 1 + n \cdot r . \end{equation*}

We prove \(S(n)\) for every natural number \(n\) by the Principle of Mathematical Induction.

The base case \(S(1)\). By definition \((1 + r)^1 = 1 + r = 1 + 1 \cdot r\), so \(S(1)\) holds, with equality.

The induction step. Suppose \(S(k)\) holds for the natural number \(k\), that is,

\begin{equation*} (1 + r)^k \geq 1 + k \cdot r . \end{equation*}

Since \(r > 0\) and \(1 > 0\) we have \(1 + r > 0\). Multiplication by the positive number \(1 + r\) preserves the inequality: if \((1 + r)^k > 1 + k r\), then \((1 + r)^k (1 + r) > (1 + k r)(1 + r)\) by Problem 2(iii), while if \((1 + r)^k = 1 + k r\) the two products are equal. In either case

\begin{equation*} (1 + r)^{k+1} = (1 + r)^k (1 + r) \geq (1 + k r)(1 + r). \end{equation*}

Expanding the right-hand side by the distributive law,

\begin{equation*} \begin{aligned} (1 + k r)(1 + r) &= 1 + r + k r + k r^2 \\ &= 1 + (k + 1) r + k r^2 . \end{aligned} \end{equation*}

Now \(k\) is a natural number, so \(k \geq 1 > 0\), and \(r \neq 0\) gives \(r^2 > 0\) (Problem 2(i)); hence \(k r^2 > 0\) and therefore

\begin{equation*} 1 + (k + 1) r + k r^2 \geq 1 + (k + 1) r . \end{equation*}

Combining the two displays,

\begin{equation*} (1 + r)^{k+1} \geq 1 + (k + 1) r , \end{equation*}

which is the assertion \(S(k + 1)\).

By the Principle of Mathematical Induction, \(S(n)\) holds for every natural number \(n\); that is, \((1 + r)^n \geq 1 + n \cdot r\) for every natural number \(n\) whenever \(r > 0\).

We remark that the same argument works for every real \(r\) with \(r > -1\), since the only properties of \(r\) used were \(1 + r > 0\) and \(k r^2 \geq 0\); this more general form is Bernoulli’s inequality.

Throughout we use the Principle of Mathematical Induction in the form stated in Section 1.2: if \(S\) is a set of natural numbers with \(1 \in S\), and with the property that \(n \in S\) implies \(n + 1 \in S\), then \(S\) is all of \(\mathbb{N}\). In each part let \(S\) be the set of natural numbers \(n\) for which the asserted identity holds; we verify that \(1 \in S\) and that \(S\) is closed under successor.

(i) For \(n = 1\) the left-hand side is \(1^2 = 1\) and the right-hand side is \(\frac{1 \cdot 2 \cdot 3}{6} = 1\), so \(1 \in S\).

Suppose \(n \in S\), that is, \(\sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6}\). Then

\begin{equation*} \begin{aligned} \sum_{j=1}^{n+1} j^2 &= \left(\sum_{j=1}^{n} j^2\right) + (n+1)^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ &= \frac{(n+1)\big[\,n(2n+1) + 6(n+1)\,\big]}{6} = \frac{(n+1)\big(2n^2 + 7n + 6\big)}{6} \\ &= \frac{(n+1)(n+2)(2n+3)}{6} = \frac{(n+1)\big((n+1)+1\big)\big(2(n+1)+1\big)}{6}, \end{aligned} \end{equation*}

where the factorization \(2n^2 + 7n + 6 = (n+2)(2n+3)\) is checked by expanding the right-hand product. Hence \(n + 1 \in S\), and by induction \(S = \mathbb{N}\).

(ii) We first record the companion identity

\begin{equation*} 1 + 2 + \ldots + n = \frac{n(n+1)}{2} \qquad \text{for all } n \in \mathbb{N}, \end{equation*}

which is itself proved by induction: it holds for \(n = 1\) since \(1 = \frac{1 \cdot 2}{2}\), and if it holds for \(n\) then

\begin{equation*} \big(1 + \ldots + n\big) + (n+1) = \frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2}, \end{equation*}

which is the same formula with \(n\) replaced by \(n+1\).

Now let \(S\) be the set of \(n\) for which \(1^3 + \ldots + n^3 = (1 + \ldots + n)^2\). For \(n = 1\) both sides equal \(1\), so \(1 \in S\). Suppose \(n \in S\). Using the induction hypothesis and then the companion identity,

\begin{equation*} \begin{aligned} 1^3 + \ldots + n^3 + (n+1)^3 &= (1 + \ldots + n)^2 + (n+1)^3 = \left(\frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ &= \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2\big[\,n^2 + 4(n+1)\,\big]}{4} \\ &= \frac{(n+1)^2(n+2)^2}{4} = \left(\frac{(n+1)\big((n+1)+1\big)}{2}\right)^2 \\ &= \big(1 + 2 + \ldots + (n+1)\big)^2 . \end{aligned} \end{equation*}

Thus \(n + 1 \in S\), and by induction the identity holds for every natural number \(n\).

(iii) As printed the exponent is a misprint: with \(1 - r^{n-1}\) in the numerator the identity already fails at \(n = 1\), where the left-hand side is \(1 + r\) while the right-hand side is \(\frac{1 - r^0}{1-r} = 0\). The intended (and correct) statement is

\begin{equation*} 1 + r + \ldots + r^n = \frac{1 - r^{n+1}}{1 - r} \qquad \text{if } r \neq 1, \end{equation*}

and this is what we prove. Equivalently, multiplying by the nonzero number \(1 - r\), the assertion is that \((1-r)(1 + r + \ldots + r^n) = 1 - r^{n+1}\).

Fix a real number \(r \neq 1\) and let \(S\) be the set of natural numbers \(n\) for which the displayed identity holds. For \(n = 1\),

\begin{equation*} \frac{1 - r^{2}}{1 - r} = \frac{(1-r)(1+r)}{1-r} = 1 + r, \end{equation*}

so \(1 \in S\). Suppose \(n \in S\). Then

\begin{equation*} \begin{aligned} 1 + r + \ldots + r^{n} + r^{n+1} &= \frac{1 - r^{n+1}}{1-r} + r^{n+1} = \frac{1 - r^{n+1} + r^{n+1}(1 - r)}{1-r} \\ &= \frac{1 - r^{n+1} + r^{n+1} - r^{n+2}}{1-r} = \frac{1 - r^{n+2}}{1-r}, \end{aligned} \end{equation*}

which is the identity with \(n\) replaced by \(n+1\). Hence \(n+1 \in S\), and by induction \(S = \mathbb{N}\). (The identity also holds for \(n = 0\), where both sides equal \(1\).)

By the definition in Section 1.3, a set is countably infinite provided it is equipotent to \(\mathbb{N}\), and a set is countable provided it is either finite or countably infinite. So it suffices to exhibit a one-to-one mapping of \(\mathbb{N}\) onto \(\mathbb{Z}\).

We first record that each natural number \(n\) is either odd or even, and not both: that is, \(n\) can be written as \(n = 2k - 1\) for a natural number \(k\), or as \(n = 2k\) for a natural number \(k\), but not both. Existence follows by induction: \(1 = 2 \cdot 1 - 1\) is odd; and if \(n = 2k - 1\) then \(n + 1 = 2k\) is even, while if \(n = 2k\) then \(n + 1 = 2(k+1) - 1\) is odd. The two alternatives are mutually exclusive: if \(2k - 1 = 2k^{\prime}\) then \(2(k - k^{\prime}) = 1\), so the integer \(j = k - k^{\prime}\) satisfies \(j = 1/2\); but \(0 < 1/2 < 1\), and by Problem 10 the only integer in the interval \([0,1)\) is \(0\), so no integer equals \(1/2\). Finally \(k\) is uniquely determined by \(n\) in either case, since \(2k = 2k^{\prime}\) forces \(k = k^{\prime}\).

Define \(f \colon \mathbb{N} \to \mathbb{Z}\) by

\begin{equation*} f(n) = \begin{cases} \dfrac{n-1}{2}, & \text{if } n \text{ is odd},\\[2mm] -\dfrac{n}{2}, & \text{if } n \text{ is even}. \end{cases} \end{equation*}

This is well defined: if \(n\) is odd then \(n - 1\) is even so \((n-1)/2\) is a nonnegative integer, and if \(n\) is even then \(n/2\) is a natural number so \(-n/2\) is a negative integer. Thus \(f\) maps the odd natural numbers into \(\{0, 1, 2, \ldots\}\) and the even natural numbers into \(\{-1, -2, \ldots\}\). Explicitly, \(f\) enumerates \(\mathbb{Z}\) as

\begin{equation*} 0,\; -1,\; 1,\; -2,\; 2,\; -3,\; 3,\; \ldots \end{equation*}

The mapping \(f\) is one-to-one. Indeed, suppose \(f(m) = f(n)\). If \(m\) and \(n\) are both odd, then \((m-1)/2 = (n-1)/2\), and multiplying by \(2\) and adding \(1\) gives \(m = n\). If \(m\) and \(n\) are both even, then \(-m/2 = -n/2\), so \(m = n\). The remaining case cannot occur: if, say, \(m\) is odd and \(n\) is even, then \(f(m) = (m-1)/2 \geq 0\) while \(f(n) = -n/2 \leq -1\), so \(f(m) \neq f(n)\).

The mapping \(f\) is onto \(\mathbb{Z}\). Let \(p\) be an integer. If \(p \geq 0\), set \(n = 2p + 1\). Then \(n\) is an odd natural number and \(f(n) = \frac{(2p+1)-1}{2} = p\). If \(p < 0\), set \(n = -2p\). Then \(-p\) is a natural number, so \(n = 2(-p)\) is an even natural number, and \(f(n) = -\frac{-2p}{2} = p\).

Therefore \(f\) is a one-to-one correspondence between \(\mathbb{N}\) and \(\mathbb{Z}\), so \(\mathbb{Z}\) is equipotent to \(\mathbb{N}\). Hence \(\mathbb{Z}\) is countably infinite and, in particular, countable.

Here “a one-to-one mapping of \(A\) to \(\mathbb{N}\)” means a one-to-one mapping of \(A\) into \(\mathbb{N}\); it is not required to be onto.

First suppose \(A\) is countable; we produce a one-to-one mapping of \(A\) into \(\mathbb{N}\). By definition \(A\) is either finite or countably infinite.

If \(A\) is countably infinite, then \(A\) is equipotent to \(\mathbb{N}\), so there is a one-to-one mapping \(f\) of \(A\) onto \(\mathbb{N}\); in particular \(f\) is a one-to-one mapping of \(A\) into \(\mathbb{N}\).

If \(A\) is finite, then either \(A = \emptyset\), in which case the empty mapping is (vacuously) a one-to-one mapping of \(A\) into \(\mathbb{N}\), or else there is a natural number \(n\) and a one-to-one correspondence \(g\) between \(A\) and \(\{1, \ldots, n\}\). Let \(\iota \colon \{1, \ldots, n\} \to \mathbb{N}\) be the inclusion mapping, which is one-to-one. Then \(\iota \circ g\) is a mapping of \(A\) into \(\mathbb{N}\), and it is one-to-one: if \((\iota \circ g)(x) = (\iota \circ g)(x^{\prime})\) then \(g(x) = g(x^{\prime})\) since \(\iota\) is one-to-one, and hence \(x = x^{\prime}\) since \(g\) is one-to-one.

Conversely, suppose \(f\) is a one-to-one mapping of \(A\) into \(\mathbb{N}\). Let

\begin{equation*} B = f(A) = \{\, f(x) \mid x \in A \,\} \subseteq \mathbb{N}. \end{equation*}

Regard \(f\) as a mapping of \(A\) to \(B\). So regarded it is onto, by the very definition of \(B\), and it remains one-to-one. Hence \(f\) is a one-to-one correspondence between \(A\) and \(B\), so \(A\) is equipotent to \(B\).

Now \(B\) is a set of natural numbers, and by Theorem 3 every set of natural numbers is countable; thus \(B\) is countable. As observed immediately after the definition of countability in Section 1.3, if a set is equipotent to a countable set then it is countable. Since \(A\) is equipotent to the countable set \(B\), we conclude that \(A\) is countable.

(For the reader who prefers to avoid the appeal to that observation: if \(B\) is finite, then \(B\) is empty, in which case \(A\) is empty and hence finite, or \(B\) is equipotent to some \(\{1, \ldots, n\}\), and composing the one-to-one correspondence \(A \to B\) with the one-to-one correspondence \(B \to \{1, \ldots, n\}\) gives a one-to-one correspondence between \(A\) and \(\{1, \ldots, n\}\), by Problem 20; so \(A\) is finite. If \(B\) is countably infinite, the same composition argument, using a one-to-one correspondence \(B \to \mathbb{N}\), shows \(A\) is equipotent to \(\mathbb{N}\), so \(A\) is countably infinite. In either case \(A\) is countable.)

Write \(\mathbb{N}^n\) for the \(n\)-fold Cartesian product \(\mathbb{N} \times \cdots \times \mathbb{N}\), whose members are the ordered \(n\)-tuples \((k_1, \ldots, k_n)\) of natural numbers. Two such \(n\)-tuples are equal exactly when their corresponding entries agree. We must show that \(\mathbb{N}^n\) is countably infinite for every natural number \(n\); that is, it is countable and it is not finite.

Step 1: \(\mathbb{N}^n\) is countable, by induction on \(n\).

Let \(S\) be the set of natural numbers \(n\) for which \(\mathbb{N}^n\) is countable.

For \(n = 1\) we have \(\mathbb{N}^1 = \mathbb{N}\), which is countable (the identity mapping is a one-to-one correspondence between \(\mathbb{N}\) and \(\mathbb{N}\)). So \(1 \in S\).

Suppose \(n \in S\), so \(\mathbb{N}^n\) is countable. By Problem 17 there is a one-to-one mapping

\begin{equation*} h \colon \mathbb{N}^n \to \mathbb{N}. \end{equation*}

Let \(g \colon \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) be the mapping \(g(m, n) = (m+n)^2 + n\) used in the proof of Corollary 4; it was shown there to be one-to-one. Define

\begin{equation*} G \colon \mathbb{N}^{n+1} \to \mathbb{N}, \qquad G(k_1, \ldots, k_n, k_{n+1}) = g\big(h(k_1, \ldots, k_n),\, k_{n+1}\big). \end{equation*}

This is well defined, since \(h(k_1, \ldots, k_n)\) and \(k_{n+1}\) are natural numbers.

We claim \(G\) is one-to-one. Suppose

\begin{equation*} G(k_1, \ldots, k_{n+1}) = G(k^{\prime}_1, \ldots, k^{\prime}_{n+1}). \end{equation*}

Then \(g\big(h(k_1, \ldots, k_n), k_{n+1}\big) = g\big(h(k^{\prime}_1, \ldots, k^{\prime}_n), k^{\prime}_{n+1}\big)\). Since \(g\) is one-to-one, the two arguments agree as ordered pairs, that is,

\begin{equation*} h(k_1, \ldots, k_n) = h(k^{\prime}_1, \ldots, k^{\prime}_n) \quad \text{and} \quad k_{n+1} = k^{\prime}_{n+1}. \end{equation*}

Since \(h\) is one-to-one, the first equality forces \((k_1, \ldots, k_n) = (k^{\prime}_1, \ldots, k^{\prime}_n)\), so \(k_j = k^{\prime}_j\) for \(1 \leq j \leq n\); together with \(k_{n+1} = k^{\prime}_{n+1}\) this gives \((k_1, \ldots, k_{n+1}) = (k^{\prime}_1, \ldots, k^{\prime}_{n+1})\). Hence \(G\) is one-to-one.

So \(G\) is a one-to-one mapping of \(\mathbb{N}^{n+1}\) into \(\mathbb{N}\), and therefore, by Problem 17 again, \(\mathbb{N}^{n+1}\) is countable. Thus \(n + 1 \in S\), and by the Principle of Mathematical Induction \(S = \mathbb{N}\): the set \(\mathbb{N}^n\) is countable for every natural number \(n\).

Step 2: \(\mathbb{N}^n\) is not finite.

We first record the following consequence of the pigeonhole principle.

Lemma. If there is a one-to-one mapping \(\varphi\) of \(\mathbb{N}\) into a set \(A\), then \(A\) is not finite.

Proof of the Lemma. Suppose, to get a contradiction, that \(A\) is finite. Since \(\varphi(\mathbb{N})\) is a nonempty subset of \(A\), \(A\) is nonempty, so there is a natural number \(p\) and a one-to-one correspondence \(\psi\) between \(A\) and \(\{1, \ldots, p\}\). The composition \(\psi \circ \varphi\) is a one-to-one mapping of \(\mathbb{N}\) into \(\{1, \ldots, p\}\) (a composition of one-to-one mappings is one-to-one, by Problem 20). Restrict it to the initial segment \(\{1, \ldots, p+1\}\): we obtain a one-to-one mapping of \(\{1, \ldots, p+1\}\) into \(\{1, \ldots, p\}\), hence a one-to-one correspondence between \(\{1, \ldots, p+1\}\) and its image \(C\), a subset of \(\{1, \ldots, p\}\). By Theorem 3, and specifically by the argument given there for a subset of a finite set, \(C\) is equipotent to \(\{1, \ldots, q\}\) for some natural number \(q \leq p\) (the set \(C\) is nonempty, since \(p + 1 \geq 1\)). Since equipotence is transitive, \(\{1, \ldots, p+1\}\) is equipotent to \(\{1, \ldots, q\}\). Since \(q \leq p < p+1\), Problem 9 tells us that \(m = (p+1) - q\) is a natural number. Writing \(p + 1 = q + m\), this says that \(\{1, \ldots, q + m\}\) is equipotent to \(\{1, \ldots, q\}\), contradicting the pigeonhole principle. Hence \(A\) is not finite, and the Lemma is proved.

Now fix a natural number \(n\) and define \(\varphi \colon \mathbb{N} \to \mathbb{N}^n\) by

\begin{equation*} \varphi(k) = (k, 1, 1, \ldots, 1), \end{equation*}

that is, first entry \(k\) and all remaining entries equal to \(1\). If \(\varphi(k) = \varphi(k^{\prime})\), then comparing first entries gives \(k = k^{\prime}\), so \(\varphi\) is one-to-one. By the Lemma, \(\mathbb{N}^n\) is not finite.

Combining the two steps: \(\mathbb{N}^n\) is countable and not finite, hence, by the definition of countability, \(\mathbb{N}^n\) is countably infinite. This is part (i) of Corollary 4.

Let \(\Lambda\) be a finite index set and for each \(\lambda \in \Lambda\) let \(E_\lambda\) be a countable set. Put

\begin{equation*} E = \bigcup_{\lambda \in \Lambda} E_\lambda . \end{equation*}

We must show \(E\) is countable.

If \(\Lambda\) is empty, then \(E\) is empty and hence finite, so it is countable. If \(E\) is empty, it is likewise countable. So assume \(\Lambda \neq \emptyset\) and \(E \neq \emptyset\). Since \(\Lambda\) is finite and nonempty, there is a natural number \(M\) and a one-to-one correspondence between \(\{1, \ldots, M\}\) and \(\Lambda\); relabelling, we may write

\begin{equation*} \Lambda = \{\lambda_1, \ldots, \lambda_M\}, \qquad E = \bigcup_{n=1}^{M} E_{\lambda_n}. \end{equation*}

Now we argue exactly as in the text’s proof of Corollary 6, with the initial segment \(\{1, \ldots, M\}\) in place of \(\mathbb{N}\) as the index range.

Fix \(n\) with \(1 \leq n \leq M\). If \(E_{\lambda_n}\) is finite and nonempty, choose a natural number \(N(n)\) and a one-to-one mapping \(f_n\) of \(\{1, \ldots, N(n)\}\) onto \(E_{\lambda_n}\); such a choice is possible by the definition of a finite nonempty set. If \(E_{\lambda_n}\) is countably infinite, choose a one-to-one mapping \(f_n\) of \(\mathbb{N}\) onto \(E_{\lambda_n}\); such a choice is possible by the definition of a countably infinite set. (These are the only two possibilities, since each \(E_{\lambda_n}\) is countable.) If \(E_{\lambda_n}\) is empty, no \(f_n\) is chosen. Define

\begin{equation*} E^{\prime} = \big\{\, (n,k) \in \mathbb{N} \times \mathbb{N} \;\big|\; 1 \leq n \leq M,\ E_{\lambda_n} \text{ is nonempty, and } 1 \leq k \leq N(n) \text{ if } E_{\lambda_n} \text{ is also finite} \,\big\} \end{equation*}

and define

\begin{equation*} f \colon E^{\prime} \to E, \qquad f(n,k) = f_n(k). \end{equation*}

The definition makes sense: for \((n,k) \in E^{\prime}\) the set \(E_{\lambda_n}\) is nonempty and \(k\) lies in the domain of \(f_n\), and \(f_n(k) \in E_{\lambda_n} \subseteq E\).

The mapping \(f\) is onto \(E\). Indeed, let \(x \in E\). Then \(x \in E_{\lambda_n}\) for some \(n\) with \(1 \leq n \leq M\), and in particular \(E_{\lambda_n}\) is nonempty. Since \(f_n\) maps its domain onto \(E_{\lambda_n}\), there is a \(k\) in the domain of \(f_n\) with \(f_n(k) = x\); that \(k\) is a natural number with \(1 \leq k \leq N(n)\) in the case \(E_{\lambda_n}\) is finite, and an arbitrary natural number in the case \(E_{\lambda_n}\) is countably infinite. In either case \((n,k) \in E^{\prime}\) and \(f(n,k) = x\).

Finally, \(E^{\prime}\) is a subset of \(\mathbb{N} \times \mathbb{N}\), which is countable by Corollary 4 (part (i) with \(n = 2\)), so \(E^{\prime}\) is countable by Theorem 3. Moreover \(E^{\prime}\) is nonempty, since \(E\) is nonempty and \(f\) maps \(E^{\prime}\) onto \(E\). Therefore \(E\) is a nonempty set that is the image of a function whose domain is a nonempty countable set, and Theorem 5 tells us that \(E\) is countable.

Remark. Equivalently one may argue by induction on \(M\): the case \(M = 1\) is trivial, and for the inductive step it suffices to know that the union of two countable sets \(A\) and \(B\) is countable. That special case follows from the argument above with \(M = 2\), or directly: choose one-to-one mappings \(g \colon A \to \mathbb{N}\) and \(h \colon B \to \mathbb{N}\) as furnished by Problem 17, and define \(\Phi \colon A \cup B \to \mathbb{N}\) by \(\Phi(x) = 2g(x)\) if \(x \in A\) and \(\Phi(x) = 2h(x) - 1\) if \(x \in B \sim A\). Then \(\Phi\) is one-to-one, since \(\Phi\) takes even values on \(A\) and odd values on \(B \sim A\) and is one-to-one on each of these two sets, and so \(A \cup B\) is countable by Problem 17.

(The target of the composition is misprinted in the statement: \(g \circ f\) maps \(A\) to \(C\), not to \(B\). We prove the assertion for \(g \circ f \colon A \to C\).)

First, the composition. By definition \(g \circ f\) is the mapping of \(A\) to \(C\) given by

\begin{equation*} (g \circ f)(x) = g\big(f(x)\big) \qquad \text{for } x \in A, \end{equation*}

which makes sense because \(f(x) \in B\) and \(g\) is defined on all of \(B\).

\(g \circ f\) is one-to-one. Let \(x, x^{\prime} \in A\) and suppose \((g \circ f)(x) = (g \circ f)(x^{\prime})\), that is, \(g(f(x)) = g(f(x^{\prime}))\). Since \(f(x)\) and \(f(x^{\prime})\) belong to \(B\) and \(g\) is one-to-one, we conclude that \(f(x) = f(x^{\prime})\). Since \(f\) is one-to-one, we conclude that \(x = x^{\prime}\).

\(g \circ f\) is onto \(C\). Let \(z \in C\). Since \(g\) maps \(B\) onto \(C\), there is a \(y \in B\) with \(g(y) = z\). Since \(f\) maps \(A\) onto \(B\), there is an \(x \in A\) with \(f(x) = y\). Then

\begin{equation*} (g \circ f)(x) = g(f(x)) = g(y) = z . \end{equation*}

Second, the inverse. We must first check that \(f^{-1}\) is a well-defined mapping of \(B\) to \(A\). Let \(y \in B\). Since \(f\) is onto, there is at least one \(x \in A\) with \(f(x) = y\); since \(f\) is one-to-one, there is at most one such \(x\). Hence there is exactly one \(x \in A\) with \(f(x) = y\), and we may define \(f^{-1}(y)\) to be that \(x\). This gives a mapping \(f^{-1} \colon B \to A\) characterized by the equivalence

\begin{equation*} f^{-1}(y) = x \iff f(x) = y \qquad (x \in A,\ y \in B). \end{equation*}

In particular \(f^{-1}(f(x)) = x\) for all \(x \in A\) and \(f(f^{-1}(y)) = y\) for all \(y \in B\).

\(f^{-1}\) is one-to-one. Let \(y, y^{\prime} \in B\) and suppose \(f^{-1}(y) = f^{-1}(y^{\prime})\); call this common value \(x \in A\). By the characterizing equivalence, \(f(x) = y\) and \(f(x) = y^{\prime}\). Since \(f\) is a mapping, it assigns to \(x\) a single value, so \(y = f(x) = y^{\prime}\).

\(f^{-1}\) is onto \(A\). Let \(x \in A\) and set \(y = f(x) \in B\). By the characterizing equivalence, \(f^{-1}(y) = x\). Thus every member of \(A\) is a value of \(f^{-1}\).

Therefore \(g \circ f \colon A \to C\) and \(f^{-1} \colon B \to A\) are both one-to-one and onto, as asserted.

Remark. These two facts are exactly what is needed to see that equipotence is an equivalence relation among sets, as asserted in Section 1.3: reflexivity is witnessed by the identity mapping, symmetry by the inverse \(f^{-1}\) of a one-to-one correspondence \(f\), and transitivity by the composition \(g \circ f\) of one-to-one correspondences \(f\) and \(g\).

For a natural number \(k\) we write, as in the text, \(\{1, \ldots, k\}\) for the initial segment \(\{ j \in \mathbb{N} \mid 1 \leq j \leq k \}\). Let

\begin{equation*} S = \big\{\, n \in \mathbb{N} \;\big|\; \text{for every } m \in \mathbb{N}, \ \{1, \ldots, n+m\} \text{ is not equipotent to } \{1, \ldots, n\} \,\big\}. \end{equation*}

We show \(1 \in S\) and that \(n \in S\) implies \(n + 1 \in S\); the Principle of Mathematical Induction then gives \(S = \mathbb{N}\), which is the pigeonhole principle. Since equipotence is symmetric (Problem 20), it does not matter in which direction we look for a one-to-one correspondence.

Base case. Let \(m\) be a natural number and suppose, to get a contradiction, that \(f\) is a one-to-one mapping of \(\{1, \ldots, 1+m\}\) onto \(\{1\}\). Since \(m \geq 1\), we have \(1 + m \geq 2\), so \(1\) and \(2\) are two distinct members of \(\{1, \ldots, 1+m\}\). But \(f(1)\) and \(f(2)\) both lie in the one-element set \(\{1\}\), so \(f(1) = 1 = f(2)\), contradicting the assumption that \(f\) is one-to-one. Hence no such \(f\) exists, and \(1 \in S\).

Inductive step. Suppose \(n \in S\). Let \(m\) be a natural number and suppose, to get a contradiction, that there is a one-to-one mapping \(f\) of \(\{1, \ldots, (n+1)+m\}\) onto \(\{1, \ldots, n+1\}\). Write

\begin{equation*} q = (n+1) + m = (n+m) + 1, \end{equation*}

so that the domain of \(f\) is \(\{1, \ldots, q\}\) and \(\{1, \ldots, q\} \sim \{q\} = \{1, \ldots, n+m\}\).

We first arrange that the largest element of the domain is sent to the largest element of the range. Since \(f\) is onto, there is a \(p \in \{1, \ldots, q\}\) with \(f(p) = n+1\). Let \(\tau\) be the transposition of \(\{1, \ldots, q\}\) that interchanges \(p\) and \(q\) and fixes every other element:

\begin{equation*} \tau(p) = q, \qquad \tau(q) = p, \qquad \tau(j) = j \ \text{ for } j \notin \{p, q\}. \end{equation*}

(If \(p = q\), then \(\tau\) is the identity mapping.) One checks directly that \(\tau \circ \tau\) is the identity mapping of \(\{1, \ldots, q\}\); consequently \(\tau\) is one-to-one, since \(\tau(i) = \tau(j)\) gives \(i = \tau(\tau(i)) = \tau(\tau(j)) = j\), and \(\tau\) is onto, since any \(j\) equals \(\tau(\tau(j))\).

By Problem 20, the composition \(h = f \circ \tau\) is a one-to-one mapping of \(\{1, \ldots, q\}\) onto \(\{1, \ldots, n+1\}\), and

\begin{equation*} h(q) = f(\tau(q)) = f(p) = n + 1 . \end{equation*}

Now let \(h_0\) be the restriction of \(h\) to \(\{1, \ldots, n+m\} = \{1, \ldots, q\} \sim \{q\}\). We claim \(h_0\) is a one-to-one mapping of \(\{1, \ldots, n+m\}\) onto \(\{1, \ldots, n\}\).

\(h_0\) takes its values in \(\{1, \ldots, n\}\): if \(1 \leq j \leq n+m\), then \(j \neq q\), so \(h(j) \neq h(q) = n+1\) because \(h\) is one-to-one; since \(h(j) \in \{1, \ldots, n+1\}\) and \(h(j) \neq n+1\), we get \(h(j) \in \{1, \ldots, n\}\).

\(h_0\) is one-to-one: it is the restriction of the one-to-one mapping \(h\) to a subset of its domain.

\(h_0\) is onto \(\{1, \ldots, n\}\): let \(i \in \{1, \ldots, n\}\). Since \(h\) is onto \(\{1, \ldots, n+1\}\), there is a \(j \in \{1, \ldots, q\}\) with \(h(j) = i\). As \(i \leq n < n+1 = h(q)\), we have \(h(j) \neq h(q)\), hence \(j \neq q\), hence \(1 \leq j \leq n+m\) and \(h_0(j) = i\).

Thus \(\{1, \ldots, n+m\}\) is equipotent to \(\{1, \ldots, n\}\), with \(m\) a natural number. This contradicts the assumption \(n \in S\).

Therefore no such \(f\) exists for any natural number \(m\), that is, \(\{1, \ldots, (n+1)+m\}\) is not equipotent to \(\{1, \ldots, n+1\}\) for any natural number \(m\). Hence \(n + 1 \in S\).

By induction \(S = \mathbb{N}\), and the pigeonhole principle is established: for all natural numbers \(n\) and \(m\), the set \(\{1, \ldots, n+m\}\) is not equipotent to \(\{1, \ldots, n\}\).

Remark. Two standard consequences follow at once. If \(n\) and \(n^{\prime}\) are natural numbers with \(n \neq n^{\prime}\), then \(\{1, \ldots, n\}\) and \(\{1, \ldots, n^{\prime}\}\) are not equipotent: for one of them, say \(n < n^{\prime}\), we may write \(n^{\prime} = n + m\) with \(m = n^{\prime} - n\), and \(m\) is a natural number by Problem 9; the principle then applies. Hence the natural number \(n\) attached to a nonempty finite set by the definition of finiteness is uniquely determined by the set.

We first record an elementary observation that will be used repeatedly.

Observation. If \(E\) is a nonempty countable set, then there is a mapping of \(\mathbb{N}\) onto \(E\). Indeed, by the definition of countability \(E\) is either finite or countably infinite. If \(E\) is countably infinite there is a one-to-one correspondence \(h\) of \(\mathbb{N}\) onto \(E\), and \(h\) itself is such a mapping. If \(E\) is finite and nonempty there is a natural number \(n\) and a one-to-one correspondence \(h\) of \(\{1,\ldots,n\}\) onto \(E\); define \(f\colon \mathbb{N}\to E\) by \(f(k)=h(k)\) for \(1\le k\le n\) and \(f(k)=h(1)\) for \(k>n\). Then \(f\) maps \(\mathbb{N}\) onto \(E\).

Now suppose, by way of contradiction, that \(2^{\mathbb{N}}\) is countable. The set \(2^{\mathbb{N}}\) is nonempty, since the empty-set belongs to it. Hence, by the Observation, there is a mapping

\begin{equation*} f\colon \mathbb{N}\longrightarrow 2^{\mathbb{N}} \end{equation*}

of \(\mathbb{N}\) onto \(2^{\mathbb{N}}\); that is, every set of natural numbers is of the form \(f(n)\) for some natural number \(n\).

Define

\begin{equation*} D=\{\, n\in\mathbb{N} \;:\; n\notin f(n) \,\}. \end{equation*}

This is a well-defined set of natural numbers, so \(D\) belongs to \(2^{\mathbb{N}}\). Since \(f\) is onto, there is a natural number \(m\) for which \(D=f(m)\). We ask whether \(m\) belongs to \(D\).

If \(m\in D\), then by the definition of \(D\) we have \(m\notin f(m)=D\), a contradiction. If \(m\notin D\), then \(m\notin f(m)\), and therefore, again by the definition of \(D\), \(m\in D\), once more a contradiction.

Both alternatives are impossible, so no mapping of \(\mathbb{N}\) onto \(2^{\mathbb{N}}\) exists. This contradicts the Observation, and therefore \(2^{\mathbb{N}}\) is not countable, that is, \(2^{\mathbb{N}}\) is uncountable.

Remark. The same argument shows, for an arbitrary set \(X\), that there is no mapping of \(X\) onto \(2^{X}\); in particular \(X\) and \(2^{X}\) are never equipotent.

Part 1: finite Cartesian products of countable sets.

Let \(A_1,\ldots,A_n\) be countable sets; we show \(A_1\times\cdots\times A_n\) is countable, arguing by induction on \(n\).

For \(n=1\) there is nothing to prove. Consider \(n=2\), so let \(A\) and \(B\) be countable. If either \(A\) or \(B\) is empty, then \(A\times B\) is empty and hence countable. So assume both are nonempty. By the Observation recorded in the solution of Problem 22, there are mappings \(f\) of \(\mathbb{N}\) onto \(A\) and \(g\) of \(\mathbb{N}\) onto \(B\). Define

\begin{equation*} h\colon \mathbb{N}\times\mathbb{N}\longrightarrow A\times B,\qquad h(m,k)=(f(m),g(k)). \end{equation*}

If \((a,b)\) belongs to \(A\times B\), choose \(m\) with \(f(m)=a\) and \(k\) with \(g(k)=b\); then \(h(m,k)=(a,b)\). Thus \(h\) maps \(\mathbb{N}\times\mathbb{N}\) onto \(A\times B\). By Corollary 4(i) the set \(\mathbb{N}\times\mathbb{N}\) is countably infinite, in particular a nonempty countable set. Since \(A\times B\) is nonempty and is the image of a function whose domain is a nonempty countable set, Theorem 5 tells us that \(A\times B\) is countable.

Now assume the assertion holds for \(n\), and let \(A_1,\ldots,A_{n+1}\) be countable. The mapping

\begin{equation*} \big((x_1,\ldots,x_n),x_{n+1}\big)\longmapsto (x_1,\ldots,x_n,x_{n+1}) \end{equation*}

is a one-to-one correspondence between \((A_1\times\cdots\times A_n)\times A_{n+1}\) and \(A_1\times\cdots\times A_{n+1}\). By the induction hypothesis \(A_1\times\cdots\times A_n\) is countable, so by the case \(n=2\) the set \((A_1\times\cdots\times A_n)\times A_{n+1}\) is countable; since a set equipotent to a countable set is countable, so is \(A_1\times\cdots\times A_{n+1}\). The induction is complete.

Part 2: \(\mathbb{N}^{\mathbb{N}}\) is uncountable.

For a set \(E\) of natural numbers define \(\chi_E\colon \mathbb{N}\to\mathbb{N}\) by

\begin{equation*} \chi_E(n)=\begin{cases} 1, & n\in E,\\ 2, & n\notin E.\end{cases} \end{equation*}

Then \(\chi_E\) is a mapping of \(\mathbb{N}\) into \(\mathbb{N}\), so \(\chi_E\) belongs to \(\mathbb{N}^{\mathbb{N}}\), and the assignment \(E\mapsto \chi_E\) is a mapping of \(2^{\mathbb{N}}\) into \(\mathbb{N}^{\mathbb{N}}\).

This assignment is one-to-one: if \(E\ne F\) are sets of natural numbers, then there is a natural number \(n\) belonging to exactly one of them, say \(n\in E\) and \(n\notin F\); then \(\chi_E(n)=1\ne 2=\chi_F(n)\), so \(\chi_E\ne\chi_F\). Consequently \(2^{\mathbb{N}}\) is equipotent to the set

\begin{equation*} S=\{\,\chi_E \;:\; E\in 2^{\mathbb{N}}\,\}\subseteq \mathbb{N}^{\mathbb{N}}. \end{equation*}

Suppose \(\mathbb{N}^{\mathbb{N}}\) were countable. Then, by Theorem 3, its subset \(S\) would be countable; and since \(2^{\mathbb{N}}\) is equipotent to \(S\), and a set equipotent to a countable set is countable, \(2^{\mathbb{N}}\) would be countable. This contradicts the preceding problem. Therefore \(\mathbb{N}^{\mathbb{N}}\) is not countable.

Recall that an interval of real numbers is called degenerate provided it is empty or contains a single member. Thus a nondegenerate interval \(I\) contains two distinct points; choose \(a,b\in I\) with \(a<b\). Since \(I\) is an interval, every real number lying between two of its members belongs to \(I\), so

\begin{equation*} [a,b]\subseteq I . \end{equation*}

Step 1: \(I\) contains a subset equipotent to \(\mathbb{N}\). For each natural number \(k\) put

\begin{equation*} x_k=a+\frac{b-a}{k}. \end{equation*}

Since \(b-a>0\) and \(0<1/k\le 1\), we have \(a<x_k\le b\), so \(x_k\) belongs to \([a,b]\) and hence to \(I\). Moreover \(k\mapsto x_k\) is one-to-one: if \(k\ne k^{\prime}\) then \(1/k\ne 1/k^{\prime}\), and since \(b-a\ne0\) we get \(x_k\ne x_{k^{\prime}}\). Therefore

\begin{equation*} X=\{\,x_k \;:\; k\in\mathbb{N}\,\} \end{equation*}

is a subset of \(I\) that is equipotent to \(\mathbb{N}\).

Step 2: a subset of a finite set is finite, and indeed corresponds to an initial segment. We use the following fact, which is exactly what the finite case of the proof of Theorem 3 establishes: if \(B\) is equipotent to \(\{1,\ldots,n\}\) and \(A\) is a nonempty subset of \(B\), then there is a natural number \(N\) with \(N\le n\) and a one-to-one correspondence between \(\{1,\ldots,N\}\) and \(A\).

Step 3: \(\mathbb{N}\) is not finite. Suppose \(\mathbb{N}\) were equipotent to \(\{1,\ldots,m\}\) for some natural number \(m\), by way of a one-to-one correspondence \(\psi\) of \(\mathbb{N}\) onto \(\{1,\ldots,m\}\). The set \(\{1,\ldots,m+1\}\) is a nonempty subset of \(\mathbb{N}\), so its image \(\psi(\{1,\ldots,m+1\})\) is a nonempty subset of \(\{1,\ldots,m\}\) equipotent to \(\{1,\ldots,m+1\}\). By Step 2 there is a natural number \(j\le m\) with \(\{1,\ldots,j\}\) equipotent to \(\psi(\{1,\ldots,m+1\})\), and hence, by transitivity of equipotence, \(\{1,\ldots,j\}\) is equipotent to \(\{1,\ldots,m+1\}\). Writing \(m+1=j+p\) with \(p=m+1-j\ge 1\) a natural number, this says that \(\{1,\ldots,j\}\) is equipotent to \(\{1,\ldots,j+p\}\), which contradicts the pigeonhole principle (Problem 21). Therefore \(\mathbb{N}\) is not finite.

Step 4: conclusion. Suppose \(I\) were finite. Since \(I\) is nonempty, there is a natural number \(n\) and a one-to-one correspondence \(\varphi\) of \(I\) onto \(\{1,\ldots,n\}\). The set \(X\) of Step 1 is a nonempty subset of \(I\), so \(\varphi(X)\) is a nonempty subset of \(\{1,\ldots,n\}\), and \(\varphi\) restricted to \(X\) is a one-to-one correspondence of \(X\) onto \(\varphi(X)\). By Step 2 there is a natural number \(N\le n\) with \(\{1,\ldots,N\}\) equipotent to \(\varphi(X)\). Since \(\mathbb{N}\) is equipotent to \(X\), \(X\) is equipotent to \(\varphi(X)\), and \(\varphi(X)\) is equipotent to \(\{1,\ldots,N\}\), transitivity of equipotence gives that \(\mathbb{N}\) is equipotent to \(\{1,\ldots,N\}\), that is, \(\mathbb{N}\) is finite. This contradicts Step 3.

Therefore a nondegenerate interval of real numbers is not finite.

Equipotence is reflexive, symmetric and transitive (Problem 20), so it suffices to prove that every nondegenerate interval is equipotent to the single interval \((0,1)\).

Lemma (absorbing a countable set). Let \(X\) be a set, let \(C\) be a countable subset of \(X\), and suppose \(X\sim C\) contains a countably infinite subset \(D\). Then \(X\) and \(X\sim C\) are equipotent.

Proof of Lemma. Here \(X\sim C\) denotes the set of points of \(X\) that do not belong to \(C\). Put \(E=C\cup D\). By Corollary 6 the set \(E\) is countable. It is not finite: it contains \(D\), and were \(E\) finite then, by Step 2 of Problem 24, \(D\) would be equipotent to an initial segment \(\{1,\ldots,N\}\), so that \(\mathbb{N}\) would be finite, contradicting Step 3 of Problem 24. Hence \(E\) is countably infinite. Since \(D\subseteq X\sim C\), the sets \(C\) and \(D\) are disjoint, so \(E\sim C=D\). Let \(\{e_k \mid k\in\mathbb{N}\}\) be an enumeration of \(E\) and \(\{d_k \mid k\in\mathbb{N}\}\) an enumeration of \(D\). Define

\begin{equation*} h\colon X\to X\sim C,\qquad h(e_k)=d_k \ \ (k\in\mathbb{N}),\qquad h(x)=x \ \ \text{for } x\in X\sim E . \end{equation*}

This is well defined, since every point of \(X\) either belongs to \(E\), and then equals exactly one \(e_k\), or belongs to \(X\sim E\); and its values lie in \(D\cup(X\sim E)\subseteq X\sim C\). It is one-to-one: \(h\) is one-to-one on \(E\) because the \(d_k\) are distinct, it is one-to-one on \(X\sim E\) because it is the identity there, and the two images \(D\) and \(X\sim E\) are disjoint because \(D\subseteq E\). It is onto \(X\sim C\): its image is \(D\cup(X\sim E)=(E\sim C)\cup(X\sim E)=X\sim C\). This proves the Lemma.

Step 1: every nondegenerate interval is equipotent to a nondegenerate open interval. Let \(I\) be a nondegenerate interval and let \(\alpha=\inf I\) and \(\beta=\sup I\) be its infimum and supremum in the extended real numbers. If \(\alpha<z<\beta\), then \(z\) is not a lower bound for \(I\), so there is \(u\in I\) with \(u<z\), and \(z\) is not an upper bound for \(I\), so there is \(v\in I\) with \(z<v\); since \(I\) is an interval, \(z\in I\). Hence, writing \(J=(\alpha,\beta)\),

\begin{equation*} J\subseteq I\subseteq J\cup F,\qquad F=\{\alpha,\beta\}\cap\mathbb{R}, \end{equation*}

so \(I=J\cup C\) where \(C=I\sim J\) is a subset of \(F\) and therefore has at most two elements. Since \(I\) is nondegenerate it contains two distinct points, so \(\alpha<\beta\) and \(J\) is a nondegenerate open interval. Now apply the Lemma with \(X=I\) and this finite (hence countable) set \(C\): the set \(X\sim C=J\) is a nondegenerate interval, so by Problem 24 it is not finite, and hence (as in Step 1 of Problem 24) it contains a countably infinite subset \(D\). The Lemma then tells us that \(I\) is equipotent to \(I\sim C=J\), that is, \(I\) is equipotent to the open interval \(J=(\alpha,\beta)\).

Step 2: the interval \(J=(\alpha,\beta)\) produced in Step 1 is equipotent to \((0,1)\). Here \(\alpha<\beta\) are extended real numbers, and we use the notation introduced in Section 1.2, namely \((a,\infty)=\{x\in\mathbb{R} \mid a<x\}\), \((-\infty,b)=\{x\in\mathbb{R} \mid x<b\}\) and \((-\infty,\infty)=\mathbb{R}\). Since \(\alpha<\beta\), we have \(\alpha\ne+\infty\) and \(\beta\ne-\infty\); hence, according to whether each of \(\alpha\), \(\beta\) is real or infinite, \(J\) is of exactly one of the four forms \((a,b)\) with \(a<b\) real, \((a,\infty)\) with \(a\) real, \((-\infty,b)\) with \(b\) real, or \(\mathbb{R}\). We treat each in turn.

If \(J=(a,b)\) with \(a<b\) real, then \(x\mapsto (x-a)/(b-a)\) is a one-to-one correspondence of \((a,b)\) onto \((0,1)\), with inverse \(s\mapsto a+s(b-a)\).

If \(J=(a,\infty)\), then \(x\mapsto x-a\) is a one-to-one correspondence of \((a,\infty)\) onto \((0,\infty)\), and

\begin{equation*} t\longmapsto \frac{t}{1+t} \end{equation*}

is a one-to-one correspondence of \((0,\infty)\) onto \((0,1)\): it carries \((0,\infty)\) into \((0,1)\) since \(0<t<1+t\), and \(s\mapsto s/(1-s)\) carries \((0,1)\) into \((0,\infty)\) and is a two-sided inverse, because if \(s=t/(1+t)\) then \(1-s=1/(1+t)\) and so \(s/(1-s)=t\), while if \(t=s/(1-s)\) then \(1+t=1/(1-s)\) and so \(t/(1+t)=s\).

If \(J=(-\infty,b)\), then \(x\mapsto b-x\) is a one-to-one correspondence of \((-\infty,b)\) onto \((0,\infty)\), and we conclude as in the previous case.

If \(J=\mathbb{R}\), then

\begin{equation*} x\longmapsto \frac{x}{1+|x|} \end{equation*}

is a one-to-one correspondence of \(\mathbb{R}\) onto \((-1,1)\): its values satisfy \(|x|/(1+|x|)<1\), and \(y\mapsto y/(1-|y|)\) maps \((-1,1)\) into \(\mathbb{R}\) and is a two-sided inverse, since for \(x\ge 0\) the value \(y=x/(1+x)\) lies in \([0,1)\) and \(y/(1-y)=x\), while both maps are odd, so the case \(x<0\) follows by symmetry. Composing with the one-to-one correspondence \(y\mapsto (y+1)/2\) of \((-1,1)\) onto \((0,1)\) gives the assertion.

Conclusion. By Step 1 and Step 2, every nondegenerate interval of real numbers is equipotent to \((0,1)\). By symmetry and transitivity of equipotence, any two nondegenerate intervals of real numbers are equipotent to each other.

Yes. We construct an explicit one-to-one correspondence, in five steps. Throughout we use two trivial remarks: equipotence is an equivalence relation (Problem 20), and if \(f\) is a one-to-one correspondence of \(X\) onto \(X^{\prime}\) and \(g\) one of \(Y\) onto \(Y^{\prime}\), then \((x,y)\mapsto (f(x),g(y))\) is a one-to-one correspondence of \(X\times Y\) onto \(X^{\prime}\times Y^{\prime}\).

Let \(\Sigma\) denote the collection of all sequences \(a=(a_n)_{n\ge 1}\) with each \(a_n\in\{0,1\}\) and with \(a_n=0\) for infinitely many indices \(n\).

Step 1: the map \(\sigma(a)=\sum_{n=1}^{\infty}a_n2^{-n}\) is a one-to-one correspondence of \(\Sigma\) onto \([0,1)\).

The series converges since it is dominated by \(\sum 2^{-n}=1\), and \(\sigma(a)\ge 0\). If \(N\) is an index with \(a_N=0\), then \(\sigma(a)\le \sum_{n\ne N}2^{-n}=1-2^{-N}<1\); such an \(N\) exists, so \(\sigma\) maps \(\Sigma\) into \([0,1)\).

\(\sigma\) is one-to-one: let \(a\ne b\) in \(\Sigma\) and let \(m\) be the least index with \(a_m\ne b_m\), say \(a_m=1\) and \(b_m=0\). Since \(b\) has infinitely many zero terms, \(\sum_{n>m}b_n2^{-n}<\sum_{n>m}2^{-n}=2^{-m}\), and therefore

\begin{equation*} \sigma(a)-\sigma(b)=2^{-m}+\sum_{n>m}(a_n-b_n)2^{-n}\ \ge\ 2^{-m}-\sum_{n>m}b_n2^{-n}\ >\ 0 . \end{equation*}

\(\sigma\) is onto \([0,1)\): given \(x\in[0,1)\), define \(a_n\) recursively by declaring \(a_n=1\) if \(\sum_{k=1}^{n-1}a_k2^{-k}+2^{-n}\le x\) and \(a_n=0\) otherwise. Put \(r_n=x-\sum_{k=1}^{n}a_k2^{-k}\), with \(r_0=x\). We claim \(0\le r_n<2^{-n}\) for every \(n\ge 0\). This holds for \(n=0\). Assume it for \(n-1\). The defining condition for \(a_n\) reads \(r_{n-1}\ge 2^{-n}\). If it holds, then \(a_n=1\) and \(r_n=r_{n-1}-2^{-n}\), so \(0\le r_n<2^{-(n-1)}-2^{-n}=2^{-n}\). If it fails, then \(a_n=0\) and \(r_n=r_{n-1}<2^{-n}\), and \(r_n\ge0\). This proves the claim, and letting \(n\to\infty\) gives \(\sigma(a)=x\). Finally \(a\in\Sigma\): if \(a_n=1\) for all \(n>N\), then for \(n>N\)

\begin{equation*} r_n=r_N-\big(2^{-N}-2^{-n}\big)=\big(r_N-2^{-N}\big)+2^{-n}, \end{equation*}

and \(r_N-2^{-N}<0\); taking \(n\) large enough that \(2^{-n}<2^{-N}-r_N\) gives \(r_n<0\), a contradiction. So \(a\) has infinitely many zero terms.

Step 2: let \(B\) be the set of \(a\in\Sigma\) having infinitely many terms equal to \(1\) as well. Then \(B\) is equipotent to \(\mathbb{R}\).

The complement \(\Sigma\sim B\) consists of the sequences in \(\Sigma\) with only finitely many terms equal to \(1\). For each natural number \(N\) there are exactly \(2^{N}\) such sequences whose terms vanish from index \(N+1\) on, so \(\Sigma\sim B\) is the union of a countable collection of finite sets and hence is countable by Corollary 6. Since \(\sigma\) is one-to-one, \(C=\sigma(\Sigma\sim B)\) is a countable subset of \([0,1)\) and \(\sigma(B)=[0,1)\sim C\). For each natural number \(m\) let \(b^{(m)}\) be the sequence whose \(n\)-th term is \(1\) exactly when \(n\) is even and \(n\ge 2m\); each \(b^{(m)}\) has infinitely many ones and infinitely many zeros, so \(b^{(m)}\in B\), and these sequences are distinct because \(2m\) is the least index carrying a \(1\) in \(b^{(m)}\). Hence \(D=\{\sigma(b^{(m)}) \mid m\in\mathbb{N}\}\) is a countably infinite subset of \([0,1)\sim C\). By the Lemma proved in Problem 25, applied with \(X=[0,1)\), the sets \([0,1)\) and \([0,1)\sim C=\sigma(B)\) are equipotent. Since \(\sigma\) restricted to \(B\) is a one-to-one correspondence of \(B\) onto \(\sigma(B)\), the set \(B\) is equipotent to \([0,1)\), which by Problem 25 is equipotent to \(\mathbb{R}\).

Step 3: \(B\) is equipotent to \(\mathbb{N}^{\mathbb{N}}\), the collection of all mappings of \(\mathbb{N}\) into \(\mathbb{N}\), which we regard as the collection of all sequences \((m_j)_{j\ge1}\) of natural numbers.

Define \(\Psi\colon \mathbb{N}^{\mathbb{N}}\to B\) as follows: \(\Psi(m)\) is the sequence obtained by concatenating the blocks

\begin{equation*} \underbrace{0\cdots 0}_{m_1-1},\ \underbrace{1\cdots 1}_{m_2},\ \underbrace{0\cdots 0}_{m_3},\ \underbrace{1\cdots 1}_{m_4},\ \ldots \end{equation*}

that is, the first block consists of \(m_1-1\ge 0\) zeros and, for \(j\ge2\), the \(j\)-th block consists of \(m_j\ge 1\) terms, all equal to \(1\) if \(j\) is even and all equal to \(0\) if \(j\) is odd. Since each block with \(j\ge 2\) is nonempty, the concatenation is an infinite sequence of \(0\)’s and \(1\)’s; it has infinitely many ones (one from each even block) and infinitely many zeros (one from each odd block with \(j\ge3\)). Hence \(\Psi(m)\in B\).

\(\Psi\) is onto \(B\) and one-to-one because a sequence in \(B\) has a unique decomposition into maximal constant runs. We first describe a recovery recipe \(\Phi\colon B\to\mathbb{N}^{\mathbb{N}}\). Let \(a\in B\) and put \(S=\{n\in\mathbb{N} \mid a_n\ne a_{n+1}\}\). If \(S\) were finite, \(a\) would be constant from some index on, contradicting the fact that both values occur infinitely often; so \(S\) is infinite and we may enumerate it increasingly as \(s_1<s_2<\cdots\). Put \(s_0=0\) and \(\ell_j=s_j-s_{j-1}\ge 1\). On each set of indices \(\{s_{j-1}+1,\ldots,s_j\}\) the sequence \(a\) is constant, and its values on consecutive such sets differ; hence they alternate, starting with the value \(a_1\). If \(a_1=0\), set \(m_1=\ell_1+1\) and \(m_j=\ell_j\) for \(j\ge2\); if \(a_1=1\), set \(m_1=1\) and \(m_{j+1}=\ell_j\) for \(j\ge1\). In either case \((m_j)\) is a sequence of natural numbers; write \(\Phi(a)=(m_j)\). Comparing blocks one at a time, \(\Psi(\Phi(a))=a\). Hence \(\Psi\) maps \(\mathbb{N}^{\mathbb{N}}\) onto \(B\).

It remains to check that \(\Phi(\Psi(m))=m\) for every \(m\in\mathbb{N}^{\mathbb{N}}\), which gives that \(\Psi\) is one-to-one. Put \(a=\Psi(m)\). Suppose first \(m_1\ge 2\). Then every block in the definition of \(\Psi(m)\) is nonempty and consecutive blocks carry different values, so the maximal constant runs of \(a\) are precisely those blocks, in order, with lengths \(m_1-1,m_2,m_3,\ldots\); that is, \(\ell_1=m_1-1\) and \(\ell_j=m_j\) for \(j\ge2\). Also \(a_1=0\), so the recipe returns \(\ell_1+1=m_1\) and \(\ell_j=m_j\) for \(j\ge2\), that is, \(\Phi(a)=m\). Suppose instead \(m_1=1\). Then the first block is empty, \(a\) begins with the second block and \(a_1=1\), and the maximal constant runs of \(a\) have lengths \(m_2,m_3,\ldots\); that is, \(\ell_j=m_{j+1}\) for \(j\ge1\). The recipe for the case \(a_1=1\) returns first coordinate \(1=m_1\) and \((j+1)\)-st coordinate \(\ell_j=m_{j+1}\), so again \(\Phi(a)=m\).

Thus \(\Psi\) is a one-to-one correspondence of \(\mathbb{N}^{\mathbb{N}}\) onto \(B\).

Step 4: \(\mathbb{N}^{\mathbb{N}}\times\mathbb{N}^{\mathbb{N}}\) is equipotent to \(\mathbb{N}^{\mathbb{N}}\).

Given \(u=(u_k)\) and \(v=(v_k)\) in \(\mathbb{N}^{\mathbb{N}}\), let \(w=\Theta(u,v)\) be the interleaved sequence defined by \(w_{2k-1}=u_k\) and \(w_{2k}=v_k\) for \(k\in\mathbb{N}\). Then \(\Theta\) maps into \(\mathbb{N}^{\mathbb{N}}\), and the assignment \(w\mapsto \big((w_{2k-1})_k,(w_{2k})_k\big)\) is a two-sided inverse. Hence \(\Theta\) is a one-to-one correspondence.

Step 5: conclusion. Combining the steps,

\begin{equation*} \mathbb{R}\times\mathbb{R}\ \sim\ B\times B\ \sim\ \mathbb{N}^{\mathbb{N}}\times\mathbb{N}^{\mathbb{N}}\ \sim\ \mathbb{N}^{\mathbb{N}}\ \sim\ B\ \sim\ \mathbb{R}, \end{equation*}

where each \(\sim\) denotes equipotence: the first by Step 2 and the product remark, the second by Step 3 and the product remark, the third by Step 4, the fourth by Step 3, the last by Step 2. By transitivity, \(\mathbb{R}\times\mathbb{R}\) is equipotent to \(\mathbb{R}\).

The set \(\mathbb{Q}\) of rational numbers is neither open nor closed.

\(\mathbb{Q}\) is not open. Let \(x\) be a rational number and let \(r>0\). The interval \((x-r,x+r)\) is a nondegenerate interval, and the irrational numbers are dense in \(\mathbb{R}\) (Problem 12), so there is an irrational number \(y\) with \(x-r<y<x+r\). Thus \((x-r,x+r)\) is not contained in \(\mathbb{Q}\). Since this holds for every \(r>0\), and since \(\mathbb{Q}\) contains a point (for instance \(0\)), the defining property of openness fails at every point of \(\mathbb{Q}\). Hence \(\mathbb{Q}\) is not open.

\(\mathbb{Q}\) is not closed. Let \(x\) be any real number and let \(I\) be an open interval containing \(x\). Then \(I\) contains an interval of the form \((x-r,x+r)\) with \(r>0\), and by the density of the rational numbers, Theorem 2, there is a rational number \(q\) with \(x-r<q<x+r\). Hence every open interval containing \(x\) contains a point of \(\mathbb{Q}\), so \(x\) is a point of closure of \(\mathbb{Q}\). Since \(x\) was arbitrary,

\begin{equation*} \overline{\mathbb{Q}}=\mathbb{R}. \end{equation*}

But \(\mathbb{R}\ne\mathbb{Q}\), since for instance \(\sqrt{2}\) is irrational. Therefore \(\mathbb{Q}\) does not contain all of its points of closure, that is, \(\mathbb{Q}\) is not closed.

Equivalently, one may argue the second assertion from the first through Proposition 11: were \(\mathbb{Q}\) closed, its complement, the set of irrational numbers, would be open; but by the density of the rationals, Theorem 2, no interval \((x-r,x+r)\) about an irrational \(x\) consists entirely of irrationals.

The only subsets of \(\mathbb{R}\) that are both open and closed are the empty-set \(\emptyset\) and \(\mathbb{R}\) itself.

First, these two sets do have the property. By Proposition 8 both \(\emptyset\) and \(\mathbb{R}\) are open, and by Proposition 12 both are closed. (Alternatively: \(\mathbb{R}\) is open and its complement \(\emptyset\) is open, so by Proposition 11 the set \(\mathbb{R}\) is closed, and symmetrically for \(\emptyset\).)

Conversely, let \(E\) be a set of real numbers that is both open and closed, and suppose \(E\) is neither empty nor all of \(\mathbb{R}\); we derive a contradiction.

Choose \(x\in E\). Since \(E\) is open and nonempty, the construction in the proof of Proposition 9 applies: with

\begin{equation*} a_x=\inf\{\,z \mid (z,x)\subseteq E\,\},\qquad b_x=\sup\{\,y \mid (x,y)\subseteq E\,\} \end{equation*}

taken in the extended real numbers, the interval \(I_x=(a_x,b_x)\) contains \(x\) and, by assertion (2) in that proof,

\begin{equation*} I_x\subseteq E,\qquad a_x\notin E,\qquad b_x\notin E . \end{equation*}

Since \(E\ne\mathbb{R}\) and \(I_x\subseteq E\), we have \(I_x\ne\mathbb{R}\), so at least one of \(a_x\), \(b_x\) is a real number. Suppose first that \(b_x\) is real. We claim \(b_x\) is a point of closure of \(E\). Indeed, let \(I\) be an open interval containing \(b_x\); then \(I\) contains an interval \((b_x-r,b_x+r)\) for some \(r>0\). Since \(a_x<x<b_x\), the number

\begin{equation*} t=\max\Big\{\,x,\ b_x-\tfrac{r}{2}\,\Big\} \end{equation*}

satisfies \(a_x<t<b_x\) and \(b_x-r<t\), so \(t\) belongs both to \(I_x\subseteq E\) and to \((b_x-r,b_x+r)\subseteq I\). Thus every open interval containing \(b_x\) contains a point of \(E\), so \(b_x\) is a point of closure of \(E\). Since \(E\) is closed, \(E=\overline{E}\), and therefore \(b_x\in E\). This contradicts \(b_x\notin E\).

If instead \(b_x=\infty\), then \(a_x\) must be a real number, and the symmetric argument applies: for any \(r>0\) the number \(t=\min\{x,\ a_x+\tfrac{r}{2}\}\) satisfies \(a_x<t<b_x\) and \(t<a_x+r\), so every open interval containing \(a_x\) contains a point of \(I_x\subseteq E\); hence \(a_x\) is a point of closure of \(E\), so \(a_x\in E\) because \(E\) is closed, contradicting \(a_x\notin E\).

In either case we reach a contradiction. Therefore a set of real numbers that is both open and closed must be \(\emptyset\) or \(\mathbb{R}\).

Take

\begin{equation*} A = (0,1), \qquad B = (1,2). \end{equation*}

These are disjoint: a real number cannot satisfy both \(x < 1\) and \(x > 1\), so \(A \cap B = \emptyset\).

Now compute the closures. I claim \(\overline{A} = [0,1]\). First, \([0,1]\) is a closed set containing \(A\): its complement \((-\infty,0) \cup (1,\infty)\) is the union of two open intervals, hence open by Proposition 8, so \([0,1]\) is closed by Proposition 11. Since \(\overline{A}\) is the smallest closed set containing \(A\) (Proposition 10), \(\overline{A} \subseteq [0,1]\). Conversely, both \(0\) and \(1\) are points of closure of \(A\): if \(I\) is an open interval containing \(0\), then \(I\) contains an interval \((-r,r)\) with \(r > 0\), and by the Archimedean Property of \(\mathbb{R}\) we may choose a natural number \(n\) with \(1/n < \min\{r,1\}\); then \(1/n \in (0,1) \cap I = A \cap I\). The same argument at the point \(1\), using the numbers \(1 - 1/n\), shows that \(1\) is a point of closure of \(A\). Since also \(A \subseteq \overline{A}\), we get \([0,1] = A \cup \{0,1\} \subseteq \overline{A}\). Hence \(\overline{A} = [0,1]\), and by the identical argument (translate by \(1\)) \(\overline{B} = [1,2]\).

Therefore

\begin{equation*} \overline{A} \cap \overline{B} = [0,1] \cap [1,2] = \{1\} \neq \emptyset . \end{equation*}

A second, more extreme example: let \(A = \mathbb{Q}\) and \(B = \mathbb{R} \sim \mathbb{Q}\) be the rationals and the irrationals. These are disjoint by definition. Fix a real number \(x\) and an open interval \(I\) containing \(x\); whatever the type of \(I\), there is an \(r > 0\) with \((x - r, x + r) \subseteq I\), so it suffices to produce points of \(A\) and of \(B\) in each bounded interval \((a,b)\) with \(a < b\). By Theorem 2 the rational numbers are dense in \(\mathbb{R}\), so \((a,b)\) contains a rational number; hence every real number is a point of closure of \(A\) and \(\overline{A} = \mathbb{R}\). The interval \((a,b)\) also contains an irrational number: by Theorem 2 pick rationals \(q\) and \(q^{\prime}\) with \(a < q < q^{\prime} < b\), and set \(z = q + (q^{\prime} - q)/\sqrt{2}\). Since \(0 < 1/\sqrt{2} < 1\) we have \(q < z < q^{\prime}\), so \(z \in (a,b)\); and \(z\) is irrational, for if \(z\) were rational then \(\sqrt{2} = (q^{\prime} - q)/(z - q)\) would be a quotient of nonzero rationals and so rational, contradicting the irrationality of \(\sqrt{2}\) established in Section 1.2. Hence \(\overline{B} = \mathbb{R}\) as well, and \(\overline{A} \cap \overline{B} = \mathbb{R} \neq \emptyset\), although \(A \cap B = \emptyset\).

The examples show that the operation \(E \mapsto \overline{E}\) does not preserve disjointness: closure can create new common points on the shared boundary of two disjoint sets.

Throughout, recall the definition: \(x\) is a point of closure of a set \(S\) provided every open interval that contains \(x\) also contains a point of \(S\). Thus \(x \in E^{\prime}\) means: every open interval containing \(x\) contains a point of \(E\) different from \(x\).

(i) \(E^{\prime}\) is closed.

By definition a set is closed provided it contains all of its points of closure, so let \(x\) be a point of closure of \(E^{\prime}\); we must show \(x \in E^{\prime}\). Let \(I\) be an open interval containing \(x\). We must produce a point of \(E \sim \{x\}\) inside \(I\).

Since \(x\) is a point of closure of \(E^{\prime}\), the interval \(I\) contains some point \(y \in E^{\prime}\).

Case 1: \(y = x\). Then \(x \in E^{\prime}\) and there is nothing more to prove.

Case 2: \(y \neq x\). The set \(I \sim \{x\}\) is open: indeed \(I\) is open and \(\mathbb{R} \sim \{x\}\) is open (its complement \(\{x\}\) is closed, since any \(z \neq x\) has the open interval \((z - |z-x|, z + |z-x|)\) about it missing \(x\), so \(z\) is not a point of closure of \(\{x\}\)), and the intersection of two open sets is open by Proposition 8. Since \(y \in I \sim \{x\}\), there is an \(r > 0\) with \(J := (y-r, y+r) \subseteq I \sim \{x\}\). Now \(y \in E^{\prime}\), so \(y\) is a point of closure of \(E \sim \{y\}\), and \(J\) is an open interval containing \(y\); hence \(J\) contains a point \(z \in E \sim \{y\}\). Then \(z \in E\), \(z \in J \subseteq I\), and \(z \neq x\) because \(J\) misses \(x\). So \(z \in (E \sim \{x\}) \cap I\).

In either case \(I\) contains a point of \(E \sim \{x\}\). As \(I\) was an arbitrary open interval containing \(x\), we conclude \(x\) is a point of closure of \(E \sim \{x\}\), that is, \(x \in E^{\prime}\). Therefore \(E^{\prime}\) contains all its points of closure and so is closed.

(ii) \(\overline{E} = E \cup E^{\prime}\).

\((\supseteq)\) We always have \(E \subseteq \overline{E}\). And \(E^{\prime} \subseteq \overline{E}\): if \(x \in E^{\prime}\), then every open interval containing \(x\) contains a point of \(E \sim \{x\}\), and such a point lies in \(E\); hence \(x\) is a point of closure of \(E\), i.e. \(x \in \overline{E}\). Thus \(E \cup E^{\prime} \subseteq \overline{E}\).

\((\subseteq)\) Let \(x \in \overline{E}\). If \(x \in E\) then \(x \in E \cup E^{\prime}\) and we are done. So suppose \(x \notin E\). Then \(E \sim \{x\} = E\). Since \(x \in \overline{E}\), every open interval containing \(x\) contains a point of \(E = E \sim \{x\}\); hence \(x\) is a point of closure of \(E \sim \{x\}\), i.e. \(x \in E^{\prime}\). In either case \(x \in E \cup E^{\prime}\).

Therefore \(\overline{E} = E \cup E^{\prime}\).

Two remarks worth recording. First, part (i) is not a corollary of Proposition 10: that proposition says \(\overline{E}\) is closed, and by part (ii) \(\overline{E} = E \cup E^{\prime}\), which carries no information about \(E^{\prime}\) by itself. Indeed \(E^{\prime}\) can be closed while \(E\) is not, and \(E^{\prime}\) can be a proper subset of the closed set \(\overline{E}\); the direct argument above is what is needed. Second, combining (i) and (ii): \(E\) is closed if and only if \(E^{\prime} \subseteq E\), since \(E\) is closed exactly when \(E = \overline{E} = E \cup E^{\prime}\), and that holds exactly when \(E^{\prime} \subseteq E\).

Assume every point of \(E\) is an isolated point of \(E\). For each \(x \in E\) fix a number \(r_x > 0\) with

\begin{equation*} (x - r_x,\, x + r_x) \cap E = \{x\}. \end{equation*}

By Theorem 2 the rational numbers are dense in \(\mathbb{R}\), so we may choose rational numbers \(p_x\) and \(q_x\) with

\begin{equation*} x - r_x < p_x < x < q_x < x + r_x . \end{equation*}

(Apply Theorem 2 once to the pair \(x - r_x < x\) and once to the pair \(x < x + r_x\).) Define a map

\begin{equation*} \varphi : E \longrightarrow \mathbb{Q} \times \mathbb{Q}, \qquad \varphi(x) = (p_x, q_x). \end{equation*}

\(\varphi\) is one-to-one. Suppose \(x, y \in E\) and \(\varphi(x) = \varphi(y)\), that is \(p_x = p_y\) and \(q_x = q_y\). Then

\begin{equation*} y \in (p_y, q_y) = (p_x, q_x) \subseteq (x - r_x,\, x + r_x), \end{equation*}

and \(y \in E\), so \(y \in (x - r_x, x + r_x) \cap E = \{x\}\). Hence \(y = x\). Thus \(\varphi\) is one-to-one.

\(\mathbb{Q} \times \mathbb{Q}\) is countable. Write

\begin{equation*} \mathbb{Q} \times \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \big( \{q\} \times \mathbb{Q} \big). \end{equation*}

Each set \(\{q\} \times \mathbb{Q}\) is equipotent to \(\mathbb{Q}\), hence countable by Corollary 4, and the index set \(\mathbb{Q}\) is countable by Corollary 4 as well. So this is the union of a countable collection of countable sets, which is countable by Corollary 6.

Conclusion. Since \(\varphi\) is one-to-one, \(E\) is equipotent to the subset \(\varphi(E)\) of the countable set \(\mathbb{Q} \times \mathbb{Q}\). By Theorem 3 a subset of a countable set is countable, so \(\varphi(E)\) is countable, and therefore \(E\) is countable.

A second proof, avoiding \(\mathbb{Q} \times \mathbb{Q}\). For each natural number \(n\) set

\begin{equation*} E_n = \Big\{ x \in E \;:\; \big(x - \tfrac1n,\, x + \tfrac1n\big) \cap E = \{x\} \Big\}. \end{equation*}

Every \(x \in E\) has some \(r_x > 0\) as above, and by the Archimedean Property there is an \(n\) with \(1/n < r_x\); then \((x - 1/n, x + 1/n) \cap E \subseteq (x - r_x, x + r_x) \cap E = \{x\}\) and this intersection contains \(x\), so \(x \in E_n\). Hence \(E = \bigcup_{n=1}^{\infty} E_n\).

Fix \(n\). Any two distinct points of \(E_n\) are at distance at least \(1/n\): if \(x, y \in E_n\) with \(x \neq y\) and \(|x - y| < 1/n\), then \(y \in (x - 1/n, x + 1/n) \cap E = \{x\}\), so \(y = x\), a contradiction. Consequently, for each natural number \(k\) the set \(E_n \cap [-k, k]\) is finite. Indeed, partition \([-k,k]\) into the \(2kn\) half-open intervals \(\big[-k + \tfrac{j-1}{n},\, -k + \tfrac{j}{n}\big)\) for \(j = 1, \ldots, 2kn\), together with the single point \(\{k\}\); two points lying in the same one of these \(2kn + 1\) pieces are at distance strictly less than \(1/n\), so by the pigeonhole principle \(E_n \cap [-k,k]\) has at most \(2kn + 1\) members.

Every real number lies in \([-k,k]\) for some natural number \(k\), by the Archimedean Property applied to the pair \(1\) and \(|x|\) when \(x \neq 0\). Therefore \(E_n = \bigcup_{k=1}^{\infty} \big( E_n \cap [-k,k] \big)\) is a countable union of finite sets, hence countable by Corollary 6, and then \(E = \bigcup_{n=1}^{\infty} E_n\) is a countable union of countable sets, hence countable by Corollary 6 again.

Recall the two definitions in force. A set \(\mathcal{O}\) of real numbers is open provided for each \(x \in \mathcal{O}\) there is an \(r > 0\) for which \((x - r, x + r)\) is contained in \(\mathcal{O}\). A set \(E\) of real numbers is dense in \(\mathbb{R}\) provided between any two real numbers there lies a member of \(E\); equivalently, every nonempty open interval \((a,b)\) with \(a < b\) contains a point of \(E\).

First note the inclusion \(\operatorname{int} E \subseteq E\), valid for every set \(E\): if \(x \in \operatorname{int} E\), choose \(r > 0\) with \((x-r, x+r) \subseteq E\); since \(x \in (x-r,x+r)\), we get \(x \in E\).

(i) Suppose \(E\) is open. Let \(x \in E\). By the definition of openness there is an \(r > 0\) with \((x-r, x+r) \subseteq E\), which is precisely the statement that \(x\) is an interior point of \(E\). Hence \(E \subseteq \operatorname{int} E\), and combined with the inclusion above, \(E = \operatorname{int} E\).

Conversely suppose \(E = \operatorname{int} E\). Let \(x \in E\). Then \(x \in \operatorname{int} E\), so there is an \(r > 0\) with \((x - r, x + r) \subseteq E\). Since \(x \in E\) was arbitrary, \(E\) is open.

(ii) Suppose first that \(E\) is dense in \(\mathbb{R}\), and suppose, to get a contradiction, that \(\operatorname{int}(\mathbb{R} \sim E) \neq \emptyset\). Choose \(x \in \operatorname{int}(\mathbb{R} \sim E)\) and \(r > 0\) with

\begin{equation*} (x - r,\, x + r) \subseteq \mathbb{R} \sim E, \qquad \text{that is} \qquad (x-r,\, x+r) \cap E = \emptyset . \end{equation*}

But \(x - r < x + r\), so by density there is a member of \(E\) lying between \(x - r\) and \(x + r\), i.e. a point of \(E\) in \((x-r, x+r)\). This contradicts the displayed line. Therefore \(\operatorname{int}(\mathbb{R} \sim E) = \emptyset\).

Conversely suppose \(\operatorname{int}(\mathbb{R} \sim E) = \emptyset\). Let \(a < b\) be real numbers; we must find a member of \(E\) strictly between them. Suppose not, so that \((a,b) \cap E = \emptyset\), i.e. \((a,b) \subseteq \mathbb{R} \sim E\). Put

\begin{equation*} x = \frac{a+b}{2}, \qquad r = \frac{b-a}{2} > 0 . \end{equation*}

Then \((x - r, x + r) = (a,b) \subseteq \mathbb{R} \sim E\), so \(x\) is an interior point of \(\mathbb{R} \sim E\) and hence \(\operatorname{int}(\mathbb{R} \sim E) \neq \emptyset\), a contradiction. Therefore \((a,b)\) contains a point of \(E\) for all \(a < b\), i.e. \(E\) is dense in \(\mathbb{R}\).

It is worth recording what falls out of the two parts. The set \(\operatorname{int} E\) is always open: it is exactly the union of all open intervals contained in \(E\), and the union of any collection of open sets is open by Proposition 8. Moreover \(\operatorname{int} E\) is the largest open set contained in \(E\), since any open \(\mathcal{O} \subseteq E\) has every one of its points interior to \(E\), so \(\mathcal{O} = \operatorname{int} \mathcal{O} \subseteq \operatorname{int} E\) by part (i). Dually, part (ii) says \(E\) is dense exactly when \(\mathbb{R} \sim E\) contains no nonempty open set.

The Nested Set Theorem reads: let \(\{F_n\}_{n=1}^{\infty}\) be a descending countable collection of nonempty closed sets of real numbers for which \(F_1\) is bounded; then \(\bigcap_{n=1}^{\infty} F_n \neq \emptyset\). We exhibit a descending countable collection of nonempty closed sets whose first member is unbounded and whose intersection is empty; this shows the boundedness hypothesis on \(F_1\) cannot be dropped.

For each natural number \(n\) define

\begin{equation*} F_n = [n, \infty) = \{ x \in \mathbb{R} \;:\; x \ge n \}. \end{equation*}

Each \(F_n\) is nonempty, since \(n \in F_n\).

Each \(F_n\) is closed. Its complement is \(\mathbb{R} \sim F_n = (-\infty, n)\), which is an open interval and hence an open set. By Proposition 11, a set of real numbers is closed if and only if its complement in \(\mathbb{R}\) is open, so \(F_n\) is closed.

The collection is descending: if \(x \ge n+1\) then \(x \ge n\), so \(F_{n+1} \subseteq F_n\) for every natural number \(n\).

\(F_1 = [1,\infty)\) is unbounded: it is not bounded above, for if \(b\) were an upper bound of \(F_1\) then \(b \ge 1 > 0\) and, by the Archimedean Property of \(\mathbb{R}\) applied to the pair \(1\) and \(b\), there is a natural number \(n\) with \(n \cdot 1 > b\); but \(n \in F_1\), so \(b\) is not an upper bound, a contradiction.

Finally the intersection is empty. Let \(x\) be any real number. If \(x \le 0\) then \(x \notin F_1\). If \(x > 0\), then by the Archimedean Property applied to the pair \(1\) and \(x\) there is a natural number \(n\) with \(n > x\), and then \(x \notin F_n\). In either case \(x \notin \bigcap_{n=1}^{\infty} F_n\). Hence

\begin{equation*} \bigcap_{n=1}^{\infty} [n, \infty) = \emptyset . \end{equation*}

So the conclusion of the Nested Set Theorem fails for this descending collection of nonempty closed sets, the only hypothesis violated being the boundedness of \(F_1\).

The failure is not an artifact of using unbounded sets at every stage: the proof of the Nested Set Theorem invokes the Heine-Borel Theorem for the closed set \(F_1\), and the Heine-Borel Theorem requires the set to be both closed and bounded. Concretely, in the proof one covers \(F_1\) by the ascending open sets \(\mathcal{O}_n = \mathbb{R} \sim F_n\) and extracts a finite subcover; for \(F_1 = [1,\infty)\) the open cover \(\{(-\infty,n)\}_{n=1}^{\infty}\) of \(F_1\) has no finite subcover, since a finite union of these is \((-\infty, N)\) for the largest index \(N\) present, and \(N \in F_1 \sim (-\infty, N)\).

Here is the precise setting. Let \(\mathbb{R}\) denote a set satisfying the Field Axioms and the Positivity Axioms, so that all the algebra of Section 1.1 and the whole vocabulary of Section 1.4 – open interval, open set, closed set, point of closure, closure, cover, open cover, finite subcover, descending collection, bounded set – is available, together with Propositions 8, 10, 11 and 12, whose proofs use only the field and order axioms. Consider the following three assertions about such an \(\mathbb{R}\).

(C) The Completeness Axiom. Every nonempty set of real numbers that is bounded above has a least upper bound.

(HB) The Heine-Borel assertion. If \(F\) is a closed and bounded set of real numbers, then every open cover of \(F\) has a finite subcover.

(NST) The Nested Set assertion. If \(\{F_n\}_{n=1}^{\infty}\) is a descending countable collection of nonempty closed sets of real numbers for which \(F_1\) is bounded, then \(\bigcap_{n=1}^{\infty} F_n \neq \emptyset\).

We prove (C) \(\Rightarrow\) (HB) \(\Rightarrow\) (C) and (C) \(\Rightarrow\) (NST) \(\Rightarrow\) (C). The two forward implications are exactly the two theorems proved in the text: the proof of the Heine-Borel Theorem uses (C) once, to form \(c = \sup E\) for the set \(E\) of numbers \(x \in [a,b]\) such that \([a,x]\) can be covered by a finite number of the sets of the cover, and otherwise uses only the definitions of open set, closed set and cover together with Proposition 11; and the proof of the Nested Set Theorem uses (C) only through its appeal to the Heine-Borel Theorem, its remaining ingredients being Proposition 11 and De Morgan’s Identities. So it remains to prove the two converses.

Part 1. (HB) implies (C).

Assume (HB). Let \(E\) be a nonempty set of real numbers that is bounded above, and suppose, to get a contradiction, that \(E\) has no least upper bound.

First, \(E\) has no largest member: a largest member of \(E\) would be an upper bound of \(E\) that is less than or equal to every other upper bound, hence a least upper bound. Consequently no member of \(E\) is an upper bound of \(E\), for such a member would be a largest member.

Fix \(a \in E\) and fix an upper bound \(b\) of \(E\). Then \(a \le b\), and \(a \neq b\) since \(b\) is an upper bound while \(a\) is not. So \(a < b\), and \([a,b]\) is a nonempty bounded set. It is closed: its complement \((-\infty, a) \cup (b, \infty)\) is a union of open intervals, hence open by Proposition 8, so \([a,b]\) is closed by Proposition 11.

Now build an open cover of \([a,b]\). Let \(x \in [a,b]\).

Case N (\(x\) is not an upper bound of \(E\)). Then there is an \(e \in E\) with \(e > x\). Set \(\mathcal{O}_x = (-\infty, e)\). This is an open interval containing \(x\), and no point of \(\mathcal{O}_x\) is an upper bound of \(E\), since every \(z \in \mathcal{O}_x\) satisfies \(z < e\) with \(e \in E\).

Case U (\(x\) is an upper bound of \(E\)). Since \(E\) has no least upper bound, \(x\) is not the least upper bound, so there is an upper bound \(y\) of \(E\) with \(y < x\). Set \(\mathcal{O}_x = (y, \infty)\). This is an open interval containing \(x\), and every point of \(\mathcal{O}_x\) is an upper bound of \(E\): if \(z > y\) then for each \(e \in E\) we have \(e \le y < z\).

The collection \(\{\mathcal{O}_x\}_{x \in [a,b]}\) is an open cover of \([a,b]\). By (HB) there is a finite subcover \(\mathcal{O}_{x_1}, \ldots, \mathcal{O}_{x_m}\) of \([a,b]\). Call the index \(i\) of type N or of type U according to which case produced \(\mathcal{O}_{x_i}\).

Suppose first that no index is of type U. Then every \(\mathcal{O}_{x_i}\) is of the form \((-\infty, e_i)\) with \(e_i \in E\). Let \(e^{*} = \max\{e_1, \ldots, e_m\}\), which exists because the list is finite and \(\mathbb{R}\) is totally ordered; then \(e^{*} \in E\) and \(\bigcup_{i=1}^{m} \mathcal{O}_{x_i} = (-\infty, e^{*})\). Since this covers \([a,b]\) and \(b \in [a,b]\), we get \(b < e^{*}\). But \(b\) is an upper bound of \(E\) and \(e^{*} \in E\), so \(e^{*} \le b\). This is a contradiction.

So at least one index is of type U. Let \(y_{j_1}, \ldots, y_{j_k}\) be the numbers produced in Case U for those indices, and let \(y^{*} = \min\{y_{j_1}, \ldots, y_{j_k}\}\); each \(y_{j}\) is an upper bound of \(E\), hence so is \(y^{*}\). Define

\begin{equation*} c = \min\{ y^{*},\, b \}. \end{equation*}

Then \(c\) is an upper bound of \(E\): for each \(e \in E\) we have \(e \le y^{*}\) and \(e \le b\), so \(e \le c\). Since \(a \in E\) and \(c\) is an upper bound, \(a \le c\); and \(c \le b\) by construction. Hence \(c \in [a,b]\), so \(c\) lies in some \(\mathcal{O}_{x_i}\) of the finite subcover.

If that index \(i\) is of type N, then \(\mathcal{O}_{x_i} = (-\infty, e_i)\) with \(e_i \in E\), so \(c < e_i\), contradicting the fact that \(c\) is an upper bound of \(E\). Therefore \(i\) is of type U, so \(\mathcal{O}_{x_i} = (y_{j}, \infty)\) for one of the numbers \(y_j\) above, whence \(c > y_{j} \ge y^{*}\). But \(c = \min\{y^{*}, b\} \le y^{*}\). This is a contradiction.

Both cases being impossible, \(E\) must have a least upper bound. Thus (HB) implies (C).

Part 2. (NST) implies (C).

Assume (NST). We argue in two steps.

Step 1: (NST) implies the Archimedean Property, equivalently that \(\mathbb{N}\) is not bounded above in \(\mathbb{R}\). (The two are equivalent without completeness: if \(\mathbb{N}\) is bounded above by \(M\), then for the positive pair \(a = 1\), \(b = \max\{M, 1\}\) we have \(n \cdot 1 \le b\) for every natural number \(n\), so the Archimedean Property fails; conversely if the Archimedean Property fails for the positive pair \(a, b\), then \(n \le b/a\) for every natural number \(n\) and \(\mathbb{N}\) is bounded above.)

Suppose, for a contradiction, that \(\mathbb{N}\) is bounded above by some \(M\); note \(M \ge 1\) since \(1 \in \mathbb{N}\). For each natural number \(n\) put

\begin{equation*} T_n = \{ m \in \mathbb{N} \;:\; m \ge n \}, \qquad F_n = \overline{T_n}. \end{equation*}

Then each \(F_n\) is nonempty (it contains \(n\)) and closed, by Proposition 10. The collection is descending: \(T_{n+1} \subseteq T_n\) implies \(\overline{T_{n+1}} \subseteq \overline{T_n}\), again by Proposition 10. Finally \(F_1\) is bounded: \(T_1 \subseteq [1, M]\), and \([1,M]\) is closed (its complement is open by Proposition 8, so Proposition 11 applies), so by the minimality clause of Proposition 10 we get \(F_1 = \overline{T_1} \subseteq [1,M]\).

By (NST) there is a point \(x \in \bigcap_{n=1}^{\infty} F_n\). Fix \(n\). Since \(x\) is a point of closure of \(T_n\) and \(\big(x - \tfrac12,\, x + \tfrac12\big)\) is an open interval containing \(x\), there is a natural number \(m \ge n\) with \(|x - m| < \tfrac12\).

Apply this with \(n = 1\) to obtain \(m_1 \in \mathbb{N}\) with \(|x - m_1| < \tfrac12\). Apply it again with \(n = m_1 + 1\) to obtain \(m_2 \in \mathbb{N}\) with \(m_2 \ge m_1 + 1\) and \(|x - m_2| < \tfrac12\). Then

\begin{equation*} 1 \le m_2 - m_1 = (m_2 - x) + (x - m_1) \le |x - m_2| + |x - m_1| < \tfrac12 + \tfrac12 = 1, \end{equation*}

which is absurd. Hence \(\mathbb{N}\) is not bounded above, i.e. the Archimedean Property holds.

Step 2: (NST) together with the Archimedean Property implies (C).

Let \(E\) be nonempty and bounded above. If \(E\) has a largest member \(M_0\), then \(M_0\) is an upper bound of \(E\) and every upper bound \(u\) of \(E\) satisfies \(u \ge M_0\) because \(M_0 \in E\); so \(M_0\) is the least upper bound and we are done. So assume \(E\) has no largest member. Then no member of \(E\) is an upper bound of \(E\), since a member of \(E\) that is an upper bound of \(E\) is a largest member of \(E\).

Choose \(a_1 \in E\) and an upper bound \(b_1\) of \(E\). Then \(a_1 \le b_1\), and \(a_1 \neq b_1\) because \(b_1\) is an upper bound of \(E\) while \(a_1\), being a member of \(E\), is not; so \(a_1 < b_1\), the number \(a_1\) is not an upper bound of \(E\), and \(b_1\) is. Define \(a_n, b_n\) recursively by bisection: given \(a_n < b_n\) with \(a_n\) not an upper bound of \(E\) and \(b_n\) an upper bound of \(E\), let \(m_n = (a_n + b_n)/2\) and set

\begin{equation*} \begin{aligned} a_{n+1} = a_n, \quad b_{n+1} = m_n \quad &\text{if } m_n \text{ is an upper bound of } E, \\ a_{n+1} = m_n, \quad b_{n+1} = b_n \quad &\text{if } m_n \text{ is not an upper bound of } E. \end{aligned} \end{equation*}

In either case \(a_{n+1}\) is not an upper bound of \(E\), \(b_{n+1}\) is an upper bound of \(E\), \(a_n \le a_{n+1} < b_{n+1} \le b_n\), and \(b_{n+1} - a_{n+1} = (b_n - a_n)/2\). An induction gives

\begin{equation*} b_n - a_n = \frac{b_1 - a_1}{2^{\,n-1}} \qquad \text{for every natural number } n. \end{equation*}

Put \(F_n = [a_n, b_n]\). Each \(F_n\) is nonempty and closed (complement open by Proposition 8, then Proposition 11), the collection is descending because \(a_n \le a_{n+1} < b_{n+1} \le b_n\), and \(F_1 = [a_1, b_1]\) is bounded. By (NST) there is a point

\begin{equation*} c \in \bigcap_{n=1}^{\infty} F_n, \qquad \text{so} \qquad a_n \le c \le b_n \text{ for every } n. \end{equation*}

We record one consequence of the Archimedean Property: for every \(\epsilon > 0\) there is an \(n\) with \(b_n - a_n < \epsilon\). Indeed, choose a natural number \(n\) with \(n\epsilon > b_1 - a_1\). An induction shows \(2^{\,n-1} \ge n\) for every natural number \(n\) (it holds at \(n = 1\), and if \(2^{\,n-1} \ge n\) then \(2^{\,n} \ge 2n \ge n+1\)). Hence

\begin{equation*} b_n - a_n = \frac{b_1 - a_1}{2^{\,n-1}} \le \frac{b_1 - a_1}{n} < \epsilon . \end{equation*}

\(c\) is an upper bound of \(E\). Suppose not; then there is \(e \in E\) with \(e > c\). Put \(\epsilon = e - c > 0\) and choose \(n\) with \(b_n - a_n < \epsilon\). Since \(a_n \le c\),

\begin{equation*} b_n = a_n + (b_n - a_n) \le c + (b_n - a_n) < c + \epsilon = e, \end{equation*}

so \(b_n < e\) with \(e \in E\), contradicting that \(b_n\) is an upper bound of \(E\).

\(c\) is the least upper bound of \(E\). Let \(u\) be any upper bound of \(E\) and suppose \(u < c\). Put \(\delta = c - u > 0\) and choose \(n\) with \(b_n - a_n < \delta\). Since \(c \le b_n\),

\begin{equation*} a_n = b_n - (b_n - a_n) \ge c - (b_n - a_n) > c - \delta = u . \end{equation*}

So \(a_n > u\). But \(a_n\) is not an upper bound of \(E\), so there is an \(e \in E\) with \(e > a_n > u\), contradicting that \(u\) is an upper bound of \(E\). Hence \(u \ge c\) for every upper bound \(u\).

Therefore \(c\) is the least upper bound of \(E\), and (C) holds.

Conclusion. Assertions (C), (HB) and (NST) are equivalent over the Field and Positivity Axioms. In particular each of the Heine-Borel Theorem and the Nested Set Theorem is a faithful substitute for the Completeness Axiom: one may take either as the completeness postulate for \(\mathbb{R}\) and recover the least upper bound property, and hence all of the analysis built on it. Note also that the equivalence gives (HB) \(\Leftrightarrow\) (NST) directly.

Write \(\mathcal{B}\) for the collection of Borel sets, that is, by definition, the smallest \(\sigma\)-algebra of sets of real numbers that contains all of the open sets of real numbers. Let \(\mathcal{F}\) denote the collection of all closed sets of real numbers, and let \(\mathcal{C}\) denote the smallest \(\sigma\)-algebra of sets of real numbers that contains \(\mathcal{F}\).

Both \(\mathcal{B}\) and \(\mathcal{C}\) exist. Indeed, the collection \(2^{\mathbb{R}}\) of all sets of real numbers is a \(\sigma\)-algebra containing every open set and every closed set, so in each case the family of \(\sigma\)-algebras being intersected is nonempty; Proposition 13 then tells us that the intersection of all \(\sigma\)-algebras of subsets of \(\mathbb{R}\) containing a given collection is itself a \(\sigma\)-algebra containing that collection, and it is the smallest such in the sense that it is contained in every \(\sigma\)-algebra containing the collection.

We must show \(\mathcal{B} = \mathcal{C}\). The whole argument rests on Proposition 11: a set of real numbers is open if and only if its complement in \(\mathbb{R}\) is closed. Equivalently, \(E\) is closed if and only if \(\mathbb{R} \sim E\) is open.

\(\mathcal{C} \subseteq \mathcal{B}\). We check that \(\mathcal{B}\) contains every closed set. Let \(F\) be a closed set of real numbers. By Proposition 11 the set \(\mathbb{R} \sim F\) is open, so \(\mathbb{R} \sim F \in \mathcal{B}\) because \(\mathcal{B}\) contains all the open sets. A \(\sigma\)-algebra is closed with respect to the formation of complements, so

\begin{equation*} F = \mathbb{R} \sim (\mathbb{R} \sim F) \in \mathcal{B}. \end{equation*}

Thus \(\mathcal{B}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) containing \(\mathcal{F}\). Since \(\mathcal{C}\) is the smallest such \(\sigma\)-algebra, \(\mathcal{C} \subseteq \mathcal{B}\).

\(\mathcal{B} \subseteq \mathcal{C}\). We check that \(\mathcal{C}\) contains every open set. Let \(\mathcal{O}\) be an open set of real numbers. By Proposition 11 the set \(\mathbb{R} \sim \mathcal{O}\) is closed, so \(\mathbb{R} \sim \mathcal{O} \in \mathcal{F} \subseteq \mathcal{C}\). Since \(\mathcal{C}\) is closed with respect to the formation of complements,

\begin{equation*} \mathcal{O} = \mathbb{R} \sim (\mathbb{R} \sim \mathcal{O}) \in \mathcal{C}. \end{equation*}

Thus \(\mathcal{C}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) containing all the open sets. Since \(\mathcal{B}\) is by definition the smallest such \(\sigma\)-algebra, \(\mathcal{B} \subseteq \mathcal{C}\).

Combining the two inclusions, \(\mathcal{B} = \mathcal{C}\): the collection of Borel sets is the smallest \(\sigma\)-algebra of sets of real numbers that contains all of the closed sets.

The point is entirely formal, and it is worth isolating it. If \(\mathcal{S}\) is any collection of subsets of a set \(X\) and \(\mathcal{S}^{c} = \{ X \sim S : S \in \mathcal{S} \}\), then a \(\sigma\)-algebra contains \(\mathcal{S}\) if and only if it contains \(\mathcal{S}^{c}\), since \(\sigma\)-algebras are closed with respect to complements. Hence the smallest \(\sigma\)-algebra containing \(\mathcal{S}\) equals the smallest \(\sigma\)-algebra containing \(\mathcal{S}^{c}\). Proposition 11 says exactly that the collection of closed sets is the collection of complements of the open sets, and the result follows.

Let \(\mathcal{I} = \{\, [a,b) \mid a < b \,\}\) and let \(\mathcal{A}\) be the smallest \(\sigma\)-algebra of sets of real numbers that contains \(\mathcal{I}\); this exists by Proposition 13, being the intersection of all \(\sigma\)-algebras containing \(\mathcal{I}\) (there is at least one, namely \(2^{\mathbb{R}}\)). Let \(\mathcal{B}\) denote the collection of Borel sets, that is, the smallest \(\sigma\)-algebra containing all the open sets. We show \(\mathcal{A} = \mathcal{B}\) by two inclusions.

Step 1. We show \(\mathcal{A} \subseteq \mathcal{B}\).

It suffices to show \(\mathcal{I} \subseteq \mathcal{B}\), for then \(\mathcal{B}\) is a \(\sigma\)-algebra containing \(\mathcal{I}\) and the minimality of \(\mathcal{A}\) forces \(\mathcal{A} \subseteq \mathcal{B}\).

Let \(a < b\). I claim

\begin{equation*} [a,b) \;=\; \bigcap_{n=1}^{\infty} \left( a - \tfrac{1}{n},\, b \right). \end{equation*}

Indeed, if \(a \le x < b\) then \(a - 1/n < a \le x < b\) for every \(n\), so \(x\) lies in the intersection. Conversely, if \(x\) belongs to every \((a - 1/n, b)\) then \(x < b\) and \(x > a - 1/n\) for all \(n\); by the Archimedean Property of \(\mathbb{R}\) the latter forces \(x \ge a\) (were \(x < a\), choose \(n\) with \(1/n < a - x\) to get \(x < a - 1/n\)). Hence \(x \in [a,b)\).

Each \((a - 1/n, b)\) is an open set, hence a Borel set, and a \(\sigma\)-algebra is closed with respect to the formation of countable intersections. Therefore \([a,b) \in \mathcal{B}\), which proves the claim.

Step 2. We show \(\mathcal{B} \subseteq \mathcal{A}\).

It suffices to show that every open set belongs to \(\mathcal{A}\), for then \(\mathcal{A}\) is a \(\sigma\)-algebra containing the open sets and the minimality of \(\mathcal{B}\) forces \(\mathcal{B} \subseteq \mathcal{A}\).

First we check that every open interval belongs to \(\mathcal{A}\).

• A bounded open interval \((a,b)\) with \(a<b\): choose, by the Archimedean Property, a natural number \(N\) with \(1/N < b - a\). Then

\begin{equation*} (a,b) \;=\; \bigcup_{n = N}^{\infty} \left[ a + \tfrac{1}{n},\, b \right). \end{equation*}

Each set on the right belongs to \(\mathcal{I}\) because \(a + 1/n \le a + 1/N < b\), and it is contained in \((a,b)\); conversely if \(a < x < b\), choose \(n \ge N\) with \(1/n < x - a\), so that \(x \in [a + 1/n, b)\). Since \(\mathcal{A}\) is closed with respect to countable unions, \((a,b) \in \mathcal{A}\).

• The interval \((a, \infty)\): it equals \(\bigcup_{n=1}^{\infty} (a, a+n)\), a countable union of bounded open intervals, hence lies in \(\mathcal{A}\).

• The interval \((-\infty, b)\): it equals \(\bigcup_{n=1}^{\infty} (b-n, b)\), hence lies in \(\mathcal{A}\).

• All of \(\mathbb{R}\): it equals \(\bigcup_{n=1}^{\infty} (-n, n)\), hence lies in \(\mathcal{A}\).

Now let \(\mathcal{O}\) be an open set of real numbers. If \(\mathcal{O} = \emptyset\) then \(\mathcal{O} \in \mathcal{A}\) by the very definition of a \(\sigma\)-algebra. Otherwise, by Proposition 9, \(\mathcal{O}\) is the disjoint union of a countable collection \(\{ I_k \}\) of open intervals. Each \(I_k\) belongs to \(\mathcal{A}\) by the preceding paragraph, and \(\mathcal{A}\) is closed with respect to the formation of countable unions, so \(\mathcal{O} \in \mathcal{A}\).

Thus every open set belongs to \(\mathcal{A}\), and Step 2 is proved.

Combining the two steps, \(\mathcal{A} = \mathcal{B}\): the collection of Borel sets is the smallest \(\sigma\)-algebra containing the intervals \([a,b)\) with \(a<b\).

Recall that a countable union of closed sets is called an \(F_\sigma\) set.

Let \(\mathcal{O}\) be an open set of real numbers and let \(C = \mathbb{R} \sim \mathcal{O}\), which is closed by Proposition 11. For each natural number \(n\) define

\begin{equation*} F_n \;=\; \left\{\, x \in \mathbb{R} \;\middle|\; |x| \le n \ \text{ and } \ |x - c| \ge \tfrac{1}{n} \ \text{ for every } c \in C \,\right\}. \end{equation*}

(If \(C = \emptyset\) the second condition is vacuous and \(F_n = [-n,n]\).)

Each \(F_n\) is closed. Indeed, let \(x\) be a point of closure of \(F_n\) and let \(\epsilon > 0\). The open interval \((x - \epsilon, x + \epsilon)\) contains a point \(x^{\prime} \in F_n\). From \(|x^{\prime}| \le n\) we get \(|x| \le |x^{\prime}| + \epsilon \le n + \epsilon\), and since \(\epsilon > 0\) is arbitrary, \(|x| \le n\). Moreover, for any \(c \in C\) we have \(|x^{\prime} - c| \ge 1/n\), so

\begin{equation*} |x - c| \;\ge\; |x^{\prime} - c| - |x - x^{\prime}| \;>\; \tfrac{1}{n} - \epsilon . \end{equation*}

Again since \(\epsilon > 0\) is arbitrary, \(|x - c| \ge 1/n\) for every \(c \in C\). Hence \(x \in F_n\), and \(F_n\) contains all of its points of closure, that is, \(F_n\) is closed.

We now claim

\begin{equation*} \mathcal{O} \;=\; \bigcup_{n=1}^{\infty} F_n . \end{equation*}

First, \(F_n \subseteq \mathcal{O}\) for each \(n\). For if \(x \in F_n\) and we had \(x \in C\), then taking \(c = x\) in the defining condition would give \(0 = |x - x| \ge 1/n\), which is false. So \(x \notin C\), that is, \(x \in \mathcal{O}\).

Conversely, let \(x \in \mathcal{O}\). Since \(\mathcal{O}\) is open there is an \(r > 0\) with \((x - r, x + r) \subseteq \mathcal{O}\), and therefore no point \(c\) of \(C\) satisfies \(|x - c| < r\); that is, \(|x - c| \ge r\) for all \(c \in C\). By the Archimedean Property of \(\mathbb{R}\) choose a natural number \(n\) with \(n \ge |x|\) and \(1/n \le r\). Then \(|x| \le n\) and \(|x - c| \ge r \ge 1/n\) for every \(c \in C\), so \(x \in F_n\).

This proves the claim, and exhibits \(\mathcal{O}\) as a countable union of closed sets. Hence every open set is an \(F_\sigma\) set.

Two remarks. First, taking complements and using De Morgan’s Identities together with Proposition 11, the result is equivalent to the dual statement that every closed set is a \(G_\delta\) set. Second, an alternative proof runs through Proposition 9: a nonempty open set is the union of a countable collection of open intervals, and each open interval is an \(F_\sigma\), since for \(a < b\)

\begin{equation*} (a,b) = \bigcup_{n \ge N} \left[ a + \tfrac{1}{n},\, b - \tfrac{1}{n} \right] \quad \text{where } \tfrac{2}{N} < b - a, \end{equation*}

while \((a,\infty) = \bigcup_{n} [a + 1/n,\, a + n]\), \((-\infty,b) = \bigcup_{n} [b - n,\, b - 1/n]\), and \(\mathbb{R} = \bigcup_n [-n,n]\); a countable union of \(F_\sigma\) sets is again an \(F_\sigma\) set.

Throughout, write

\begin{equation*} s_n = \sup\{ a_k \mid k \ge n \}, \qquad i_n = \inf\{ a_k \mid k \ge n \}, \end{equation*}

these being extended real numbers. The sequence \(\{s_n\}\) is decreasing and \(\{i_n\}\) is increasing, since the sets over which the supremum and infimum are taken shrink as \(n\) increases. By definition \(L := \limsup\{a_n\} = \lim_{n \to \infty} s_n\) and \(\ell := \liminf\{a_n\} = \lim_{n \to \infty} i_n\), the limits being taken in the extended sense; because \(\{s_n\}\) is decreasing, \(L = \inf_n s_n\), and because \(\{i_n\}\) is increasing, \(\ell = \sup_n i_n\). This last identification is exactly the Lemma established in the solution to Problem 39 below (a decreasing sequence of extended real numbers converges in the extended sense to its infimum, an increasing one to its supremum); for a bounded monotone sequence of real numbers it is Theorem 15 together with its proof, and the unbounded and infinite cases are checked there directly. Nothing in Problem 39 uses the present problem, so there is no circularity.

We prove the assertion about \(\limsup\) in full; the assertion about \(\liminf\) then follows by a reflection argument given at the end.

Part 1. \(L\) is a cluster point of \(\{a_n\}\).

Case (a): \(L = \infty\). Then \(\{a_n\}\) is not bounded above, since a bound \(M\) would give \(s_n \le M\) for all \(n\) and hence \(L \le M\). Deleting finitely many terms from a sequence that is unbounded above leaves a sequence that is still unbounded above, so for every index \(N\) and every real \(c\) there is an index \(k \ge N\) with \(a_k > c\). Choose \(n_1\) with \(a_{n_1} > 1\); having chosen \(n_1 < \dots < n_{k-1}\), choose \(n_k > n_{k-1}\) with \(a_{n_k} > k\). Then \(\{n_k\}\) is strictly increasing and \(\{a_{n_k}\} \to \infty\), so \(L = \infty\) is a cluster point.

Case (b): \(L = -\infty\). Then \(\{s_n\} \to -\infty\), so for each natural number \(k\) there is an index \(N_k\) with \(s_{N_k} < -k\). Choose \(n_1 \ge N_1\); having chosen \(n_{k-1}\), let \(n_k\) be any index with \(n_k > n_{k-1}\) and \(n_k \ge N_k\). Since \(\{s_n\}\) is decreasing and \(n_k \ge N_k\),

\begin{equation*} a_{n_k} \;\le\; s_{n_k} \;\le\; s_{N_k} \;<\; -k . \end{equation*}

Hence \(\{a_{n_k}\} \to -\infty\), and \(L = -\infty\) is a cluster point.

Case (c): \(L \in \mathbb{R}\). We claim that for every \(\epsilon > 0\) and every index \(N\) there is an index \(k \ge N\) with \(|a_k - L| < \epsilon\). Indeed, since \(\{s_n\}\) decreases to \(L\), choose \(n \ge N\) with \(s_n < L + \epsilon\); also \(s_n \ge L\), because \(\{s_n\}\) is decreasing with limit \(L\). From \(s_n \ge L > L - \epsilon\) and the definition of supremum there is an index \(k \ge n\) with \(a_k > s_n - \epsilon \ge L - \epsilon\). On the other hand \(a_k \le s_n < L + \epsilon\) since \(k \ge n\). As \(k \ge n \ge N\), the claim is proved.

Now build a subsequence: apply the claim with \(\epsilon = 1\) and \(N = 1\) to get \(n_1\) with \(|a_{n_1} - L| < 1\); having chosen \(n_{k-1}\), apply the claim with \(\epsilon = 1/k\) and \(N = n_{k-1} + 1\) to get \(n_k > n_{k-1}\) with \(|a_{n_k} - L| < 1/k\). By the Archimedean Property, \(\{a_{n_k}\} \to L\). So \(L\) is a cluster point.

Part 2. Every cluster point \(c\) of \(\{a_n\}\) satisfies \(c \le L\).

If \(L = \infty\) there is nothing to prove, so assume \(L < \infty\) and let \(t\) be any real number with \(t > L\). Since \(L = \inf_n s_n < t\), there is an index \(N\) with \(s_N < t\), and therefore

\begin{equation*} a_n \le s_N < t \quad \text{for all } n \ge N . \end{equation*}

Let \(c\) be a cluster point, say \(\{a_{n_k}\} \to c\) where \(\{n_k\}\) is strictly increasing. A strictly increasing sequence of natural numbers satisfies \(n_k \ge k\) for every \(k\), by induction: \(n_1 \ge 1\), and \(n_k > n_{k-1} \ge k-1\) forces \(n_k \ge k\). Hence \(n_k \ge k \ge N\) for \(k \ge N\), and so \(a_{n_k} < t\) for all \(k \ge N\). If \(c = -\infty\) then certainly \(c \le t\). If \(c\) were \(+\infty\) then \(a_{n_k}\) would eventually exceed \(t\), a contradiction. If \(c \in \mathbb{R}\), then \(c \le t\) by Proposition 14 applied to the convergent sequence \(\{a_{n_k}\}_{k \ge N}\), all of whose terms are at most \(t\). In every case \(c \le t\).

This holds for every real \(t > L\). If \(L \in \mathbb{R}\), letting \(t\) decrease to \(L\) gives \(c \le L\). If \(L = -\infty\), then \(c \le t\) for every real number \(t\), which forces \(c = -\infty = L\).

By Parts 1 and 2, the set of cluster points of \(\{a_n\}\) is nonempty, contains \(L\), and has \(L\) as an upper bound; hence \(L = \limsup\{a_n\}\) is its largest element.

Part 3. The statement for \(\liminf\).

Consider the sequence \(\{-a_n\}\). For a strictly increasing sequence \(\{n_k\}\) of indices and an extended real number \(c\), we have \(\{a_{n_k}\} \to c\) if and only if \(\{-a_{n_k}\} \to -c\): for finite \(c\) this is Theorem 18 with \(\alpha = -1\), \(\beta = 0\) (or directly, since \(|(-a_{n_k}) - (-c)| = |a_{n_k} - c|\)), and for \(c = \pm\infty\) it is immediate from the definition of convergence to infinity. Consequently the set of cluster points of \(\{-a_n\}\) is exactly the set of negatives of the cluster points of \(\{a_n\}\).

By Part 1 and Part 2 applied to \(\{-a_n\}\), the largest cluster point of \(\{-a_n\}\) is \(\limsup\{-a_n\}\). By Proposition 19(iii) (with \(\{-a_n\}\) in place of \(\{a_n\}\)),

\begin{equation*} \limsup\{-a_n\} \;=\; -\liminf\{a_n\}. \end{equation*}

Since negation reverses order among extended real numbers and carries cluster points of \(\{a_n\}\) onto cluster points of \(\{-a_n\}\), the number \(-\limsup\{-a_n\} = \liminf\{a_n\}\) is the smallest cluster point of \(\{a_n\}\). This completes the proof.

Throughout put

\begin{equation*} s_n = \sup\{ a_k \mid k \ge n\}, \qquad i_n = \inf\{a_k \mid k \ge n\}, \end{equation*}

so that \(\limsup\{a_n\} = \lim_{n \to \infty} s_n\) and \(\liminf\{a_n\} = \lim_{n \to \infty} i_n\). The sets \(\{a_k \mid k \ge n\}\) decrease as \(n\) increases, so \(\{s_n\}\) is decreasing and \(\{i_n\}\) is increasing, and clearly \(i_n \le a_n \le s_n\) for every \(n\).

Lemma. A decreasing sequence \(\{s_n\}\) of extended real numbers converges in the extended sense to \(\inf_n s_n\), and an increasing sequence \(\{i_n\}\) converges to \(\sup_n i_n\).

Proof of the Lemma. Consider the decreasing case. If \(s_n = \infty\) for all \(n\), both sides equal \(\infty\). Otherwise fix \(N\) with \(s_N < \infty\); then \(\{s_n\}_{n \ge N}\) is a decreasing sequence of extended reals none of which is \(\infty\), and \(\inf_n s_n = \inf_{n \ge N} s_n\). If some \(s_m = -\infty\) then \(s_n = -\infty\) for all \(n \ge m\) and both sides are \(-\infty\). So assume \(s_n \in \mathbb{R}\) for \(n \ge N\). If this real sequence is bounded below it converges to its infimum by Theorem 15 (as in the proof there, the limit of a decreasing bounded sequence is its infimum); if it is not bounded below, then for each real \(c\) there is \(M \ge N\) with \(s_M < c\), hence \(s_n \le s_M < c\) for \(n \ge M\), so \(\{s_n\} \to -\infty = \inf_n s_n\). In each case the conclusion was reached for the tail \(\{s_n\}_{n \ge N}\), and altering or ignoring finitely many initial terms affects neither convergence nor the value of the limit, since every assertion of convergence quantifies only over indices beyond some threshold. The increasing case follows by the same argument, or by applying the decreasing case to \(\{-i_n\}\). \(\square\)

So \(\limsup\{a_n\} = \inf_n s_n\) and \(\liminf\{a_n\} = \sup_n i_n\); we use this freely.

Proof of (i).

Suppose first \(\limsup\{a_n\} = \ell \in \mathbb{R}\), and let \(\epsilon > 0\). Since \(\ell = \inf_n s_n < \ell + \epsilon\), there is an index \(N\) with \(s_N < \ell + \epsilon\), and therefore \(a_n \le s_N < \ell + \epsilon\) for every \(n \ge N\). Thus at most the indices \(n < N\) can satisfy \(a_n > \ell + \epsilon\): only finitely many. On the other hand, for every index \(N\) we have \(s_N \ge \inf_n s_n = \ell > \ell - \epsilon\), and since \(s_N\) is the supremum of \(\{a_k \mid k \ge N\}\) there is some \(k \ge N\) with \(a_k > \ell - \epsilon\). As \(N\) was arbitrary, the set of indices \(n\) with \(a_n > \ell - \epsilon\) is not contained in any finite initial segment, hence is infinite.

Conversely, suppose that for each \(\epsilon > 0\) there are infinitely many indices \(n\) with \(a_n > \ell - \epsilon\) and only finitely many with \(a_n > \ell + \epsilon\). Taking \(\epsilon = 1\), only finitely many terms exceed \(\ell + 1\), so \(\{a_n\}\) is bounded above and each \(s_n\) is a real number. Let \(\epsilon > 0\). Let \(N\) be an index strictly greater than each of the finitely many indices \(n\) with \(a_n > \ell + \epsilon\). Then \(a_k \le \ell + \epsilon\) for all \(k \ge N\), so \(s_n \le s_N \le \ell + \epsilon\) for \(n \ge N\). Also, for every \(n\) there are infinitely many indices \(k\) with \(a_k > \ell - \epsilon\), hence some such \(k\) with \(k \ge n\), so \(s_n \ge a_k > \ell - \epsilon\). Therefore

\begin{equation*} \ell - \epsilon \;<\; s_n \;\le\; \ell + \epsilon \qquad \text{for all } n \ge N, \end{equation*}

that is, \(|s_n - \ell| \le \epsilon\) for \(n \ge N\). Given \(\epsilon^{\prime} > 0\), running this argument with \(\epsilon = \epsilon^{\prime}/2\) produces an \(N\) with \(|s_n - \ell| \le \epsilon^{\prime}/2 < \epsilon^{\prime}\) for all \(n \ge N\). Hence \(\{s_n\} \to \ell\), that is, \(\limsup\{a_n\} = \ell\).

Proof of (ii).

If \(\{a_n\}\) is not bounded above, then for each \(n\) the set \(\{a_k \mid k \ge n\}\) is also not bounded above (it differs from \(\{a_k \mid k \ge 1\}\) by the finitely many terms \(a_1, \dots, a_{n-1}\), and adjoining finitely many real numbers to a set cannot turn an unbounded-above set into a bounded one). Hence \(s_n = \infty\) for every \(n\), and \(\limsup\{a_n\} = \inf_n s_n = \infty\).

Conversely, if \(\{a_n\}\) is bounded above by a real number \(M\), then \(M\) is an upper bound for each set \(\{a_k \mid k \ge n\}\), so \(s_n \le M\) for all \(n\) and therefore \(\limsup\{a_n\} = \inf_n s_n \le M < \infty\).

Proof of (iii).

For a nonempty set \(S\) of real numbers, \(\sup S = -\inf(-S)\), where \(-S = \{-x \mid x \in S\}\), with the convention that this holds in the extended sense (if \(S\) is unbounded above then \(-S\) is unbounded below, and both sides are \(\infty\)). Applying this to \(S = \{a_k \mid k \ge n\}\), whose negative is \(\{-a_k \mid k \ge n\}\), we get

\begin{equation*} s_n \;=\; -\, i_n^{\,-}, \qquad \text{where } i_n^{\,-} = \inf\{-a_k \mid k \ge n\}. \end{equation*}

Now \(\{i_n^{\,-}\}\) is increasing, so by the Lemma \(\liminf\{-a_n\} = \sup_n i_n^{\,-}\), while \(\limsup\{a_n\} = \inf_n s_n = \inf_n (- i_n^{\,-}) = -\sup_n i_n^{\,-}\), using that negation reverses order and interchanges suprema and infima of a set of extended real numbers. Hence \(\limsup\{a_n\} = -\liminf\{-a_n\}\).

Proof of (iv).

Suppose first that \(\liminf\{a_n\} = \limsup\{a_n\} = a\).

If \(a \in \mathbb{R}\): let \(\epsilon > 0\) and choose \(N\) so large that \(|s_n - a| < \epsilon\) and \(|i_n - a| < \epsilon\) for \(n \ge N\). Then for \(n \ge N\),

\begin{equation*} a - \epsilon \;<\; i_n \;\le\; a_n \;\le\; s_n \;<\; a + \epsilon, \end{equation*}

so \(|a_n - a| < \epsilon\). Thus \(\{a_n\} \to a\).

If \(a = \infty\): then \(\{i_n\} \to \infty\), so for each real \(c\) there is \(N\) with \(i_n > c\) for \(n \ge N\); since \(a_n \ge i_n\), also \(a_n > c\) for \(n \ge N\). Thus \(\{a_n\} \to \infty\).

If \(a = -\infty\): then \(\{s_n\} \to -\infty\), and \(a_n \le s_n\), so \(\{a_n\} \to -\infty\) by the same reasoning.

Conversely, suppose \(\{a_n\} \to a\).

If \(a \in \mathbb{R}\): let \(\epsilon > 0\) and choose \(N\) with \(a - \epsilon < a_k < a + \epsilon\) for all \(k \ge N\). Then for \(n \ge N\) the number \(a + \epsilon\) is an upper bound and \(a - \epsilon\) a lower bound for \(\{a_k \mid k \ge n\}\), so \(a - \epsilon \le i_n \le s_n \le a + \epsilon\). Since \(\{s_n\}\) and \(\{i_n\}\) are monotone with these bounds holding from \(N\) on, their limits also lie in \([a - \epsilon, a + \epsilon]\). As \(\epsilon > 0\) was arbitrary, \(\liminf\{a_n\} = \limsup\{a_n\} = a\).

If \(a = \infty\): given a real number \(c\), choose \(N\) with \(a_n > c\) for \(n \ge N\). Then \(c\) is a lower bound for \(\{a_k \mid k \ge n\}\) whenever \(n \ge N\), so \(i_n \ge c\) for \(n \ge N\). Hence \(\{i_n\} \to \infty\), and since \(s_n \ge i_n\) also \(\{s_n\} \to \infty\). Thus both equal \(\infty = a\).

If \(a = -\infty\): symmetrically, for each real \(c\) there is \(N\) with \(a_n < c\) for \(n \ge N\), so \(s_n \le c\) for \(n \ge N\), giving \(\{s_n\} \to -\infty\), and \(i_n \le s_n\) gives \(\{i_n\} \to -\infty\). Thus both equal \(-\infty = a\).

Proof of (v).

As printed, part (v) asserts that \(a_n \le b_n\) for all \(n\) implies \(\limsup\{a_n\} \le \liminf\{b_n\}\). This is a misprint: the assertion is false. Take \(a_n = b_n = (-1)^n\). Then \(a_n \le b_n\) for all \(n\), while \(s_n = 1\) for all \(n\) and \(i_n = -1\) for all \(n\), so \(\limsup\{a_n\} = 1\) and \(\liminf\{b_n\} = -1\), and \(1 \le -1\) fails.

The correct (and intended) statement is that the limit superior and the limit inferior are each monotone:

\begin{equation*} \text{if } a_n \le b_n \text{ for all } n, \text{ then } \limsup\{a_n\} \le \limsup\{b_n\} \text{ and } \liminf\{a_n\} \le \liminf\{b_n\}. \end{equation*}

We prove this. Write \(s_n^a, i_n^a\) and \(s_n^b, i_n^b\) for the associated suprema and infima. Fix \(n\). For every \(k \ge n\) we have \(a_k \le b_k \le s_n^b\), so \(s_n^b\) is an upper bound for \(\{a_k \mid k \ge n\}\) and therefore \(s_n^a \le s_n^b\). Taking infima over \(n\) and using the Lemma,

\begin{equation*} \limsup\{a_n\} = \inf_n s_n^a \;\le\; \inf_n s_n^b = \limsup\{b_n\}. \end{equation*}

Similarly, for every \(k \ge n\) we have \(i_n^a \le a_k \le b_k\), so \(i_n^a\) is a lower bound for \(\{b_k \mid k \ge n\}\) and therefore \(i_n^a \le i_n^b\). Taking suprema over \(n\),

\begin{equation*} \liminf\{a_n\} = \sup_n i_n^a \;\le\; \sup_n i_n^b = \liminf\{b_n\}. \end{equation*}

(One may also note that the printed inequality does hold under the stronger hypothesis that \(\{b_n\}\) converges, since then \(\limsup\{b_n\} = \liminf\{b_n\}\) by part (iv).)

Recall from Problem 38 that an extended real number is a cluster point of \(\{a_n\}\) provided some subsequence of \(\{a_n\}\) converges to it.

Suppose first that \(\{a_n\}\) converges to the extended real number \(a\).

We check that every subsequence \(\{a_{n_k}\}\) also converges to \(a\). Recall that a strictly increasing sequence \(\{n_k\}\) of natural numbers satisfies \(n_k \ge k\) for all \(k\) (an easy induction: \(n_1 \ge 1\), and \(n_k > n_{k-1} \ge k-1\) forces \(n_k \ge k\)). If \(a \in \mathbb{R}\) and \(\epsilon > 0\), choose \(N\) with \(|a_n - a| < \epsilon\) for \(n \ge N\); then for \(k \ge N\) we have \(n_k \ge k \ge N\), so \(|a_{n_k} - a| < \epsilon\). If \(a = \infty\) and \(c\) is real, choose \(N\) with \(a_n > c\) for \(n \ge N\); then \(a_{n_k} > c\) for \(k \ge N\). The case \(a = -\infty\) is symmetric. Hence every subsequence converges to \(a\).

In particular \(a\) is a cluster point of \(\{a_n\}\), since \(\{a_n\}\) is a subsequence of itself. And \(a\) is the only one: if \(c\) is a cluster point, then \(\{a_{n_k}\} \to c\) for some subsequence, while we have just shown \(\{a_{n_k}\} \to a\); so it suffices to know that a sequence cannot converge to two distinct extended real numbers. If both limits are real this is the uniqueness of the limit in Proposition 14. A sequence converging to a real number is bounded (Proposition 14), whereas a sequence converging to \(\infty\) or to \(-\infty\) is respectively unbounded above or unbounded below, so a real limit cannot coexist with an infinite one. Finally, \(\{a_{n_k}\} \to \infty\) gives \(a_{n_k} > 0\) for all large \(k\) while \(\{a_{n_k}\} \to -\infty\) gives \(a_{n_k} < 0\) for all large \(k\), so these cannot both hold. Thus \(c = a\), and \(a\) is the unique cluster point.

Conversely, suppose \(\{a_n\}\) has exactly one cluster point, say \(c\). By Problem 38, \(\liminf\{a_n\}\) is the smallest cluster point of \(\{a_n\}\) and \(\limsup\{a_n\}\) is the largest; in particular both are cluster points. Since there is only one cluster point,

\begin{equation*} \liminf\{a_n\} \;=\; c \;=\; \limsup\{a_n\}. \end{equation*}

By Proposition 19(iv), \(\{a_n\}\) converges to the extended real number \(c\).

As above put \(i_n = \inf\{a_k \mid k \ge n\}\) and \(s_n = \sup\{a_k \mid k \ge n\}\), so that \(\{i_n\}\) is increasing, \(\{s_n\}\) is decreasing, and (by the Lemma proved in Problem 39, which is just Theorem 15 extended to the extended real numbers)

\begin{equation*} \liminf a_n = \sup_n i_n, \qquad \limsup a_n = \inf_n s_n . \end{equation*}

Fix indices \(m\) and \(n\) and put \(p = \max\{m, n\}\). Since \(p \ge m\) and \(\{i_n\}\) is increasing, \(i_m \le i_p\). Since \(a_p\) belongs to the nonempty set \(\{a_k \mid k \ge p\}\), we have \(i_p \le a_p \le s_p\). Since \(p \ge n\) and \(\{s_n\}\) is decreasing, \(s_p \le s_n\). Chaining these,

\begin{equation*} i_m \;\le\; i_p \;\le\; a_p \;\le\; s_p \;\le\; s_n . \end{equation*}

Thus \(s_n\) is an upper bound for \(\{ i_m \mid m \in \mathbb{N}\}\), whence \(\sup_m i_m \le s_n\). As this holds for every \(n\), the extended real number \(\sup_m i_m\) is a lower bound for \(\{ s_n \mid n \in \mathbb{N}\}\), whence \(\sup_m i_m \le \inf_n s_n\). That is,

\begin{equation*} \liminf a_n \;\le\; \limsup a_n . \end{equation*}

An alternative argument: by Problem 38 the set of cluster points of \(\{a_n\}\) is nonempty, \(\liminf a_n\) is its smallest element and \(\limsup a_n\) is its largest element, and the smallest element of a nonempty set is at most its largest element.

Write \(A = \limsup a_n\) and \(B = \limsup b_n\), and as before put

\begin{equation*} s_n^a = \sup\{a_k \mid k \ge n\}, \qquad s_n^b = \sup\{b_k \mid k \ge n\}, \qquad s_n^{ab} = \sup\{a_k b_k \mid k \ge n\}, \end{equation*}

so that, by the Lemma of Problem 39, \(A = \inf_n s_n^a\), \(B = \inf_n s_n^b\) and \(\limsup[a_n b_n] = \inf_n s_n^{ab}\).

First note that \(A\) and \(B\) are nonnegative. Indeed, for every \(n\) we have \(s_n^a \ge a_n > 0\), so every \(s_n^a\) is positive and hence \(A = \inf_n s_n^a \ge 0\); likewise \(s_n^b \ge b_n \ge 0\), so \(B = \inf_n s_n^b \ge 0\). Thus \(A, B \in [0, \infty]\). Also \(a_n b_n \ge 0\) for all \(n\), so by the same reasoning \(\limsup[a_n b_n] \ge 0\); in particular the left-hand side is never \(-\infty\), so no indeterminacy arises there.

Since \(A\) and \(B\) are nonnegative, the hypothesis that the right-hand product is not of the form \(0 \cdot \infty\) says exactly that neither of the two combinations \(A = 0\) with \(B = \infty\), nor \(A = \infty\) with \(B = 0\), occurs.

Case 1: \(A = \infty\) or \(B = \infty\).

If \(A = \infty\) then by hypothesis \(B \ne 0\), so \(B > 0\) and \(A \cdot B = \infty\). If \(B = \infty\) then by hypothesis \(A \ne 0\), so \(A > 0\) and \(A \cdot B = \infty\). In either case the right-hand side is \(\infty\) and the asserted inequality holds trivially.

Case 2: \(A\) and \(B\) are both finite.

Then \(A, B \in [0, \infty)\) and \(A \cdot B\) is a nonnegative real number. Let \(\epsilon > 0\).

Since \(A = \inf_n s_n^a < A + \epsilon\), there is an index \(N_1\) with \(s_{N_1}^a < A + \epsilon\), and hence

\begin{equation*} 0 \;<\; a_k \;\le\; s_{N_1}^a \;<\; A + \epsilon \qquad \text{for every } k \ge N_1 . \end{equation*}

Similarly there is an index \(N_2\) with

\begin{equation*} 0 \;\le\; b_k \;\le\; s_{N_2}^b \;<\; B + \epsilon \qquad \text{for every } k \ge N_2 . \end{equation*}

Let \(N = \max\{N_1, N_2\}\). Note \(A + \epsilon > 0\) and \(B + \epsilon > 0\). For \(k \ge N\) we have \(b_k < B + \epsilon\) and \(a_k > 0\), so \(a_k b_k < a_k (B + \epsilon)\); and \(a_k < A + \epsilon\) with \(B + \epsilon > 0\) gives \(a_k (B + \epsilon) < (A + \epsilon)(B + \epsilon)\). Hence

\begin{equation*} a_k b_k \;<\; (A + \epsilon)(B + \epsilon) \qquad \text{for every } k \ge N . \end{equation*}

Thus \((A+\epsilon)(B+\epsilon)\) is an upper bound for \(\{a_k b_k \mid k \ge N\}\), so \(s_N^{ab} \le (A+\epsilon)(B+\epsilon)\), and therefore

\begin{equation*} \limsup [a_n b_n] \;=\; \inf_n s_n^{ab} \;\le\; s_N^{ab} \;\le\; (A + \epsilon)(B + \epsilon) \;=\; AB + \epsilon(A + B) + \epsilon^2 . \end{equation*}

This holds for every \(\epsilon > 0\). Since \(A + B\) and \(AB\) are real numbers, letting \(\epsilon\) run through the values \(1/m\), \(m \in \mathbb{N}\), we obtain a sequence of real bounds converging to \(AB\); hence \(\limsup[a_n b_n] \le AB\) by Proposition 14 (or directly: if \(\limsup[a_n b_n] > AB\), choose \(\epsilon > 0\) small enough that \(AB + \epsilon(A+B) + \epsilon^2 < \limsup[a_n b_n]\), a contradiction).

In both cases

\begin{equation*} \limsup\, [a_n \cdot b_n] \;\le\; (\limsup a_n) \cdot (\limsup b_n), \end{equation*}

as required.

Two remarks. The inequality can be strict: with \(a_n = 2 + (-1)^n\) and \(b_n = 2 - (-1)^n\) we have \(a_n > 0\), \(b_n \ge 0\), \(a_n b_n = 4 - 1 = 3\) for every \(n\), so the left side is \(3\) while \(\limsup a_n = \limsup b_n = 3\) and the right side is \(9\). Also, the proviso is genuinely needed, and not merely because \(0 \cdot \infty\) is undefined: take \(a_n = n\) and \(b_n = 1/n\), so that \(a_n > 0\) and \(b_n \ge 0\) for every \(n\). Then \(a_n b_n = 1\) for all \(n\), so \(\limsup [a_n b_n] = 1\), while \(\limsup a_n = \infty\) and \(\limsup b_n = 0\). Under any convention that assigns the value \(0\) to \(0 \cdot \infty\) the asserted inequality would read \(1 \le 0\), which is false. So the excluded case is one in which the conclusion genuinely fails, not merely one in which the right-hand side is unassigned.

Let \(\{a_n\}\) be a sequence of real numbers. Call an index \(n\) a peak index provided

\begin{equation*} a_m \le a_n \quad \text{for every } m \ge n . \end{equation*}

Let \(P\) be the set of peak indices. There are two cases.

Case 1: \(P\) is infinite. List the members of \(P\) in increasing order as \(n_1 < n_2 < n_3 < \cdots\); this is possible because every nonempty set of natural numbers has a smallest member (Theorem 1), so we may choose \(n_1 = \min P\) and, inductively, \(n_{k+1} = \min\{n \in P \mid n > n_k\}\), which is a nonempty set since \(P\) is infinite. Fix \(k\). Since \(n_k\) is a peak index and \(n_{k+1} > n_k\), we have \(a_{n_{k+1}} \le a_{n_k}\). Hence \(\{a_{n_k}\}\) is a decreasing subsequence of \(\{a_n\}\).

Case 2: \(P\) is finite. Choose a natural number \(N\) that exceeds every member of \(P\) (if \(P = \emptyset\) take \(N = 1\)). Define \(n_1 = N\). Suppose \(n_1 < n_2 < \cdots < n_k\) have been chosen with \(n_k \ge N\). Since \(n_k \ge N\), the index \(n_k\) is not a peak index, so there is an index \(m > n_k\) for which \(a_m > a_{n_k}\); define \(n_{k+1} = m\). Then \(\{n_k\}\) is a strictly increasing sequence of natural numbers and \(a_{n_k} < a_{n_{k+1}}\) for every \(k\), so \(\{a_{n_k}\}\) is an increasing subsequence of \(\{a_n\}\).

In either case \(\{a_n\}\) has a monotone subsequence.

Proof of the Bolzano-Weierstrass Theorem. Let \(\{a_n\}\) be a bounded sequence of real numbers, say \(|a_n| \le M\) for all \(n\). By what we just proved, \(\{a_n\}\) has a monotone subsequence \(\{a_{n_k}\}\). Every term of the subsequence is a term of the original sequence, so \(|a_{n_k}| \le M\) for all \(k\); that is, \(\{a_{n_k}\}\) is a bounded monotone sequence. By the Monotone Convergence Criterion for Real Sequences (Theorem 15), \(\{a_{n_k}\}\) converges. Thus \(\{a_n\}\) has a convergent subsequence.

Observe that this argument is not circular: the peak construction uses nothing but the order properties of \(\mathbb{R}\) and Theorem 1, while Theorem 15 is deduced directly from the Completeness Axiom without appeal to the Bolzano-Weierstrass Theorem.

Throughout, for a real number \(y \ge 0\) let \(\lfloor y \rfloor\) denote the largest integer not exceeding \(y\). This exists: by the Archimedean Property the set of natural numbers \(k\) with \(k > y\) is nonempty, so by Theorem 1 it has a least member \(k_0\), and \(\lfloor y \rfloor = k_0 - 1\). It satisfies \(\lfloor y \rfloor \le y < \lfloor y \rfloor + 1\).

Convergence of any such series. Let \(\{a_n\}\) be any sequence of integers with \(0 \le a_n < p\), that is, \(0 \le a_n \le p-1\). The terms \(a_n/p^n\) are nonnegative, so the partial sums \(s_N = \sum_{n=1}^N a_n/p^n\) form an increasing sequence, and

\begin{equation*} s_N \le \sum_{n=1}^{N} \frac{p-1}{p^n} = (p-1) \cdot \frac{1/p - 1/p^{N+1}}{1 - 1/p} = 1 - \frac{1}{p^N} < 1 . \end{equation*}

Thus the sequence of partial sums is bounded, and by Proposition 20(iii) (Problem 45) the series is summable. Let \(x\) be its sum. Since \(s_N \le 1\) for every \(N\), Proposition 14 gives \(x \le 1\); since \(-s_N \le 0\) for every \(N\), the same proposition applied to \(\{-s_N\}\), whose limit is \(-x\) by Theorem 18, gives \(-x \le 0\), that is \(x \ge 0\). Hence \(0 \le x \le 1\).

Existence of an expansion. Let \(0 < x < 1\). Define \(x_0 = x\) and, inductively,

\begin{equation*} a_n = \lfloor p\, x_{n-1} \rfloor, \qquad x_n = p\, x_{n-1} - a_n \qquad (n \ge 1). \end{equation*}

We verify by induction that \(0 \le x_n < 1\) for every \(n \ge 0\). This holds for \(n = 0\). If \(0 \le x_{n-1} < 1\), then \(0 \le p\,x_{n-1} < p\), so \(a_n = \lfloor p\,x_{n-1}\rfloor\) is an integer with \(0 \le a_n \le p-1\), and \(x_n = p\,x_{n-1} - a_n\) satisfies \(0 \le x_n < 1\) by the defining property of the floor. So each \(a_n\) is an integer with \(0 \le a_n < p\).

From \(x_{n-1} = (a_n + x_n)/p\) an induction gives, for every \(N \ge 1\),

\begin{equation*} x = \sum_{n=1}^{N} \frac{a_n}{p^n} + \frac{x_N}{p^N}. \end{equation*}

Since \(0 \le x_N < 1\) we get \(0 \le x - s_N < 1/p^N\). Now \(p^N \ge 2^N \ge N\) for every \(N\), both inequalities by induction on \(N\) (using \(p \ge 2\) for the first, and \(2^{N+1} = 2^N + 2^N \ge N + 1\) for the second). Given \(\epsilon > 0\), the Archimedean Property provides a natural number \(N_0\) with \(1/N_0 < \epsilon\), and then \(0 \le x - s_N < 1/p^N \le 1/N \le 1/N_0 < \epsilon\) for all \(N \ge N_0\). Hence \(s_N \to x\), that is, \(x = \sum_{n=1}^{\infty} a_n/p^n\).

The structure of two distinct expansions. Suppose \(\{a_n\}\) and \(\{b_n\}\) are sequences of integers in \([0, p)\) with

\begin{equation*} \sum_{n=1}^{\infty} \frac{a_n}{p^n} = \sum_{n=1}^{\infty} \frac{b_n}{p^n} = x , \end{equation*}

and suppose they are not identical. Let \(N\) be the smallest index with \(a_N \ne b_N\) (Theorem 1); relabelling if necessary, assume \(a_N > b_N\), so \(a_N - b_N \ge 1\). Subtracting the two series (legitimate by Theorem 18, applied to the partial sums) and using \(a_n = b_n\) for \(n < N\),

\begin{equation*} \frac{a_N - b_N}{p^N} = \sum_{n = N+1}^{\infty} \frac{b_n - a_n}{p^n} . \end{equation*}

Now \(b_n - a_n \le p-1\) for every \(n\), so the right-hand side is at most \(\sum_{n>N}(p-1)/p^n = 1/p^N\), while the left-hand side is at least \(1/p^N\). Hence both sides equal \(1/p^N\), which forces

\begin{equation*} a_N - b_N = 1, \qquad b_n - a_n = p - 1 \ \text{ for all } n > N , \end{equation*}

the latter because a series of nonnegative terms \(\bigl((p-1) - (b_n - a_n)\bigr)/p^n\) with sum zero must have every term zero. Since \(0 \le a_n, b_n \le p-1\), this says

\begin{equation*} b_n = p-1 \quad\text{and}\quad a_n = 0 \qquad \text{for all } n > N . \end{equation*}

In particular, of any two distinct expansions of the same \(x\), one terminates (is eventually \(0\)) and the other is eventually \(p-1\).

Uniqueness except for \(p\)-adic fractions. Suppose \(x \in (0,1)\) has two distinct expansions \(\{a_n\}, \{b_n\}\) as above. Then \(x = \sum_{n=1}^{N} a_n/p^n = q/p^N\) where \(q = \sum_{n=1}^N a_n p^{N-n}\) is an integer. So \(x\) has the form \(q/p^N\).

Conversely, let \(x = q/p^N\) with \(q\) an integer and \(0 < x < 1\); then \(1 \le q \le p^N - 1\). We claim that every integer \(q\) with \(0 \le q \le p^N - 1\) can be written as \(q = \sum_{n=1}^{N} c_n p^{N-n}\) with each \(c_n\) an integer, \(0 \le c_n \le p-1\). Argue by induction on \(N\). For \(N = 1\) take \(c_1 = q\). Assume the claim for \(N\) and let \(0 \le q \le p^{N+1} - 1\). Put \(c_1 = \lfloor q/p^N \rfloor\); then \(0 \le c_1 \le p - 1\), since \(q/p^N < p\), and \(q^{\prime} = q - c_1 p^N\) satisfies \(0 \le q^{\prime} \le p^N - 1\) by the defining property of the floor. Applying the inductive hypothesis to \(q^{\prime}\) and shifting indices gives the representation of \(q\) with \(N+1\) digits. This proves the claim, and dividing by \(p^N\) gives \(x = \sum_{n=1}^{N} c_n/p^n\). Since \(x > 0\), not all \(c_n\) vanish; let \(M \le N\) be the largest index with \(c_M \ne 0\). Then

\begin{equation*} x = \sum_{n=1}^{M} \frac{c_n}{p^n}, \end{equation*}

and the sequence \(a_n = c_n\) for \(n \le M\), \(a_n = 0\) for \(n > M\), is one expansion. Define a second by

\begin{equation*} b_n = c_n \ (n < M), \qquad b_M = c_M - 1 \ (\ge 0), \qquad b_n = p-1 \ (n > M). \end{equation*}

These are integers in \([0,p)\), and since \(\sum_{n>M}(p-1)/p^n = 1/p^M\),

\begin{equation*} \sum_{n=1}^{\infty} \frac{b_n}{p^n} = \sum_{n<M} \frac{c_n}{p^n} + \frac{c_M - 1}{p^M} + \frac{1}{p^M} = x . \end{equation*}

The two sequences differ at index \(M\), so \(x\) has at least two expansions.

There are never more than two. Suppose \(\{a_n\}, \{b_n\}, \{c_n\}\) were three pairwise distinct expansions of the same \(x\). By the structural result above, each pair consists of one sequence that is eventually \(0\) and one that is eventually \(p-1\); note that since \(p \ge 2\) we have \(p - 1 \ne 0\), so no single sequence can be both. Say \(\{a_n\}\) is eventually \(0\) and \(\{b_n\}\) eventually \(p-1\). Applying the result to the pair \(\{a_n\}, \{c_n\}\) shows \(\{c_n\}\) is eventually \(p-1\) (it cannot be the eventually-\(0\) one, since then \(\{a_n\}\) would have to be eventually \(p-1\)). But then \(\{b_n\}\) and \(\{c_n\}\) are both eventually \(p-1\), contradicting the structural result for that pair. Hence there are at most two expansions.

Combining: every \(x \in (0,1)\) has an expansion; it is unique unless \(x = q/p^N\) for integers \(q, N\), in which case there are exactly two.

Write \(s_n = \sum_{k=1}^{n} a_k\) for the \(n\)-th partial sum and set \(s_0 = 0\), so that for all \(j < i\),

\begin{equation*} s_i - s_j = \sum_{k=j+1}^{i} a_k . \end{equation*}

By definition the series is summable exactly when the sequence \(\{s_n\}\) converges.

(i). Suppose first that the series is summable, so \(\{s_n\}\) converges. By the Cauchy Convergence Criterion (Theorem 17), \(\{s_n\}\) is Cauchy. Let \(\epsilon > 0\) and choose an index \(N^{\prime}\) such that \(|s_i - s_j| < \epsilon\) whenever \(i, j \ge N^{\prime}\). Put \(N = N^{\prime} + 1\). If \(n \ge N\) and \(m\) is a natural number, then \(n - 1 \ge N^{\prime}\) and \(n + m \ge N^{\prime}\), so

\begin{equation*} \left| \sum_{k=n}^{n+m} a_k \right| = |s_{n+m} - s_{n-1}| < \epsilon . \end{equation*}

Conversely, suppose the stated condition holds. Let \(\epsilon > 0\) and choose \(N\) so that \(\bigl|\sum_{k=n}^{n+m} a_k\bigr| < \epsilon/2\) for all \(n \ge N\) and all natural numbers \(m\). We first observe that the condition controls individual terms: for \(n \ge N\),

\begin{equation*} a_{n+2} = \sum_{k=n}^{n+2} a_k - \sum_{k=n}^{n+1} a_k , \end{equation*}

and both sums on the right are of the permitted form (with \(m = 2\) and \(m = 1\)), so \(|a_{n+2}| < \epsilon/2 + \epsilon/2 = \epsilon\). Hence \(|a_k| < \epsilon\) for all \(k \ge N+2\).

Now set \(N^{\prime\prime} = N + 1\) and let \(i > j \ge N^{\prime\prime}\). If \(i \ge j+2\), then

\begin{equation*} |s_i - s_j| = \left|\sum_{k=j+1}^{i} a_k\right| = \left|\sum_{k=n}^{n+m} a_k\right| \quad\text{with } n = j+1 \ge N,\ m = i - j - 1 \ge 1, \end{equation*}

so \(|s_i - s_j| < \epsilon/2 < \epsilon\). If \(i = j+1\), then \(|s_i - s_j| = |a_{j+1}|\) and \(j + 1 \ge N + 2\), so \(|s_i - s_j| < \epsilon\). Thus \(|s_i - s_j| < \epsilon\) whenever \(i, j \ge N^{\prime\prime}\) (the case \(i = j\) being trivial), so \(\{s_n\}\) is Cauchy and, by Theorem 17, converges. The series is summable.

(ii). Suppose \(\sum_{k=1}^{\infty} |a_k|\) is summable. Let \(\epsilon > 0\). Applying part (i) to the series \(\sum |a_k|\), there is an index \(N\) such that for \(n \ge N\) and every natural number \(m\),

\begin{equation*} \sum_{k=n}^{n+m} |a_k| = \left| \sum_{k=n}^{n+m} |a_k| \right| < \epsilon . \end{equation*}

By the triangle inequality (Problem 7),

\begin{equation*} \left| \sum_{k=n}^{n+m} a_k \right| \le \sum_{k=n}^{n+m} |a_k| < \epsilon \end{equation*}

for all \(n \ge N\) and all natural numbers \(m\). By part (i), applied in the other direction to \(\sum a_k\), the series \(\sum_{k=1}^{\infty} a_k\) is summable.

(iii). Suppose each \(a_k \ge 0\). Then \(s_{n+1} - s_n = a_{n+1} \ge 0\), so \(\{s_n\}\) is an increasing sequence. By the Monotone Convergence Criterion (Theorem 15), a monotone sequence converges if and only if it is bounded. Hence the series is summable, that is \(\{s_n\}\) converges, if and only if \(\{s_n\}\) is bounded. (For an increasing sequence, boundedness is the same as boundedness above, since \(s_n \ge s_1\) for all \(n\).)

Throughout, we work with an ordered field \(F\) satisfying the field axioms and the positivity axioms for \(\mathbb{R}\), and we ask which of the following three assertions imply which:

• (C) Completeness Axiom. Every nonempty subset of \(F\) that is bounded above has a least upper bound in \(F\).
• (BW) Bolzano-Weierstrass. Every bounded sequence in \(F\) has a convergent subsequence.
• (MC) Monotone Convergence. Every monotone bounded sequence in \(F\) converges.

(The natural numbers \(\mathbb{N}\) are defined inside \(F\) as the intersection of all inductive subsets, exactly as in Section 1.2; this definition and the Principle of Mathematical Induction use no completeness. Note also that the proofs of Proposition 14 and Theorem 18 — uniqueness of limits, boundedness of convergent sequences, and linearity and monotonicity of convergence — use only the definition of convergence and the ordered-field axioms, so they are available to us here.)

(C) implies (BW). This is exactly the Bolzano-Weierstrass Theorem (Theorem 16), proved in the text from the Completeness Axiom (via the Nested Set Theorem).

(C) implies (MC). This is exactly the Monotone Convergence Criterion (Theorem 15), proved in the text from the Completeness Axiom.

It remains to prove the two converses. Both use the following two lemmas.

Lemma A. Each of (BW) and (MC) implies the Archimedean Property, that is, \(\mathbb{N}\) is not bounded above in \(F\). Suppose \(\mathbb{N}\) were bounded above. Consider the sequence \(a_n = n\). It is bounded (below by \(1\), above by the assumed bound) and increasing.

Assume (BW). Then \(\{a_n\}\) has a convergent subsequence \(\{n_k\}\), where \(n_1 < n_2 < \cdots\). A convergent sequence is Cauchy (the first half of the proof of Theorem 17 uses only the definition of convergence), so there is an index \(K\) with \(|n_{k+1} - n_k| < 1\) for \(k \ge K\). But \(n_{k+1} > n_k\) are natural numbers, so \(n_{k+1} - n_k\) is itself a natural number (a routine induction from Problem 9) and hence \(n_{k+1} - n_k \ge 1\), a contradiction.

Assume (MC). Then \(\{a_n\}\), being increasing and bounded, converges to some \(a \in F\). Choosing \(N\) with \(|a_n - a| < 1/2\) for \(n \ge N\) gives \(|a_{N+1} - a_N| < 1\), that is \(1 < 1\), a contradiction.

Hence \(\mathbb{N}\) is unbounded above. Consequently, for each \(\epsilon > 0\) there is a natural number \(n\) with \(1/n < \epsilon\), and since \(2^n \ge n\) for all \(n\) (induction), also \(1/2^n \to 0\).

Lemma B (bisection). Let \(E \subseteq F\) be nonempty and bounded above by \(b_0\). Choose \(e \in E\) and put \(a_0 = e - 1\); then \(a_0\) is not an upper bound for \(E\), and \(a_0 < b_0\). Define sequences \(\{a_n\}\), \(\{b_n\}\) inductively: given \(a_n\) not an upper bound for \(E\) and \(b_n\) an upper bound for \(E\) with \(a_n < b_n\), let \(c = (a_n + b_n)/2\) and set

\begin{equation*} (a_{n+1}, b_{n+1}) = \begin{cases} (a_n, c) & \text{if } c \text{ is an upper bound for } E,\\ (c, b_n) & \text{if } c \text{ is not an upper bound for } E.\end{cases} \end{equation*}

By induction, for every \(n\): \(a_n\) is not an upper bound for \(E\), \(b_n\) is an upper bound for \(E\), \(\{a_n\}\) is increasing, \(\{b_n\}\) is decreasing, \(a_n < b_n\), and

\begin{equation*} b_n - a_n = \frac{b_0 - a_0}{2^n}. \end{equation*}

In particular \(a_0 \le a_n < b_n \le b_0\), so both sequences are bounded, and by Lemma A, \(b_n - a_n \to 0\).

(BW) implies (C). Assume (BW), and let \(E\) be nonempty and bounded above. Form \(\{a_n\}, \{b_n\}\) as in Lemma B. By (BW) the bounded sequence \(\{a_n\}\) has a subsequence \(\{a_{n_k}\}\) converging to some \(L \in F\). Since \(\{b_n - a_n\}\) converges to \(0\), so does its subsequence \(\{b_{n_k} - a_{n_k}\}\) (immediate from the definition of convergence, since \(n_k \ge k\)). As \(b_{n_k} = a_{n_k} + (b_{n_k} - a_{n_k})\), Theorem 18 gives \(b_{n_k} \to L\) as well.

\(L\) is an upper bound for \(E\): if \(x \in E\), then \(x \le b_{n_k}\) for every \(k\), since each \(b_n\) is an upper bound. Comparing the constant sequence with constant value \(x\), which converges to \(x\), against \(\{b_{n_k}\}\), which converges to \(L\), the monotonicity of convergence (Theorem 18, formula (7)) gives \(x \le L\).

\(L\) is the least upper bound: suppose \(c\) is an upper bound for \(E\) with \(c < L\). Since \(a_{n_k} \to L\), there is \(k\) with \(a_{n_k} > c\). But \(a_{n_k}\) is not an upper bound for \(E\), so there is \(y \in E\) with \(y > a_{n_k} > c\), contradicting that \(c\) is an upper bound. Hence no upper bound is smaller than \(L\), so \(L = \sup E\) exists in \(F\). This is (C).

(MC) implies (C). Assume (MC), and let \(E\) be nonempty and bounded above. Form \(\{a_n\}, \{b_n\}\) as in Lemma B. The sequence \(\{a_n\}\) is increasing and bounded, so by (MC) it converges to some \(L \in F\); as above \(b_n = a_n + (b_n - a_n) \to L\) by Lemma A and Theorem 18. The very same two paragraphs as in the preceding proof (with \(a_n, b_n\) in place of \(a_{n_k}, b_{n_k}\)) show that \(L\) is an upper bound for \(E\) and that no smaller number is an upper bound. Hence \(\sup E\) exists, which is (C).

Therefore (C), (BW) and (MC) are mutually equivalent: the assertion of the Bolzano-Weierstrass Theorem is equivalent to the Completeness Axiom, and so is the assertion of the Monotone Convergence Theorem.

If \(E = \emptyset\) take \(g \equiv 0\); if \(E = \mathbb{R}\) take \(g = f\). So assume \(E\) is a nonempty proper closed subset of \(\mathbb{R}\).

Construction of \(g\). Since \(E\) is closed, \(\mathcal{O} = \mathbb{R} \sim E\) is open and nonempty (Proposition 11). By Proposition 9, \(\mathcal{O}\) is the disjoint union of a countable collection \(\mathcal{I}\) of open intervals. Each member of \(\mathcal{I}\) is of one of the forms \((a,b)\), \((-\infty, b)\), \((a, \infty)\) with \(a,b\) real; none is all of \(\mathbb{R}\), since \(E \ne \emptyset\). Call the members of \(\mathcal{I}\) the components of \(\mathcal{O}\).

We first record that a finite endpoint of a component belongs to \(E\). Let \(I \in \mathcal{I}\) have right endpoint \(b \in \mathbb{R}\), and suppose \(b \in \mathcal{O}\). Then \(b\) lies in some \(J = (c,d) \in \mathcal{I}\), and \(J \ne I\) because \(b \notin I\). Write \(I = (a,b)\), where \(a\) is real or \(a = -\infty\). Put \(z_0 = c\) if \(a = -\infty\), and \(z_0 = \max\{a, c\}\) if \(a\) is real; in either case \(z_0\) is real and \(z_0 < b\), so we may choose \(z\) with \(z_0 < z < b\). Then \(a < z < b\), so \(z \in I\), and \(c < z < b < d\), so \(z \in J\), contradicting the disjointness of \(\mathcal{I}\). Hence \(b \in E\); the argument for a finite left endpoint is symmetric. (This is also exactly what is verified in the proof of Proposition 9.)

Define \(g : \mathbb{R} \to \mathbb{R}\) by:

• \(g(x) = f(x)\) for \(x \in E\);
• on a bounded component \((a,b)\) (so \(a, b \in E\)), let \(g\) be the affine function determined by \(g(a) = f(a)\), \(g(b) = f(b)\), that is,

\begin{equation*} g(x) = f(a) + \frac{f(b) - f(a)}{b-a}\,(x - a) = \frac{b-x}{b-a} f(a) + \frac{x-a}{b-a} f(b); \end{equation*}

• on a component \((a, \infty)\) (so \(a \in E\)), let \(g(x) = f(a)\);
• on a component \((-\infty, b)\) (so \(b \in E\)), let \(g(x) = f(b)\).

This defines \(g\) unambiguously on all of \(\mathbb{R}\), and \(g = f\) on \(E\). Note that on a bounded component \((a,b)\) the value \(g(x)\) is a convex combination of \(f(a)\) and \(f(b)\), since \(\frac{b-x}{b-a}\) and \(\frac{x-a}{b-a}\) are nonnegative and sum to \(1\). Consequently, if \(|f(a) - c| < \epsilon\) and \(|f(b) - c| < \epsilon\) for some real \(c\), then for \(x \in (a,b)\),

\begin{equation*} |g(x) - c| = \left| \frac{b-x}{b-a}\bigl(f(a)-c\bigr) + \frac{x-a}{b-a}\bigl(f(b)-c\bigr) \right| \le \frac{b-x}{b-a}\,\epsilon + \frac{x-a}{b-a}\,\epsilon = \epsilon. \tag{\(\ast\)} \end{equation*}

Continuity at points of \(\mathcal{O}\). Let \(x_0 \in \mathcal{O}\) and let \(I\) be its component. Since \(I\) is open, \(x_0\) is an interior point of \(I\), and on \(I\) the function \(g\) agrees with an affine function (a constant, in the unbounded cases), which is continuous. So \(g\) is continuous at \(x_0\).

Continuity at points of \(E\). Let \(x_0 \in E\) and let \(\epsilon > 0\). By continuity of \(f\) at \(x_0\) relative to \(E\), there is a \(\delta > 0\) such that

\begin{equation*} |f(y) - f(x_0)| < \epsilon \quad \text{whenever } y \in E \text{ and } |y - x_0| < \delta. \tag{\(\ast\ast\)} \end{equation*}

We produce \(\delta_+ > 0\) such that \(|g(x) - g(x_0)| \le \epsilon\) for \(x_0 \le x < x_0 + \delta_+\); the argument on the left is symmetric, and taking the minimum of the two numbers finishes the proof.

Case 1: there is \(\eta > 0\) with \((x_0, x_0 + \eta) \cap E = \emptyset\). Then \((x_0, x_0+\eta) \subseteq \mathcal{O}\). Let \(I\) be the component of \(\mathcal{O}\) that contains the point \(x_0 + \eta/2\), and let \(a\) be its left endpoint. If \(a < x_0\) (this includes \(a = -\infty\)), then \(x_0\) would lie strictly between \(a\) and \(x_0 + \eta/2\), hence in \(I \subseteq \mathcal{O}\), contradicting \(x_0 \in E\). So \(a\) is real and \(a \ge x_0\). If \(a > x_0\), then \(a\) is a finite endpoint of a component, so \(a \in E\) by the fact recorded above, while \(x_0 < a < x_0 + \eta/2 < x_0 + \eta\); this contradicts \((x_0, x_0+\eta) \cap E = \emptyset\). Hence \(a = x_0\), so \(I = (x_0, b)\) or \(I = (x_0, \infty)\). On \(I\), \(g\) agrees with an affine function \(\ell\) (in the unbounded case, the constant \(f(x_0)\)), and in the bounded case \(\ell(x_0) = f(a) = f(x_0)\) by construction, while in the unbounded case \(\ell \equiv f(x_0)\). Since \(\ell\) is continuous at \(x_0\) and \(\ell(x_0) = f(x_0) = g(x_0)\), there is \(\delta_+ > 0\) with \(|\ell(x) - g(x_0)| < \epsilon\) for \(|x - x_0| < \delta_+\); shrinking \(\delta_+\) so that \((x_0, x_0+\delta_+) \subseteq I\), we get \(|g(x) - g(x_0)| < \epsilon\) for \(x_0 \le x < x_0 + \delta_+\).

Case 2: every interval \((x_0, x_0 + \eta)\) meets \(E\). Choose \(x_1 \in E\) with \(x_0 < x_1 < x_0 + \delta\), and set \(\delta_+ = x_1 - x_0 > 0\). Let \(x_0 \le x < x_0 + \delta_+\), so \(x_0 \le x < x_1 < x_0 + \delta\). If \(x \in E\), then \(|x - x_0| < \delta\) and (\(\ast\ast\)) gives \(|g(x) - g(x_0)| = |f(x) - f(x_0)| < \epsilon\). If \(x \notin E\), let \(I\) be the component of \(\mathcal{O}\) containing \(x\). Since \(x_0 \in E\) and \(x_0 \le x\), and since \(x_1 \in E\) with \(x < x_1\), the interval \(I\) contains neither \(x_0\) nor \(x_1\), so \(I\) is bounded, \(I = (a,b)\) with

\begin{equation*} x_0 \le a < x < b \le x_1 . \end{equation*}

Both \(a\) and \(b\) lie in \(E\) and satisfy \(|a - x_0| < \delta\), \(|b - x_0| < \delta\), so by (\(\ast\ast\)) both \(|f(a) - f(x_0)| < \epsilon\) and \(|f(b) - f(x_0)| < \epsilon\). By (\(\ast\)) with \(c = f(x_0) = g(x_0)\) we conclude \(|g(x) - g(x_0)| \le \epsilon\).

Thus in both cases \(|g(x) - g(x_0)| \le \epsilon\) for all \(x\) with \(x_0 \le x < x_0 + \delta_+\). The mirror-image argument on the left of \(x_0\) produces \(\delta_- > 0\) with \(|g(x) - g(x_0)| \le \epsilon\) for \(x_0 - \delta_- < x \le x_0\). With \(\delta_0 = \min\{\delta_+, \delta_-\}\) we have \(|g(x) - g(x_0)| \le \epsilon\) whenever \(|x - x_0| < \delta_0\). Since \(\epsilon > 0\) was arbitrary (apply the above to \(\epsilon/2\) to obtain a strict inequality), \(g\) is continuous at \(x_0\).

Hence \(g\) is continuous on all of \(\mathbb{R}\) and extends \(f\).

Answer: \(f\) is continuous exactly at the irrational numbers and at \(x = 0\); it is discontinuous at every nonzero rational.

Here “in lowest terms” means \(x = p/q\) with \(p\) an integer, \(q\) a natural number, and \(p\) and \(q\) having no common factor; this representation is unique, and \(0 = 0/1\), so \(f(0) = 0 \cdot \sin 1 = 0\).

We use three standard facts about the sine function from elementary calculus: \(|\sin t| \le |t|\) for all real \(t\); \(\sin t < t\) for \(t > 0\); and \(\sin t / t \to 1\) as \(t \to 0\). Throughout, we use Proposition 21: \(f\) is continuous at \(x_*\) if and only if \(f(x_n) \to f(x_*)\) whenever \(\{x_n\} \to x_*\).

A useful rewriting. If \(x = p/q\) in lowest terms, then \(p = qx\), so

\begin{equation*} f(x) = q x \sin\frac{1}{q} = x \cdot \frac{\sin(1/q)}{1/q}. \tag{1} \end{equation*}

Lemma. If \(\{x_n\}\) is a sequence of rationals converging to \(x_*\), and \(x_n = p_n/q_n\) in lowest terms, then either the denominators \(q_n\) tend to infinity, or some subsequence of \(\{x_n\}\) is constant with value a rational number equal to \(x_*\). Indeed, suppose \(\{q_n\}\) does not tend to infinity. Then there is a natural number \(Q\) and infinitely many indices \(n\) with \(q_n \le Q\). A convergent sequence is bounded (Proposition 14), say \(|x_n| \le M\) for all \(n\); then for those indices \(|p_n| = q_n |x_n| \le QM\). So along those indices the pairs \((p_n, q_n)\) take only finitely many values, hence \(x_n\) takes only finitely many values; consequently some value is attained infinitely often, giving a constant subsequence. A constant subsequence of a sequence converging to \(x_*\) has value \(x_*\), so \(x_*\) is that rational number.

Continuity at every irrational \(x_*\). Let \(\{x_n\} \to x_*\); we must show \(f(x_n) \to f(x_*) = x_*\). Consider

\begin{equation*} |f(x_n) - x_n| = \begin{cases} 0 & \text{if } x_n \text{ irrational},\\[2pt] |x_n| \left| \dfrac{\sin(1/q_n)}{1/q_n} - 1 \right| & \text{if } x_n = p_n/q_n \text{ in lowest terms,} \end{cases} \end{equation*}

using (1). Let \(\epsilon > 0\). Since \(\{x_n\}\) is bounded, say \(|x_n| \le M\) with \(M > 0\), and since \(\sin t / t \to 1\) as \(t \to 0\), there is \(\eta > 0\) such that \(|\sin t / t - 1| < \epsilon/M\) whenever \(0 < t < \eta\). Now consider the rational terms of \(\{x_n\}\). By the Lemma, since \(x_*\) is irrational no subsequence of rationals can have bounded denominators (a constant subsequence would force \(x_*\) to be rational); more precisely, if there were a natural number \(Q\) and infinitely many rational terms with \(q_n \le Q\), the Lemma applied to that subsequence would make \(x_*\) rational. Hence \(q_n \to \infty\) through the rational terms: there is an index \(N_1\) such that every rational term \(x_n\) with \(n \ge N_1\) has \(1/q_n < \eta\). For such \(n\),

\begin{equation*} |f(x_n) - x_n| = |x_n| \left| \frac{\sin(1/q_n)}{1/q_n} - 1\right| \le M \cdot \frac{\epsilon}{M} = \epsilon . \end{equation*}

For irrational terms \(|f(x_n) - x_n| = 0 \le \epsilon\). Therefore \(|f(x_n) - x_n| \le \epsilon\) for all \(n \ge N_1\). Since \(\epsilon > 0\) was arbitrary, \(f(x_n) - x_n \to 0\), and since \(x_n \to x_*\), Theorem 18 gives \(f(x_n) \to x_* = f(x_*)\). By Proposition 21, \(f\) is continuous at \(x_*\).

Continuity at \(0\). For every real \(x\) we claim \(|f(x)| \le |x|\). If \(x\) is irrational this is an equality. If \(x = p/q\) in lowest terms, then, using \(|\sin t| \le |t|\),

\begin{equation*} |f(x)| = |p| \left| \sin \frac{1}{q}\right| \le |p| \cdot \frac{1}{q} = |x| . \end{equation*}

Hence if \(\{x_n\} \to 0\), then \(|f(x_n)| \le |x_n| \to 0\), so \(f(x_n) \to 0 = f(0)\). By Proposition 21, \(f\) is continuous at \(0\).

Discontinuity at every nonzero rational. Let \(x_* = p/q\) in lowest terms with \(x_* \ne 0\), so \(p \ne 0\) and \(q \ge 1\). Then \(0 < 1/q \le 1\), so \(\sin(1/q) < 1/q\) and hence \(\dfrac{\sin(1/q)}{1/q} < 1\). By (1),

\begin{equation*} f(x_*) = x_* \cdot \frac{\sin(1/q)}{1/q} \ne x_*, \end{equation*}

since \(x_* \ne 0\) and the factor is different from \(1\). Next, the irrationals are dense in \(\mathbb{R}\): given \(a < b\), Theorem 2 provides a rational \(r\) with \(a - \sqrt{2} < r < b - \sqrt{2}\), and then \(r + \sqrt{2} \in (a,b)\) is irrational, for otherwise \(\sqrt{2} = (r + \sqrt{2}) - r\) would be rational, which Section 1.2 shows it is not. So for each natural number \(n\) choose an irrational \(x_n\) with \(|x_n - x_*| < 1/n\); by the Archimedean Property \(\{x_n\} \to x_*\), while \(f(x_n) = x_n \to x_* \ne f(x_*)\). By Proposition 21, \(f\) is not continuous at \(x_*\).

Thus the set of points of continuity of \(f\) is \(\{0\} \cup \{x \in \mathbb{R} \mid x \text{ irrational}\}\).

We use Proposition 21 throughout: a real-valued function \(\varphi\) defined on a set \(E\) of real numbers is continuous at \(x_* \in E\) if and only if \(\varphi(x_n) \to \varphi(x_*)\) whenever \(\{x_n\}\) is a sequence in \(E\) converging to \(x_*\).

(i). Fix \(x_* \in E\) and let \(\{x_n\}\) be a sequence in \(E\) with \(\{x_n\} \to x_*\). Since \(f\) and \(g\) are continuous at \(x_*\), Proposition 21 gives \(f(x_n) \to f(x_*)\) and \(g(x_n) \to g(x_*)\).

Sum: by Theorem 18 (linearity of convergence, with \(\alpha = \beta = 1\)),

\begin{equation*} (f+g)(x_n) = f(x_n) + g(x_n) \to f(x_*) + g(x_*) = (f+g)(x_*). \end{equation*}

By Proposition 21, \(f + g\) is continuous at \(x_*\), and since \(x_*\) was arbitrary, on \(E\).

Product: we first check that the product of convergent sequences converges to the product of the limits. Let \(a_n \to a\) and \(b_n \to b\). By Proposition 14 the sequence \(\{a_n\}\) is bounded; choose \(M > 0\) with \(|a_n| \le M\) for all \(n\), and enlarge \(M\) if necessary so that \(M \ge |b|\). Then for every \(n\),

\begin{equation*} |a_n b_n - ab| = |a_n(b_n - b) + b(a_n - a)| \le M|b_n - b| + M|a_n - a| . \end{equation*}

Given \(\epsilon > 0\), choose \(N\) so large that \(|a_n - a| < \epsilon/(2M)\) and \(|b_n - b| < \epsilon/(2M)\) for \(n \ge N\); then \(|a_n b_n - ab| < \epsilon\) for \(n \ge N\). So \(a_n b_n \to ab\). Applying this with \(a_n = f(x_n)\), \(b_n = g(x_n)\),

\begin{equation*} (fg)(x_n) = f(x_n) g(x_n) \to f(x_*) g(x_*) = (fg)(x_*), \end{equation*}

so by Proposition 21, \(fg\) is continuous on \(E\).

(ii). Let \(h\) be a continuous real-valued function defined on a set \(D\) of real numbers with \(h(D) \subseteq E\). Then \(f \circ h\) is defined on \(D\). Fix \(x_* \in D\) and let \(\{x_n\}\) be a sequence in \(D\) with \(\{x_n\} \to x_*\). Since \(h\) is continuous at \(x_*\), Proposition 21 gives \(h(x_n) \to h(x_*)\). Now \(\{h(x_n)\}\) is a sequence in \(E\) converging to the point \(h(x_*)\) of \(E\), so continuity of \(f\) at \(h(x_*)\) and Proposition 21 give

\begin{equation*} (f \circ h)(x_n) = f(h(x_n)) \to f(h(x_*)) = (f \circ h)(x_*). \end{equation*}

By Proposition 21 again, \(f \circ h\) is continuous at \(x_*\), hence on \(D\).

(iii). We first record an elementary inequality: for real numbers \(u, v, u^{\prime}, v^{\prime}\),

\begin{equation*} \bigl| \max\{u,v\} - \max\{u^{\prime},v^{\prime}\} \bigr| \le \max \bigl\{ |u - u^{\prime}|,\ |v - v^{\prime}| \bigr\}. \tag{2} \end{equation*}

To see this, assume without loss of generality that \(\max\{u,v\} \ge \max\{u^{\prime},v^{\prime}\}\) (otherwise interchange the roles of the two pairs, which does not change either side). Say \(\max\{u,v\} = u\), the case \(\max\{u,v\} = v\) being identical with \(v\) in place of \(u\). Then, since \(\max\{u^{\prime},v^{\prime}\} \ge u^{\prime}\),

\begin{equation*} 0 \le \max\{u,v\} - \max\{u^{\prime},v^{\prime}\} \le u - u^{\prime} \le |u - u^{\prime}| \le \max\{|u-u^{\prime}|, |v-v^{\prime}|\}, \end{equation*}

which is (2).

Now let \(x_* \in E\) and \(\epsilon > 0\). By continuity of \(f\) and of \(g\) at \(x_*\), there are \(\delta_1 > 0\) and \(\delta_2 > 0\) such that for \(x \in E\): \(|x - x_*| < \delta_1\) implies \(|f(x) - f(x_*)| < \epsilon\), and \(|x - x_*| < \delta_2\) implies \(|g(x) - g(x_*)| < \epsilon\). Put \(\delta = \min\{\delta_1, \delta_2\} > 0\). If \(x \in E\) and \(|x - x_*| < \delta\), then by (2) with \(u = f(x)\), \(v = g(x)\), \(u^{\prime} = f(x_*)\), \(v^{\prime} = g(x_*)\),

\begin{equation*} \bigl| \max\{f,g\}(x) - \max\{f,g\}(x_*) \bigr| \le \max\bigl\{ |f(x) - f(x_*)|,\ |g(x) - g(x_*)| \bigr\} < \epsilon . \end{equation*}

So \(\max\{f,g\}\) is continuous at \(x_*\), and hence on \(E\).

(Alternatively, once (i) and (iv) are known one may use the identity \(\max\{f,g\} = \tfrac{1}{2}\bigl(f + g + |f-g|\bigr)\).)

(iv). By the triangle inequality (Problem 7), for all real \(s,t\) we have \(|s| \le |s - t| + |t|\) and \(|t| \le |t-s| + |s|\), whence

\begin{equation*} \bigl| |s| - |t| \bigr| \le |s - t| . \end{equation*}

Let \(x_* \in E\) and \(\epsilon > 0\). By continuity of \(f\) at \(x_*\) there is \(\delta > 0\) such that \(|f(x) - f(x_*)| < \epsilon\) whenever \(x \in E\) and \(|x - x_*| < \delta\). For such \(x\),

\begin{equation*} \bigl| |f|(x) - |f|(x_*) \bigr| = \bigl| |f(x)| - |f(x_*)| \bigr| \le |f(x) - f(x_*)| < \epsilon . \end{equation*}

Hence \(|f|\) is continuous at \(x_*\), and therefore on \(E\).

Recall the definitions in force. A real-valued function \(f\) on a set \(E\) of real numbers is Lipschitz provided there is a \(c \ge 0\) for which

\begin{equation*} |f(x^{\prime}) - f(x)| \le c\,|x^{\prime} - x| \quad \text{for all } x^{\prime}, x \in E , \end{equation*}

and \(f\) is uniformly continuous on \(E\) provided for each \(\epsilon > 0\) there is a \(\delta > 0\) such that for all \(x, x^{\prime}\) in \(E\), if \(|x - x^{\prime}| < \delta\) then \(|f(x) - f(x^{\prime})| < \epsilon\).

Lipschitz implies uniformly continuous. Let \(f\) be Lipschitz on \(E\) with constant \(c \ge 0\). Let \(\epsilon > 0\). Define

\begin{equation*} \delta = \frac{\epsilon}{c + 1} > 0 . \end{equation*}

This \(\delta\) responds to the \(\epsilon\) challenge. Indeed, let \(x, x^{\prime}\) belong to \(E\) with \(|x - x^{\prime}| < \delta\). Then

\begin{equation*} |f(x) - f(x^{\prime})| \le c\,|x - x^{\prime}| \le (c+1)\,|x - x^{\prime}| < (c+1)\cdot \frac{\epsilon}{c+1} = \epsilon . \end{equation*}

Since \(\delta\) was chosen independently of the points \(x, x^{\prime}\), the function \(f\) is uniformly continuous on \(E\). Observe that the same \(\delta\) works for every \(\epsilon\)-challenge at every point, which is exactly the extra strength of uniform continuity over continuity.

A uniformly continuous function that is not Lipschitz. Define \(f(x) = \sqrt{x}\) for \(x\) in \(E = [0,1]\). The function \(f\) is continuous on \(E\). Indeed, at a point \(x > 0\), for \(x^{\prime} \in [0,1]\) we have \(\sqrt{x^{\prime}} + \sqrt{x} \ge \sqrt{x} > 0\) and hence

\begin{equation*} |\sqrt{x^{\prime}} - \sqrt{x}| = \frac{|x^{\prime} - x|}{\sqrt{x^{\prime}} + \sqrt{x}} \le \frac{|x^{\prime} - x|}{\sqrt{x}} , \end{equation*}

so \(\delta = \epsilon \sqrt{x}\) responds to the \(\epsilon\) challenge at \(x\); and at \(x = 0\) we have \(|f(x^{\prime}) - f(0)| = \sqrt{x^{\prime}} < \epsilon\) whenever \(0 \le x^{\prime} < \epsilon^2\). The set \([0,1]\) is closed and bounded, so by Theorem 23 the continuous function \(f\) is uniformly continuous on \([0,1]\).

However \(f\) is not Lipschitz. Suppose it were, with constant \(c \ge 0\). Taking \(x = 0\) and \(x^{\prime} = t\) with \(0 < t \le 1\) gives

\begin{equation*} \sqrt{t} = |f(t) - f(0)| \le c\,|t - 0| = c\,t , \end{equation*}

so that, dividing by \(\sqrt{t} > 0\),

\begin{equation*} \frac{1}{\sqrt{t}} \le c \quad \text{for all } 0 < t \le 1 . \end{equation*}

But \(1/\sqrt{t}\) is unbounded above as \(t\) decreases to \(0\): choosing \(t = 1/(c+1)^2\), which lies in \((0,1]\), yields \(c + 1 \le c\), a contradiction. Hence no such \(c\) exists and \(f\) is not Lipschitz.

Therefore the class of Lipschitz functions is properly contained in the class of uniformly continuous functions.

We may assume \(a < b\), since if \(a = b\) there is nothing to prove.

The interval \([a,b]\) is a closed, bounded set of real numbers and \(f\) is continuous on it, so by Theorem 23 the function \(f\) is uniformly continuous on \([a,b]\). Let \(\epsilon > 0\) be given and choose \(\delta > 0\) so that

\begin{equation*} \text{if } x, x^{\prime} \in [a,b] \text{ and } |x - x^{\prime}| < \delta, \text{ then } |f(x) - f(x^{\prime})| < \epsilon . \end{equation*}

By the Archimedean Property of \(\mathbb{R}\) choose a natural number \(n\) with \(n > (b-a)/\delta\), so that \((b-a)/n < \delta\). Define the uniform partition

\begin{equation*} x_i = a + i\cdot \frac{b-a}{n} \quad \text{for } i = 0, 1, \dots, n , \end{equation*}

so \(a = x_0 < x_1 < \cdots < x_n = b\) and each subinterval \([x_i, x_{i+1}]\) has length \((b-a)/n < \delta\).

Define \(\varphi \colon [a,b] \to \mathbb{R}\) by prescribing it on each subinterval to be the linear function agreeing with \(f\) at the two endpoints: for \(0 \le i \le n-1\) and \(x \in [x_i, x_{i+1}]\), set

\begin{equation*} \varphi(x) = f(x_i) + \frac{f(x_{i+1}) - f(x_i)}{x_{i+1} - x_i}\,(x - x_i) . \end{equation*}

This is well defined. At an interior partition point \(x_i\) with \(1 \le i \le n-1\) the formula on \([x_{i-1}, x_i]\) gives the value \(f(x_i)\) and the formula on \([x_i, x_{i+1}]\) also gives \(f(x_i)\), so the two prescriptions agree. Thus \(\varphi(x_i) = f(x_i)\) for every \(i\). Each piece is a linear, hence continuous, function on a closed interval and consecutive pieces agree at the shared endpoint, so \(\varphi\) is continuous on \([a,b]\) and is linear on each \([x_i, x_{i+1}]\). Hence \(\varphi\) is piecewise linear.

It remains to estimate \(|f - \varphi|\). Let \(x \in [a,b]\) and choose \(i\) with \(x \in [x_i, x_{i+1}]\). Write

\begin{equation*} \lambda = \frac{x_{i+1} - x}{x_{i+1} - x_i} \in [0,1], \qquad 1 - \lambda = \frac{x - x_i}{x_{i+1} - x_i} . \end{equation*}

A direct computation from the defining formula shows that \(\varphi(x)\) is the corresponding convex combination of the endpoint values:

\begin{equation*} \varphi(x) = \lambda\, f(x_i) + (1-\lambda)\, f(x_{i+1}) . \end{equation*}

Since \(\lambda + (1-\lambda) = 1\) we may write \(f(x) = \lambda f(x) + (1-\lambda) f(x)\), and therefore

\begin{equation*} |f(x) - \varphi(x)| = \bigl| \lambda\,(f(x) - f(x_i)) + (1-\lambda)\,(f(x) - f(x_{i+1})) \bigr| \le \lambda\,|f(x) - f(x_i)| + (1-\lambda)\,|f(x) - f(x_{i+1})| . \end{equation*}

Now \(x\) and \(x_i\) both lie in \([x_i, x_{i+1}]\), an interval of length \((b-a)/n < \delta\), so \(|x - x_i| < \delta\) and hence \(|f(x) - f(x_i)| < \epsilon\); for the same reason \(|f(x) - f(x_{i+1})| < \epsilon\). Consequently

\begin{equation*} |f(x) - \varphi(x)| < \lambda\,\epsilon + (1-\lambda)\,\epsilon = \epsilon . \end{equation*}

(If \(\lambda\) is \(0\) or \(1\) the corresponding term is absent and the strict inequality persists, since then \(|f(x) - \varphi(x)|\) equals \(|f(x) - f(x_{i+1})|\) or \(|f(x) - f(x_i)|\), each less than \(\epsilon\).)

Since \(x \in [a,b]\) was arbitrary, \(|f(x) - \varphi(x)| < \epsilon\) for all \(x \in [a,b]\), as required.

Let \(E\) be a nonempty set of real numbers. We say that a real-valued function \(f\) on \(E\) takes a maximum value provided there is a point \(x_{*}\) in \(E\) with \(f(x) \le f(x_{*})\) for all \(x\) in \(E\).

Necessity. Suppose \(E\) is closed and bounded. Let \(f\) be any continuous real-valued function on \(E\). Since \(E\) is nonempty, closed and bounded, the Extreme Value Theorem asserts that \(f\) takes a maximum value on \(E\). This is exactly the assertion to be proved.

Sufficiency. Suppose every continuous real-valued function on \(E\) takes a maximum value. We prove \(E\) is bounded and closed.

\(E\) is bounded. Consider \(f(x) = |x|\) for \(x\) in \(E\). This function is continuous on \(E\): for \(x, x^{\prime}\) in \(E\) the reverse triangle inequality gives \(\bigl| |x^{\prime}| - |x| \bigr| \le |x^{\prime} - x|\), so \(f\) is Lipschitz with constant \(1\) and hence continuous. By hypothesis \(f\) takes a maximum value, say at \(x_{*} \in E\). Then \(|x| \le |x_{*}|\) for all \(x \in E\), so \(E \subseteq [-|x_{*}|, |x_{*}|]\) and \(E\) is bounded.

\(E\) is closed. By definition \(E\) is closed provided it contains all of its points of closure. Argue by contradiction: suppose there is a point of closure \(x_0\) of \(E\) with \(x_0 \notin E\). Define

\begin{equation*} f(x) = \frac{1}{|x - x_0|} \quad \text{for } x \in E . \end{equation*}

This is well defined, since \(x_0 \notin E\) forces \(|x - x_0| > 0\) for every \(x \in E\). It is continuous on \(E\): the function \(x \mapsto |x - x_0|\) is continuous and nowhere zero on \(E\), and \(t \mapsto 1/t\) is continuous on \((0, \infty)\), so by the continuity of compositions the quotient is continuous at each point of \(E\). (Concretely, fix \(x \in E\) and put \(r = |x - x_0| > 0\); if \(x^{\prime} \in E\) and \(|x^{\prime} - x| < r/2\) then \(|x^{\prime} - x_0| > r/2\) and

\begin{equation*} |f(x^{\prime}) - f(x)| = \frac{\bigl| |x - x_0| - |x^{\prime} - x_0| \bigr|}{|x^{\prime} - x_0|\,|x - x_0|} \le \frac{|x^{\prime} - x|}{(r/2)\, r} , \end{equation*}

which is less than any prescribed \(\epsilon > 0\) once \(|x^{\prime} - x|\) is small enough.)

By hypothesis \(f\) takes a maximum value on \(E\), say \(f(x) \le M\) for all \(x \in E\), where \(M = f(x_{*}) > 0\). Thus \(|x - x_0| \ge 1/M\) for every \(x \in E\). But \(x_0\) is a point of closure of \(E\), so every open interval about \(x_0\) contains a point of \(E\); in particular there is an \(x \in E\) with \(|x - x_0| < 1/M\). This contradiction shows that every point of closure of \(E\) belongs to \(E\), that is, \(E\) is closed.

Hence \(E\) is closed and bounded, and the equivalence is established.

Recall that a collection \(\mathcal{F}\) of sets is a cover of \(E\) provided \(E \subseteq \bigcup_{F \in \mathcal{F}} F\), and it is an open cover provided each member is open. A finite subcover is a finite subcollection of \(\mathcal{F}\) that still covers \(E\).

Necessity. Suppose \(E\) is closed and bounded. Then the Heine-Borel Theorem asserts precisely that every open cover of \(E\) has a finite subcover.

Sufficiency. Suppose every open cover of \(E\) has a finite subcover. (If \(E\) is empty it is trivially closed and bounded, so assume \(E \neq \emptyset\).)

\(E\) is bounded. For each natural number \(n\) let \(\mathcal{O}_n = (-n, n)\), an open interval. Every real number \(x\) satisfies \(|x| < n\) for some natural number \(n\), by the Archimedean Property, so \(\{\mathcal{O}_n\}_{n=1}^{\infty}\) is an open cover of \(E\). By hypothesis there are indices \(n_1, \dots, n_k\) with

\begin{equation*} E \subseteq \mathcal{O}_{n_1} \cup \cdots \cup \mathcal{O}_{n_k} . \end{equation*}

The intervals \(\mathcal{O}_n\) are ascending, so putting \(N = \max\{n_1, \dots, n_k\}\) the union equals \(\mathcal{O}_N = (-N, N)\). Thus \(E \subseteq (-N, N)\) and \(E\) is bounded.

\(E\) is closed. We must show \(E\) contains each of its points of closure. Argue by contradiction: let \(x_0\) be a point of closure of \(E\) with \(x_0 \notin E\). For each natural number \(n\) define

\begin{equation*} \mathcal{O}_n = \Bigl\{ x \in \mathbb{R} : |x - x_0| > \tfrac{1}{n} \Bigr\} = \Bigl(-\infty,\ x_0 - \tfrac{1}{n}\Bigr) \cup \Bigl(x_0 + \tfrac{1}{n},\ \infty\Bigr) , \end{equation*}

which, being the union of two open intervals, is open. The collection \(\{\mathcal{O}_n\}_{n=1}^{\infty}\) covers \(E\): if \(x \in E\) then \(x \neq x_0\) since \(x_0 \notin E\), hence \(|x - x_0| > 0\), and by the Archimedean Property there is an \(n\) with \(1/n < |x - x_0|\), so \(x \in \mathcal{O}_n\).

By hypothesis finitely many of these sets cover \(E\). The sets \(\mathcal{O}_n\) ascend with \(n\), so as above there is a single natural number \(N\) with \(E \subseteq \mathcal{O}_N\), that is,

\begin{equation*} |x - x_0| > \frac{1}{N} \quad \text{for all } x \in E . \end{equation*}

Therefore the open interval \(\bigl(x_0 - 1/N,\ x_0 + 1/N\bigr)\) contains no point of \(E\). But \(x_0\) is a point of closure of \(E\), so every open interval containing \(x_0\) must contain a point of \(E\). This contradiction shows no such \(x_0\) exists, so \(E\) contains all its points of closure and hence is closed.

Thus \(E\) is closed and bounded, completing the proof of the equivalence.

It is convenient first to record the internal characterization of intervals.

Lemma. A nonempty set \(I\) of real numbers is an interval if and only if it has the betweenness property: whenever \(x, y \in I\) and \(x < z < y\), then \(z \in I\).

Proof of the Lemma. Every interval visibly has the betweenness property, since each of the sets \((a,b)\), \([a,b)\), \((a,b]\), \([a,b]\), \((-\infty, b)\), \((-\infty, b]\), \((a, \infty)\), \([a, \infty)\), \(\mathbb{R}\) is described by inequalities on the point itself. Conversely, let \(I\) be nonempty with the betweenness property. Put \(a = \inf I\) and \(b = \sup I\), allowing the values \(-\infty\) and \(\infty\) if \(I\) fails to be bounded below or above. Then \(I \subseteq [a,b]\) (interpreted appropriately when an endpoint is infinite). We claim \((a,b) \subseteq I\). Let \(a < z < b\). Since \(z > a = \inf I\), the number \(z\) is not a lower bound for \(I\), so there is \(x \in I\) with \(x < z\); since \(z < b = \sup I\), the number \(z\) is not an upper bound for \(I\), so there is \(y \in I\) with \(z < y\). By the betweenness property \(z \in I\). Hence \((a,b) \subseteq I \subseteq [a,b]\), so \(I\) is obtained from \((a,b)\) by possibly adjoining the finite endpoints \(a\) and \(b\); in every case \(I\) is an interval. (If \(a = b\) then \(I = \{a\} = [a,a]\), a degenerate closed interval.) \(\square\)

Necessity. Suppose \(E\) is an interval and let \(f\) be a continuous real-valued function on \(E\). The image \(f(E)\) is nonempty since \(E\) is. We verify the betweenness property for \(f(E)\); the Lemma then shows \(f(E)\) is an interval.

Let \(u, v \in f(E)\) and \(u < w < v\). Choose \(x, y \in E\) with \(f(x) = u\) and \(f(y) = v\). Since \(u \neq v\) we have \(x \neq y\).

Case \(x < y\). Because \(E\) is an interval containing \(x\) and \(y\), the betweenness property gives \([x,y] \subseteq E\). The restriction of \(f\) to the closed, bounded interval \([x,y]\) is continuous and \(f(x) = u < w < v = f(y)\), so by the Intermediate Value Theorem there is a point \(x_0 \in (x,y)\) with \(f(x_0) = w\). Hence \(w \in f(E)\).

Case \(y < x\). Again \([y,x] \subseteq E\). Consider \(g = -f\) on \([y,x]\), which is continuous, and note

\begin{equation*} g(y) = -v < -w < -u = g(x) . \end{equation*}

By the Intermediate Value Theorem applied to \(g\) on \([y,x]\) there is \(x_0 \in (y,x)\) with \(g(x_0) = -w\), that is, \(f(x_0) = w\). Hence \(w \in f(E)\).

In either case \(w \in f(E)\), so \(f(E)\) has the betweenness property and is therefore an interval.

Sufficiency. Suppose every continuous real-valued function on \(E\) has an interval as its image. Consider the identity function \(f(x) = x\) for \(x \in E\). It is Lipschitz with constant \(1\), hence continuous on \(E\), and its image is \(f(E) = E\). By hypothesis this image is an interval, so \(E\) is an interval.

This proves the asserted equivalence.

Let \(I\) be an open interval and \(f\) a monotone real-valued function on \(I\).

Throughout we use the following characterization, established in the solution to Problem 54: a nonempty set \(S\) of real numbers is an interval if and only if it has the betweenness property, namely that \(u, v \in S\) and \(u < w < v\) imply \(w \in S\).

Reduction to the increasing case. If \(f\) is decreasing then \(g = -f\) is increasing on \(I\); moreover \(f\) is continuous at a point if and only if \(g\) is, and \(g(I) = \{-y : y \in f(I)\}\). A set \(S\) has the betweenness property if and only if \(-S = \{-y : y \in S\}\) does, since \(u < w < v\) holds exactly when \(-v < -w < -u\) holds. Hence \(f(I)\) is an interval exactly when \(g(I)\) is. So the equivalence for \(g\) yields the equivalence for \(f\), and we may assume throughout that \(f\) is increasing, that is, \(f(x) \le f(x^{\prime})\) whenever \(x, x^{\prime} \in I\) and \(x \le x^{\prime}\).

Continuity implies the image is an interval. This half needs no monotonicity. An open interval \(I\) is an interval, so by the necessity half of Problem 54 (whose proof rests on the Intermediate Value Theorem) the image under a continuous real-valued function of an interval is again an interval. Explicitly: if \(u = f(x)\) and \(v = f(y)\) lie in \(f(I)\) and \(u < w < v\), then \([x,y] \subseteq I\) or \([y,x] \subseteq I\) according to the order of \(x\) and \(y\), and the Intermediate Value Theorem applied to \(f\) (or to \(-f\)) on that closed, bounded subinterval produces a point \(x_0\) between \(x\) and \(y\) with \(f(x_0) = w\). So \(f(I)\) has the betweenness property and hence is an interval.

The image is an interval implies continuity. Assume \(f(I)\) is an interval, and let \(x_0 \in I\). As in the discussion preceding these problems, monotonicity together with the Monotone Convergence Theorem for real sequences guarantees that the one-sided limits

\begin{equation*} f(x_0^-) = \lim_{x \to x_0^-} f(x), \qquad f(x_0^+) = \lim_{x \to x_0^+} f(x) \end{equation*}

exist and are independent of the approximating sequence, and \(f\) is continuous at \(x_0\) if and only if \(f(x_0^-) = f(x_0) = f(x_0^+)\). Since \(I\) is open, both one-sided limits are genuinely available: there are points of \(I\) on each side of \(x_0\).

For an increasing \(f\) these limits are given by

\begin{equation*} f(x_0^-) = \sup\{ f(x) : x \in I,\ x < x_0 \}, \qquad f(x_0^+) = \inf\{ f(x) : x \in I,\ x > x_0 \} . \end{equation*}

Indeed, since \(I\) is open there are points of \(I\) strictly to the left of \(x_0\), so the set \(S^{-} = \{f(x) : x \in I,\ x < x_0\}\) is nonempty; as \(f\) is increasing, \(S^{-}\) is bounded above by \(f(x_0)\), and hence \(s = \sup S^{-}\) is a real number. Choose a sequence \(\{x_n\}\) in \(I\) with \(x_n < x_0\) for all \(n\) and \(x_n \to x_0\); by the discussion above \(f(x_n) \to f(x_0^-)\). Each \(f(x_n)\) belongs to \(S^{-}\), so \(f(x_n) \le s\) for every \(n\) and therefore \(f(x_0^-) \le s\). Conversely, fix \(x \in I\) with \(x < x_0\). Since \(x_n \to x_0\) and \(x < x_0\), we have \(x_n > x\) for all sufficiently large \(n\), and then \(f(x_n) \ge f(x)\); letting \(n \to \infty\) gives \(f(x_0^-) \ge f(x)\). As \(x\) was an arbitrary point of \(I\) to the left of \(x_0\), taking the supremum yields \(f(x_0^-) \ge s\). Therefore \(f(x_0^-) = s\). The formula for \(f(x_0^+)\) is proved symmetrically. In particular

\begin{equation*} f(x_0^-) \le f(x_0) \le f(x_0^+) . \end{equation*}

Suppose, to get a contradiction, that \(f\) is not continuous at \(x_0\). Then \(f(x_0^-) < f(x_0^+)\). Two consequences of the sup/inf formulas are:

\begin{equation*} f(x) \le f(x_0^-) \ \text{ for all } x \in I \text{ with } x < x_0, \qquad f(x) \ge f(x_0^+) \ \text{ for all } x \in I \text{ with } x > x_0 . \end{equation*}

Hence the only value of \(f\) that can lie in the open interval \(\bigl(f(x_0^-),\, f(x_0^+)\bigr)\) is \(f(x_0)\) itself.

The open interval \(\bigl(f(x_0^-),\, f(x_0^+)\bigr)\) is nonempty and contains more than one point, so we may choose a number \(c\) with

\begin{equation*} f(x_0^-) < c < f(x_0^+) \quad \text{and} \quad c \neq f(x_0) . \end{equation*}

By the preceding paragraph \(c \notin f(I)\).

Since \(I\) is open and \(x_0 \in I\), choose \(x_1, x_2 \in I\) with \(x_1 < x_0 < x_2\). Then

\begin{equation*} f(x_1) \le f(x_0^-) < c < f(x_0^+) \le f(x_2) , \end{equation*}

so \(f(x_1)\) and \(f(x_2)\) belong to \(f(I)\) and \(c\) lies strictly between them, yet \(c \notin f(I)\). This contradicts the betweenness property of the interval \(f(I)\).

Therefore \(f(x_0^-) = f(x_0) = f(x_0^+)\) and \(f\) is continuous at \(x_0\). As \(x_0 \in I\) was arbitrary, \(f\) is continuous on \(I\).

Recall that a set of real numbers is a \(G_\delta\) set provided it is the intersection of a countable collection of open sets. Let

\begin{equation*} C = \{ x \in \mathbb{R} : f \text{ is continuous at } x \} . \end{equation*}

We exhibit \(C\) as a countable intersection of open sets by means of the oscillation of \(f\).

The oscillation. For \(x \in \mathbb{R}\) and \(\delta > 0\) define

\begin{equation*} \operatorname{osc}(f, x, \delta) = \sup \bigl\{ |f(y) - f(z)| : y, z \in (x - \delta,\ x + \delta) \bigr\} , \end{equation*}

an extended real number in \([0, \infty]\) (the supremum may be \(+\infty\); nothing below is affected). Since the interval shrinks as \(\delta\) decreases, \(\delta \mapsto \operatorname{osc}(f, x, \delta)\) is a nondecreasing function of \(\delta\), and we may define the oscillation of \(f\) at \(x\) by

\begin{equation*} \omega(x) = \inf_{\delta > 0} \operatorname{osc}(f, x, \delta) \in [0, \infty] . \end{equation*}

Step 1: \(f\) is continuous at \(x\) if and only if \(\omega(x) = 0\).

Suppose \(f\) is continuous at \(x\) and let \(\epsilon > 0\). There is \(\delta > 0\) such that \(|f(y) - f(x)| < \epsilon/3\) whenever \(|y - x| < \delta\). Then for any \(y, z \in (x - \delta, x + \delta)\),

\begin{equation*} |f(y) - f(z)| \le |f(y) - f(x)| + |f(x) - f(z)| < \frac{2\epsilon}{3} , \end{equation*}

so \(\operatorname{osc}(f, x, \delta) \le 2\epsilon/3 < \epsilon\) and hence \(\omega(x) < \epsilon\). As \(\epsilon > 0\) was arbitrary and \(\omega(x) \ge 0\), we get \(\omega(x) = 0\).

Conversely suppose \(\omega(x) = 0\) and let \(\epsilon > 0\). By definition of the infimum there is \(\delta > 0\) with \(\operatorname{osc}(f, x, \delta) < \epsilon\). If \(|y - x| < \delta\), then \(y\) and \(x\) both lie in \((x - \delta, x + \delta)\), so

\begin{equation*} |f(y) - f(x)| \le \operatorname{osc}(f, x, \delta) < \epsilon . \end{equation*}

Thus \(f\) is continuous at \(x\). This proves \(C = \{x \in \mathbb{R} : \omega(x) = 0\}\).

Step 2: for each natural number \(n\), the set

\begin{equation*} \mathcal{O}_n = \Bigl\{ x \in \mathbb{R} : \omega(x) < \tfrac{1}{n} \Bigr\} \end{equation*}

is open.

Let \(x \in \mathcal{O}_n\). Since \(\omega(x) < 1/n\), there is \(\delta > 0\) with \(\operatorname{osc}(f, x, \delta) < 1/n\). We claim the open interval \(\bigl(x - \delta/2,\ x + \delta/2\bigr)\) is contained in \(\mathcal{O}_n\). Indeed, let \(x^{\prime}\) satisfy \(|x^{\prime} - x| < \delta/2\). If \(y \in \bigl(x^{\prime} - \delta/2,\ x^{\prime} + \delta/2\bigr)\) then

\begin{equation*} |y - x| \le |y - x^{\prime}| + |x^{\prime} - x| < \frac{\delta}{2} + \frac{\delta}{2} = \delta , \end{equation*}

so \(\bigl(x^{\prime} - \delta/2,\ x^{\prime} + \delta/2\bigr) \subseteq (x - \delta,\ x + \delta)\). The supremum defining the oscillation is therefore taken over a smaller collection of pairs, whence

\begin{equation*} \operatorname{osc}\Bigl(f, x^{\prime}, \frac{\delta}{2}\Bigr) \le \operatorname{osc}(f, x, \delta) < \frac{1}{n} , \end{equation*}

and consequently \(\omega(x^{\prime}) \le \operatorname{osc}(f, x^{\prime}, \delta/2) < 1/n\), that is, \(x^{\prime} \in \mathcal{O}_n\). So every point of \(\mathcal{O}_n\) is contained in an open interval lying in \(\mathcal{O}_n\), and \(\mathcal{O}_n\) is open.

Step 3: conclusion.

We claim

\begin{equation*} C = \bigcap_{n=1}^{\infty} \mathcal{O}_n . \end{equation*}

If \(x \in C\) then \(\omega(x) = 0 < 1/n\) for every \(n\), so \(x\) lies in every \(\mathcal{O}_n\). Conversely, if \(x \in \mathcal{O}_n\) for every natural number \(n\), then \(0 \le \omega(x) < 1/n\) for all \(n\); by the Archimedean Property this forces \(\omega(x) = 0\), so \(x \in C\) by Step 1.

Thus \(C\) is the intersection of the countable collection \(\{\mathcal{O}_n\}_{n=1}^{\infty}\) of open sets, that is, the set of points at which \(f\) is continuous is a \(G_\delta\) set.

Write

\begin{equation*} X = \{\, x \in \mathbb{R} : \{f_n(x)\}_{n=1}^{\infty} \text{ converges to a real number} \,\}. \end{equation*}

Step 1: a Cauchy description of \(X\). By Theorem 17 (the Cauchy Convergence Criterion for Real Sequences), a sequence of real numbers converges to a real number if and only if it is Cauchy. Hence \(x\) belongs to \(X\) if and only if the sequence \(\{f_n(x)\}\) is Cauchy, that is, if and only if

for every \(k \in \mathbb{N}\) there is an \(N \in \mathbb{N}\) such that \(|f_n(x) - f_m(x)| \le 1/k\) for all \(n, m \ge N\).

(Using the closed inequality \(\le 1/k\) rather than \(< 1/k\) is harmless: if the Cauchy condition holds with strict inequality for every positive \(\epsilon\) it certainly holds with \(\le 1/k\) for every \(k\), and conversely, given \(\epsilon > 0\) choose \(k\) with \(1/k < \epsilon\).)

Translating the three quantifiers into set operations, this says exactly that

\begin{equation*} X \;=\; \bigcap_{k=1}^{\infty} \; \bigcup_{N=1}^{\infty} \; \bigcap_{n \ge N} \; \bigcap_{m \ge N} E_{n,m,k}, \qquad\text{where}\qquad E_{n,m,k} = \Big\{\, x \in \mathbb{R} : |f_n(x) - f_m(x)| \le \tfrac{1}{k} \,\Big\}. \end{equation*}

Step 2: each \(E_{n,m,k}\) is closed. Fix \(n\), \(m\), \(k\) and set \(g = |f_n - f_m|\). First, \(-f_m\) is continuous on \(\mathbb{R}\): for any point \(x\) and any \(\epsilon > 0\), a \(\delta > 0\) that responds to the \(\epsilon\) challenge for the continuity of \(f_m\) at \(x\) also responds for \(-f_m\), since \(|(-f_m)(x^{\prime}) - (-f_m)(x)| = |f_m(x^{\prime}) - f_m(x)|\). Next, by Problem 49 of this chapter, the sum of two continuous real-valued functions with a common domain is continuous, so \(f_n + (-f_m) = f_n - f_m\) is continuous on \(\mathbb{R}\); and by the same problem the absolute value of a continuous function is continuous, so \(g = |f_n - f_m|\) is continuous on \(\mathbb{R}\). The complement of \(E_{n,m,k}\) is

\begin{equation*} \mathbb{R} \sim E_{n,m,k} = \Big\{\, x \in \mathbb{R} : g(x) > \tfrac1k \,\Big\} = g^{-1}\big( (\tfrac1k, \infty) \big). \end{equation*}

The set \((1/k, \infty)\) is open, so by Proposition 22, applied with \(E = \mathbb{R}\), we have \(g^{-1}((1/k,\infty)) = \mathbb{R} \cap U = U\) for some open set \(U\); thus \(\mathbb{R} \sim E_{n,m,k}\) is open, and by Proposition 11 the set \(E_{n,m,k}\) is closed.

Step 3: assembling. For fixed \(k\) and \(N\) put

\begin{equation*} A_{N,k} \;=\; \bigcap_{n \ge N} \bigcap_{m \ge N} E_{n,m,k}. \end{equation*}

This is an intersection of a collection of closed sets, hence is closed by Proposition 12, which states that the intersection of any collection of closed sets is closed.

Now define

\begin{equation*} F_k \;=\; \bigcup_{N=1}^{\infty} A_{N,k}. \end{equation*}

This is a countable union of closed sets, so \(F_k\) is an \(F_\sigma\) set.

By Step 1,

\begin{equation*} X \;=\; \bigcap_{k=1}^{\infty} F_k , \end{equation*}

which exhibits \(X\) as the intersection of the countable collection \(\{F_k\}_{k=1}^{\infty}\) of \(F_\sigma\) sets.

Remark. Since a \(\sigma\)-algebra is closed with respect to the formation of countable unions and countable intersections, each \(F_\sigma\) set is a Borel set, and therefore \(X\) itself is a Borel set.

Throughout we use the elementary identities, valid for any function \(f\) from \(\mathbb{R}\) to \(\mathbb{R}\) and any sets \(A\), \(A_n\) of real numbers:

\begin{equation*} f^{-1}(\mathbb{R} \sim A) = \mathbb{R} \sim f^{-1}(A), \qquad f^{-1}\Big( \bigcup_{n=1}^{\infty} A_n \Big) = \bigcup_{n=1}^{\infty} f^{-1}(A_n). \end{equation*}

Both are immediate from the definition \(f^{-1}(A) = \{x \in \mathbb{R} : f(x) \in A\}\): for the first, \(x \in f^{-1}(\mathbb{R}\sim A)\) means \(f(x) \notin A\), which means \(x \notin f^{-1}(A)\); for the second, \(f(x)\) lies in some \(A_n\) if and only if \(x\) lies in some \(f^{-1}(A_n)\).

Inverse images of open sets. Let \(O\) be an open set of real numbers. Since \(f\) is continuous on \(E = \mathbb{R}\), Proposition 22 tells us that \(f^{-1}(O) = E \cap U = \mathbb{R} \cap U = U\) for some open set \(U\). Hence \(f^{-1}(O)\) is open.

Inverse images of closed sets. Let \(C\) be a closed set of real numbers. By Proposition 11 the set \(\mathbb{R} \sim C\) is open, so by the preceding paragraph \(f^{-1}(\mathbb{R} \sim C)\) is open. But

\begin{equation*} f^{-1}(\mathbb{R} \sim C) = \mathbb{R} \sim f^{-1}( C), \end{equation*}

so \(\mathbb{R} \sim f^{-1}( C)\) is open, and therefore, again by Proposition 11, \(f^{-1}( C)\) is closed.

Inverse images of Borel sets. Define

\begin{equation*} \mathcal{A} \;=\; \{\, A \subseteq \mathbb{R} : f^{-1}(A) \text{ is a Borel set} \,\}. \end{equation*}

We claim \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\).

(i) The empty-set belongs to \(\mathcal{A}\), since \(f^{-1}(\emptyset) = \emptyset\), and \(\emptyset\) is open, hence Borel.

(ii) If \(A\) belongs to \(\mathcal{A}\), then \(f^{-1}(\mathbb{R} \sim A) = \mathbb{R} \sim f^{-1}(A)\). Since \(f^{-1}(A)\) is Borel and the collection \(\mathcal{B}\) of Borel sets is a \(\sigma\)-algebra, hence closed with respect to the formation of complements, \(\mathbb{R} \sim f^{-1}(A)\) is Borel. Thus \(\mathbb{R} \sim A\) belongs to \(\mathcal{A}\).

(iii) If \(\{A_n\}_{n=1}^{\infty}\) is a countable collection of sets in \(\mathcal{A}\), then

\begin{equation*} f^{-1}\Big( \bigcup_{n=1}^{\infty} A_n \Big) = \bigcup_{n=1}^{\infty} f^{-1}(A_n), \end{equation*}

a countable union of Borel sets, which is Borel because \(\mathcal{B}\) is closed with respect to the formation of countable unions. Thus \(\bigcup_{n=1}^{\infty} A_n\) belongs to \(\mathcal{A}\).

So \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\). Moreover \(\mathcal{A}\) contains every open set: if \(O\) is open then, by the first paragraph, \(f^{-1}(O)\) is open and hence is a Borel set, so \(O\) belongs to \(\mathcal{A}\).

By definition, the collection \(\mathcal{B}\) of Borel sets is the smallest \(\sigma\)-algebra of sets of real numbers that contains all of the open sets; by Proposition 13 this means precisely that any \(\sigma\)-algebra of subsets of \(\mathbb{R}\) containing the open sets also contains \(\mathcal{B}\). Since \(\mathcal{A}\) is such a \(\sigma\)-algebra, \(\mathcal{B} \subseteq \mathcal{A}\). That is, for every Borel set \(A\) the set \(f^{-1}(A)\) is a Borel set, which is what we wished to show.

Let \(x\) be a point of \(E\); we show that \(f\) is continuous at \(x\). Since \(x\) is an arbitrary point of the domain \(E\), this establishes that \(f\) is continuous on \(E\).

Let \(\epsilon > 0\). Apply the definition of uniform convergence with the positive number \(\epsilon/3\): there is an index \(N\) such that

\begin{equation*} |f_n(y) - f(y)| < \frac{\epsilon}{3} \quad \text{for all } y \in E \text{ and all } n \ge N. \end{equation*}

Fix this index \(N\). The crucial point is that the single index \(N\) works simultaneously at every point of \(E\); this is exactly what uniformity provides, and it is what makes the following estimate possible.

The function \(f_N\) is continuous on \(E\), hence continuous at the point \(x\). Therefore there is a \(\delta > 0\) such that

\begin{equation*} \text{if } y \in E \text{ and } |y - x| < \delta, \text{ then } |f_N(y) - f_N(x)| < \frac{\epsilon}{3}. \end{equation*}

We claim this \(\delta > 0\) responds to the \(\epsilon > 0\) challenge regarding the criterion for the continuity of \(f\) at \(x\). Indeed, let \(y\) belong to \(E\) with \(|y - x| < \delta\). Adding and subtracting and using the triangle inequality twice,

\begin{equation*} \begin{aligned} |f(y) - f(x)| &= |\,[f(y) - f_N(y)] + [f_N(y) - f_N(x)] + [f_N(x) - f(x)]\,| \\ &\le |f(y) - f_N(y)| + |f_N(y) - f_N(x)| + |f_N(x) - f(x)|. \end{aligned} \end{equation*}

The first term is less than \(\epsilon/3\) by the choice of \(N\): the uniform estimate holds for every point of \(E\) and every index at least \(N\), and here the point is \(y \in E\) and the index is \(n = N\). The third term is less than \(\epsilon/3\) for the same reason, with the point \(x \in E\). The middle term is less than \(\epsilon/3\) by the choice of \(\delta\). Therefore

\begin{equation*} |f(y) - f(x)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon. \end{equation*}

Thus for each \(\epsilon > 0\) there is a \(\delta > 0\) for which \(y \in E\) and \(|y - x| < \delta\) imply \(|f(y) - f(x)| < \epsilon\); that is, \(f\) is continuous at \(x\). Since \(x \in E\) was arbitrary, \(f\) is continuous on \(E\).

Part 1: proof of Proposition 21.

Suppose first that \(f\) is continuous at \(x_*\), and let \(\{x_n\}\) be a sequence in \(E\) that converges to \(x_*\). Let \(\epsilon > 0\). By continuity of \(f\) at \(x_*\) there is a \(\delta > 0\) such that

\begin{equation*} \text{if } x^{\prime} \in E \text{ and } |x^{\prime} - x_*| < \delta, \text{ then } |f(x^{\prime}) - f(x_*)| < \epsilon . \end{equation*}

Since \(\{x_n\} \to x_*\), applied with the positive number \(\delta\), there is an index \(N\) such that \(|x_n - x_*| < \delta\) for all \(n \ge N\). Each \(x_n\) belongs to \(E\), so for all \(n \ge N\) we have \(|f(x_n) - f(x_*)| < \epsilon\). As \(\epsilon > 0\) was arbitrary, \(\{f(x_n)\} \to f(x_*)\).

Conversely, suppose \(f\) is not continuous at \(x_*\). Negating the definition of continuity at a point, there is an \(\epsilon_0 > 0\) with the property that no \(\delta > 0\) responds to the \(\epsilon_0\) challenge: for every \(\delta > 0\) there is a point \(x^{\prime} \in E\) with \(|x^{\prime} - x_*| < \delta\) and \(|f(x^{\prime}) - f(x_*)| \ge \epsilon_0\). For each natural number \(n\), apply this with \(\delta = 1/n\) to select a point \(x_n \in E\) with

\begin{equation*} |x_n - x_*| < \frac{1}{n} \qquad \text{and} \qquad |f(x_n) - f(x_*)| \ge \epsilon_0 . \end{equation*}

The sequence \(\{x_n\}\) lies in \(E\) and converges to \(x_*\): given \(\epsilon > 0\), the Archimedean Property of \(\mathbb{R}\) provides a natural number \(N\) with \(1/N < \epsilon\), and then \(|x_n - x_*| < 1/n \le 1/N < \epsilon\) for all \(n \ge N\). But \(\{f(x_n)\}\) does not converge to \(f(x_*)\), since no term of the sequence \(\{f(x_n)\}\) lies within \(\epsilon_0\) of \(f(x_*)\). Hence the sequential condition fails.

Contrapositively, if the sequential condition holds then \(f\) is continuous at \(x_*\). This proves Proposition 21.

A lemma on closed sets. Before turning to the Extreme Value Theorem we record the following, which we use twice. If \(E\) is a closed set of real numbers, \(\{x_n\}\) is a sequence in \(E\), and \(\{x_n\} \to x_*\), then \(x_* \in E\). Indeed, let \(I\) be an open interval containing \(x_*\); choose \(r > 0\) with \((x_* - r, x_* + r) \subseteq I\). Since \(\{x_n\} \to x_*\) there is an index \(n\) with \(|x_n - x_*| < r\), so \(x_n \in I\), and \(x_n \in E\). Thus every open interval containing \(x_*\) contains a point of \(E\), that is, \(x_*\) is a point of closure of \(E\). Since \(E\) is closed it contains all of its points of closure, so \(x_* \in E\).

Part 2: another proof of the Extreme Value Theorem.

Let \(E\) be a nonempty closed, bounded set of real numbers and let \(f\) be a continuous real-valued function on \(E\). We must show that \(f\) takes a maximum and a minimum value on \(E\).

Claim A: \(f(E)\) is bounded above. Suppose not. Then for each natural number \(n\) the number \(n\) fails to be an upper bound for \(f(E)\), so we may choose \(x_n \in E\) with \(f(x_n) > n\). The sequence \(\{x_n\}\) lies in the bounded set \(E\) and hence is bounded. By Theorem 16 (the Bolzano-Weierstrass Theorem) it has a convergent subsequence \(\{x_{n_k}\}\), say \(\{x_{n_k}\} \to x_*\). By the lemma, \(x_* \in E\). Since \(f\) is continuous at \(x_*\), Proposition 21 tells us that \(\{f(x_{n_k})\} \to f(x_*)\), a real number. By Proposition 14, a convergent sequence of real numbers is bounded, so \(\{f(x_{n_k})\}\) is bounded. But \(\{n_k\}\) is a strictly increasing sequence of natural numbers, so \(n_k \ge k\) for every \(k\) (an immediate induction), and therefore \(f(x_{n_k}) > n_k \ge k\) for all \(k\), which makes \(\{f(x_{n_k})\}\) unbounded. This contradiction proves Claim A.

Since \(E\) is nonempty, \(f(E)\) is a nonempty set of real numbers which by Claim A is bounded above. By the Completeness Axiom it has a supremum; define

\begin{equation*} m = \sup f(E) \in \mathbb{R}. \end{equation*}

Claim B: \(f\) takes the value \(m\). For each natural number \(n\), the number \(m - 1/n\) is smaller than \(m\) and hence is not an upper bound for \(f(E)\), so we may choose \(x_n \in E\) with

\begin{equation*} m - \frac{1}{n} < f(x_n) \le m . \end{equation*}

Again \(\{x_n\}\) is a bounded sequence, so by the Bolzano-Weierstrass Theorem there is a subsequence \(\{x_{n_k}\}\) converging to some real number \(x_*\), and by the lemma \(x_* \in E\). Since \(f\) is continuous at \(x_*\), Proposition 21 gives \(\{f(x_{n_k})\} \to f(x_*)\).

On the other hand \(\{f(x_{n_k})\} \to m\). Indeed, let \(\epsilon > 0\) and use the Archimedean Property to choose a natural number \(K\) with \(1/K < \epsilon\). For \(k \ge K\) we have \(n_k \ge k \ge K\), so

\begin{equation*} |f(x_{n_k}) - m| = m - f(x_{n_k}) < \frac{1}{n_k} \le \frac{1}{K} < \epsilon . \end{equation*}

By Proposition 14 the limit of a convergent sequence is unique, so \(f(x_*) = m\). Thus \(f(x) \le m = f(x_*)\) for every \(x \in E\): the function \(f\) takes the maximum value \(m\) on \(E\), attained at the point \(x_* \in E\).

Finally, the function \(-f\) is continuous on \(E\) (given \(\epsilon > 0\), any \(\delta\) that responds to the \(\epsilon\) challenge for \(f\) at a point also responds for \(-f\), since \(|-f(x^{\prime}) + f(x)| = |f(x^{\prime}) - f(x)|\)). Applying what has just been proved to \(-f\), there is a point \(x_{**} \in E\) with \(-f(x) \le -f(x_{**})\) for all \(x \in E\), that is, \(f(x_{**}) \le f(x)\) for all \(x \in E\). Hence \(f\) takes a minimum value on \(E\), attained at \(x_{**}\).

This completes the proof of the Extreme Value Theorem using only Proposition 21, the Bolzano-Weierstrass Theorem, Proposition 14, and the Completeness Axiom; in particular the Heine-Borel Theorem, used in the text’s proof, is not needed.

Since \(\mathcal{A}\) is a \(\sigma\)-algebra it contains \(\mathbb{R}\) and is closed with respect to the formation of complements and countable unions; by De Morgan’s Identities it is therefore also closed with respect to the formation of countable intersections. In particular \(\emptyset = \mathbb{R} \setminus \mathbb{R}\) belongs to \(\mathcal{A}\), and since

\begin{equation*} B \setminus A = B \cap (\mathbb{R} \setminus A), \end{equation*}

the set \(B \setminus A\) also belongs to \(\mathcal{A}\).

Define a countable collection \(\{E_k\}_{k=1}^{\infty}\) of sets in \(\mathcal{A}\) by

\begin{equation*} E_1 = A, \qquad E_2 = B \setminus A, \qquad E_k = \emptyset \ \text{ for } k \ge 3 . \end{equation*}

This collection is disjoint: \(E_1 \cap E_2 = A \cap (B \setminus A) = \emptyset\), and every intersection involving one of the sets \(E_k\) with \(k \ge 3\) is empty because those sets are empty. Moreover, since \(A \subseteq B\),

\begin{equation*} \bigcup_{k=1}^{\infty} E_k = A \cup (B \setminus A) = B . \end{equation*}

By the assumed countable additivity of \(m\) over countable disjoint collections of sets in \(\mathcal{A}\),

\begin{equation*} m(B) = m\left(\bigcup_{k=1}^{\infty} E_k\right) = \sum_{k=1}^{\infty} m(E_k) = m(A) + m(B \setminus A) + \sum_{k=3}^{\infty} m(\emptyset). \end{equation*}

The set function \(m\) takes values in \([0, \infty]\), so every term on the right-hand side is nonnegative and the series of nonnegative terms is at least its first term. Therefore

\begin{equation*} m(B) \ge m(A), \end{equation*}

which is the asserted monotonicity. Observe that this argument does not require knowing in advance that \(m(\emptyset) = 0\); only the nonnegativity of \(m\) is used.

Since \(\mathcal{A}\) is a \(\sigma\)-algebra it contains \(\mathbb{R}\) and is closed with respect to complementation, so \(\emptyset = \mathbb{R} \setminus \mathbb{R}\) belongs to \(\mathcal{A}\) and \(m(\emptyset)\) is defined.

Let \(A \in \mathcal{A}\) satisfy \(m(A) < \infty\). Define a countable collection \(\{E_k\}_{k=1}^{\infty}\) of sets in \(\mathcal{A}\) by

\begin{equation*} E_1 = A, \qquad E_k = \emptyset \ \text{ for } k \ge 2 . \end{equation*}

The collection is disjoint, since the intersection of any two distinct members is empty (indeed \(A \cap \emptyset = \emptyset\) and \(\emptyset \cap \emptyset = \emptyset\)), and its union is \(A\). Countable additivity of \(m\) therefore gives

\begin{equation*} m(A) = \sum_{k=1}^{\infty} m(E_k) = m(A) + \sum_{k=2}^{\infty} m(\emptyset). \end{equation*}

Suppose, to obtain a contradiction, that \(m(\emptyset) = c > 0\). Then \(\sum_{k=2}^{\infty} m(\emptyset) = \sum_{k=2}^{\infty} c = \infty\), so the right-hand side of the displayed identity equals \(\infty\), while the left-hand side equals \(m(A) < \infty\). This is a contradiction. Since \(m\) takes values in \([0, \infty]\), the only remaining possibility is

\begin{equation*} m(\emptyset) = 0 . \end{equation*}

Equivalently and without contradiction: because \(m(A) < \infty\) we may subtract the finite number \(m(A)\) from both sides of the identity above to obtain \(\sum_{k=2}^{\infty} m(\emptyset) = 0\); a series of identical nonnegative terms vanishes only if each term is \(0\), so \(m(\emptyset) = 0\).

The hypothesis that some set has finite measure cannot be dropped: the set function defined on all of \(\mathcal{A}\) by \(m(E) = \infty\) for every \(E \in \mathcal{A}\), including \(E = \emptyset\), is countably additive in the stated sense yet has \(m(\emptyset) \ne 0\).

The strategy is to replace the collection \(\{E_k\}_{k=1}^{\infty}\) by a disjoint collection with the same union, each member of which is contained in the corresponding \(E_k\), and then to apply countable additivity together with the monotonicity established in Problem 1.

Define

\begin{equation*} F_1 = E_1, \qquad F_k = E_k \setminus \bigcup_{j=1}^{k-1} E_j \quad \text{for } k \ge 2 . \end{equation*}

Each \(F_k\) belongs to \(\mathcal{A}\): the union \(\bigcup_{j=1}^{k-1} E_j\) is a finite, hence countable, union of members of \(\mathcal{A}\) and so lies in \(\mathcal{A}\); its complement lies in \(\mathcal{A}\); and \(F_k\) is the intersection of \(E_k\) with that complement, which lies in \(\mathcal{A}\) because a \(\sigma\)-algebra is closed with respect to countable intersections by De Morgan’s Identities.

The collection \(\{F_k\}_{k=1}^{\infty}\) is disjoint. Indeed, let \(j < k\). Every point of \(F_k\) fails to belong to \(E_j\), whereas every point of \(F_j\) belongs to \(E_j\) (since \(F_j \subseteq E_j\)); hence \(F_j \cap F_k = \emptyset\).

The two collections have the same union. Since \(F_k \subseteq E_k\) for every \(k\), we have \(\bigcup_{k=1}^{\infty} F_k \subseteq \bigcup_{k=1}^{\infty} E_k\). Conversely, let \(x \in \bigcup_{k=1}^{\infty} E_k\). The set of indices \(k\) with \(x \in E_k\) is a nonempty set of natural numbers, so by the Well-Ordering Principle it has a smallest member \(k_0\). Then \(x \in E_{k_0}\) while \(x \notin E_j\) for \(1 \le j < k_0\), that is, \(x \in F_{k_0}\). Therefore

\begin{equation*} \bigcup_{k=1}^{\infty} F_k = \bigcup_{k=1}^{\infty} E_k . \end{equation*}

Now apply countable additivity of \(m\) to the disjoint collection \(\{F_k\}_{k=1}^{\infty}\) of sets in \(\mathcal{A}\):

\begin{equation*} m\left(\bigcup_{k=1}^{\infty} E_k\right) = m\left(\bigcup_{k=1}^{\infty} F_k\right) = \sum_{k=1}^{\infty} m(F_k). \end{equation*}

Since \(F_k \subseteq E_k\) and both sets belong to \(\mathcal{A}\), monotonicity (Problem 1) gives \(m(F_k) \le m(E_k)\) for each \(k\). Comparison of series of nonnegative terms, term by term, yields

\begin{equation*} \sum_{k=1}^{\infty} m(F_k) \le \sum_{k=1}^{\infty} m(E_k). \end{equation*}

Combining the last two displays,

\begin{equation*} m\left(\bigcup_{k=1}^{\infty} E_k\right) \le \sum_{k=1}^{\infty} m(E_k), \end{equation*}

as required. This property is called countable monotonicity (or countable subadditivity). The corresponding statement for a finite collection \(\{E_k\}_{k=1}^{n}\) follows by setting \(E_k = \emptyset\) for \(k > n\), provided \(m(\emptyset) = 0\); and if instead \(m(\emptyset) > 0\), then applying countable additivity to the disjoint collection \(E, \emptyset, \emptyset, \ldots\) shows \(m(E) = \infty\) for every \(E \in \mathcal{A}\), so every such inequality holds trivially.

Throughout, for a finite set \(F\) we write \(\#F\) for the number of its elements, so that \(c(F) = \#F\) for finite \(F\) and \(c(E) = \infty\) for infinite \(E\). Note first the elementary monotonicity property: if \(F \subseteq E\) and \(F\) is finite, then \(c(E) \ge \#F\). Indeed, if \(E\) is infinite this reads \(\infty \ge \#F\), and if \(E\) is finite then \(\#E \ge \#F\) because a subset of a finite set has no more elements than the set.

Countable additivity. Let \(\{E_k\}_{k=1}^{\infty}\) be a countable disjoint collection of subsets of \(\mathbb{R}\) and put \(E = \bigcup_{k=1}^{\infty} E_k\). We show

\begin{equation*} c(E) = \sum_{k=1}^{\infty} c(E_k). \end{equation*}

First we prove \(c(E) \ge \sum_{k=1}^{\infty} c(E_k)\). Since the right-hand side is the supremum of its partial sums, it suffices to show that \(c(E) \ge \sum_{k=1}^{n} c(E_k)\) for every natural number \(n\), and, when some \(c(E_k) = \infty\), that \(c(E) = \infty\). In the latter case some \(E_{k_0}\) is infinite; since \(E_{k_0} \subseteq E\), the set \(E\) is infinite as well and \(c(E) = \infty\), so the inequality holds. So suppose every \(E_k\) is finite and fix \(n\). The sets \(E_1, \ldots, E_n\) are pairwise disjoint finite sets, so their union \(F = \bigcup_{k=1}^{n} E_k\) is finite with

\begin{equation*} \#F = \sum_{k=1}^{n} \#E_k = \sum_{k=1}^{n} c(E_k), \end{equation*}

the first equality being the additivity of cardinality over disjoint finite sets. Since \(F \subseteq E\), the monotonicity noted above gives \(c(E) \ge \#F = \sum_{k=1}^{n} c(E_k)\). Letting \(n \to \infty\),

\begin{equation*} c(E) \ge \sum_{k=1}^{\infty} c(E_k). \end{equation*}

Next we prove \(c(E) \le \sum_{k=1}^{\infty} c(E_k)\). If the right-hand side is \(\infty\) there is nothing to prove, so assume \(\sum_{k=1}^{\infty} c(E_k) = M < \infty\). Then each \(c(E_k)\) is finite, that is, each \(E_k\) is a finite set, and \(M\) is a finite nonnegative number (in fact a nonnegative integer, since a convergent series of nonnegative integers has all but finitely many terms equal to zero). We claim \(E\) is finite with \(\#E \le M\). Suppose otherwise. Then \(E\) contains at least \(M + 1\) distinct points; more precisely, if \(E\) were infinite we could choose \(M+1\) distinct points of \(E\), and if \(E\) were finite with \(\#E > M\) we could likewise choose \(M+1\) distinct points. Call them \(x_1, \ldots, x_{M+1}\). Each \(x_i\) lies in \(E\), hence in \(E_{k_i}\) for some index \(k_i\). Let \(n = \max\{k_1, \ldots, k_{M+1}\}\), a finite number since only finitely many indices occur. Then the \(M+1\) distinct points all lie in the finite set \(\bigcup_{k=1}^{n} E_k\), whence, using disjointness again,

\begin{equation*} M + 1 \le \#\left(\bigcup_{k=1}^{n} E_k\right) = \sum_{k=1}^{n} c(E_k) \le \sum_{k=1}^{\infty} c(E_k) = M, \end{equation*}

a contradiction. Therefore \(E\) is finite and \(c(E) = \#E \le M = \sum_{k=1}^{\infty} c(E_k)\).

The two inequalities give \(c(E) = \sum_{k=1}^{\infty} c(E_k)\), so \(c\) is countably additive. (In particular, taking all \(E_k\) empty is consistent with the stipulation \(c(\emptyset) = 0\).)

Translation invariance. Let \(E \subseteq \mathbb{R}\) and let \(y\) be a real number, and consider \(E + y = \{x + y \mid x \in E\}\). The map \(\tau_y : \mathbb{R} \to \mathbb{R}\) defined by \(\tau_y(x) = x + y\) is a bijection of \(\mathbb{R}\) onto \(\mathbb{R}\), with inverse \(\tau_{-y}\), and \(\tau_y(E) = E + y\). A bijection restricted to \(E\) is a one-to-one correspondence between \(E\) and \(E + y\). Consequently \(E\) is finite if and only if \(E + y\) is finite, and in that case \(\#(E + y) = \#E\); likewise \(E\) is infinite if and only if \(E + y\) is infinite. Also \(\emptyset + y = \emptyset\). Examining the three clauses of the definition of \(c\), we conclude in every case that

\begin{equation*} c(E + y) = c(E), \end{equation*}

so \(c\) is translation invariant.

We use two facts about Lebesgue outer measure \(m^{*}\) established in Section 2.2: the outer measure of an interval is its length (Proposition 1), and a countable set has outer measure zero (the Example preceding Proposition 1).

For completeness we recall the proof of the second fact. Let \(C\) be a countable set, enumerated as \(C = \{c_k\}_{k=1}^{\infty}\) (a finite set is handled the same way, repeating a point). Let \(\epsilon > 0\). For each natural number \(k\) define the open bounded interval

\begin{equation*} I_k = \left(c_k - \frac{\epsilon}{2^{k+1}},\ c_k + \frac{\epsilon}{2^{k+1}}\right), \end{equation*}

which has length \(\epsilon / 2^{k}\). The countable collection \(\{I_k\}_{k=1}^{\infty}\) covers \(C\), since \(c_k \in I_k\) for each \(k\). Hence, by the definition of outer measure as an infimum over such covers,

\begin{equation*} 0 \le m^{*}( C) \le \sum_{k=1}^{\infty} \ell(I_k) = \sum_{k=1}^{\infty} \frac{\epsilon}{2^{k}} = \epsilon . \end{equation*}

This inequality holds for every \(\epsilon > 0\), so \(m^{*}( C) = 0\).

Now suppose, to obtain a contradiction, that \([0, 1]\) is countable. Then by the fact just proved,

\begin{equation*} m^{*}([0, 1]) = 0 . \end{equation*}

On the other hand, \([0,1]\) is an interval of length \(1\), so by Proposition 1,

\begin{equation*} m^{*}([0, 1]) = \ell([0, 1]) = 1 - 0 = 1 . \end{equation*}

Thus \(0 = 1\), which is absurd. Therefore \([0, 1]\) is not countable.

(The set \([0,1]\) is certainly not finite, so the conclusion is that \([0,1]\) is uncountable. The same argument shows that no interval of positive length is countable.)

Write \(B = [0,1] \setminus A\) for the set of rational numbers in \([0,1]\), so that

\begin{equation*} [0, 1] = A \cup B . \end{equation*}

The set \(B\) is countable, being a subset of the countable set \(\mathbb{Q}\) of rational numbers. Hence, by the Example of Section 2.2 (a countable set has outer measure zero),

\begin{equation*} m^{*}(B) = 0 . \end{equation*}

By Proposition 1, the outer measure of an interval is its length, so

\begin{equation*} m^{*}([0,1]) = \ell([0,1]) = 1 . \end{equation*}

Now use the finite subadditivity of outer measure, which is the special case of Proposition 3 (countable subadditivity) obtained by taking all but finitely many sets in the collection to be empty:

\begin{equation*} 1 = m^{*}([0,1]) = m^{*}(A \cup B) \le m^{*}(A) + m^{*}(B) = m^{*}(A) + 0 = m^{*}(A). \end{equation*}

Therefore \(m^{*}(A) \ge 1\).

For the reverse inequality, observe that \(A \subseteq [0,1]\), so by the monotonicity of outer measure, which follows immediately from the definition since any cover of \([0,1]\) is a cover of \(A\),

\begin{equation*} m^{*}(A) \le m^{*}([0,1]) = 1 . \end{equation*}

Combining the two inequalities, \(m^{*}(A) = 1\).

Thus \([0,1]\) is the disjoint union of the rationals in \([0,1]\), which carry outer measure \(0\), and the irrationals in \([0,1]\), which carry all of the outer measure \(1\).

Let \(E\) be a bounded set of real numbers.

Step 1: the outer measure of \(E\) is finite. Since \(E\) is bounded there is a number \(M > 0\) with \(E \subseteq (-M, M)\). By the monotonicity of outer measure and Proposition 1 (the outer measure of an interval is its length),

\begin{equation*} m^{*}(E) \le m^{*}((-M, M)) = \ell((-M,M)) = 2M < \infty . \end{equation*}

Step 2: construction of open sets of nearly minimal outer measure containing \(E\). Fix a natural number \(n\). By the definition of \(m^{*}(E)\) as the infimum of the sums \(\sum_{k=1}^{\infty} \ell(I_k)\) taken over all countable collections \(\{I_k\}_{k=1}^{\infty}\) of nonempty open, bounded intervals that cover \(E\), and since \(m^{*}(E)\) is finite so that \(m^{*}(E) + 1/n\) is not a lower bound for these sums, there is such a collection \(\{I_{n,k}\}_{k=1}^{\infty}\) with

\begin{equation*} E \subseteq \bigcup_{k=1}^{\infty} I_{n,k} \quad \text{and} \quad \sum_{k=1}^{\infty} \ell(I_{n,k}) < m^{*}(E) + \frac{1}{n}. \end{equation*}

Define

\begin{equation*} \mathcal{O}_n = \bigcup_{k=1}^{\infty} I_{n,k}. \end{equation*}

Then \(\mathcal{O}_n\) is open, being a union of open intervals, and \(E \subseteq \mathcal{O}_n\). Moreover, by the countable subadditivity of outer measure (Proposition 3) together with Proposition 1,

\begin{equation*} m^{*}(\mathcal{O}_n) \le \sum_{k=1}^{\infty} m^{*}(I_{n,k}) = \sum_{k=1}^{\infty} \ell(I_{n,k}) < m^{*}(E) + \frac{1}{n}. \end{equation*}

Step 3: intersection of the open sets. Define

\begin{equation*} G = \bigcap_{n=1}^{\infty} \mathcal{O}_n . \end{equation*}

As the intersection of a countable collection of open sets, \(G\) is a \(G_{\delta}\) set. Since \(E \subseteq \mathcal{O}_n\) for every \(n\), we have \(E \subseteq G\).

Step 4: equality of the outer measures. By monotonicity of outer measure, \(E \subseteq G\) gives

\begin{equation*} m^{*}(E) \le m^{*}(G). \end{equation*}

On the other hand, for each natural number \(n\) we have \(G \subseteq \mathcal{O}_n\), so monotonicity and Step 2 give

\begin{equation*} m^{*}(G) \le m^{*}(\mathcal{O}_n) < m^{*}(E) + \frac{1}{n}. \end{equation*}

Thus \(m^{*}(E) \le m^{*}(G) < m^{*}(E) + 1/n\) for every natural number \(n\). Letting \(n \to \infty\) (equivalently, since \(m^{*}(G) - m^{*}(E)\) is a nonnegative number smaller than every \(1/n\), it must be zero; here finiteness of \(m^{*}(E)\) from Step 1 permits the subtraction), we conclude

\begin{equation*} m^{*}(G) = m^{*}(E). \end{equation*}

Hence \(G\) is a \(G_{\delta}\) set with \(E \subseteq G\) and \(m^{*}(G) = m^{*}(E)\), as required.

Remark. Boundedness of \(E\) was used only to guarantee \(m^{*}(E) < \infty\). If \(E\) is any set with \(m^{*}(E) = \infty\), one may simply take \(G = \mathbb{R}\), which is open and hence a \(G_{\delta}\) set, and then \(m^{*}(G) = \infty = m^{*}(E)\) by monotonicity. So the conclusion in fact holds for every set of real numbers.

If some \(I_k\) is unbounded then \(m^*(I_k) = \infty\) by Proposition 1 (the outer measure of an interval is its length), and the asserted inequality is trivial. So assume every \(I_k\) is a bounded open interval, say \(I_k = (a_k, b_k)\) with \(a_k \le b_k\).

The essential point is that the finiteness of the collection lets us pass to closures. Write \(\bar{S}\) for the closure of a set \(S\). For a finite family of sets the closure of the union is the union of the closures:

\begin{equation*} \overline{\bigcup_{k=1}^{n} I_k} \;=\; \bigcup_{k=1}^{n} \overline{I_k}. \end{equation*}

Indeed, \(\bigcup_{k=1}^n \overline{I_k}\) is a finite union of closed sets, hence closed, and it contains \(\bigcup_{k=1}^n I_k\); therefore it contains the closure of that union. Conversely each \(\overline{I_k}\) is contained in the closure of the union by monotonicity of closure. (This is exactly where finiteness is used: an infinite union of closed sets need not be closed.)

Now \(B \subseteq \bigcup_{k=1}^{n} I_k\) by hypothesis, so

\begin{equation*} [0,1] \;=\; \bar{B} \;\subseteq\; \overline{\bigcup_{k=1}^{n} I_k} \;=\; \bigcup_{k=1}^{n} \overline{I_k}, \end{equation*}

the first equality holding because the rationals in \([0,1]\) are dense in \([0,1]\): every point of \([0,1]\) is a limit of rationals lying in \([0,1]\), and \([0,1]\) is closed, so \(\bar{B} = [0,1]\).

Apply monotonicity of outer measure and then finite subadditivity (the finite form of Proposition 3):

\begin{equation*} 1 \;=\; m^*([0,1]) \;\le\; m^*\Big( \bigcup_{k=1}^{n} \overline{I_k} \Big) \;\le\; \sum_{k=1}^{n} m^*\big( \overline{I_k} \big). \end{equation*}

Here \(m^*([0,1]) = \ell([0,1]) = 1\) by Proposition 1. Finally, \(\overline{I_k} = [a_k, b_k]\) is a closed bounded interval of length \(b_k - a_k = \ell(I_k)\), so by Proposition 1 again

\begin{equation*} m^*\big( \overline{I_k} \big) \;=\; \ell\big( \overline{I_k} \big) \;=\; \ell(I_k) \;=\; m^*(I_k). \end{equation*}

(If \(I_k\) is empty this reads \(0 = 0\).) Substituting,

\begin{equation*} 1 \;\le\; \sum_{k=1}^{n} m^*(I_k), \end{equation*}

which is the desired inequality.

A remark on why the hypothesis of finiteness cannot be dropped: \(B\) is countable, say \(B = \{q_1, q_2, \dots\}\), and for any \(\epsilon > 0\) the countable collection \(I_k = (q_k - \epsilon/2^{k+1},\, q_k + \epsilon/2^{k+1})\) is a collection of open intervals covering \(B\) with \(\sum_{k=1}^{\infty} \ell(I_k) = \epsilon\). Thus \(m^*(B) = 0\), and the conclusion \(\sum_k m^*(I_k) \ge 1\) genuinely depends on the cover being finite.

Both inequalities are immediate from the basic properties of outer measure established in Section 2.2.

First, \(B \subseteq A \cup B\), so by the monotonicity of outer measure

\begin{equation*} m^*(B) \;\le\; m^*(A \cup B). \end{equation*}

Monotonicity holds because every countable collection of open bounded intervals covering \(A \cup B\) also covers \(B\), so the infimum defining \(m^*(B)\) is taken over a larger collection of admissible sums than the infimum defining \(m^*(A \cup B)\).

Second, by the finite subadditivity of outer measure (the finite form of Proposition 3, obtained by taking \(E_1 = A\), \(E_2 = B\) and \(E_k = \emptyset\) for \(k > 2\)),

\begin{equation*} m^*(A \cup B) \;\le\; m^*(A) + m^*(B) \;=\; 0 + m^*(B) \;=\; m^*(B). \end{equation*}

Combining the two displays gives \(m^*(A \cup B) = m^*(B)\). Note that the argument is valid whether \(m^*(B)\) is finite or infinite: if \(m^*(B) = \infty\) the first inequality already forces \(m^*(A \cup B) = \infty\).

The inequality \(m^*(A \cup B) \le m^*(A) + m^*(B)\) is finite subadditivity (Proposition 3), and requires no hypothesis. The content of the problem is the reverse inequality, and the separation hypothesis is what supplies it. Since \(A \cup B\) is bounded it is contained in some bounded interval, so \(m^*(A \cup B) < \infty\) by monotonicity and Proposition 1.

Step 1 (a subdivision lemma). Let \(I\) be a nonempty bounded open interval, and let \(\eta > 0\). Then there are finitely many open intervals \(J_1, \dots, J_N\), each of length less than \(\alpha\), such that

\begin{equation*} I \subseteq \bigcup_{i=1}^{N} J_i \quad\text{and}\quad \sum_{i=1}^{N} \ell(J_i) < \ell(I) + \eta . \end{equation*}

To see this, write \(I = (c,d)\) and \(L = d - c > 0\). Choose a natural number \(N\) with \(L/N < \alpha/2\), put \(h = L/N\) and \(t_i = c + ih\) for \(i = 0, 1, \dots, N\), so that \(t_0 = c\), \(t_N = d\) and \(\bigcup_{i=1}^{N}[t_{i-1}, t_i] = [c,d]\). Choose

\begin{equation*} \delta = \min\Big\{ \frac{\alpha}{8}, \; \frac{\eta}{4N} \Big\} > 0 \end{equation*}

and set \(J_i = (t_{i-1} - \delta,\; t_i + \delta)\). Then \(I \subseteq [c,d] \subseteq \bigcup_{i=1}^N J_i\); each \(J_i\) has length \(h + 2\delta < \alpha/2 + \alpha/4 < \alpha\); and

\begin{equation*} \sum_{i=1}^{N} \ell(J_i) = Nh + 2N\delta = L + 2N\delta \le \ell(I) + \frac{\eta}{2} < \ell(I) + \eta . \end{equation*}

Step 2 (refining an arbitrary cover). Let \(\epsilon > 0\). By the definition of outer measure there is a countable collection \(\{I_k\}_{k=1}^{\infty}\) of open bounded intervals with

\begin{equation*} A \cup B \subseteq \bigcup_{k=1}^{\infty} I_k \quad\text{and}\quad \sum_{k=1}^{\infty} \ell(I_k) < m^*(A \cup B) + \epsilon . \end{equation*}

Discard any empty \(I_k\). For each \(k\) apply Step 1 with \(\eta = \epsilon/2^k\) to obtain finitely many open intervals \(\{J_{k,1}, \dots, J_{k,N_k}\}\), each of length less than \(\alpha\), which cover \(I_k\) and satisfy \(\sum_{i=1}^{N_k} \ell(J_{k,i}) < \ell(I_k) + \epsilon/2^k\). The collection \(\mathcal{J} = \{J_{k,i} : k \ge 1,\; 1 \le i \le N_k\}\) is countable (a countable union of finite collections), consists of open bounded intervals each of length less than \(\alpha\), covers \(\bigcup_k I_k\) and hence covers \(A \cup B\), and satisfies

\begin{equation*} \sum_{J \in \mathcal{J}} \ell(J) \;<\; \sum_{k=1}^{\infty} \Big( \ell(I_k) + \frac{\epsilon}{2^k} \Big) \;<\; m^*(A \cup B) + 2\epsilon . \end{equation*}

Step 3 (splitting the refined cover). If \(J\) is an interval with \(\ell(J) < \alpha\), then any two points \(x, y \in J\) satisfy \(|x - y| \le \ell(J) < \alpha\). Consequently no member of \(\mathcal{J}\) can meet both \(A\) and \(B\): if \(a \in A \cap J\) and \(b \in B \cap J\) we would get \(|a - b| < \alpha\), contradicting the hypothesis \(|a-b| \ge \alpha\).

Therefore we may split \(\mathcal{J}\) into two disjoint subcollections

\begin{equation*} \mathcal{J}_A = \{ J \in \mathcal{J} : J \cap A \ne \emptyset \}, \qquad \mathcal{J}_B = \mathcal{J} \setminus \mathcal{J}_A , \end{equation*}

and every \(J \in \mathcal{J}\) that meets \(B\) lies in \(\mathcal{J}_B\), by the previous paragraph. Since \(\mathcal{J}\) covers \(A\), and a member of \(\mathcal{J}\) contributes a point of \(A\) only if it meets \(A\), the subcollection \(\mathcal{J}_A\) covers \(A\); likewise \(\mathcal{J}_B\) contains every member of \(\mathcal{J}\) meeting \(B\) and hence covers \(B\). Both subcollections are countable collections of open bounded intervals, so by the definition of outer measure

\begin{equation*} m^*(A) \le \sum_{J \in \mathcal{J}_A} \ell(J), \qquad m^*(B) \le \sum_{J \in \mathcal{J}_B} \ell(J). \end{equation*}

(If one of the subcollections is empty, the corresponding set is empty and its outer measure is \(0\), so the inequality still holds.)

Step 4 (conclusion). Since \(\mathcal{J}_A\) and \(\mathcal{J}_B\) are disjoint subcollections of \(\mathcal{J}\) and all terms are nonnegative,

\begin{equation*} m^*(A) + m^*(B) \;\le\; \sum_{J \in \mathcal{J}_A} \ell(J) + \sum_{J \in \mathcal{J}_B} \ell(J) \;=\; \sum_{J \in \mathcal{J}} \ell(J) \;<\; m^*(A \cup B) + 2\epsilon . \end{equation*}

This holds for every \(\epsilon > 0\), so \(m^*(A) + m^*(B) \le m^*(A \cup B)\). Together with subadditivity,

\begin{equation*} m^*(A \cup B) = m^*(A) + m^*(B). \end{equation*}

Let \(\mathcal{A}\) be a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) with \((a, \infty) \in \mathcal{A}\) for every \(a \in \mathbb{R}\). Recall that a \(\sigma\)-algebra contains \(\emptyset\) and \(\mathbb{R}\), is closed under complements and under countable unions, and hence, by De Morgan’s Identities, is also closed under countable intersections: if \(\{E_k\}_{k=1}^{\infty} \subseteq \mathcal{A}\) then

\begin{equation*} \bigcap_{k=1}^{\infty} E_k = \Big( \bigcup_{k=1}^{\infty} E_k^{\,C} \Big)^{C} \in \mathcal{A}. \end{equation*}

Being closed under countable unions and intersections, \(\mathcal{A}\) is in particular closed under finite unions and intersections (pad a finite family with \(\emptyset\) or with \(\mathbb{R}\)).

We build up all interval types in turn. Fix real numbers \(a < b\).

Rays of the form \((-\infty, a]\). Since \((a, \infty) \in \mathcal{A}\) and \((-\infty, a] = (a, \infty)^{C}\), closure under complements gives \((-\infty, a] \in \mathcal{A}\).

Rays of the form \([a, \infty)\). Observe that

\begin{equation*} [a, \infty) \;=\; \bigcap_{n=1}^{\infty} \Big( a - \frac{1}{n},\, \infty \Big). \end{equation*}

Indeed, if \(x \ge a\) then \(x > a - 1/n\) for every \(n\); conversely if \(x > a - 1/n\) for every \(n\) then letting \(n \to \infty\) gives \(x \ge a\). Each set on the right lies in \(\mathcal{A}\), and \(\mathcal{A}\) is closed under countable intersections, so \([a, \infty) \in \mathcal{A}\).

Rays of the form \((-\infty, a)\). This is \([a, \infty)^{C}\), hence in \(\mathcal{A}\).

Bounded intervals. Using closure under finite intersections,

\begin{equation*} (a,b) = (a, \infty) \cap (-\infty, b), \qquad [a,b] = [a, \infty) \cap (-\infty, b], \end{equation*}

\begin{equation*} [a,b) = [a, \infty) \cap (-\infty, b), \qquad (a,b] = (a, \infty) \cap (-\infty, b], \end{equation*}

and all four right-hand sides are intersections of two members of \(\mathcal{A}\). Hence every bounded interval with endpoints \(a < b\), of any of the four types, belongs to \(\mathcal{A}\).

Degenerate intervals. The empty set is in \(\mathcal{A}\) by definition of a \(\sigma\)-algebra, and a singleton is

\begin{equation*} \{a\} = [a, \infty) \cap (-\infty, a] \in \mathcal{A}. \end{equation*}

The whole line and half-lines already covered. We have \(\mathbb{R} \in \mathcal{A}\), and the four unbounded types \((a,\infty)\), \([a,\infty)\), \((-\infty,a)\), \((-\infty,a]\) were treated above.

Every interval of real numbers is of one of the listed forms: it is empty, a singleton, a bounded interval with two distinct endpoints (open, closed, or half-open), a half-line of one of the four types, or all of \(\mathbb{R}\). Therefore \(\mathcal{A}\) contains all intervals.

Recall the definition given in Section 2.3: the Borel \(\sigma\)-algebra \(\mathcal{B}\) is the intersection of all the \(\sigma\)-algebras of subsets of \(\mathbb{R}\) that contain the open sets, and its members are the Borel sets. Two facts follow immediately from this definition.

First, \(\mathcal{B}\) really is a \(\sigma\)-algebra. An intersection of a family of \(\sigma\)-algebras of subsets of \(\mathbb{R}\) is again a \(\sigma\)-algebra: \(\emptyset\) and \(\mathbb{R}\) belong to every member of the family and hence to the intersection; if \(E\) belongs to the intersection then \(E^{C}\) belongs to each \(\sigma\)-algebra in the family and hence to the intersection; and if \(\{E_k\}_{k=1}^{\infty}\) lies in the intersection then \(\bigcup_{k} E_k\) lies in each \(\sigma\)-algebra of the family and hence in the intersection. (The family is nonempty, since the collection of all subsets of \(\mathbb{R}\) is a \(\sigma\)-algebra containing the open sets.)

Second, \(\mathcal{B}\) contains every open set, since every \(\sigma\)-algebra in the defining family does.

Now let \(a \in \mathbb{R}\). The ray \((a, \infty)\) is an open subset of \(\mathbb{R}\): if \(x > a\), then with \(r = x - a > 0\) we have \((x - r, x + r) \subseteq (a, \infty)\). Hence \((a, \infty) \in \mathcal{B}\) for every real \(a\).

So \(\mathcal{B}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) containing every interval of the form \((a, \infty)\). By Problem 11 of this chapter, \(\mathcal{B}\) contains all intervals. That is, every interval is a Borel set.

(For completeness, this also recovers part of Theorem 9: since the measurable sets form a \(\sigma\)-algebra containing all the open sets, \(\mathcal{B} \subseteq \mathcal{M}\), so every interval is measurable as well.)

Throughout, for a set \(E \subseteq \mathbb{R}\) and \(y \in \mathbb{R}\) we write \(E + y = \{x + y : x \in E\}\), and we use the elementary equivalence

\begin{equation*} x \in E + y \iff x - y \in E . \end{equation*}

From this equivalence, translation commutes with arbitrary unions and intersections: for any family \(\{E_\lambda\}\),

\begin{equation*} \Big( \bigcup_\lambda E_\lambda \Big) + y = \bigcup_\lambda (E_\lambda + y), \qquad \Big( \bigcap_\lambda E_\lambda \Big) + y = \bigcap_\lambda (E_\lambda + y), \end{equation*}

since in each case \(x\) belongs to the left-hand side if and only if \(x - y\) belongs to some (respectively, to every) \(E_\lambda\), which is exactly membership in the right-hand side.

We also record two topological facts.

The translate of an open set is open. Let \(\mathcal{O}\) be open and \(x \in \mathcal{O} + y\). Then \(x - y \in \mathcal{O}\), so there is \(r > 0\) with \((x - y - r,\, x - y + r) \subseteq \mathcal{O}\). Then \((x - r, x + r) \subseteq \mathcal{O} + y\), since \(|t - x| < r\) implies \(|(t-y) - (x-y)| < r\), so \(t - y \in \mathcal{O}\). Hence \(\mathcal{O} + y\) is open.

The translate of a closed set is closed. If \(C\) is closed then \(C^{C}\) is open, and complementation commutes with translation: \(x \in (C + y)^{C}\) if and only if \(x - y \notin C\), i.e. \(x - y \in C^{C}\), i.e. \(x \in C^{C} + y\). So \((C + y)^{C} = C^{C} + y\) is open by the previous paragraph, and therefore \(C + y\) is closed.

(i) Let \(F\) be an \(F_\sigma\) set, that is, \(F = \bigcup_{k=1}^{\infty} C_k\) with each \(C_k\) closed. Then

\begin{equation*} F + y = \Big( \bigcup_{k=1}^{\infty} C_k \Big) + y = \bigcup_{k=1}^{\infty} (C_k + y), \end{equation*}

and each \(C_k + y\) is closed. Thus \(F + y\) is a countable union of closed sets, i.e. an \(F_\sigma\) set.

(ii) Let \(G\) be a \(G_\delta\) set, that is, \(G = \bigcap_{k=1}^{\infty} \mathcal{O}_k\) with each \(\mathcal{O}_k\) open. Then

\begin{equation*} G + y = \Big( \bigcap_{k=1}^{\infty} \mathcal{O}_k \Big) + y = \bigcap_{k=1}^{\infty} (\mathcal{O}_k + y), \end{equation*}

and each \(\mathcal{O}_k + y\) is open. Thus \(G + y\) is a countable intersection of open sets, i.e. a \(G_\delta\) set.

(iii) Suppose \(m^*(E) = 0\). By Proposition 2, outer measure is translation invariant, so

\begin{equation*} m^*(E + y) = m^*(E) = 0 . \end{equation*}

By Proposition 4 a set of outer measure zero is measurable, so \(E + y\) is measurable with measure \(m(E + y) = 0\). (Alternatively, one may argue directly, as in the proof of Proposition 2: if \(\{I_k\}\) is a countable collection of open bounded intervals covering \(E\) with \(\sum_k \ell(I_k) < \epsilon\), then \(\{I_k + y\}\) is a countable collection of open bounded intervals covering \(E + y\) with the same total length, since \(I_k + y\) is an open interval of the same length as \(I_k\). Hence \(m^*(E + y) < \epsilon\) for every \(\epsilon > 0\).)

Let \(E \subseteq \mathbb{R}\) with \(m^*(E) > 0\). For each natural number \(n\) define

\begin{equation*} E_n = E \cap [-n, n]. \end{equation*}

Each \(E_n\) is a subset of \(E\) and is bounded, being contained in \([-n,n]\). Moreover

\begin{equation*} E = \bigcup_{n=1}^{\infty} E_n , \end{equation*}

since every real number \(x \in E\) satisfies \(|x| \le n\) for some natural number \(n\) (by the Archimedean property of \(\mathbb{R}\)), and hence lies in \(E_n\); the reverse inclusion is clear because each \(E_n \subseteq E\).

Suppose, for the sake of contradiction, that every bounded subset of \(E\) had outer measure zero. In particular \(m^*(E_n) = 0\) for every \(n\). By the countable subadditivity of outer measure (Proposition 3),

\begin{equation*} m^*(E) = m^*\Big( \bigcup_{n=1}^{\infty} E_n \Big) \;\le\; \sum_{n=1}^{\infty} m^*(E_n) = \sum_{n=1}^{\infty} 0 = 0 . \end{equation*}

Since outer measure is nonnegative, this gives \(m^*(E) = 0\), contradicting the hypothesis \(m^*(E) > 0\).

Therefore \(m^*(E_n) > 0\) for at least one natural number \(n\), and this \(E_n\) is a bounded subset of \(E\) with positive outer measure.

One can say slightly more: the sets \(E_n\) increase to \(E\), and by monotonicity \(m^*(E_1) \le m^*(E_2) \le \dots \le m^*(E)\). The argument above shows the limit of this increasing sequence cannot be \(0\) when \(m^*(E) > 0\); thus in fact, if \(N\) is an index with \(m^*(E_N) > 0\), then \(m^*(E_n) \ge m^*(E_N) > 0\) for every \(n \ge N\). Note also that the same reasoning works with any countable collection of bounded sets whose union is \(\mathbb{R}\) in place of \(\{[-n,n]\}_{n=1}^{\infty}\).

Let \(E\) be measurable with \(m(E) = m^*(E) < \infty\) and let \(\epsilon > 0\).

Slice \(E\) by the half-open intervals of length \(\epsilon\). For \(k \in \mathbb{Z}\) put

\begin{equation*} A_k = E \cap \big( k\epsilon,\ (k+1)\epsilon \big]. \end{equation*}

Each half-open interval is measurable (Proposition 8 gives measurability of every interval; alternatively \((k\epsilon,(k+1)\epsilon] = (k\epsilon, (k+1)\epsilon) \cup \{(k+1)\epsilon\}\) and singletons have outer measure zero, hence are measurable by Proposition 4), so each \(A_k\) is measurable, being the intersection of two measurable sets. The intervals \(\big(k\epsilon, (k+1)\epsilon\big]\), \(k \in \mathbb{Z}\), are pairwise disjoint and their union is all of \(\mathbb{R}\); hence the sets \(A_k\) are pairwise disjoint and

\begin{equation*} E = \bigcup_{k \in \mathbb{Z}} A_k . \end{equation*}

By monotonicity of outer measure and Proposition 1,

\begin{equation*} m(A_k) \le m^*\big( (k\epsilon, (k+1)\epsilon] \big) = \epsilon \quad \text{for every } k . \end{equation*}

Since \(\mathbb{Z}\) is countably infinite, fix a bijection \(\varphi \colon \mathbb{N} \to \mathbb{Z}\) and put \(B_k = A_{\varphi(k)}\) for each natural number \(k\). Then \(\{B_k\}_{k=1}^{\infty}\) is a countable disjoint collection of measurable sets whose union is \(E\), and \(m(B_k) \le \epsilon\) for every \(k\). I claim

\begin{equation*} \sum_{k=1}^{\infty} m(B_k) \le m(E) < \infty . \end{equation*}

Indeed, fix \(n\). Applying Proposition 6 with \(A = \mathbb{R}\) to the finite disjoint collection \(\{B_k\}_{k=1}^n\) of measurable sets gives

\begin{equation*} \sum_{k=1}^{n} m^*(B_k) = m^*\Big( \bigcup_{k=1}^{n} B_k \Big) \le m^*(E), \end{equation*}

the last inequality by monotonicity, since \(\bigcup_{k=1}^n B_k \subseteq E\). The bound \(m^*(E)\) does not depend on \(n\), so the series of nonnegative terms converges and its sum is at most \(m(E) < \infty\).

Since the series converges, choose a natural number \(N\) so large that

\begin{equation*} \sum_{k=N+1}^{\infty} m(B_k) \le \epsilon . \end{equation*}

Define

\begin{equation*} C = \bigcup_{k=N+1}^{\infty} B_k . \end{equation*}

Then \(C\) is measurable, being a countable union of measurable sets (Proposition 7), and by the countable subadditivity of outer measure (Proposition 3),

\begin{equation*} m( C) = m^*( C) \le \sum_{k=N+1}^{\infty} m(B_k) \le \epsilon . \end{equation*}

Finally,

\begin{equation*} E = B_1 \cup B_2 \cup \cdots \cup B_N \cup C \end{equation*}

is a disjoint union of finitely many measurable sets: the \(B_k\) are pairwise disjoint by construction, and \(C\) is disjoint from each of \(B_1, \dots, B_N\) because it is the union of the remaining members of that disjoint collection. Each of these \(N+1\) sets has measure at most \(\epsilon\), since \(m(B_k) \le \epsilon\) for every \(k\) and \(m( C) \le \epsilon\). This is the required decomposition.

Recall the four assertions of Theorem 11 for a set \(E\) of real numbers:

(i) for each \(\epsilon > 0\) there is an open set \(\mathcal{O}\) containing \(E\) for which \(m^*(\mathcal{O} \sim E) < \epsilon\);

(ii) there is a \(G_\delta\) set \(G\) containing \(E\) for which \(m^*(G \sim E) = 0\);

(iii) for each \(\epsilon > 0\) there is a closed set \(F\) contained in \(E\) for which \(m^*(E \sim F) < \epsilon\);

(iv) there is an \(F_\sigma\) set \(F\) contained in \(E\) for which \(m^*(E \sim F) = 0\).

The text established that measurability of \(E\) is equivalent to (i) and to (ii). We deduce the remaining two equivalences by complementation, using three facts: a set is measurable if and only if its complement is measurable (immediate from the Carathéodory definition, which is symmetric in \(E\) and \(E^C\)); a set is open if and only if its complement is closed; and a set is \(F_\sigma\) if and only if its complement is \(G_\delta\) (De Morgan’s Identities, since a countable union of closed sets has as complement the countable intersection of their open complements, and conversely).

The computational heart of the argument is the following identity. Let \(A\) and \(B\) be sets with \(B = A^C\). If \(F = \mathcal{O}^C\), then

\begin{equation*} \mathcal{O} \sim A = \mathcal{O} \cap A^{C} = F^{C} \cap B = B \cap F^{C} = B \sim F . \end{equation*}

So passing to complements interchanges an “outer” difference for \(A\) with an “inner” difference for \(B = A^C\), and the two differences are literally the same set. We use this repeatedly.

Measurability is equivalent to (iii).

Suppose first that \(E\) is measurable. Then \(E^C\) is measurable, so by the already proved equivalence of measurability with (i), applied to the set \(E^C\): given \(\epsilon > 0\) there is an open set \(\mathcal{O}\) with \(E^C \subseteq \mathcal{O}\) and \(m^*(\mathcal{O} \sim E^C) < \epsilon\). Put \(F = \mathcal{O}^C\). Then \(F\) is closed, and \(E^C \subseteq \mathcal{O}\) gives \(F = \mathcal{O}^C \subseteq E\). Moreover, by the identity above (with \(A = E^C\), \(B = E\)),

\begin{equation*} \mathcal{O} \sim E^{C} = \mathcal{O} \cap E = E \cap F^{C} = E \sim F , \end{equation*}

so \(m^*(E \sim F) < \epsilon\). Thus (iii) holds.

Conversely, suppose (iii) holds for \(E\). Let \(\epsilon > 0\) and choose a closed set \(F \subseteq E\) with \(m^*(E \sim F) < \epsilon\). Put \(\mathcal{O} = F^C\). Then \(\mathcal{O}\) is open and \(F \subseteq E\) gives \(E^C \subseteq F^C = \mathcal{O}\). Again

\begin{equation*} \mathcal{O} \sim E^{C} = \mathcal{O} \cap E = E \cap F^{C} = E \sim F , \end{equation*}

so \(m^*(\mathcal{O} \sim E^C) < \epsilon\). As \(\epsilon > 0\) was arbitrary, \(E^C\) satisfies (i) and is therefore measurable; hence \(E\) is measurable.

Measurability is equivalent to (iv).

Suppose \(E\) is measurable. Then \(E^C\) is measurable, so by the equivalence of measurability with (ii), applied to \(E^C\), there is a \(G_\delta\) set \(G\) containing \(E^C\) with \(m^*(G \sim E^C) = 0\). Put \(F = G^C\). Then \(F\) is an \(F_\sigma\) set, \(E^C \subseteq G\) gives \(F \subseteq E\), and

\begin{equation*} G \sim E^{C} = G \cap E = E \cap F^{C} = E \sim F , \end{equation*}

so \(m^*(E \sim F) = 0\). Thus (iv) holds.

Conversely, suppose (iv) holds for \(E\): there is an \(F_\sigma\) set \(F \subseteq E\) with \(m^*(E \sim F) = 0\). Put \(G = F^C\), a \(G_\delta\) set. From \(F \subseteq E\) we get \(E^C \subseteq G\), and as computed above \(G \sim E^C = E \sim F\), so \(m^*(G \sim E^C) = 0\). Hence \(E^C\) satisfies (ii) and is measurable, and therefore \(E\) is measurable.

(Alternatively, for this last step one may argue directly, without complementation: with \(F \subseteq E\) an \(F_\sigma\) set for which \(m^*(E \sim F) = 0\), the set \(E \sim F\) has outer measure zero and so is measurable by Proposition 4, while \(F\) is measurable by Theorem 9, since every \(F_\sigma\) set is a Borel set. Hence

\begin{equation*} E = F \cup [E \sim F] \end{equation*}

is measurable, being the union of two measurable sets, by Proposition 5.)

Suppose \(E\) is measurable. Let \(\epsilon > 0\). By assertion (i) of Theorem 11 there is an open set \(\mathcal{O}\) containing \(E\) with

\begin{equation*} m^*(\mathcal{O} \sim E) < \epsilon/2 , \end{equation*}

and by assertion (iii) of Theorem 11 (proved in Problem 16) there is a closed set \(F\) contained in \(E\) with

\begin{equation*} m^*(E \sim F) < \epsilon/2 . \end{equation*}

Then \(F \subseteq E \subseteq \mathcal{O}\). I claim

\begin{equation*} \mathcal{O} \sim F = [\mathcal{O} \sim E] \cup [E \sim F]. \end{equation*}

Indeed, if \(x \in \mathcal{O} \sim F\) then \(x \in \mathcal{O}\) and \(x \notin F\); either \(x \notin E\), in which case \(x \in \mathcal{O} \sim E\), or \(x \in E\), in which case \(x \in E \sim F\). Conversely \(\mathcal{O} \sim E \subseteq \mathcal{O} \sim F\) because \(F \subseteq E\), and \(E \sim F \subseteq \mathcal{O} \sim F\) because \(E \subseteq \mathcal{O}\). Hence, by the countable (in particular finite) subadditivity of outer measure, Proposition 3,

\begin{equation*} m^*(\mathcal{O} \sim F) \le m^*(\mathcal{O} \sim E) + m^*(E \sim F) < \epsilon/2 + \epsilon/2 = \epsilon . \end{equation*}

Conversely, suppose that for each \(\epsilon > 0\) there are a closed \(F\) and an open \(\mathcal{O}\) with \(F \subseteq E \subseteq \mathcal{O}\) and \(m^*(\mathcal{O} \sim F) < \epsilon\). Fix \(\epsilon > 0\) and take such \(F\) and \(\mathcal{O}\). Since \(F \subseteq E\),

\begin{equation*} \mathcal{O} \sim E = \mathcal{O} \cap E^{C} \subseteq \mathcal{O} \cap F^{C} = \mathcal{O} \sim F , \end{equation*}

so by the monotonicity of outer measure

\begin{equation*} m^*(\mathcal{O} \sim E) \le m^*(\mathcal{O} \sim F) < \epsilon . \end{equation*}

Thus for each \(\epsilon > 0\) there is an open set \(\mathcal{O}\) containing \(E\) with \(m^*(\mathcal{O} \sim E) < \epsilon\); that is, \(E\) satisfies assertion (i) of Theorem 11, and therefore \(E\) is measurable.

(One could equally well have used the inner half of the hypothesis in the converse: \(E \sim F \subseteq \mathcal{O} \sim F\) gives \(m^*(E \sim F) < \epsilon\), which is assertion (iii) of Theorem 11 and again forces \(E\) to be measurable.)

A word on the hypotheses before the proof. The \(G_\delta\) half of the conclusion holds for every set of finite outer measure. The \(F_\sigma\) half does not: as the Remark at the end shows, it holds precisely for the measurable sets, so the exercise is to be read within the standing context of this section, namely for a measurable set \(E\) of finite outer measure. We prove each half under the hypothesis it actually needs.

The outer half, for an arbitrary set \(E\) of finite outer measure. Let \(k\) be a natural number. Since \(m^*(E) < \infty\), by the definition of outer measure there is a countable collection \(\{I_j^{k}\}_{j=1}^{\infty}\) of open, bounded intervals covering \(E\) with

\begin{equation*} \sum_{j=1}^{\infty} \ell(I_j^{k}) < m^*(E) + 1/k . \end{equation*}

Set \(\mathcal{O}_k = \bigcup_{j=1}^{\infty} I_j^{k}\), an open set containing \(E\). By the definition of outer measure (or by Propositions 1 and 3),

\begin{equation*} m^*(\mathcal{O}_k) \le \sum_{j=1}^{\infty} \ell(I_j^{k}) < m^*(E) + 1/k . \end{equation*}

Define \(G = \bigcap_{k=1}^{\infty} \mathcal{O}_k\). Then \(G\) is a \(G_\delta\) set, \(E \subseteq G\), and for every \(k\), by monotonicity, \(m^*(G) \le m^*(\mathcal{O}_k) < m^*(E) + 1/k\). Letting \(k \to \infty\) gives \(m^*(G) \le m^*(E)\), while \(E \subseteq G\) gives \(m^*(E) \le m^*(G)\). Hence \(m^*(G) = m^*(E)\). This half used nothing about \(E\) beyond \(m^*(E) < \infty\).

The inner half. Assume \(E\) is measurable (see the remark below, which shows this hypothesis cannot be dropped). By assertion (iv) of Theorem 11 (proved in Problem 16) there is an \(F_\sigma\) set \(F\) contained in \(E\) with \(m^*(E \sim F) = 0\). Since \(E = F \cup [E \sim F]\), the countable subadditivity of outer measure (Proposition 3) gives

\begin{equation*} m^*(E) \le m^*(F) + m^*(E \sim F) = m^*(F), \end{equation*}

while monotonicity gives \(m^*(F) \le m^*(E)\) because \(F \subseteq E\). Hence \(m^*(F) = m^*(E)\).

Combining the two halves: for measurable \(E\) of finite outer measure there are an \(F_\sigma\) set \(F\) and a \(G_\delta\) set \(G\) with \(F \subseteq E \subseteq G\) and \(m^*(F) = m^*(E) = m^*(G)\).

Remark: measurability is exactly what is needed. The conclusion is in fact equivalent to the measurability of \(E\), so it fails for every nonmeasurable set of finite outer measure.

First, suppose that for a set \(E\) with \(m^*(E) < \infty\) there are an \(F_\sigma\) set \(F\) and a \(G_\delta\) set \(G\) with \(F \subseteq E \subseteq G\) and \(m^*(F) = m^*(E) = m^*(G)\). Both \(F\) and \(G\) are Borel sets, hence measurable by Theorem 9, and \(m(F) = m^*(E) < \infty\). Since \(F\) is a measurable set of finite outer measure contained in \(G\), the excision property (7) gives

\begin{equation*} m^*(G \sim F) = m^*(G) - m^*(F) = 0 . \end{equation*}

Now \(E \sim F \subseteq G \sim F\), so by monotonicity \(m^*(E \sim F) = 0\), and therefore \(E \sim F\) is measurable by Proposition 4. Consequently

\begin{equation*} E = F \cup [E \sim F] \end{equation*}

is measurable by Proposition 5.

Second, a concrete failure for a nonmeasurable set. This part draws on Section 2.6, which lies ahead of the present section; nothing in the proof above depends on it. Let \(V \subseteq [0,1]\) be a choice set for the rational equivalence relation on \([0,1]\), that is, a set consisting of exactly one member of each rational equivalence class of \([0,1]\); such a set exists by the Axiom of Choice, as in Section 2.6, and \(m^*(V) \le 1 < \infty\). Let \(\Lambda\) be the set of rational numbers in \([-1,1]\), a bounded, countably infinite set. If \(x \in [0,1]\), then by the second characteristic property of a choice set there is a \(c \in V\) with \(x = c + q\) for some rational \(q\), and \(q = x - c \in [-1,1]\) since both \(x\) and \(c\) lie in \([0,1]\); hence

\begin{equation*} [0,1] \subseteq \bigcup_{\lambda \in \Lambda} [\lambda + V]. \end{equation*}

By countable subadditivity (Proposition 3), the translation invariance of outer measure (Proposition 2), and Proposition 1, \(1 = m^*([0,1]) \le \sum_{\lambda \in \Lambda} m^*(V)\), which forces \(m^*(V) > 0\). On the other hand, let \(F_0\) be any closed subset of \(V\). Then \(F_0\) is bounded and measurable, and by the first characteristic property of a choice set, recorded as (16) in Section 2.6, the translates \(\{\lambda + V\}_{\lambda \in \Lambda}\) are disjoint; since \(F_0 \subseteq V\), the translates \(\{\lambda + F_0\}_{\lambda \in \Lambda}\) are disjoint as well. By Lemma 16, \(m(F_0) = 0\). Hence every \(F_\sigma\) set contained in \(V\), being a countable union of closed subsets of \(V\), has outer measure zero by Proposition 3. So no \(F_\sigma\) set \(F \subseteq V\) satisfies \(m^*(F) = m^*(V) > 0\).

Thus the exercise, read literally for an arbitrary set of finite outer measure, is false; the \(G_\delta\) statement holds in general, and the \(F_\sigma\) statement holds precisely for the measurable sets.

Let \(E\) be a nonmeasurable set with \(m^*(E) < \infty\).

Step 1: a uniform gap. By Theorem 11, measurability of \(E\) is equivalent to assertion (i): for each \(\epsilon > 0\) there is an open set containing \(E\) whose difference with \(E\) has outer measure less than \(\epsilon\). Since \(E\) is not measurable, (i) fails. The negation of (i) is precisely: there is an \(\epsilon_0 > 0\) such that

\begin{equation*} m^*(\mathcal{U} \sim E) \ge \epsilon_0 \quad \text{for every open set } \mathcal{U} \text{ containing } E. \end{equation*}

Call this property \((\ast)\).

Step 2: a tight open cover. Since \(m^*(E) < \infty\), by the definition of outer measure there is a countable collection \(\{I_k\}_{k=1}^{\infty}\) of open, bounded intervals covering \(E\) for which

\begin{equation*} \sum_{k=1}^{\infty} \ell(I_k) < m^*(E) + \epsilon_0 . \end{equation*}

Define \(\mathcal{O} = \bigcup_{k=1}^{\infty} I_k\). Then \(\mathcal{O}\) is an open set containing \(E\), and by the countable subadditivity of outer measure (Proposition 3) together with the equality of the outer measure of an interval and its length (Proposition 1),

\begin{equation*} m^*(\mathcal{O}) \le \sum_{k=1}^{\infty} \ell(I_k) < m^*(E) + \epsilon_0 < \infty . \end{equation*}

In particular \(\mathcal{O}\) has finite outer measure.

Step 3: the strict inequality. From the display of Step 2,

\begin{equation*} m^*(\mathcal{O}) - m^*(E) < \epsilon_0 , \end{equation*}

the subtraction being legitimate because \(m^*(E)\) is finite. On the other hand, \(\mathcal{O}\) is an open set containing \(E\), so \((\ast)\) applies to it and gives

\begin{equation*} m^*(\mathcal{O} \sim E) \ge \epsilon_0 . \end{equation*}

Combining the two,

\begin{equation*} m^*(\mathcal{O} \sim E) \ge \epsilon_0 > m^*(\mathcal{O}) - m^*(E), \end{equation*}

which is the required strict inequality. Note also that \(\mathcal{O} \sim E\) has finite outer measure, since \(\mathcal{O} \sim E \subseteq \mathcal{O}\).

This is exactly the point made in the Remark following Theorem 12: for any bounded set \(E\), measurable or not, one can always find an open \(\mathcal{O} \supseteq E\) with \(m^*(\mathcal{O}) - m^*(E) < \epsilon\); but the excision identity \(m^*(\mathcal{O} \sim E) = m^*(\mathcal{O}) - m^*(E)\) requires the measurability of \(E\), and for a nonmeasurable \(E\) the left-hand side stays bounded away from zero no matter how tightly \(\mathcal{O}\) is chosen.

Necessity. Suppose \(E\) is measurable. Let \((a,b)\) be an open, bounded interval. Taking \(A = (a,b)\) in the Carathéodory definition of measurability,

\begin{equation*} m^*\big( (a,b) \big) = m^*\big( (a,b) \cap E \big) + m^*\big( (a,b) \cap E^{C} \big) = m^*\big( (a,b) \cap E \big) + m^*\big( (a,b) \sim E \big). \end{equation*}

By Proposition 1, \(m^*((a,b)) = \ell((a,b)) = b - a\). This is the asserted identity, and it holds for measurable \(E\) whether or not \(m^*(E)\) is finite.

Sufficiency. Suppose \(m^*(E) < \infty\) and that

\begin{equation*} \ell(I) = m^*(I \cap E) + m^*(I \sim E) \end{equation*}

for every nonempty open, bounded interval \(I\); call this hypothesis \((\ast)\). We verify assertion (i) of Theorem 11.

Let \(\epsilon > 0\). Since \(m^*(E) < \infty\), the definition of outer measure provides a countable collection of open, bounded intervals covering \(E\) the sum of whose lengths is less than \(m^*(E) + \epsilon\); let \(\mathcal{O}\) be the union of that collection. Then \(\mathcal{O}\) is an open set containing \(E\) and, by Propositions 1 and 3,

\begin{equation*} m^*(\mathcal{O}) < m^*(E) + \epsilon < \infty . \end{equation*}

Every open set of real numbers is the disjoint union of a countable collection of open intervals (page 17). Write \(\mathcal{O} = \bigcup_{k} I_k\) with \(\{I_k\}\) a countable disjoint collection of open intervals; discarding empty ones, we may assume each \(I_k\) is nonempty. Each \(I_k\) is bounded: \(I_k \subseteq \mathcal{O}\), so by Proposition 1 and monotonicity \(\ell(I_k) = m^*(I_k) \le m^*(\mathcal{O}) < \infty\), and an unbounded interval has infinite length. So each \(I_k\) is an open, bounded interval and \((\ast)\) applies to it.

Two summation facts. First,

\begin{equation*} \sum_{k} \ell(I_k) \le m^*(\mathcal{O}). \end{equation*}

Indeed, each \(I_k\) is measurable (Proposition 8) with \(m^*(I_k) = \ell(I_k)\), so for each \(n\), Proposition 6 applied with \(A = \mathbb{R}\) to the finite disjoint collection \(\{I_k\}_{k=1}^{n}\) together with monotonicity gives

\begin{equation*} \sum_{k=1}^{n} \ell(I_k) = m^*\Big( \bigcup_{k=1}^{n} I_k \Big) \le m^*(\mathcal{O}), \end{equation*}

and the bound is independent of \(n\). Second, since \(E \subseteq \mathcal{O}\) we have \(E = \bigcup_k [I_k \cap E]\), so by countable subadditivity (Proposition 3),

\begin{equation*} m^*(E) \le \sum_{k} m^*(I_k \cap E). \end{equation*}

All the terms \(m^*(I_k \cap E) \le \ell(I_k)\) are finite, and both series converge, being dominated by \(m^*(\mathcal{O}) < \infty\).

Now \(\mathcal{O} \sim E = \bigcup_k [I_k \sim E]\), so by countable subadditivity and then \((\ast)\),

\begin{equation*} m^*(\mathcal{O} \sim E) \le \sum_{k} m^*(I_k \sim E) = \sum_{k} \big[ \ell(I_k) - m^*(I_k \cap E) \big] = \sum_{k} \ell(I_k) - \sum_{k} m^*(I_k \cap E), \end{equation*}

the splitting of the last series being legitimate since both series converge. Therefore, by the two summation facts,

\begin{equation*} m^*(\mathcal{O} \sim E) \le m^*(\mathcal{O}) - m^*(E) < \big[ m^*(E) + \epsilon \big] - m^*(E) = \epsilon . \end{equation*}

Thus for each \(\epsilon > 0\) there is an open set \(\mathcal{O}\) containing \(E\) with \(m^*(\mathcal{O} \sim E) < \epsilon\). By assertion (i) of Theorem 11, \(E\) is measurable.

Observe where the hypotheses were used: the finiteness of \(m^*(E)\) made the subtraction \(m^*(\mathcal{O}) - m^*(E)\) meaningful and produced the tight open cover, while \((\ast)\) converted the Carathéodory splitting condition, a priori demanded of every test set \(A\), into a condition on the single family of bounded open intervals. The content of the exercise is that intervals are enough test sets.

Throughout, the only properties of outer measure used are its monotonicity and its countable subadditivity (Proposition 3); no part of the Carathéodory theory is invoked, which is the point of the exercise.

Part 1: property (ii) as the definition.

Definition: a set \(E\) is measurable provided there is a \(G_\delta\) set \(G\) containing \(E\) for which \(m^*(G \sim E) = 0\).

We need one topological lemma.

Lemma A. The union of two \(G_\delta\) sets is a \(G_\delta\) set.

Proof of Lemma A. Let \(G_1 = \bigcap_{n=1}^{\infty} U_n\) and \(G_2 = \bigcap_{n=1}^{\infty} V_n\) with each \(U_n\), \(V_n\) open. Consider the countable collection of open sets \(\{U_n \cup V_m\}_{n,m \in \mathbb{N}}\) (a countable family, being indexed by \(\mathbb{N} \times \mathbb{N}\)), and set

\begin{equation*} H = \bigcap_{n,m} \big[ U_n \cup V_m \big], \end{equation*}

a \(G_\delta\) set. If \(x \in G_1\) then \(x \in U_n\) for all \(n\), so \(x \in U_n \cup V_m\) for all \(n, m\); hence \(G_1 \subseteq H\), and symmetrically \(G_2 \subseteq H\), so \(G_1 \cup G_2 \subseteq H\). Conversely, let \(x \in H\) and suppose \(x \notin G_1\). Then \(x \notin U_{n_0}\) for some index \(n_0\). Since \(x \in U_{n_0} \cup V_m\) for every \(m\), it follows that \(x \in V_m\) for every \(m\), that is, \(x \in G_2\). Hence \(H \subseteq G_1 \cup G_2\), and \(G_1 \cup G_2 = H\) is a \(G_\delta\) set.

Now let \(E_1\) and \(E_2\) be measurable in the sense of (ii). Choose \(G_\delta\) sets \(G_1 \supseteq E_1\) and \(G_2 \supseteq E_2\) with

\begin{equation*} m^*(G_1 \sim E_1) = m^*(G_2 \sim E_2) = 0 . \end{equation*}

Put \(G = G_1 \cup G_2\), a \(G_\delta\) set by Lemma A, and clearly \(E_1 \cup E_2 \subseteq G\). I claim

\begin{equation*} G \sim [E_1 \cup E_2] \subseteq [G_1 \sim E_1] \cup [G_2 \sim E_2]. \end{equation*}

Indeed, let \(x \in G\) with \(x \notin E_1 \cup E_2\). Then \(x \in G_1\) or \(x \in G_2\). In the first case \(x \in G_1\) and \(x \notin E_1\), so \(x \in G_1 \sim E_1\); in the second case, likewise, \(x \in G_2 \sim E_2\). This proves the inclusion.

By monotonicity and the subadditivity of outer measure,

\begin{equation*} m^*\big( G \sim [E_1 \cup E_2] \big) \le m^*(G_1 \sim E_1) + m^*(G_2 \sim E_2) = 0 . \end{equation*}

Thus \(G\) is a \(G_\delta\) set containing \(E_1 \cup E_2\) with \(m^*(G \sim [E_1 \cup E_2]) = 0\), so \(E_1 \cup E_2\) is measurable.

Part 2: property (iv) as the definition.

Definition: a set \(E\) is measurable provided there is an \(F_\sigma\) set \(F\) contained in \(E\) for which \(m^*(E \sim F) = 0\).

Lemma B. The union of two \(F_\sigma\) sets is an \(F_\sigma\) set.

Proof of Lemma B. If \(F_1 = \bigcup_{n=1}^{\infty} C_n\) and \(F_2 = \bigcup_{n=1}^{\infty} D_n\) with each \(C_n\), \(D_n\) closed, then

\begin{equation*} F_1 \cup F_2 = \bigcup_{n=1}^{\infty} \big[ C_n \cup D_n \big], \end{equation*}

a countable union of closed sets, since the union of two closed sets is closed. (Equivalently, interlace the two sequences.) Hence \(F_1 \cup F_2\) is \(F_\sigma\).

Now let \(E_1\) and \(E_2\) be measurable in the sense of (iv). Choose \(F_\sigma\) sets \(F_1 \subseteq E_1\) and \(F_2 \subseteq E_2\) with

\begin{equation*} m^*(E_1 \sim F_1) = m^*(E_2 \sim F_2) = 0 . \end{equation*}

Put \(F = F_1 \cup F_2\), an \(F_\sigma\) set by Lemma B, with \(F \subseteq E_1 \cup E_2\). I claim

\begin{equation*} [E_1 \cup E_2] \sim F \subseteq [E_1 \sim F_1] \cup [E_2 \sim F_2]. \end{equation*}

Indeed, let \(x \in E_1 \cup E_2\) with \(x \notin F = F_1 \cup F_2\); then \(x \notin F_1\) and \(x \notin F_2\). If \(x \in E_1\) then \(x \in E_1 \sim F_1\); otherwise \(x \in E_2\) and then \(x \in E_2 \sim F_2\). This proves the inclusion.

By monotonicity and subadditivity of outer measure,

\begin{equation*} m^*\big( [E_1 \cup E_2] \sim F \big) \le m^*(E_1 \sim F_1) + m^*(E_2 \sim F_2) = 0 . \end{equation*}

So \(F\) is an \(F_\sigma\) set contained in \(E_1 \cup E_2\) whose complement in \(E_1 \cup E_2\) has outer measure zero, and therefore \(E_1 \cup E_2\) is measurable.

By induction, both arguments extend at once to the union of any finite collection of measurable sets.

They are the same set function: \(m^{**}(A) = m^*(A)\) for every set \(A\) of real numbers. (That is, outer measure is outer regular: it is already the infimum of the outer measures of the open sets that contain the given set.)

First observe that \(m^{**}\) is well defined, since \(\mathbb{R}\) is an open set containing \(A\), so the set over which the infimum is taken is nonempty.

One inequality is immediate from monotonicity of outer measure. If \(\mathcal{O}\) is any open set with \(A \subseteq \mathcal{O}\), then \(m^*(A) \le m^*(\mathcal{O})\). Taking the infimum over all such \(\mathcal{O}\),

\begin{equation*} m^*(A) \le m^{**}(A). \end{equation*}

For the reverse inequality we use the definition of outer measure directly. If \(m^*(A) = \infty\) there is nothing to prove, since then \(m^{**}(A) \ge m^*(A) = \infty\) forces \(m^{**}(A) = \infty = m^*(A)\). So suppose \(m^*(A) < \infty\) and let \(\epsilon > 0\). By the definition of \(m^*(A)\) as an infimum, there is a countable collection \(\{ I_k \}_{k=1}^{\infty}\) of open, bounded intervals which covers \(A\) and for which

\begin{equation*} \sum_{k=1}^{\infty} \ell(I_k) < m^*(A) + \epsilon . \end{equation*}

Define \(\mathcal{O} = \bigcup_{k=1}^{\infty} I_k\). Then \(\mathcal{O}\) is open, being a union of open intervals, and \(A \subseteq \mathcal{O}\). By the countable subadditivity of outer measure (Proposition 3) together with the fact that the outer measure of an interval is its length (Proposition 1),

\begin{equation*} m^*(\mathcal{O}) \le \sum_{k=1}^{\infty} m^*(I_k) = \sum_{k=1}^{\infty} \ell(I_k) < m^*(A) + \epsilon . \end{equation*}

Since \(\mathcal{O}\) is one of the competitors in the infimum defining \(m^{**}(A)\),

\begin{equation*} m^{**}(A) \le m^*(\mathcal{O}) < m^*(A) + \epsilon . \end{equation*}

As \(\epsilon > 0\) was arbitrary, \(m^{**}(A) \le m^*(A)\).

Combining the two inequalities, \(m^{**}(A) = m^*(A)\) for every set \(A\).

Two remarks are worth recording. First, the argument used no measurability whatsoever: the equality holds for every set \(A\), measurable or not. So outer regularity, by itself, detects nothing about measurability.

Second, the equality must not be confused with the stronger assertion that the difference \(\mathcal{O} \sim A\) can be made small. That assertion is exactly assertion (i) of Theorem 11, and it is equivalent to the measurability of \(A\). The inference from one to the other would require the excision formula \(m^*(\mathcal{O} \sim A) = m^*(\mathcal{O}) - m^*(A)\), and excision is available only when \(A\) is measurable and of finite outer measure. Indeed Problem 19 shows that if \(A\) has finite outer measure and is not measurable, then there is an open set \(\mathcal{O}\) containing \(A\) of finite outer measure with \(m^*(\mathcal{O} \sim A) > m^*(\mathcal{O}) - m^*(A)\), even though, by what we have just proved, \(m^*(\mathcal{O}) - m^*(A)\) may be made arbitrarily small.

The set function \(m^{***}\) is the inner measure determined by closed sets. It never exceeds \(m^*\), and it agrees with \(m^*\) on every measurable set; conversely, for sets of finite outer measure, agreement forces measurability. Thus \(m^{***}\) and \(m^*\) are genuinely different set functions, and the sets on which they disagree are non-measurable. Precisely, we prove:

(1) \(m^{***}(A) \le m^*(A)\) for every set \(A\);

(2) if \(A\) is measurable, then \(m^{***}(A) = m^*(A)\);

(3) if \(m^*(A) < \infty\) and \(m^{***}(A) = m^*(A)\), then \(A\) is measurable;

(4) neither the finiteness hypothesis in (3) nor the converse in general can be dropped, and there are sets for which the inequality in (1) is strict.

Note first that \(m^{***}\) is well defined: the empty set is a closed subset of \(A\), so the supremum is over a nonempty set of numbers.

Proof of (1). If \(F\) is closed and \(F \subseteq A\), then by the monotonicity of outer measure \(m^*(F) \le m^*(A)\). Taking the supremum over all such \(F\) gives \(m^{***}(A) \le m^*(A)\).

Proof of (2). Let \(A\) be measurable and let \(\epsilon > 0\). By assertion (iii) of Theorem 11 there is a closed set \(F\) contained in \(A\) for which \(m^*(A \sim F) < \epsilon\). Since \(A = F \cup (A \sim F)\), the finite subadditivity of outer measure gives

\begin{equation*} m^*(A) \le m^*(F) + m^*(A \sim F) < m^*(F) + \epsilon \le m^{***}(A) + \epsilon . \end{equation*}

This holds for every \(\epsilon > 0\), whence \(m^*(A) \le m^{***}(A)\); with (1) this gives equality. The argument covers the case \(m^*(A) = \infty\) as well: the displayed inequality then forces \(m^*(F) = \infty\), so \(m^{***}(A) = \infty = m^*(A)\).

Proof of (3). Suppose \(m^*(A) < \infty\) and \(m^{***}(A) = m^*(A)\). By the definition of the supremum, for each natural number \(n\) there is a closed set \(F_n \subseteq A\) with

\begin{equation*} m(F_n) > m^*(A) - 1/n . \end{equation*}

Closed sets are measurable, so each \(F_n\) is measurable, and hence so is \(F = \bigcup_{n=1}^{\infty} F_n\), an \(F_{\sigma}\) set contained in \(A\). Since \(F_n \subseteq F \subseteq A\), monotonicity gives

\begin{equation*} m^*(A) - 1/n < m(F_n) \le m(F) \le m^*(A) \quad \text{for all } n, \end{equation*}

so \(m(F) = m^*(A)\).

Next we produce a measurable set containing \(A\) of the same measure. By Problem 22 (equivalently, by the definition of outer measure), for each \(n\) there is an open set \(\mathcal{O}_n \supseteq A\) with \(m^*(\mathcal{O}_n) < m^*(A) + 1/n\). Put \(G = \bigcap_{n=1}^{\infty} \mathcal{O}_n\), a \(G_{\delta}\) set containing \(A\); it is measurable, and by monotonicity \(m(G) \le m^*(\mathcal{O}_n) < m^*(A) + 1/n\) for every \(n\), while \(m(G) \ge m^*(A)\) since \(A \subseteq G\). Hence \(m(G) = m^*(A)\).

Now \(F \subseteq A \subseteq G\) with \(F\) and \(G\) measurable and \(m(F) = m(G) = m^*(A) < \infty\). By the excision property of measurable sets,

\begin{equation*} m(G \sim F) = m(G) - m(F) = 0 . \end{equation*}

Since \(A \sim F \subseteq G \sim F\), monotonicity gives \(m^*(A \sim F) = 0\), and a set of outer measure zero is measurable (Proposition 4). Therefore

\begin{equation*} A = F \cup (A \sim F) \end{equation*}

is the union of two measurable sets and so is measurable.

Proof of (4). Let \(V = \mathcal{C}_{[0,1]}\) be a choice set for the rational equivalence relation on \([0, 1]\), as constructed in Section 2.6. Then \(V \subseteq [0, 1]\), and \(V\) is not measurable: this is precisely the content of the proof of Theorem 17 (Vitali), applied to the set \(E = [0,1]\), which has positive outer measure. Moreover, by the first characteristic property of a choice set, recorded as (16) in Section 2.6, for any set \(\Lambda \subseteq \mathbb{Q}\) the collection of translates \(\{ \lambda + V \}_{\lambda \in \Lambda}\) is disjoint.

Now let \(F\) be any closed set with \(F \subseteq V\), and put \(\Lambda = [-1, 1] \cap \mathbb{Q}\), a bounded, countably infinite set of real numbers. Since \(F \subseteq V\), the translates \(\{ \lambda + F \}_{\lambda \in \Lambda}\) are contained in the corresponding translates of \(V\) and hence are disjoint. Also \(F\) is bounded, being a subset of \([0,1]\), and \(F\) is measurable, being closed. Lemma 16 therefore gives \(m(F) = 0\). As \(F\) was an arbitrary closed subset of \(V\),

\begin{equation*} m^{***}(V) = 0 < m^*(V), \end{equation*}

the last inequality because \(m^*(V) = 0\) would make \(V\) measurable. So the inequality in (1) can be strict.

Finally, the finiteness hypothesis in (3) cannot be omitted. Put \(A = \mathbb{R} \sim V\). Then \(A\) is not measurable, for otherwise its complement \(V\) would be measurable. But \(A\) contains the closed set \([2, \infty)\), so

\begin{equation*} m^{***}(A) = \infty = m^*(A), \end{equation*}

the right-hand equality holding since \(A \supseteq [2, \infty)\). Thus a non-measurable set of infinite outer measure may still satisfy \(m^{***}(A) = m^*(A)\).

In summary: \(m^{***} \le m^*\) always, with equality on all measurable sets; among sets of finite outer measure the two set functions agree exactly on the measurable sets. In particular, \(m^{***}\) is not countably additive on all sets, and it is the natural “inner” counterpart of the “outer” description of \(m^*\) obtained in Problem 22.

All four sets appearing in the identity are measurable: by Theorem 9 the collection \(\mathcal{M}\) of measurable sets is a \(\sigma\)-algebra, so it is closed with respect to the formation of complements and countable unions, and hence, by De Morgan’s Identities, also with respect to finite unions, finite intersections and relative complements. In particular \(E_2 \sim E_1 = E_2 \cap E_1^{C}\) is measurable. Every term is therefore a well-defined element of \([0, \infty]\), and we prove the identity as an identity in \([0, \infty]\), where addition is commutative, associative, and satisfies \(a + \infty = \infty\). Notice that no subtraction occurs anywhere in the argument, so no finiteness hypothesis is needed.

Decompose the union. The sets \(E_1\) and \(E_2 \sim E_1\) are disjoint measurable sets whose union is \(E_1 \cup E_2\). Indeed, they are disjoint because \(E_2 \sim E_1\) misses \(E_1\), and their union is

\begin{equation*} E_1 \cup (E_2 \sim E_1) = E_1 \cup (E_2 \cap E_1^{C}) = (E_1 \cup E_2) \cap (E_1 \cup E_1^{C}) = E_1 \cup E_2 . \end{equation*}

Hence, by the finite additivity of Lebesgue measure over disjoint measurable sets (Proposition 6 with \(A = \mathbb{R}\), or Proposition 13), we obtain the first decomposition

\begin{equation*} m(E_1 \cup E_2) = m(E_1) + m(E_2 \sim E_1). \end{equation*}

Decompose \(E_2\). The sets \(E_1 \cap E_2\) and \(E_2 \sim E_1\) are disjoint measurable sets, and

\begin{equation*} (E_1 \cap E_2) \cup (E_2 \sim E_1) = E_2 \cap (E_1 \cup E_1^{C}) = E_2 . \end{equation*}

Hence, again by finite additivity, we obtain the second decomposition

\begin{equation*} m(E_2) = m(E_1 \cap E_2) + m(E_2 \sim E_1). \end{equation*}

Now add \(m(E_1 \cap E_2)\) to both sides of the first decomposition and substitute the second:

\begin{equation*} \begin{aligned} m(E_1 \cup E_2) + m(E_1 \cap E_2) &= m(E_1) + m(E_2 \sim E_1) + m(E_1 \cap E_2) \\ &= m(E_1) + \big[ m(E_1 \cap E_2) + m(E_2 \sim E_1) \big] \\ &= m(E_1) + m(E_2). \end{aligned} \end{equation*}

This is the asserted identity.

Two consequences are worth noting. If \(m(E_1 \cap E_2) < \infty\) we may cancel it and obtain the inclusion-exclusion formula

\begin{equation*} m(E_1 \cup E_2) = m(E_1) + m(E_2) - m(E_1 \cap E_2), \end{equation*}

and in particular, whether or not the measures are finite, \(m(E_1 \cup E_2) \le m(E_1) + m(E_2)\), with equality precisely when \(m(E_1 \cap E_2) = 0\) or both sides are infinite.

Part (ii) of Theorem 15 asserts that if \(\{ B_k \}_{k=1}^{\infty}\) is a descending collection of measurable sets and \(m(B_1) < \infty\), then

\begin{equation*} m \left( \bigcap_{k=1}^{\infty} B_k \right) = \lim_{k \to \infty} m(B_k). \end{equation*}

We exhibit a descending collection of measurable sets for which this conclusion fails; since the only hypothesis used besides descent is the finiteness of \(m(B_1)\), this shows that hypothesis cannot be omitted.

For each natural number \(k\) define

\begin{equation*} B_k = [k, \infty). \end{equation*}

Each \(B_k\) is an interval, hence measurable by Proposition 8, and the collection is descending: \(B_{k+1} = [k+1, \infty) \subseteq [k, \infty) = B_k\). Each \(B_k\) has infinite measure. Indeed, for every natural number \(n\) the interval \([k, k+n]\) is contained in \(B_k\), so by the monotonicity of outer measure and Proposition 1,

\begin{equation*} m(B_k) \ge m^*([k, k+n]) = \ell([k, k+n]) = n , \end{equation*}

and since \(n\) is arbitrary, \(m(B_k) = \infty\). (This is the assertion of Proposition 1 for unbounded intervals.) In particular \(m(B_1) = \infty\), so the hypothesis in question fails. Consequently

\begin{equation*} \lim_{k \to \infty} m(B_k) = \infty . \end{equation*}

On the other hand,

\begin{equation*} \bigcap_{k=1}^{\infty} B_k = \bigcap_{k=1}^{\infty} [k, \infty) = \emptyset , \end{equation*}

because a real number \(x\) belongs to this intersection only if \(x \ge k\) for every natural number \(k\), which contradicts the Archimedean property of \(\mathbb{R}\). Since \(m(\emptyset) = 0\),

\begin{equation*} m \left( \bigcap_{k=1}^{\infty} B_k \right) = 0 \ne \infty = \lim_{k \to \infty} m(B_k), \end{equation*}

and the conclusion of part (ii) fails.

No counterexample of finite measure exists, of course: if \(m(B_1) < \infty\) then Theorem 15 (ii) applies and the conclusion holds. The failure above is caused precisely by the infinitude of the measures involved, and it is instructive to see where the proof of Theorem 15 (ii) breaks down. That proof sets \(D_k = B_1 \sim B_k\), applies part (i) to the ascending collection \(\{ D_k \}\), and then cancels the term \(m(B_1)\) from the identity

\begin{equation*} m(B_1) - m \left( \bigcap_{k=1}^{\infty} B_k \right) = \lim_{k \to \infty} \big[ m(B_1) - m(B_k) \big] . \end{equation*}

Both the excisions producing this identity and the cancellation require \(m(B_1) < \infty\). With \(B_k = [k, \infty)\) the identity degenerates to the vacuous statement \(\infty - 0 = \infty - \infty\).

It is worth recording exactly how much of the hypothesis is needed. What the proof of Theorem 15 (ii) actually uses is that \(m(B_{k})\) be finite for the sets it excises, and a descending collection with \(m(B_{k_0}) < \infty\) for some index \(k_0\) may be replaced by the descending collection \(\{ B_k \}_{k \ge k_0}\), which has the same intersection and the same limit of measures. So the correct sharp hypothesis is that \(m(B_{k_0}) < \infty\) for some index \(k_0\); the example \(B_k = [k, \infty)\) shows that even this weaker hypothesis cannot be dropped, since there every \(B_k\) has infinite measure.

Note that \(A\) is an arbitrary set of real numbers; no measurability is assumed of \(A\). This is the countable version of Proposition 6, which asserts the corresponding identity for a finite disjoint collection of measurable sets. We prove the two inequalities.

First, the inequality \(\le\). Since intersection distributes over union,

\begin{equation*} A \cap \bigcup_{k=1}^{\infty} E_k = \bigcup_{k=1}^{\infty} (A \cap E_k), \end{equation*}

so the countable subadditivity of outer measure (Proposition 3) gives

\begin{equation*} m^* \left( A \cap \bigcup_{k=1}^{\infty} E_k \right) \le \sum_{k=1}^{\infty} m^*(A \cap E_k). \end{equation*}

Observe that this half needed neither the measurability of the \(E_k\) nor their disjointness.

Now the inequality \(\ge\). Fix a natural number \(n\). The collection \(\{ E_k \}_{k=1}^{n}\) is a finite disjoint collection of measurable sets, so by Proposition 6, applied to the set \(A\),

\begin{equation*} m^* \left( A \cap \bigcup_{k=1}^{n} E_k \right) = \sum_{k=1}^{n} m^*(A \cap E_k). \end{equation*}

On the other hand,

\begin{equation*} A \cap \bigcup_{k=1}^{n} E_k \subseteq A \cap \bigcup_{k=1}^{\infty} E_k , \end{equation*}

so by the monotonicity of outer measure,

\begin{equation*} \sum_{k=1}^{n} m^*(A \cap E_k) = m^* \left( A \cap \bigcup_{k=1}^{n} E_k \right) \le m^* \left( A \cap \bigcup_{k=1}^{\infty} E_k \right). \end{equation*}

The right-hand side is independent of \(n\). Every partial sum of the series of nonnegative terms \(\sum_{k=1}^{\infty} m^*(A \cap E_k)\) is therefore bounded above by the quantity \(m^*(A \cap \bigcup_{k=1}^{\infty} E_k)\). Since the sum of a series of nonnegative terms is by definition the supremum in \([0, \infty]\) of its partial sums, we conclude

\begin{equation*} \sum_{k=1}^{\infty} m^*(A \cap E_k) \le m^* \left( A \cap \bigcup_{k=1}^{\infty} E_k \right). \end{equation*}

This argument is valid in \([0, \infty]\) without any finiteness assumption: if some term \(m^*(A \cap E_k)\) is infinite, or if the partial sums are unbounded, the displayed bound simply forces the right-hand side to equal \(\infty\) as well.

The two inequalities together give the asserted equality.

Taking \(A = \mathbb{R}\) recovers the countable additivity of Lebesgue measure, Proposition 13. Conversely, the identity just proved is the natural strengthening of that proposition: the countable additivity of \(m^*\) persists after intersecting with an entirely arbitrary “test set” \(A\), so long as the pieces \(E_k\) are disjoint and measurable.

Throughout, all sets named belong to \(\mathcal{M}^{\prime}\). Recall that a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) contains \(\mathbb{R}\) and is closed with respect to the formation of complements and countable unions; by De Morgan’s Identities it is closed with respect to countable intersections, and hence also with respect to finite unions, finite intersections and relative complements \(B \sim A = B \cap A^{C}\). Note also that \(\emptyset = \mathbb{R}^{C} \in \mathcal{M}^{\prime}\). Countable additivity means: if \(\{ E_k \}_{k=1}^{\infty}\) is a countable disjoint collection of sets in \(\mathcal{M}^{\prime}\), then

\begin{equation*} m^{\prime} \left( \bigcup_{k=1}^{\infty} E_k \right) = \sum_{k=1}^{\infty} m^{\prime}(E_k). \end{equation*}

Since the terms are nonnegative, such a series has a well-defined sum in \([0, \infty]\) independent of the order of summation.

Part (i).

Finite additivity. Let \(\{ E_k \}_{k=1}^{n}\) be a finite disjoint collection in \(\mathcal{M}^{\prime}\). Define \(E_k = \emptyset\) for \(k > n\). Then \(\{ E_k \}_{k=1}^{\infty}\) is a countable disjoint collection in \(\mathcal{M}^{\prime}\) whose union is \(\bigcup_{k=1}^{n} E_k\), so countable additivity and \(m^{\prime}(\emptyset) = 0\) give

\begin{equation*} m^{\prime} \left( \bigcup_{k=1}^{n} E_k \right) = \sum_{k=1}^{\infty} m^{\prime}(E_k) = \sum_{k=1}^{n} m^{\prime}(E_k) + \sum_{k=n+1}^{\infty} 0 = \sum_{k=1}^{n} m^{\prime}(E_k). \end{equation*}

This is where the hypothesis \(m^{\prime}(\emptyset) = 0\) is used; without it, countable additivity applied to the collection all of whose members are \(\emptyset\) would force \(m^{\prime}(\emptyset) \in \{ 0, \infty \}\), and the value \(\infty\) would make \(m^{\prime}\) identically \(\infty\).

Monotonicity. Let \(A, B \in \mathcal{M}^{\prime}\) with \(A \subseteq B\). Then \(B \sim A \in \mathcal{M}^{\prime}\), the sets \(A\) and \(B \sim A\) are disjoint, and their union is \(B\). By finite additivity,

\begin{equation*} m^{\prime}(B) = m^{\prime}(A) + m^{\prime}(B \sim A) \ge m^{\prime}(A), \end{equation*}

since \(m^{\prime}(B \sim A) \ge 0\).

Excision. Retain the notation of the previous paragraph and suppose in addition that \(m^{\prime}(A) < \infty\). Then the identity \(m^{\prime}(B) = m^{\prime}(A) + m^{\prime}(B \sim A)\) may be rearranged, the subtraction being legitimate because \(m^{\prime}(A)\) is finite:

\begin{equation*} m^{\prime}(B \sim A) = m^{\prime}(B) - m^{\prime}(A). \end{equation*}

(If \(m^{\prime}(B) = \infty\) this reads \(m^{\prime}(B \sim A) = \infty\), which is correct.)

Countable monotonicity. Suppose \(E \in \mathcal{M}^{\prime}\) and \(\{ E_k \}_{k=1}^{\infty}\) is a countable collection in \(\mathcal{M}^{\prime}\) with \(E \subseteq \bigcup_{k=1}^{\infty} E_k\). We must show

\begin{equation*} m^{\prime}(E) \le \sum_{k=1}^{\infty} m^{\prime}(E_k). \end{equation*}

Disjointify the covering exactly as in the discussion preceding Proposition 7: define \(E_1^{\prime} = E_1\) and, for \(k \ge 2\),

\begin{equation*} E_k^{\prime} = E_k \sim \bigcup_{i=1}^{k-1} E_i . \end{equation*}

Because \(\mathcal{M}^{\prime}\) is closed under finite unions and relative complements, each \(E_k^{\prime}\) belongs to \(\mathcal{M}^{\prime}\); the collection \(\{ E_k^{\prime} \}_{k=1}^{\infty}\) is disjoint, \(E_k^{\prime} \subseteq E_k\), and \(\bigcup_{k=1}^{\infty} E_k^{\prime} = \bigcup_{k=1}^{\infty} E_k\). Now set \(F_k = E \cap E_k^{\prime} \in \mathcal{M}^{\prime}\). The collection \(\{ F_k \}_{k=1}^{\infty}\) is disjoint, and since \(E\) is contained in the union of the \(E_k^{\prime}\),

\begin{equation*} \bigcup_{k=1}^{\infty} F_k = E \cap \bigcup_{k=1}^{\infty} E_k^{\prime} = E . \end{equation*}

By countable additivity and then by monotonicity, using \(F_k \subseteq E_k^{\prime} \subseteq E_k\),

\begin{equation*} m^{\prime}(E) = \sum_{k=1}^{\infty} m^{\prime}(F_k) \le \sum_{k=1}^{\infty} m^{\prime}(E_k). \end{equation*}

Part (ii). We show that \(m^{\prime}\) has both continuity properties of Theorem 15. The proofs are those of Theorem 15 verbatim, and they use only countable additivity, monotonicity and excision, all of which were established in part (i).

Continuity for ascending collections. Let \(\{ A_k \}_{k=1}^{\infty}\) be an ascending collection in \(\mathcal{M}^{\prime}\), that is, \(A_k \subseteq A_{k+1}\) for each \(k\). We claim

\begin{equation*} m^{\prime} \left( \bigcup_{k=1}^{\infty} A_k \right) = \lim_{k \to \infty} m^{\prime}(A_k). \end{equation*}

By monotonicity the sequence \(\{ m^{\prime}(A_k) \}\) is nondecreasing, so the limit exists in \([0, \infty]\).

Suppose first that \(m^{\prime}(A_{k_0}) = \infty\) for some index \(k_0\). Then by monotonicity \(m^{\prime}(A_k) = \infty\) for all \(k \ge k_0\), so the right-hand side is \(\infty\); and \(A_{k_0} \subseteq \bigcup_{k=1}^{\infty} A_k\), so the left-hand side is \(\infty\) too. The claim holds.

Now suppose \(m^{\prime}(A_k) < \infty\) for every \(k\). Define \(A_0 = \emptyset\) and, for \(k \ge 1\), \(C_k = A_k \sim A_{k-1} \in \mathcal{M}^{\prime}\). Because the collection is ascending, \(\{ C_k \}_{k=1}^{\infty}\) is disjoint and

\begin{equation*} \bigcup_{k=1}^{\infty} C_k = \bigcup_{k=1}^{\infty} A_k . \end{equation*}

Indeed, if \(x\) lies in the union of the \(A_k\), there is a smallest index \(k\) with \(x \in A_k\), and then \(x \in C_k\); the reverse inclusion is clear since \(C_k \subseteq A_k\). Disjointness holds because for \(j < k\) we have \(C_j \subseteq A_j \subseteq A_{k-1}\) while \(C_k\) misses \(A_{k-1}\). By countable additivity and then excision (legitimate since \(A_{k-1} \subseteq A_k\) and \(m^{\prime}(A_{k-1}) < \infty\)),

\begin{equation*} \begin{aligned} m^{\prime} \left( \bigcup_{k=1}^{\infty} A_k \right) &= \sum_{k=1}^{\infty} m^{\prime}(A_k \sim A_{k-1}) = \sum_{k=1}^{\infty} \big[ m^{\prime}(A_k) - m^{\prime}(A_{k-1}) \big] \\ &= \lim_{n \to \infty} \sum_{k=1}^{n} \big[ m^{\prime}(A_k) - m^{\prime}(A_{k-1}) \big] = \lim_{n \to \infty} \big[ m^{\prime}(A_n) - m^{\prime}(A_0) \big] = \lim_{n \to \infty} m^{\prime}(A_n), \end{aligned} \end{equation*}

the telescoping being valid because every term is a difference of finite numbers, and \(m^{\prime}(A_0) = m^{\prime}(\emptyset) = 0\).

Continuity for descending collections. Let \(\{ B_k \}_{k=1}^{\infty}\) be a descending collection in \(\mathcal{M}^{\prime}\), that is, \(B_{k+1} \subseteq B_k\) for each \(k\), and suppose \(m^{\prime}(B_1) < \infty\). We claim

\begin{equation*} m^{\prime} \left( \bigcap_{k=1}^{\infty} B_k \right) = \lim_{k \to \infty} m^{\prime}(B_k). \end{equation*}

The intersection belongs to \(\mathcal{M}^{\prime}\), and by monotonicity \(\{ m^{\prime}(B_k) \}\) is nonincreasing and bounded below by \(0\), so the limit exists. Also \(m^{\prime}(B_k) \le m^{\prime}(B_1) < \infty\) for every \(k\), and likewise \(m^{\prime}(\bigcap_{k} B_k) < \infty\); every excision performed below is therefore legitimate.

Define \(D_k = B_1 \sim B_k \in \mathcal{M}^{\prime}\). Since \(\{ B_k \}\) is descending, \(\{ D_k \}\) is ascending, so by the case already proved,

\begin{equation*} m^{\prime} \left( \bigcup_{k=1}^{\infty} D_k \right) = \lim_{k \to \infty} m^{\prime}(D_k). \end{equation*}

By De Morgan’s Identities,

\begin{equation*} \bigcup_{k=1}^{\infty} D_k = \bigcup_{k=1}^{\infty} \left[ B_1 \sim B_k \right] = B_1 \sim \bigcap_{k=1}^{\infty} B_k . \end{equation*}

By excision, \(m^{\prime}(D_k) = m^{\prime}(B_1) - m^{\prime}(B_k)\) for each \(k\), and \(m^{\prime}(B_1 \sim \bigcap_{k} B_k) = m^{\prime}(B_1) - m^{\prime}(\bigcap_{k} B_k)\). Substituting,

\begin{equation*} m^{\prime}(B_1) - m^{\prime} \left( \bigcap_{k=1}^{\infty} B_k \right) = \lim_{k \to \infty} \big[ m^{\prime}(B_1) - m^{\prime}(B_k) \big] = m^{\prime}(B_1) - \lim_{k \to \infty} m^{\prime}(B_k). \end{equation*}

Since \(m^{\prime}(B_1)\) is finite it may be cancelled from both sides, and the claim follows.

Thus every countably additive \([0, \infty]\)-valued set function on a \(\sigma\)-algebra which vanishes on the empty set, that is, every measure, enjoys all the elementary properties of Lebesgue measure established in Sections 2.4 and 2.5. Nothing particular to \(\mathbb{R}\), to intervals, or to the construction of \(m^*\) was used; this is the observation that will be taken as the starting point of the general theory of measure.

The precise assertion to be proved is the converse of the implication established in Problem 27. Let \(\mathcal{M}^{\prime}\) be a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) and let \(m^{\prime}\) be a set function on \(\mathcal{M}^{\prime}\) taking values in \([0, \infty]\) with \(m^{\prime}(\emptyset) = 0\). Assume:

(a) \(m^{\prime}\) is finitely additive, that is, if \(\{ E_k \}_{k=1}^{n}\) is a finite disjoint collection in \(\mathcal{M}^{\prime}\), then \(m^{\prime}(\bigcup_{k=1}^{n} E_k) = \sum_{k=1}^{n} m^{\prime}(E_k)\); and

(b) \(m^{\prime}\) is continuous, that is, it possesses the two continuity properties of Theorem 15; in fact only the property for ascending collections is required: if \(\{ A_n \}_{n=1}^{\infty}\) is an ascending collection in \(\mathcal{M}^{\prime}\), then \(m^{\prime}(\bigcup_{n=1}^{\infty} A_n) = \lim_{n \to \infty} m^{\prime}(A_n)\).

Then \(m^{\prime}\) is countably additive.

Proof. Let \(\{ E_k \}_{k=1}^{\infty}\) be a countable disjoint collection of sets in \(\mathcal{M}^{\prime}\), and set

\begin{equation*} E = \bigcup_{k=1}^{\infty} E_k . \end{equation*}

Since \(\mathcal{M}^{\prime}\) is a \(\sigma\)-algebra, \(E \in \mathcal{M}^{\prime}\). For each natural number \(n\) define the partial union

\begin{equation*} A_n = \bigcup_{k=1}^{n} E_k \in \mathcal{M}^{\prime} . \end{equation*}

The collection \(\{ A_n \}_{n=1}^{\infty}\) is ascending, since \(A_n \subseteq A_{n+1}\), and

\begin{equation*} \bigcup_{n=1}^{\infty} A_n = \bigcup_{k=1}^{\infty} E_k = E . \end{equation*}

Indeed, each \(A_n \subseteq E\), and conversely any \(x \in E\) lies in some \(E_k\) and hence in \(A_k\).

Because \(\{ E_k \}_{k=1}^{n}\) is a finite disjoint collection in \(\mathcal{M}^{\prime}\), hypothesis (a) gives

\begin{equation*} m^{\prime}(A_n) = \sum_{k=1}^{n} m^{\prime}(E_k) \quad \text{for every } n . \end{equation*}

Applying hypothesis (b) to the ascending collection \(\{ A_n \}\) and then using the display above,

\begin{equation*} m^{\prime}(E) = m^{\prime} \left( \bigcup_{n=1}^{\infty} A_n \right) = \lim_{n \to \infty} m^{\prime}(A_n) = \lim_{n \to \infty} \sum_{k=1}^{n} m^{\prime}(E_k) = \sum_{k=1}^{\infty} m^{\prime}(E_k), \end{equation*}

the last equality being the definition of the sum of a series of nonnegative terms; the limit exists in \([0, \infty]\) because the partial sums are nondecreasing. That is,

\begin{equation*} m^{\prime} \left( \bigcup_{k=1}^{\infty} E_k \right) = \sum_{k=1}^{\infty} m^{\prime}(E_k), \end{equation*}

which is the countable additivity of \(m^{\prime}\).

A remark on the descending form of continuity. One may instead assume finite additivity together with continuity for descending collections, in the form: if \(\{ B_n \}\) is a descending collection in \(\mathcal{M}^{\prime}\) with \(m^{\prime}(B_1) < \infty\) and \(\bigcap_{n=1}^{\infty} B_n = \emptyset\), then \(\lim_{n \to \infty} m^{\prime}(B_n) = 0\). This yields countable additivity for disjoint collections whose union has finite measure. Indeed, with \(E\) and \(E_k\) as above and \(m^{\prime}(E) < \infty\), put

\begin{equation*} B_n = E \sim \bigcup_{k=1}^{n} E_k \in \mathcal{M}^{\prime} . \end{equation*}

The collection \(\{ B_n \}\) is descending, \(m^{\prime}(B_1) \le m^{\prime}(E) < \infty\) by the monotonicity that follows from finite additivity, and \(\bigcap_{n=1}^{\infty} B_n = \emptyset\) because each point of \(E\) lies in some \(E_k\) and so is excluded from \(B_k\). Finite additivity applied to the disjoint decomposition \(E = E_1 \cup \cdots \cup E_n \cup B_n\) gives

\begin{equation*} m^{\prime}(E) = \sum_{k=1}^{n} m^{\prime}(E_k) + m^{\prime}(B_n), \end{equation*}

and letting \(n \to \infty\), with \(m^{\prime}(B_n) \to 0\), yields \(m^{\prime}(E) = \sum_{k=1}^{\infty} m^{\prime}(E_k)\). This second route needs the finiteness of \(m^{\prime}(E)\), which is why the ascending form of continuity is the one that gives the theorem in full generality.

Taken together with Problem 27, the conclusion is that for a \([0, \infty]\)-valued set function on a \(\sigma\)-algebra which vanishes on the empty set, countable additivity is equivalent to the conjunction of finite additivity and continuity. This is the sense in which Theorem 15 and Proposition 13 carry the same information about Lebesgue measure.

Throughout, for a nonempty set \(E\) of real numbers, two points \(x, y \in E\) are called rationally equivalent provided \(x - y \in \mathbb{Q}\); write \(x \sim y\).

Part (i). Let \(E\) be any nonempty set of real numbers.

Reflexivity: for \(x \in E\) we have \(x - x = 0\), and \(0 \in \mathbb{Q}\). Hence \(x \sim x\).

Symmetry: suppose \(x \sim y\), so that \(q := x - y \in \mathbb{Q}\). Then \(y - x = -q\), and \(\mathbb{Q}\) is closed under negation, so \(y - x \in \mathbb{Q}\) and therefore \(y \sim x\).

Transitivity: suppose \(x \sim y\) and \(y \sim z\), say \(x - y = q \in \mathbb{Q}\) and \(y - z = r \in \mathbb{Q}\). Then

\begin{equation*} x - z = (x-y) + (y-z) = q + r , \end{equation*}

and \(\mathbb{Q}\) is closed under addition, so \(x - z \in \mathbb{Q}\) and therefore \(x \sim z\).

Thus \(\sim\) is reflexive, symmetric and transitive on \(E\); that is, it is an equivalence relation on \(E\). Observe that the only properties of \(\mathbb{Q}\) used are that it is a subgroup of the additive group \(\mathbb{R}\): it contains \(0\) and is closed under negation and addition.

Part (ii). Take \(E = \mathbb{Q}\). If \(x, y \in \mathbb{Q}\) then \(x - y \in \mathbb{Q}\), so \(x \sim y\). Hence any two rationals are rationally equivalent, and the rational equivalence relation on \(\mathbb{Q}\) has exactly one equivalence class, namely \(\mathbb{Q}\) itself. A choice set consists of exactly one member of each equivalence class, so

\begin{equation*} \mathcal{C}_{\mathbb{Q}} = \{0\} \end{equation*}

is a choice set (as is \(\{q\}\) for any single rational \(q\); these singletons are precisely the choice sets).

It is worth checking the two characteristic properties listed in the text. Property (i), that the difference of two (distinct) points of the choice set is not rational, holds vacuously, since \(\{0\}\) contains no two distinct points. Property (ii), that each \(x \in \mathbb{Q}\) may be written \(x = c + q\) with \(c \in \mathcal{C}_{\mathbb{Q}}\) and \(q\) rational, holds with \(c = 0\) and \(q = x\).

Note that no appeal to the Axiom of Choice is needed here: there is only one equivalence class, and we have named a member of it explicitly.

Part (iii). Define \(x \approx y\) provided \(x - y\) is irrational.

On \(\mathbb{R}\) this is not an equivalence relation. It fails reflexivity at every point: for each \(x \in \mathbb{R}\), \(x - x = 0\) is rational, so \(x \not\approx x\). (The relation is symmetric, since \(x - y\) is irrational if and only if \(y - x = -(x-y)\) is irrational. But it also fails transitivity: take \(x = 0\), \(y = \sqrt{2}\), \(z = 2\). Then \(x - y = -\sqrt{2}\) and \(y - z = \sqrt{2} - 2\) are both irrational, so \(x \approx y\) and \(y \approx z\); but \(x - z = -2\) is rational, so \(x \not\approx z\).)

On \(\mathbb{Q}\) this is not an equivalence relation either. Indeed, if \(x, y \in \mathbb{Q}\) then \(x - y \in \mathbb{Q}\) is rational, so no pair of rationals is irrationally equivalent: the relation \(\approx\) restricted to \(\mathbb{Q}\) is the empty relation. It is symmetric and vacuously transitive, but it is not reflexive, since \(\mathbb{Q}\) is nonempty and \(x \not\approx x\) for every \(x \in \mathbb{Q}\). Hence it is not an equivalence relation.

Let \(E\) be a set of real numbers with \(m^*(E) > 0\), and let \(\mathcal{C}_E\) be a choice set for the rational equivalence relation on \(E\). We argue by contradiction: suppose \(\mathcal{C}_E\) is countable (recall that in this book “countable” includes “finite”, so this covers both the finite and the countably infinite case).

We first observe that

\begin{equation*} E \subseteq \bigcup_{q \in \mathbb{Q}} (q + \mathcal{C}_E). \end{equation*}

Indeed, let \(x \in E\). By the second characteristic property of a choice set, there is a point \(c \in \mathcal{C}_E\) and a rational number \(q\) with \(x = c + q\). Thus \(x \in q + \mathcal{C}_E\), which proves the inclusion.

Now \(\mathbb{Q}\) is countable, so the union above is a countable union of sets. Each set \(q + \mathcal{C}_E\) is a translate of the countable set \(\mathcal{C}_E\), hence is itself countable, and a countable set of real numbers has outer measure zero (the Example following the statement of monotonicity in Section 2.2). Therefore, by the countable subadditivity of outer measure (Proposition 3 of this chapter) together with its monotonicity,

\begin{equation*} m^*(E) \le m^*\left( \bigcup_{q \in \mathbb{Q}} (q + \mathcal{C}_E) \right) \le \sum_{q \in \mathbb{Q}} m^*(q + \mathcal{C}_E) = \sum_{q \in \mathbb{Q}} 0 = 0 . \end{equation*}

Hence \(m^*(E) = 0\), contradicting the assumption that \(m^*(E) > 0\).

We conclude that \(\mathcal{C}_E\) is not countable. In particular \(\mathcal{C}_E\) is not finite, so it is uncountably infinite, as asserted.

Two remarks. First, no measurability of \(\mathcal{C}_E\) is needed anywhere in this argument; only outer measure, which is defined for every set, is used. Second, translation invariance of outer measure was not even required, since a translate of a countable set is countable, and countable sets have outer measure zero directly.

The assertion to be justified is the opening sentence of the proof of Theorem 17: “By the countable subadditivity of outer measure, we may suppose \(E\) is bounded.” Precisely, what must be shown is the following implication.

Claim. Suppose it has been proved that every bounded set of positive outer measure contains a nonmeasurable subset. Then every set of positive outer measure contains a nonmeasurable subset.

Proof of the claim. Let \(E\) be a set of real numbers with \(m^*(E) > 0\). For each natural number \(n\) put

\begin{equation*} E_n = E \cap [-n, n] . \end{equation*}

Each \(E_n\) is bounded, and since every real number lies in \([-n,n]\) for \(n\) sufficiently large,

\begin{equation*} E = \bigcup_{n=1}^{\infty} E_n . \end{equation*}

We assert that \(m^*(E_n) > 0\) for some \(n\). Otherwise \(m^*(E_n) = 0\) for every \(n\), and then, by the countable subadditivity of outer measure (Proposition 3 of this chapter),

\begin{equation*} m^*(E) \le \sum_{n=1}^{\infty} m^*(E_n) = 0 , \end{equation*}

so that \(m^*(E) = 0\), contrary to hypothesis.

Fix an index \(n\) with \(m^*(E_n) > 0\). Then \(E_n\) is a bounded set of positive outer measure, so by the assumed bounded case, \(E_n\) contains a subset \(W\) that fails to be measurable. Since \(W \subseteq E_n \subseteq E\), the set \(E\) contains the nonmeasurable subset \(W\). This proves the claim, and hence justifies the reduction.

It is worth recording explicitly why the reduction is desirable: boundedness of \(E\) is used twice in the proof of Theorem 17. It guarantees that \(\mathcal{C}_E\) is bounded, so that Lemma 16 (which requires \(E\) bounded) may be applied to \(\mathcal{C}_E\); and it supplies an interval \([-b,b] \supseteq E\), which is what allows the bounded, countably infinite set of rationals \(\Lambda_0 = [-2b, 2b] \cap \mathbb{Q}\) to be chosen so that the inclusion \(E \subseteq \bigcup_{\lambda \in \Lambda_0} (\lambda + \mathcal{C}_E)\) holds. As Problem 32 shows, Lemma 16 genuinely fails without boundedness of \(\Lambda\), so this reduction is not a mere convenience.

Recall Lemma 16: if \(E\) is a bounded measurable set of real numbers and there is a bounded, countably infinite set \(\Lambda\) of real numbers for which the collection of translates \(\{\lambda + E\}_{\lambda \in \Lambda}\) is disjoint, then \(m(E) = 0\). We examine the three proposed relaxations in turn. The answers are: no for finite \(\Lambda\), yes for uncountably infinite (bounded) \(\Lambda\), and no for unbounded \(\Lambda\).

Finite \(\Lambda\): the lemma fails. Let \(E = [0,1)\) and \(\Lambda = \{0, 1\}\). Then \(E\) is bounded and measurable, \(\Lambda\) is bounded and finite, and the translates are \(0 + E = [0,1)\) and \(1 + E = [1,2)\), which are disjoint. Yet \(m(E) = 1 \neq 0\). Even more simply, one may take \(\Lambda = \{0\}\): the collection consisting of the single set \(E\) is trivially disjoint, so with \(E = [0,1]\) we again get a bounded measurable set of positive measure. The reason the proof of Lemma 16 breaks down is transparent from equation (15): with \(\Lambda\) finite, the right-hand sum \(\sum_{\lambda \in \Lambda} m(\lambda + E) = (\#\Lambda) \cdot m(E)\) is finite whatever the value of \(m(E)\), so no contradiction arises. It is exactly the infinitude of \(\Lambda\) that forces the sum of infinitely many equal positive numbers to be infinite.

Uncountably infinite \(\Lambda\) (still bounded): the lemma remains true. Suppose \(E\) is bounded and measurable, \(\Lambda\) is bounded and uncountably infinite, and \(\{\lambda + E\}_{\lambda \in \Lambda}\) is disjoint. If \(E = \emptyset\) then \(m(E) = 0\) and there is nothing to prove, so assume \(E \neq \emptyset\). Since \(\Lambda\) is infinite, it possesses a countably infinite subset \(\Lambda_0 = \{\lambda_1, \lambda_2, \dots\}\): choose \(\lambda_1 \in \Lambda\), and having chosen the distinct points \(\lambda_1, \dots, \lambda_n\), the set \(\Lambda \sim \{\lambda_1, \dots, \lambda_n\}\) is nonempty because \(\Lambda\) is infinite, so a point \(\lambda_{n+1}\) may be chosen from it. Now \(\Lambda_0\) is countably infinite, and it is bounded because \(\Lambda\) is. Moreover \(\{\lambda + E\}_{\lambda \in \Lambda_0}\) is a subcollection of a disjoint collection and hence is disjoint. Lemma 16, applied to \(E\) and \(\Lambda_0\), yields \(m(E) = 0\).

So allowing \(\Lambda\) to be uncountable does not change the conclusion; it merely weakens the hypothesis in an inessential way, since only a countably infinite bounded piece of \(\Lambda\) is ever used.

Unbounded \(\Lambda\): the lemma fails. Let \(E = [0,1)\) and \(\Lambda = \mathbb{Z}\). Then \(E\) is bounded and measurable, and \(\Lambda\) is countably infinite but unbounded. The translates are \(n + E = [n, n+1)\) for \(n \in \mathbb{Z}\), and these are pairwise disjoint. Yet \(m(E) = 1 \neq 0\). Here the proof of Lemma 16 breaks down at the following point: the argument used boundedness of \(\Lambda\) (together with boundedness of \(E\)) to conclude that \(\bigcup_{\lambda \in \Lambda} (\lambda + E)\) is bounded and therefore of finite measure, which is what makes the left-hand side of (15) finite. With \(\Lambda = \mathbb{Z}\) the union is all of \(\mathbb{R}\), of infinite measure, and (15) reads \(\infty = \sum_{n \in \mathbb{Z}} 1\), a true statement carrying no contradiction.

In summary, the two hypotheses on \(\Lambda\) play different roles and only one of them can be relaxed: \(\Lambda\) must be infinite (uncountable is fine), and \(\Lambda\) must be bounded.

Write \(\alpha = m^*(E)\), so that \(0 \le \alpha < \infty\) by hypothesis.

Construction of \(G\). Fix a natural number \(n\). By the definition of outer measure as an infimum, there is a countable collection \(\{I_{n,k}\}_{k}\) of nonempty open, bounded intervals which covers \(E\) and for which

\begin{equation*} \sum_k \ell(I_{n,k}) < \alpha + \frac{1}{n} . \end{equation*}

(The infimum is finite since \(\alpha < \infty\), so such a collection exists.) Set \(\mathcal{O}_n = \bigcup_k I_{n,k}\). Then \(\mathcal{O}_n\) is an open set containing \(E\), and by the countable subadditivity and monotonicity of measure together with \(m(I) = \ell(I)\) for an interval \(I\),

\begin{equation*} m(\mathcal{O}_n) \le \sum_k m(I_{n,k}) = \sum_k \ell(I_{n,k}) < \alpha + \frac{1}{n} . \end{equation*}

Now define

\begin{equation*} G = \bigcap_{n=1}^{\infty} \mathcal{O}_n . \end{equation*}

Being a countable intersection of open sets, \(G\) is a \(G_\delta\) set; in particular \(G\) is measurable, since by Theorem 9 of this chapter the measurable sets form a \(\sigma\)-algebra containing the Borel sets, and every \(G_\delta\) set is measurable. Each \(\mathcal{O}_n\) contains \(E\), so \(E \subseteq G\).

Verification that \(m^*(E) = m^*(G)\). By monotonicity of outer measure and \(E \subseteq G\) we have \(\alpha = m^*(E) \le m^*(G)\). On the other hand, \(G \subseteq \mathcal{O}_n\) for each \(n\), so by monotonicity

\begin{equation*} m^*(G) \le m(\mathcal{O}_n) < \alpha + \frac{1}{n} \quad \text{for every } n . \end{equation*}

Letting \(n \to \infty\) gives \(m^*(G) \le \alpha\). Hence \(m^*(G) = \alpha = m^*(E)\), as required. Note in particular that \(m(G) = \alpha < \infty\).

Verification that \(m^*(G \sim E) > 0\). Suppose, to the contrary, that \(m^*(G \sim E) = 0\). A set of outer measure zero is measurable (Proposition 4 of this chapter), so \(G \sim E\) is measurable. But \(G\) is measurable, and the measurable sets form a \(\sigma\)-algebra, so

\begin{equation*} E = G \sim (G \sim E) \end{equation*}

is measurable. (The displayed identity holds because \(E \subseteq G\): a point of \(G\) lies outside \(G \sim E\) precisely when it lies in \(E\).) This contradicts the hypothesis that \(E\) fails to be measurable. Therefore \(m^*(G \sim E) > 0\).

Thus \(G\) is a \(G_\delta\) set containing \(E\) with \(m^*(G) = m^*(E)\) and \(m^*(G \sim E) > 0\), which is what was to be shown.

A remark on what this exhibits. Since \(E\) and \(G \sim E\) are disjoint with union \(G\), we have

\begin{equation*} m^*(E) + m^*(G \sim E) > m^*(E) = m(G) = m^*\big( E \cup (G \sim E) \big), \end{equation*}

where the strict inequality uses \(m^*(G \sim E) > 0\) together with the finiteness of \(m^*(E)\). So the pair \(E\), \(G \sim E\) is an explicit disjoint pair for which the subadditivity inequality \(m^*(A \cup B) \le m^*(A) + m^*(B)\) is strict, exactly as promised by Theorem 18. The hypothesis \(m^*(E) < \infty\) enters twice: it is what makes the covers used in the construction of \(G\) available at all, and it is what makes the last display a strict inequality rather than the vacuous \(\infty > \infty\).

Let \(\varphi\) denote the Cantor-Lebesgue function on \([0,1]\), let \(\mathbf{C}\) be the Cantor set, and let \(\mathcal{O} = [0,1] \sim \mathbf{C}\) be the union of the countable disjoint collection \(\{I_k\}_{k=1}^{\infty}\) of open intervals removed in the Cantor deletion process. By Proposition 19, \(m(\mathbf{C}) = 0\), and hence \(m(\mathcal{O}) = 1\); since the \(I_k\) are disjoint open intervals, countable additivity gives

\begin{equation*} \sum_{k=1}^{\infty} \ell(I_k) = m(\mathcal{O}) = 1 . \end{equation*}

By Proposition 20, \(\varphi\) is increasing and continuous and maps \([0,1]\) onto \([0,1]\); in particular \(\varphi(0) = 0\) and \(\varphi(1) = 1\). Moreover, by the very definition of \(\varphi\) on the removed intervals, \(\varphi\) is constant on each \(I_k\).

Define \(\psi\) on \([0,1]\) by

\begin{equation*} \psi(x) = \frac{\varphi(x) + x}{2} . \end{equation*}

Step 1: \(\psi\) is a continuous, strictly increasing map of \([0,1]\) onto \([0,1]\). It is continuous as a linear combination of the continuous functions \(\varphi\) and the identity. If \(0 \le u < v \le 1\) then \(\varphi(u) \le \varphi(v)\) because \(\varphi\) is increasing, while \(u < v\); adding and halving gives \(\psi(u) < \psi(v)\). Thus \(\psi\) is strictly increasing, and in particular one-to-one. Finally \(\psi(0) = (\varphi(0) + 0)/2 = 0\) and \(\psi(1) = (1+1)/2 = 1\); since \(\psi\) is increasing, its values on \([0,1]\) lie in \([0,1]\), and since \(\psi\) is continuous, the Intermediate Value Theorem shows every point of \([0,1]\) is attained. Hence \(\psi\) is a one-to-one map of \([0,1]\) onto \([0,1]\).

Step 2: \(m(\psi(\mathcal{O})) = 1/2\). Fix \(k\) and write \(I_k = (a_k, b_k)\). Since \(\varphi\) is constant on \(I_k\), say \(\varphi \equiv c_k\) there, we have \(\psi(x) = (c_k + x)/2\) for \(x \in I_k\), so

\begin{equation*} \psi(I_k) = \left( \frac{c_k + a_k}{2}, \; \frac{c_k + b_k}{2} \right), \end{equation*}

an open interval of length \(\ell(I_k)/2\). Because \(\psi\) is one-to-one and the \(I_k\) are disjoint, the collection \(\{\psi(I_k)\}_{k=1}^{\infty}\) is disjoint. Therefore \(\psi(\mathcal{O}) = \bigcup_{k=1}^{\infty} \psi(I_k)\) is open, hence measurable, and by countable additivity

\begin{equation*} m(\psi(\mathcal{O})) = \sum_{k=1}^{\infty} \ell(\psi(I_k)) = \frac{1}{2} \sum_{k=1}^{\infty} \ell(I_k) = \frac{1}{2} . \end{equation*}

Step 3: \(m(\psi(\mathbf{C})) = 1/2\). The set \(\mathbf{C}\) is closed and bounded, and \(\psi\) is continuous, so \(\psi(\mathbf{C})\) is compact, hence closed, hence measurable. Since \(\psi\) is a one-to-one map of \([0,1]\) onto \([0,1]\) and \([0,1] = \mathbf{C} \cup \mathcal{O}\) is a disjoint decomposition, its image

\begin{equation*} [0,1] = \psi(\mathbf{C}) \cup \psi(\mathcal{O}) \end{equation*}

is also a disjoint decomposition. By additivity of measure and Step 2,

\begin{equation*} m(\psi(\mathbf{C})) = m([0,1]) - m(\psi(\mathcal{O})) = 1 - \frac{1}{2} = \frac{1}{2} . \end{equation*}

Step 4: the function. Let \(f = \psi^{-1}\), defined on \([0,1]\) with values in \([0,1]\); this makes sense by Step 1. As noted in the proof of Proposition 21 (and established in Problem 45), a strictly increasing continuous function defined on an interval has a continuous inverse, so \(f\) is continuous on \([0,1]\). Moreover \(f\) is strictly increasing: if \(0 \le u < v \le 1\), write \(u = \psi(s)\) and \(v = \psi(t)\) with \(s, t \in [0,1]\); were \(s \ge t\) we would get \(u = \psi(s) \ge \psi(t) = v\), a contradiction, so \(s < t\), that is, \(f(u) < f(v)\).

Now set \(A = \psi(\mathbf{C}) \subseteq [0,1]\). By Step 3, \(A\) is measurable with \(m(A) = 1/2 > 0\), and since \(\psi\) is one-to-one and onto \([0,1]\),

\begin{equation*} f(A) = \psi^{-1}(\psi(\mathbf{C})) = \mathbf{C}, \end{equation*}

which has measure zero by Proposition 19.

Thus \(f\) is a continuous, strictly increasing function on \([0,1]\) carrying the set \(A\) of measure \(1/2\) onto the set \(\mathbf{C}\) of measure zero, as required.

Recall that \(f\) increasing means \(f(u) \le f(v)\) whenever \(u \le v\) in \(I\).

Necessity. Suppose \(f\) is continuous at \(x_0\). Since \(I\) is open and \(x_0 \in I\), there is an \(r > 0\) with \((x_0 - r, x_0 + r) \subseteq I\). For each natural number \(n\) define

\begin{equation*} a_n = x_0 - \frac{r}{n+1}, \qquad b_n = x_0 + \frac{r}{n+1} . \end{equation*}

Then \(a_n\) and \(b_n\) belong to \(I\) and \(a_n < x_0 < b_n\) for every \(n\), while \(a_n \to x_0\) and \(b_n \to x_0\). Continuity of \(f\) at \(x_0\) (in its sequential form) gives \(f(a_n) \to f(x_0)\) and \(f(b_n) \to f(x_0)\), and therefore

\begin{equation*} \lim_{n \to \infty} [f(b_n) - f(a_n)] = f(x_0) - f(x_0) = 0 . \end{equation*}

Sufficiency. Suppose there are sequences \(\{a_n\}\) and \(\{b_n\}\) in \(I\) with \(a_n < x_0 < b_n\) for each \(n\) and \(f(b_n) - f(a_n) \to 0\). Note first that, since \(f\) is increasing and \(a_n < x_0 < b_n\),

\begin{equation*} f(a_n) \le f(x_0) \le f(b_n) \quad \text{for every } n , \end{equation*}

so each difference \(f(b_n) - f(a_n)\) is nonnegative.

Let \(\varepsilon > 0\). Choose an index \(N\) for which

\begin{equation*} f(b_N) - f(a_N) < \varepsilon , \end{equation*}

and define

\begin{equation*} \delta = \min \{ x_0 - a_N, \; b_N - x_0 \} > 0 . \end{equation*}

Since \(a_N\) and \(b_N\) lie in the interval \(I\), the whole interval \((a_N, b_N)\) is contained in \(I\); and \((x_0 - \delta, x_0 + \delta) \subseteq (a_N, b_N)\) by the choice of \(\delta\). Hence every \(x\) with \(|x - x_0| < \delta\) belongs to \(I\) and satisfies \(a_N < x < b_N\).

For such an \(x\), monotonicity of \(f\) gives

\begin{equation*} f(a_N) \le f(x) \le f(b_N) . \end{equation*}

Since also \(f(a_N) \le f(x_0) \le f(b_N)\), both \(f(x)\) and \(f(x_0)\) lie in the interval \([f(a_N), f(b_N)]\), whose length is less than \(\varepsilon\). Consequently

\begin{equation*} |f(x) - f(x_0)| \le f(b_N) - f(a_N) < \varepsilon . \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, \(f\) is continuous at \(x_0\).

Remark. The hypothesis that \(f\) be increasing is essential in the sufficiency direction: it is what forces \(f(x)\) to be trapped between \(f(a_N)\) and \(f(b_N)\) for all \(x\) between \(a_N\) and \(b_N\), so that the smallness of a single difference \(f(b_N) - f(a_N)\) controls the oscillation of \(f\) on an entire neighborhood of \(x_0\). The hypothesis that \(I\) be open is what guarantees, in the necessity direction, that points of \(I\) strictly on both sides of \(x_0\) exist.

Write \(\mathbf{C}\) for the Cantor set and \(\mathcal{O} = [0,1] \sim \mathbf{C}\) for its complement in \([0,1]\). We are given that \(f\) is increasing on \([0,1]\) and that \(f = \varphi\) on \(\mathcal{O}\).

Step 1: \(\mathcal{O}\) is dense in \([0,1]\). By Proposition 19, \(m(\mathbf{C}) = 0\). If some nondegenerate interval \(J \subseteq [0,1]\) were contained in \(\mathbf{C}\), then by monotonicity of measure and Proposition 1 we would have \(0 < \ell(J) = m(J) \le m(\mathbf{C}) = 0\), a contradiction. Hence every nondegenerate subinterval of \([0,1]\) contains a point of \(\mathcal{O}\).

(Alternatively and without measure theory: \(\mathbf{C} \subseteq C_k\), and \(C_k\) is a disjoint union of closed intervals of length \(3^{-k}\), so \(\mathbf{C}\) contains no interval of length exceeding \(3^{-k}\) for any \(k\).)

Step 2: \(f = \varphi\) at every interior point. Let \(x \in \mathbf{C}\) with \(0 < x < 1\), and let \(N\) be a natural number with \(1/N < \min\{x,\, 1-x\}\). For each \(n \ge N\) the intervals \((x - 1/n,\, x)\) and \((x,\, x + 1/n)\) are nondegenerate subintervals of \([0,1]\), so by Step 1 we may select points

\begin{equation*} u_n \in \mathcal{O} \cap \left(x - \tfrac{1}{n},\, x\right), \qquad v_n \in \mathcal{O} \cap \left(x,\, x + \tfrac{1}{n}\right). \end{equation*}

Then \(u_n < x < v_n\) and \(u_n \to x\), \(v_n \to x\).

Since \(f\) is increasing and agrees with \(\varphi\) on \(\mathcal{O}\),

\begin{equation*} \varphi(u_n) = f(u_n) \le f(x) \le f(v_n) = \varphi(v_n). \end{equation*}

Since \(\varphi\) is increasing (Proposition 20),

\begin{equation*} \varphi(u_n) \le \varphi(x) \le \varphi(v_n). \end{equation*}

Thus both \(f(x)\) and \(\varphi(x)\) lie in the interval \([\varphi(u_n), \varphi(v_n)]\), so

\begin{equation*} |f(x) - \varphi(x)| \le \varphi(v_n) - \varphi(u_n). \end{equation*}

By Proposition 20 the function \(\varphi\) is continuous, so \(\varphi(u_n) \to \varphi(x)\) and \(\varphi(v_n) \to \varphi(x)\), whence the right-hand side tends to \(0\). Therefore \(f(x) = \varphi(x)\).

Since \(f = \varphi\) on \(\mathcal{O}\) by hypothesis, and every point of \(\mathbf{C} \cap (0,1)\) has just been handled, we conclude that \(f = \varphi\) on all of \((0,1)\).

Step 3: the endpoints. Since \(\varphi\) is increasing and maps \([0,1]\) onto \([0,1]\) (Proposition 20), \(\varphi(0) = 0\) and \(\varphi(1) = 1\). The two points \(0\) and \(1\) lie in \(\mathbf{C}\), and the argument of Step 2 is only one-sided there, so it yields inequalities rather than equalities. By Step 1 we may choose \(v_n \in \mathcal{O} \cap (0, 1/n)\), so that \(v_n \downarrow 0\), and then

\begin{equation*} f(0) \le f(v_n) = \varphi(v_n) \to \varphi(0) = 0, \end{equation*}

so \(f(0) \le 0 = \varphi(0)\); symmetrically \(f(1) \ge 1 = \varphi(1)\). With the natural normalization that \(f\), like \(\varphi\), takes its values in \([0,1]\) (equivalently, that \(f(0) \ge 0\) and \(f(1) \le 1\)), these inequalities become equalities and \(f = \varphi\) on all of \([0,1]\).

Some such normalization is genuinely needed: the function \(g\) defined by \(g(0) = -1\) and \(g = \varphi\) on \((0,1]\) is increasing on \([0,1]\) and agrees with \(\varphi\) on \(\mathcal{O}\) (note \(0 \notin \mathcal{O}\), since \(0 \in \mathbf{C}\)), yet \(g \ne \varphi\). The substance of the problem, namely that the values of \(\varphi\) on the measure-zero set \(\mathbf{C}\) are completely determined by its values on the complement \(\mathcal{O}\) together with monotonicity, is exactly what Step 2 establishes.

No. We exhibit a continuous function \(f\) defined on a closed bounded interval \(E\) and a measurable set \(A\) for which \(f^{-1}(A)\) is not measurable.

Let \(\varphi\) be the Cantor-Lebesgue function and, as in Proposition 21, define

\begin{equation*} \psi(x) = \varphi(x) + x \qquad \text{for } x \in [0,1]. \end{equation*}

By Proposition 21, \(\psi\) is a strictly increasing continuous function that maps \([0,1]\) onto \([0,2]\), and, by assertion (ii) of that proposition, there is a measurable set \(A\), a subset of the Cantor set \(\mathbf{C}\), for which \(\psi(A)\) is nonmeasurable.

Set \(E = [0,2]\) and let

\begin{equation*} f = \psi^{-1} : [0,2] \to [0,1]. \end{equation*}

This is well defined because \(\psi\), being strictly increasing, is one-to-one, and it maps \([0,1]\) onto \([0,2]\).

The function \(f\) is continuous. This is Problem 45, and the proof is short: \(f\) is strictly increasing on \([0,2]\) and its range is all of \([0,1]\). If \(f\) failed to be continuous at some \(y_0 \in [0,2]\), then, \(f\) being increasing and bounded, the one-sided limits

\begin{equation*} \lambda = \sup_{y < y_0} f(y) \quad \text{and} \quad \mu = \inf_{y > y_0} f(y) \end{equation*}

exist (interpreting the sup as \(f(y_0)\) if \(y_0 = 0\), and the inf as \(f(y_0)\) if \(y_0 = 2\)) and satisfy \(\lambda \le f(y_0) \le \mu\) with \(\lambda < \mu\). But then \(f\) takes no value in \((\lambda, \mu)\) other than possibly \(f(y_0)\), so its range omits points of the interval \((\lambda, \mu) \subseteq [0,1]\), contradicting the fact that the range of \(f\) is all of \([0,1]\).

Now compute the inverse image of \(A\) under \(f\). For \(y \in [0,2]\),

\begin{equation*} y \in f^{-1}(A) \iff f(y) = \psi^{-1}(y) \in A \iff y \in \psi(A), \end{equation*}

the last equivalence because \(\psi\) is a bijection of \([0,1]\) onto \([0,2]\). Hence

\begin{equation*} f^{-1}(A) = \psi(A). \end{equation*}

The set \(A\) is measurable: it is a subset of \(\mathbf{C}\), which has measure zero by Proposition 19, so \(m^{*}(A) = 0\) by the monotonicity of outer measure, and therefore \(A\) is measurable by Proposition 4. But \(\psi(A)\) is nonmeasurable by the choice of \(A\). Therefore \(f\) is continuous on \(E = [0,2]\), the set \(A\) is measurable, and \(f^{-1}(A)\) is not measurable.

Remark. What is true is that \(f^{-1}(A)\) is measurable whenever \(f\) is continuous and \(A\) is a Borel set: the collection of sets \(B\) for which \(f^{-1}(B)\) is measurable is a \(\sigma\)-algebra (inverse images preserve complements and countable unions), and it contains the open sets because \(f\) is continuous, hence it contains the smallest \(\sigma\)-algebra containing the open sets, namely the Borel sets. Compare Problem 46, where the same argument gives the sharper conclusion that \(f^{-1}(A)\) is itself Borel. The example above shows this cannot be improved to Lebesgue measurable sets \(A\); this is consistent with Proposition 22, which asserts that there are Lebesgue measurable sets which are not Borel.

Throughout, all sets considered are subsets of \([a,b]\). Observe first that \(f\) is continuous, since \(|f(u) - f(v)| \le c\,|u-v|\) forces \(f(v) \to f(u)\) as \(v \to u\).

A basic estimate. Let \(J\) be any subset of \([a,b]\) that is contained in an interval of length \(\ell\). If \(J\) is empty the estimate below is trivial, so assume \(J \ne \emptyset\). For all \(u, v \in J\) we have \(|f(u) - f(v)| \le c\,|u - v| \le c\,\ell\), so the set \(f(J)\) is bounded and

\begin{equation*} \sup f(J) - \inf f(J) = \sup_{u,v \in J} |f(u) - f(v)| \le c\,\ell . \end{equation*}

Consequently \(f(J)\) is contained in the closed bounded interval \([\,\inf f(J),\ \inf f(J) + c\,\ell\,]\), whose length is \(c\,\ell\) (a degenerate one-point interval if \(c = 0\) or \(\ell = 0\)). By the monotonicity of outer measure and Proposition 1,

\begin{equation*} m^{*}\big(f(J)\big) \le c\,\ell . \end{equation*}

Call this estimate \((\ast)\).

Part 1: \(f\) maps a set of measure zero onto a set of measure zero.

Let \(E \subseteq [a,b]\) with \(m^{*}(E) = 0\), and let \(\epsilon > 0\). By the definition of outer measure there is a countable collection \(\{I_k\}_{k=1}^{\infty}\) of open intervals which covers \(E\) and for which

\begin{equation*} \sum_{k=1}^{\infty} \ell(I_k) < \epsilon . \end{equation*}

For each \(k\) put \(J_k = I_k \cap [a,b]\). Then \(\{J_k\}\) covers \(E\), each \(J_k\) is contained in an interval of length at most \(\ell(I_k)\), and

\begin{equation*} f(E) \subseteq \bigcup_{k=1}^{\infty} f(J_k). \end{equation*}

By the countable subadditivity of outer measure (Proposition 3), monotonicity, and the estimate \((\ast)\),

\begin{equation*} m^{*}\big(f(E)\big) \le \sum_{k=1}^{\infty} m^{*}\big(f(J_k)\big) \le \sum_{k=1}^{\infty} c\,\ell(I_k) = c \sum_{k=1}^{\infty} \ell(I_k) \le c\,\epsilon . \end{equation*}

Since \(\epsilon > 0\) was arbitrary, \(m^{*}(f(E)) = 0\). In particular, by Proposition 4, \(f(E)\) is measurable and \(m(f(E)) = 0\).

Part 2: \(f\) maps an \(F_\sigma\) set onto an \(F_\sigma\) set.

Let \(F \subseteq [a,b]\) be an \(F_\sigma\) set, say \(F = \bigcup_{k=1}^{\infty} F_k\) with each \(F_k\) closed. Replacing \(F_k\) by \(F_k \cap [a,b]\) (which is again closed, and does not change the union since \(F \subseteq [a,b]\)), we may assume each \(F_k\) is a closed subset of \([a,b]\). Then each \(F_k\) is closed and bounded, and \(f\) is continuous, so \(f(F_k)\) is a compact, hence closed, subset of \(\mathbb{R}\). (Concretely: \(f(F_k)\) is bounded by the estimate \((\ast)\) applied to \([a,b]\), and it is closed because if \(y_n = f(x_n) \to y\) with \(x_n \in F_k\), then by the Bolzano-Weierstrass Theorem a subsequence \(x_{n_j}\) converges to some \(x\), which lies in \(F_k\) since \(F_k\) is closed, and then \(y = \lim f(x_{n_j}) = f(x)\) by continuity, so \(y \in f(F_k)\).)

Since images commute with unions,

\begin{equation*} f(F) = \bigcup_{k=1}^{\infty} f(F_k), \end{equation*}

a countable union of closed sets, that is, an \(F_\sigma\) set.

Part 3: \(f\) maps a measurable set to a measurable set.

Let \(E \subseteq [a,b]\) be measurable. By assertion (iv) of Theorem 11 there is an \(F_\sigma\) set \(F\) contained in \(E\) for which \(m^{*}(E \sim F) = 0\). Write

\begin{equation*} E = F \cup (E \sim F), \qquad \text{so} \qquad f(E) = f(F) \cup f(E \sim F). \end{equation*}

By Part 2, \(f(F)\) is an \(F_\sigma\) set; an \(F_\sigma\) set is a Borel set and therefore measurable by Theorem 9. By Part 1, \(m^{*}(f(E \sim F)) = 0\), so \(f(E \sim F)\) is measurable by Proposition 4. Since the measurable sets form a \(\sigma\)-algebra (Theorem 9), the union \(f(E)\) of these two measurable sets is measurable.

The construction. Set \(F_0 = [0,1]\). Having obtained \(F_{n-1}\) as a disjoint union of \(2^{n-1}\) closed intervals, each of the same length \(\ell_{n-1}\), form \(F_n\) by removing from each of these \(2^{n-1}\) intervals the open middle interval of length \(\alpha 3^{-n}\); what remains is a disjoint union of \(2^{n}\) closed intervals, each of the common length

\begin{equation*} \ell_n = \frac{\ell_{n-1} - \alpha 3^{-n}}{2}, \qquad \ell_0 = 1 . \end{equation*}

Then \(F = \bigcap_{n=0}^{\infty} F_n\).

The lengths \(\ell_n\). We claim

\begin{equation*} \ell_n = (1 - \alpha)\,2^{-n} + \alpha\,3^{-n} \qquad \text{for all } n \ge 0 . \end{equation*}

Indeed this holds for \(n = 0\), since \((1-\alpha) + \alpha = 1\); and if it holds for \(n-1\), then

\begin{equation*} \frac{\ell_{n-1} - \alpha 3^{-n}}{2} = \frac{(1-\alpha)2^{-(n-1)} + \alpha 3^{-(n-1)} - \alpha 3^{-n}}{2} = (1-\alpha)2^{-n} + \frac{\alpha\, 3^{-n}(3 - 1)}{2} = (1-\alpha)2^{-n} + \alpha 3^{-n}, \end{equation*}

which is \(\ell_n\). In particular \(\ell_n > 0\) for every \(n\), since \(0 < \alpha < 1\); so the construction never runs out of room and is legitimate. Also \(\ell_n \to 0\).

\(F\) is closed. Each \(F_n\) is a finite union of closed intervals, hence closed. Therefore \(F = \bigcap_{n=0}^\infty F_n\), being an intersection of closed sets, is closed.

\(m(F) = 1 - \alpha\). Each \(F_n\) is the disjoint union of \(2^n\) closed intervals of length \(\ell_n\), so by the finite additivity of Lebesgue measure and Proposition 1,

\begin{equation*} m(F_n) = 2^n \ell_n = 2^n \big[(1-\alpha)2^{-n} + \alpha 3^{-n}\big] = (1 - \alpha) + \alpha \left(\frac{2}{3}\right)^{\! n}. \end{equation*}

The collection \(\{F_n\}_{n=0}^{\infty}\) is a descending sequence of measurable sets with \(m(F_0) = 1 < \infty\). Hence by the continuity of measure (Theorem 15),

\begin{equation*} m(F) = m\left( \bigcap_{n=0}^{\infty} F_n \right) = \lim_{n \to \infty} m(F_n) = \lim_{n \to \infty}\left[ (1-\alpha) + \alpha (2/3)^n \right] = 1 - \alpha . \end{equation*}

(As a check, the total length removed is \(\sum_{n=1}^{\infty} 2^{n-1} \cdot \alpha 3^{-n} = \frac{\alpha}{2}\sum_{n=1}^{\infty}(2/3)^n = \frac{\alpha}{2} \cdot 2 = \alpha\).)

\([0,1] \sim F\) is dense in \([0,1]\). It suffices to show that every nondegenerate interval \(J \subseteq [0,1]\) contains a point of \([0,1] \sim F\); equivalently, that \(F\) contains no nondegenerate interval.

Suppose \(J \subseteq F\) is an interval with \(\ell(J) > 0\). Since \(\ell_n \to 0\), choose \(n\) with \(\ell_n < \ell(J)\). Then \(J \subseteq F \subseteq F_n\). The set \(F_n\) is a disjoint union of \(2^n\) closed intervals separated by nonempty open gaps, so a connected subset of \(F_n\) must lie inside a single one of those intervals: indeed, if \(J\) met two distinct components of \(F_n\) it would, by the intermediate value property of intervals, contain a point of one of the removed open gaps lying between them, contradicting \(J \subseteq F_n\). Hence \(J\) is contained in a single component of \(F_n\), and therefore

\begin{equation*} \ell(J) \le \ell_n < \ell(J), \end{equation*}

a contradiction. So \(F\) contains no nondegenerate interval, and every nondegenerate subinterval of \([0,1]\) meets \([0,1] \sim F\). Thus \([0,1] \sim F\) is dense in \([0,1]\).

(Note that this density statement is not a consequence of \(m(F) = 1 - \alpha < 1\) alone; it uses the geometry of the construction. Conversely, \(F\) is a closed set with empty interior that nonetheless has positive measure \(1 - \alpha > 0\), that is, a nowhere dense set of positive measure.)

Fix \(\alpha\) with \(0 < \alpha < 1\) and let \(F \subseteq [0,1]\) be the generalized Cantor set of Problem 39, so that \(F\) is closed, \([0,1] \sim F\) is dense in \([0,1]\), and \(m(F) = 1 - \alpha > 0\). Define

\begin{equation*} \mathcal{O} = (0,1) \sim F = (0,1) \cap (\mathbb{R} \sim F). \end{equation*}

Since \(F\) is closed, \(\mathbb{R} \sim F\) is open, and therefore \(\mathcal{O}\), an intersection of two open sets, is an open set of real numbers.

Claim: \(F \cap (0,1) \subseteq \partial\mathcal{O}\).

Recall that \(\partial\mathcal{O} = \overline{\mathcal{O}} \sim \operatorname{int}\mathcal{O}\), and that \(\operatorname{int}\mathcal{O} = \mathcal{O}\) because \(\mathcal{O}\) is open. So we must show that each \(x \in F \cap (0,1)\) is a point of closure of \(\mathcal{O}\) but does not belong to \(\mathcal{O}\).

That \(x \notin \mathcal{O}\) is immediate from \(x \in F\).

That \(x\) is a point of closure of \(\mathcal{O}\): let \(\epsilon > 0\). Since \(0 < x < 1\), we may shrink \(\epsilon\) so that \((x - \epsilon, x + \epsilon) \subseteq (0,1)\). This is a nondegenerate subinterval of \([0,1]\), so by Problem 39 the dense set \([0,1] \sim F\) meets it; that is, there is a point

\begin{equation*} y \in (x - \epsilon, x + \epsilon) \quad\text{with}\quad y \notin F . \end{equation*}

Since also \(y \in (0,1)\), we have \(y \in \mathcal{O}\). Thus every neighborhood of \(x\) contains a point of \(\mathcal{O}\), so \(x \in \overline{\mathcal{O}}\). This proves the claim.

Conclusion. The boundary \(\partial\mathcal{O}\) is a closed set, hence measurable by Theorem 9. Since \(0\) and \(1\) are never removed in the construction of \(F\), we have \(F \cap (0,1) = F \sim \{0,1\}\), and a two-point set has measure zero by Proposition 4, so by the excision property of measure

\begin{equation*} m\big(F \cap (0,1)\big) = m(F) = 1 - \alpha . \end{equation*}

Therefore, by the monotonicity of measure and the claim,

\begin{equation*} m(\partial\mathcal{O}) \ge m\big(F \cap (0,1)\big) = 1 - \alpha > 0 . \end{equation*}

So \(\mathcal{O}\) is an open set whose boundary has positive measure. (Taking \(\alpha\) small makes \(m(\partial\mathcal{O})\) as close to \(1\) as we please, while \(\mathcal{O}\) itself is an open subset of \((0,1)\) of measure \(\alpha\); in fact here \(\partial \mathcal{O} = F\), since \(\overline{\mathcal{O}} = [0,1]\).)

Recall the construction: \(C_0 = [0,1]\), and \(C_k\) is the disjoint union of \(2^k\) closed intervals each of length \(3^{-k}\), obtained from \(C_{k-1}\) by removing the open middle third of each of its components; \(\mathbf{C} = \bigcap_{k=1}^{\infty} C_k\).

By Proposition 19 the set \(\mathbf{C}\) is closed, and it is nonempty since (as we verify below) \(0 \in \mathbf{C}\). It remains to show that each neighborhood of each point of \(\mathbf{C}\) contains infinitely many points of \(\mathbf{C}\).

Step 1: every endpoint of every component of every \(C_k\) belongs to \(\mathbf{C}\).

Let \(e\) be an endpoint of a component \(J\) of \(C_k\). We show by induction on \(n \ge k\) that \(e\) is an endpoint of a component of \(C_n\). This holds for \(n = k\). Suppose \(e\) is an endpoint of a component \(J^{\prime}\) of \(C_n\). Passing to \(C_{n+1}\) deletes the open middle third of \(J^{\prime}\) and leaves the two closed outer thirds of \(J^{\prime}\) as components of \(C_{n+1}\); the two endpoints of \(J^{\prime}\) are endpoints of these two outer thirds. Hence \(e\) is an endpoint of a component of \(C_{n+1}\), completing the induction.

In particular \(e \in C_n\) for every \(n \ge k\), and since \(C_0 \supseteq C_1 \supseteq \cdots\) we also have \(e \in C_n\) for \(n < k\). Therefore \(e \in \bigcap_{n=1}^{\infty} C_n = \mathbf{C}\). (Applied to \(C_0 = [0,1]\), this gives \(0, 1 \in \mathbf{C}\).)

Step 2: for each \(k\), each component \(J\) of \(C_k\) meets \(\mathbf{C}\) in an infinite set.

Fix \(j \ge 1\). The set \(J \cap C_{k+j}\) consists of exactly \(2^{j}\) pairwise disjoint closed intervals of length \(3^{-(k+j)}\): this is immediate by induction on \(j\), since each of the components of \(C_{k+j-1}\) lying inside \(J\) splits into two components of \(C_{k+j}\) lying inside \(J\).

Each of these \(2^j\) intervals is a component of \(C_{k+j}\), so by Step 1 both of its endpoints lie in \(\mathbf{C}\); and both endpoints lie in \(J\). Since the \(2^j\) intervals are pairwise disjoint and nondegenerate, these \(2^{j+1}\) endpoints are distinct. Hence \(J \cap \mathbf{C}\) contains at least \(2^{j+1}\) points. As \(j \ge 1\) was arbitrary, \(J \cap \mathbf{C}\) is infinite.

Step 3: conclusion.

Let \(x \in \mathbf{C}\) and let \(\mathcal{U}\) be a neighborhood of \(x\); choose \(\epsilon > 0\) with \((x - \epsilon, x + \epsilon) \subseteq \mathcal{U}\). Choose a natural number \(k\) with \(3^{-k} < \epsilon\). Since \(x \in \mathbf{C} \subseteq C_k\), the point \(x\) belongs to some component \(J\) of \(C_k\), and \(\ell(J) = 3^{-k}\). For every \(y \in J\) we then have

\begin{equation*} |y - x| \le \ell(J) = 3^{-k} < \epsilon, \end{equation*}

so \(J \subseteq (x - \epsilon, x + \epsilon) \subseteq \mathcal{U}\).

By Step 2 the set \(J \cap \mathbf{C}\) is infinite, and it is contained in \(\mathcal{U} \cap \mathbf{C}\). Hence \(\mathcal{U}\) contains infinitely many points of \(\mathbf{C}\).

Since \(\mathbf{C}\) is nonempty and closed, and every neighborhood of every point of \(\mathbf{C}\) contains infinitely many points of \(\mathbf{C}\), the Cantor set is perfect.

Let \(X \subseteq \mathbb{R}\) be perfect: nonempty, closed, and such that every neighborhood of every point of \(X\) contains infinitely many points of \(X\).

First note that \(X\) is infinite. Indeed, pick \(x \in X\); the neighborhood \((x-1, x+1)\) of \(x\) already contains infinitely many points of \(X\).

Suppose, to get a contradiction, that \(X\) is countable. Being infinite and countable, it may be enumerated as \(X = \{x_n\}_{n=1}^{\infty}\). For \(y \in \mathbb{R}\) and \(r > 0\) write \(B(y,r) = (y - r, y + r)\) and \(\bar B(y,r) = [y-r, y+r]\).

Construction. We construct points \(y_n \in X\) and radii \(r_n > 0\) such that, setting

\begin{equation*} F_n = X \cap \bar B(y_n, r_n), \end{equation*}

we have for every \(n\):

1. \(y_n \in X\), so \(F_n \ne \emptyset\);
2. \(x_n \notin F_n\);
3. \(\bar B(y_{n+1}, r_{n+1}) \subseteq B(y_n, r_n)\), so that \(F_{n+1} \subseteq F_n\).

Base step. Since \(X\) is infinite, choose \(y_1 \in X\) with \(y_1 \ne x_1\), and put \(r_1 = \tfrac{1}{2}|y_1 - x_1| > 0\). Then \(y_1 \in F_1\) and \(x_1 \notin \bar B(y_1, r_1)\), so \(x_1 \notin F_1\).

Inductive step. Suppose \(y_n \in X\) and \(r_n > 0\) have been chosen. Since \(y_n \in X\), the neighborhood \(B(y_n, r_n)\) of \(y_n\) contains infinitely many points of \(X\). At most one of them equals \(x_{n+1}\), so we may choose

\begin{equation*} y_{n+1} \in X \cap B(y_n, r_n) \quad\text{with}\quad y_{n+1} \ne x_{n+1}. \end{equation*}

Now choose \(r_{n+1} > 0\) with

\begin{equation*} r_{n+1} < \min\Big\{\, r_n - |y_{n+1} - y_n| \,,\ \tfrac{1}{2}|y_{n+1} - x_{n+1}| \,\Big\}, \end{equation*}

both quantities being strictly positive by the choice of \(y_{n+1}\). The first constraint gives \(\bar B(y_{n+1}, r_{n+1}) \subseteq B(y_n, r_n)\): if \(|z - y_{n+1}| \le r_{n+1}\) then

\begin{equation*} |z - y_n| \le |z - y_{n+1}| + |y_{n+1} - y_n| \le r_{n+1} + |y_{n+1} - y_n| < r_n . \end{equation*}

The second gives \(x_{n+1} \notin \bar B(y_{n+1}, r_{n+1})\). Properties 1, 2, 3 hold, and the construction proceeds.

The contradiction. Each \(F_n = X \cap \bar B(y_n, r_n)\) is the intersection of two closed sets (\(X\) is closed by hypothesis, \(\bar B(y_n, r_n)\) is closed), hence closed; it is bounded, being contained in \(\bar B(y_n, r_n)\); and it is nonempty, since \(y_n \in X \cap \bar B(y_n, r_n)\). By property 3,

\begin{equation*} F_1 \supseteq F_2 \supseteq F_3 \supseteq \cdots , \end{equation*}

so \(\{F_n\}_{n=1}^{\infty}\) is a descending countable collection of nonempty closed sets of real numbers with \(F_1\) bounded. By the Nested Set Theorem,

\begin{equation*} \bigcap_{n=1}^{\infty} F_n \ne \emptyset . \end{equation*}

On the other hand, let \(z \in \bigcap_{n=1}^{\infty} F_n\). Then \(z \in F_1 \subseteq X\), so \(z = x_N\) for some natural number \(N\). But \(z \in F_N\), while property 2 says \(x_N \notin F_N\) - a contradiction. Hence the intersection is empty.

These two conclusions are incompatible, so \(X\) cannot be countable. Therefore every perfect subset of \(\mathbb{R}\) is uncountable.

The two preceding problems are the following. Problem 41: a nonempty subset \(X\) of \(\mathbb{R}\) is called perfect provided it is closed and each neighborhood of any point in \(X\) contains infinitely many points of \(X\); the Cantor set \(\mathbf{C}\) is perfect. Problem 42: every perfect subset of \(\mathbb{R}\) is uncountable. Granting these two assertions, the Cantor set is uncountable, since by Problem 41 it is perfect and by Problem 42 every perfect set is uncountable. For completeness we carry out both arguments, using only the description of the Cantor set given in Section 2.7 and results from Chapter 1.

Recall the construction: \(C_0 = [0,1]\), and \(C_{k+1}\) is obtained from \(C_k\) by removing the open middle third of each of the \(2^k\) disjoint closed intervals of length \(3^{-k}\) whose union is \(C_k\); then

\begin{equation*} \mathbf{C} = \bigcap_{k=1}^{\infty} C_k . \end{equation*}

By Proposition 19 the set \(\mathbf{C}\) is closed, and it is nonempty since \(0 \in C_k\) for every \(k\).

Step 1: the endpoints of every interval occurring in the construction belong to \(\mathbf{C}\).

Let \(J = [a,b]\) be one of the \(2^k\) closed intervals whose union is \(C_k\). In passing from \(C_k\) to \(C_{k+1}\) the open middle third of \(J\) is deleted, so the two intervals of \(C_{k+1}\) contained in \(J\) are \([a, a + (b-a)/3]\) and \([b - (b-a)/3, b]\). Thus \(a\) is the left endpoint of an interval of \(C_{k+1}\) and \(b\) is the right endpoint of an interval of \(C_{k+1}\). Iterating this observation, for every \(j \geq k\) the points \(a\) and \(b\) are endpoints of intervals of \(C_j\), and in particular \(a, b \in C_j\). Since the collection \(\{C_j\}\) is descending, \(a, b \in C_j\) for \(j < k\) as well. Hence \(a, b \in \bigcap_{j \geq 1} C_j = \mathbf{C}\).

Step 2: \(\mathbf{C}\) is perfect.

The set \(\mathbf{C}\) is nonempty and closed. Let \(x \in \mathbf{C}\) and let \(\mathcal{O}\) be a neighborhood of \(x\); choose \(r > 0\) with \((x - r, x + r) \subseteq \mathcal{O}\). Choose an index \(k\) with \(3^{-k} < r\). Since \(x \in \mathbf{C} \subseteq C_k\) and \(C_k\) is the disjoint union of \(2^k\) closed intervals of length \(3^{-k}\), the point \(x\) belongs to one of them, say \(J\). Every point \(y\) of \(J\) satisfies \(|y - x| \leq 3^{-k} < r\), so

\begin{equation*} J \subseteq (x - r, x + r) \subseteq \mathcal{O} . \end{equation*}

Now fix \(j > k\). Exactly \(2^{\,j-k}\) of the intervals whose disjoint union is \(C_j\) are contained in \(J\) (each interval of \(C_i\) inside \(J\) splits into two intervals of \(C_{i+1}\) inside \(J\), and this is an induction on \(i\) starting at \(i = k\)). These \(2^{\,j-k}\) intervals are pairwise disjoint, and by Step 1 the left endpoint of each of them belongs to \(\mathbf{C}\). Hence \(J \cap \mathbf{C}\) contains at least \(2^{\,j-k}\) points. As \(j\) may be taken arbitrarily large, \(J \cap \mathbf{C}\) is infinite, and therefore \(\mathcal{O} \cap \mathbf{C}\) is infinite. Thus \(\mathbf{C}\) is perfect.

Step 3: a perfect subset of \(\mathbb{R}\) is uncountable.

Let \(X\) be perfect and suppose, to get a contradiction, that \(X\) is countable. Since \(X\) is nonempty it may be enumerated as \(X = \{x_1, x_2, x_3, \dots\}\) (a finite list is handled by repetition, but in fact the perfectness of \(X\) already forces \(X\) to be infinite). We construct inductively points \(y_n \in X\) and radii \(r_n > 0\) such that, writing

\begin{equation*} I_n = [\,y_n - r_n,\ y_n + r_n\,], \qquad F_n = X \cap I_n , \end{equation*}

we have for every \(n\):

(i) \(F_n \neq \emptyset\); (ii) \(I_{n+1} \subseteq (y_n - r_n, y_n + r_n)\), hence \(F_{n+1} \subseteq F_n\); (iii) \(x_n \notin I_n\), hence \(x_n \notin F_n\).

To begin, pick any point \(y_0 \in X\) and put \(r_0 = 1\). Suppose \(y_{n-1} \in X\) and \(r_{n-1} > 0\) have been chosen (\(n \geq 1\)). The interval \((y_{n-1} - r_{n-1},\, y_{n-1} + r_{n-1})\) is a neighborhood of the point \(y_{n-1}\) of \(X\), so it contains infinitely many points of \(X\); in particular we may choose

\begin{equation*} y_n \in X \cap (y_{n-1} - r_{n-1},\, y_{n-1} + r_{n-1}) \quad \text{with} \quad y_n \neq x_n . \end{equation*}

Since \(y_n \neq x_n\) and \(y_n\) lies in the open interval \((y_{n-1} - r_{n-1}, y_{n-1} + r_{n-1})\), we may choose \(r_n > 0\) so small that

\begin{equation*} r_n < |y_n - x_n| \quad \text{and} \quad [\,y_n - r_n, y_n + r_n\,] \subseteq (y_{n-1} - r_{n-1},\, y_{n-1} + r_{n-1}) . \end{equation*}

Then (iii) holds because \(|x_n - y_n| > r_n\), and (ii) holds by the second requirement; (i) holds because \(y_n \in F_n\).

Each \(F_n = X \cap I_n\) is the intersection of two closed sets, hence closed, and it is bounded since \(I_n\) is bounded. By (i) and (ii), \(\{F_n\}_{n=1}^{\infty}\) is a descending countable collection of nonempty bounded closed sets of real numbers. By the Nested Set Theorem there is a point

\begin{equation*} x \in \bigcap_{n=1}^{\infty} F_n . \end{equation*}

Since \(F_1 \subseteq X\), we have \(x \in X\), so \(x = x_N\) for some index \(N\). But \(x \in F_N\), contradicting (iii), which says \(x_N \notin F_N\). This contradiction shows \(X\) is uncountable.

Conclusion. By Step 2 the Cantor set \(\mathbf{C}\) is a perfect subset of \(\mathbb{R}\), and by Step 3 every perfect subset of \(\mathbb{R}\) is uncountable. Therefore \(\mathbf{C}\) is uncountable, which is a second proof of the uncountability asserted in Proposition 19.

Of course the empty set is an open subset of every open set and is disjoint from \(A\), so the definition is to be read as requiring a nonempty open subset; we prove this stronger statement for the Cantor set \(\mathbf{C}\).

Let \(\mathcal{O}\) be a nonempty open set of real numbers. Choose a point \(x \in \mathcal{O}\). Since \(\mathcal{O}\) is open there is an \(r > 0\) for which

\begin{equation*} (x - r, x + r) \subseteq \mathcal{O} . \end{equation*}

We first observe that the open interval \((x-r, x+r)\) is not contained in \(\mathbf{C}\). Indeed, by Proposition 19 the Cantor set has measure zero, \(m(\mathbf{C}) = 0\), whereas the measure of an interval is its length, so

\begin{equation*} m\big((x - r, x + r)\big) = 2r > 0 . \end{equation*}

If we had \((x-r, x+r) \subseteq \mathbf{C}\), then by the monotonicity of Lebesgue measure we would get \(2r = m((x-r,x+r)) \leq m(\mathbf{C}) = 0\), a contradiction. Hence there is a point

\begin{equation*} z \in (x - r, x + r) \quad \text{with} \quad z \notin \mathbf{C}. \end{equation*}

By Proposition 19 the set \(\mathbf{C}\) is closed, so its complement \(\mathbb{R} \sim \mathbf{C}\) is open. Define

\begin{equation*} \mathcal{U} = (x - r, x + r) \cap (\mathbb{R} \sim \mathbf{C}) . \end{equation*}

Then \(\mathcal{U}\) is open, being the intersection of two open sets; it is nonempty, since \(z \in \mathcal{U}\); it is contained in \(\mathcal{O}\), since \(\mathcal{U} \subseteq (x-r, x+r) \subseteq \mathcal{O}\); and it is disjoint from \(\mathbf{C}\) by its very definition. Thus every nonempty open set \(\mathcal{O}\) has a nonempty open subset disjoint from \(\mathbf{C}\), that is, \(\mathbf{C}\) is nowhere dense in \(\mathbb{R}\).

Two remarks. First, closedness of \(\mathbf{C}\) was used only to make \(\mathcal{U}\) open. The same two paragraphs prove, for an arbitrary closed set \(A\), that \(A\) is nowhere dense if and only if \(A\) has empty interior: if \(A\) contains no nonempty open set, then for a given nonempty open \(\mathcal{O}\) we may choose \(z \in \mathcal{O} \sim A\), and \(\mathcal{O} \cap (\mathbb{R} \sim A)\) is then a nonempty open subset of \(\mathcal{O}\) disjoint from \(A\); conversely, if some nonempty open \(\mathcal{O}_0\) is contained in \(A\), then every nonempty open subset of \(\mathcal{O}_0\) meets \(A\), so \(A\) is not nowhere dense. Since \(\mathbf{C}\) is closed and, as shown above, contains no open interval, \(\mathbf{C}\) indeed has empty interior.

Second, the appeal to measure can be replaced by a direct appeal to the construction. Given the interval \((x-r, x+r)\), if it fails to meet \([0,1]\) it is itself a nonempty open subset of \(\mathcal{O}\) that is disjoint from \(\mathbf{C} \subseteq [0,1]\), and we are done. Otherwise pick \(y \in (x-r,x+r) \cap [0,1]\); since \(|y - x| < r\) we may then choose an index \(k\) with

\begin{equation*} 3^{-k} < r - |y - x| . \end{equation*}

If \(y \notin \mathbf{C}\) we may take \(z = y\) in the argument above. If \(y \in \mathbf{C}\), then \(y\) lies in one of the \(2^{k}\) closed intervals \(J\) of length \(3^{-k}\) whose disjoint union is \(C_{k}\), and every point \(w\) of \(J\) satisfies \(|w - x| \leq |w - y| + |y - x| \leq 3^{-k} + |y-x| < r\), so \(J \subseteq (x-r, x+r) \subseteq \mathcal{O}\). The open middle third of \(J\) is deleted in passing from \(C_k\) to \(C_{k+1}\) and so is disjoint from \(\mathbf{C}\); it is therefore a nonempty open subset of \(\mathcal{O}\) disjoint from \(\mathbf{C}\).

Let \(I\) be an interval of real numbers and let \(f \colon I \to \mathbb{R}\) be strictly increasing, that is, \(f(u) < f(v)\) whenever \(u, v \in I\) and \(u < v\). Set \(E = f(I)\), the image of \(I\). No continuity is assumed of \(f\), and \(E\) need not be an interval.

The function \(f\) is one-to-one: if \(u, v \in I\) and \(u \neq v\), then either \(u < v\) or \(v < u\), and in either case strict monotonicity gives \(f(u) \neq f(v)\). Hence \(f\) is a one-to-one map of \(I\) onto \(E\) and the inverse function

\begin{equation*} g = f^{-1} \colon E \to I, \qquad g(f(x)) = x \ \ (x \in I), \qquad f(g(y)) = y \ \ (y \in E) \end{equation*}

is well defined. We must show that \(g\) is continuous on \(E\), continuity being understood, as always, relative to the domain \(E\): \(g\) is continuous at \(y_0 \in E\) provided for each \(\varepsilon > 0\) there is a \(\delta > 0\) such that \(|g(y) - g(y_0)| < \varepsilon\) for all \(y \in E\) with \(|y - y_0| < \delta\).

First observe that \(g\) is strictly increasing on \(E\). Indeed, let \(y_1, y_2 \in E\) with \(y_1 < y_2\) and suppose \(g(y_1) \geq g(y_2)\). If \(g(y_1) = g(y_2)\) then \(y_1 = f(g(y_1)) = f(g(y_2)) = y_2\), a contradiction; if \(g(y_1) > g(y_2)\) then, since \(f\) is strictly increasing, \(y_1 = f(g(y_1)) > f(g(y_2)) = y_2\), again a contradiction. Hence \(g(y_1) < g(y_2)\).

Now fix \(y_0 \in E\), put \(x_0 = g(y_0) \in I\), and let \(\varepsilon > 0\). If \(I\) consists of the single point \(x_0\), then \(E = \{y_0\}\) and continuity at \(y_0\) is trivial (any \(\delta\) works). So assume \(I\) has more than one point. We treat the two sides of \(x_0\) separately.

Left side. Suppose first that there is a point \(a \in I\) with \(a < x_0\). Since \(I\) is an interval containing \(a\) and \(x_0\), we have \([a, x_0] \subseteq I\), and this interval contains points strictly between \(\max\{a,\ x_0 - \varepsilon\}\) and \(x_0\); choose such a point \(u\), so that

\begin{equation*} u \in I, \qquad x_0 - \varepsilon < u < x_0 . \end{equation*}

Since \(f\) is strictly increasing, \(f(u) < f(x_0) = y_0\), so \(\delta_1 := y_0 - f(u) > 0\). If instead there is no point of \(I\) strictly below \(x_0\), put \(\delta_1 := 1\).

Right side. Symmetrically, if there is a point \(b \in I\) with \(b > x_0\), choose \(v \in I\) with \(x_0 < v < \min\{b,\ x_0 + \varepsilon\}\); then \(y_0 = f(x_0) < f(v)\), and we put \(\delta_2 := f(v) - y_0 > 0\). If there is no point of \(I\) strictly above \(x_0\), put \(\delta_2 := 1\).

Define \(\delta = \min\{\delta_1, \delta_2\} > 0\), and let \(y \in E\) satisfy \(|y - y_0| < \delta\). We claim \(|g(y) - x_0| < \varepsilon\).

Consider the lower bound. If \(I\) has no point strictly below \(x_0\), then \(x_0\) is the smallest element of \(I\), so \(g(y) \in I\) forces \(g(y) \geq x_0 > x_0 - \varepsilon\). Otherwise \(u\) was chosen as above and

\begin{equation*} y > y_0 - \delta \geq y_0 - \delta_1 = f(u) , \end{equation*}

so, \(g\) being strictly increasing on \(E\) and \(f(u) \in E\),

\begin{equation*} g(y) > g(f(u)) = u > x_0 - \varepsilon . \end{equation*}

In either case \(g(y) > x_0 - \varepsilon\).

Consider the upper bound. If \(I\) has no point strictly above \(x_0\), then \(x_0\) is the largest element of \(I\) and \(g(y) \leq x_0 < x_0 + \varepsilon\). Otherwise \(v\) was chosen as above and

\begin{equation*} y < y_0 + \delta \leq y_0 + \delta_2 = f(v), \end{equation*}

so \(g(y) < g(f(v)) = v < x_0 + \varepsilon\). In either case \(g(y) < x_0 + \varepsilon\).

Therefore \(x_0 - \varepsilon < g(y) < x_0 + \varepsilon\), that is, \(|g(y) - g(y_0)| < \varepsilon\), for every \(y \in E\) with \(|y - y_0| < \delta\). Since \(y_0 \in E\) was arbitrary, the inverse function \(g = f^{-1}\) is continuous on \(E = f(I)\).

Let \(f\) be a real-valued continuous function whose domain \(D\) is a Borel set of real numbers; the case \(D = \mathbb{R}\) is the one usually intended, and the case in which \(D\) is an interval is the one needed in the next problem. Some such assumption on the domain is necessary: if \(D\) is not Borel then already \(f^{-1}(\mathbb{R}) = D\) fails to be Borel. Denote by \(\mathcal{B}\) the collection of Borel sets, that is, by definition, the smallest \(\sigma\)-algebra of subsets of \(\mathbb{R}\) that contains all of the open sets.

Following the hint, define

\begin{equation*} \mathcal{A} = \{\, E \subseteq \mathbb{R} \ : \ f^{-1}(E) \in \mathcal{B} \,\}, \end{equation*}

where \(f^{-1}(E) = \{x \in D : f(x) \in E\}\).

Step 1: \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\).

The set \(\mathbb{R}\) belongs to \(\mathcal{A}\), since \(f^{-1}(\mathbb{R}) = D\) and \(D\) is Borel by assumption. Next, let \(E \in \mathcal{A}\). Since \(f\) is a function defined on \(D\), a point \(x \in D\) lies in \(f^{-1}(\mathbb{R} \sim E)\) if and only if \(f(x) \notin E\), that is, if and only if \(x \notin f^{-1}(E)\). Hence

\begin{equation*} f^{-1}(\mathbb{R} \sim E) = D \sim f^{-1}(E) = D \cap \big(\mathbb{R} \sim f^{-1}(E)\big). \end{equation*}

Since \(\mathcal{B}\) is a \(\sigma\)-algebra containing \(D\) and \(f^{-1}(E)\), the right-hand side belongs to \(\mathcal{B}\). Thus \(\mathbb{R} \sim E \in \mathcal{A}\), so \(\mathcal{A}\) is closed under the formation of complements. Finally, let \(\{E_k\}_{k=1}^{\infty}\) be a countable collection of sets in \(\mathcal{A}\). For \(x \in D\) we have \(f(x) \in \bigcup_k E_k\) if and only if \(f(x) \in E_k\) for some \(k\), so

\begin{equation*} f^{-1}\left(\bigcup_{k=1}^{\infty} E_k\right) = \bigcup_{k=1}^{\infty} f^{-1}(E_k), \end{equation*}

which is a countable union of Borel sets and hence Borel. Thus \(\bigcup_k E_k \in \mathcal{A}\). Therefore \(\mathcal{A}\) is a \(\sigma\)-algebra.

Step 2: \(\mathcal{A}\) contains every open set.

Let \(\mathcal{O}\) be an open set of real numbers. Since \(f\) is continuous on \(D\), Proposition 22 of Chapter 1 tells us that there is an open set \(U\) for which

\begin{equation*} f^{-1}(\mathcal{O}) = D \cap U . \end{equation*}

Both \(D\) and \(U\) are Borel sets (\(U\) is open, and the Borel sets contain the open sets), and \(\mathcal{B}\) is closed under finite intersections, being a \(\sigma\)-algebra. Hence \(f^{-1}(\mathcal{O}) \in \mathcal{B}\), that is, \(\mathcal{O} \in \mathcal{A}\).

Step 3: conclusion.

By Steps 1 and 2, \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\) that contains all the open sets. Since \(\mathcal{B}\) is the smallest such \(\sigma\)-algebra, we conclude

\begin{equation*} \mathcal{B} \subseteq \mathcal{A} . \end{equation*}

Thus if \(B\) is a Borel set, then \(B \in \mathcal{A}\), which is precisely the assertion that \(f^{-1}(B)\) is a Borel set.

In particular, taking \(D = \mathbb{R}\): the inverse image under a continuous real-valued function on \(\mathbb{R}\) of an open set is open, of a closed set is closed (take complements), and of a Borel set is Borel.

Let \(I\) be an interval of real numbers and let \(f \colon I \to \mathbb{R}\) be continuous and strictly increasing. Let \(B\) be a Borel set. We show that \(f(B \cap I)\) is a Borel set; in particular, if \(B\) is a Borel subset of \(I\), then \(f(B)\) is a Borel set. We may assume \(I\) is nonempty.

Step 1: the image \(E = f(I)\) is an interval, and hence a Borel set.

We check that \(E\) has the intermediate value property. Let \(y_1, y_2 \in E\) with \(y_1 < y_2\), and let \(y\) satisfy \(y_1 < y < y_2\). Write \(y_1 = f(x_1)\) and \(y_2 = f(x_2)\) with \(x_1, x_2 \in I\). Since \(f\) is strictly increasing and \(f(x_1) < f(x_2)\), we must have \(x_1 < x_2\). As \(I\) is an interval, \([x_1, x_2] \subseteq I\), and \(f\) is continuous on \([x_1, x_2]\). By the Intermediate Value Theorem there is a point \(x \in (x_1, x_2)\) with \(f(x) = y\). Since \(x \in I\), we conclude \(y \in E\).

A nonempty set \(E\) of real numbers with this property is an interval: let \(\alpha = \inf E\) and \(\beta = \sup E\) in the extended reals. If \(\alpha < y < \beta\), then by the definitions of infimum and supremum there are \(y_1, y_2 \in E\) with \(y_1 < y < y_2\), so \(y \in E\); thus \((\alpha, \beta) \subseteq E \subseteq [\alpha, \beta]\), and \(E\) is an interval (possibly degenerate, if \(I\) is a single point). Every interval is a Borel set, being open, closed, or the union of an open interval with one or two of its endpoints. Hence \(E\) is a Borel set.

Step 2: the inverse function is continuous on \(E\).

The function \(f\) is strictly increasing on the interval \(I\), so by Problem 45 it is one-to-one and its inverse

\begin{equation*} g = f^{-1} \colon E \to I \end{equation*}

is a continuous function defined on \(E = f(I)\).

Step 3: apply Problem 46 to \(g\).

The function \(g\) is continuous and its domain \(E\) is a Borel set by Step 1. Therefore, by Problem 46, for every Borel set \(B\) the inverse image \(g^{-1}(B)\) is a Borel set.

Step 4: identify \(g^{-1}(B)\) as an image under \(f\).

We claim

\begin{equation*} g^{-1}(B) = f(B \cap I). \end{equation*}

Indeed, suppose \(y \in g^{-1}(B)\). Then \(y \in E\) and \(g(y) \in B\). Put \(x = g(y)\); then \(x \in I\), \(x \in B\), and \(f(x) = f(g(y)) = y\). Hence \(y \in f(B \cap I)\). Conversely, suppose \(y \in f(B \cap I)\), say \(y = f(x)\) with \(x \in B \cap I\). Then \(y \in f(I) = E\) and \(g(y) = g(f(x)) = x \in B\), so \(y \in g^{-1}(B)\). This proves the claim.

Conclusion. Combining Steps 3 and 4, for every Borel set \(B\) the set

\begin{equation*} f(B \cap I) = g^{-1}(B) \end{equation*}

is a Borel set. In particular a continuous, strictly increasing function defined on an interval maps Borel subsets of that interval onto Borel sets. Since such an \(f\) is one-to-one, it maps \(I\) onto the interval \(E\) and carries Borel sets onto Borel sets, which is exactly the fact invoked in the proof of Proposition 22 of Chapter 2 to produce a measurable subset of the Cantor set that is not Borel.

A remark on where the hypotheses entered. Continuity of \(f\) was used only in Step 1, to know that \(E = f(I)\) is an interval and therefore a Borel set, which is what makes Problem 46 applicable to \(g\). Strict monotonicity was used twice: to know that \(f\) is one-to-one, so that the identification \(f(B \cap I) = g^{-1}(B)\) of Step 4 is valid, and to invoke Problem 45 for the continuity of \(g\). Note that no continuity of \(g\) beyond that supplied by Problem 45 is needed, and no appeal to measurability of \(f\) is made anywhere.

Throughout, \([a,b]\) is a nondegenerate closed bounded interval, so \(a < b\).

Put \(h = f - g\) on \([a,b]\). Since \(f\) and \(g\) are continuous, so is \(h\), and the hypothesis that \(f = g\) a.e. means precisely that there is a measurable set \(Z \subseteq [a,b]\) with \(m(Z) = 0\) such that \(f = g\), that is \(h = 0\), on \([a,b] \sim Z\). Hence

\begin{equation*} E_0 = \{x \in [a,b] \mid h(x) \neq 0\} \subseteq Z , \end{equation*}

so by the monotonicity of Lebesgue outer measure \(m^*(E_0) = 0\); every set of outer measure zero is measurable (Chapter 2), so \(E_0\) is measurable and \(m(E_0) = 0\).

Suppose, to get a contradiction, that \(h(x_0) \neq 0\) for some \(x_0 \in [a,b]\). Set \(\varepsilon = |h(x_0)|/2 > 0\). By the continuity of \(h\) at \(x_0\) there is a \(\delta > 0\) such that

\begin{equation*} |h(x) - h(x_0)| < \varepsilon \quad \text{for all } x \in [a,b] \text{ with } |x - x_0| < \delta . \end{equation*}

For such \(x\) the triangle inequality gives

\begin{equation*} |h(x)| \geq |h(x_0)| - |h(x) - h(x_0)| > 2\varepsilon - \varepsilon = \varepsilon > 0 , \end{equation*}

so \(h(x) \neq 0\). Therefore

\begin{equation*} I = (x_0 - \delta,\, x_0 + \delta) \cap [a,b] \subseteq E_0 . \end{equation*}

Now \(I\) is the intersection of two intervals, hence an interval, and it is nondegenerate: if \(a < x_0 < b\) it contains the interval \((x_0 - \delta^{\prime}, x_0 + \delta^{\prime})\) with \(\delta^{\prime} = \min\{\delta, x_0 - a, b - x_0\} > 0\); if \(x_0 = a\) it contains \([a, a + \delta^{\prime\prime})\) with \(\delta^{\prime\prime} = \min\{\delta, b-a\} > 0\); and if \(x_0 = b\) it contains \((b - \delta^{\prime\prime}, b]\). In each case \(I\) contains an interval of positive length, so by the monotonicity of outer measure and the fact that the measure of an interval is its length (Chapter 2),

\begin{equation*} m(E_0) \geq m(I) > 0 , \end{equation*}

contradicting \(m(E_0) = 0\). Hence \(h(x) = 0\) for every \(x \in [a,b]\), that is, \(f = g\) on all of \([a,b]\).

Now replace \([a,b]\) by a general measurable set \(E\). The assertion is false in general. Take

\begin{equation*} E = [0,1] \cup \{2\}, \qquad f \equiv 0 \text{ on } E, \qquad g(x) = \begin{cases} 0 & \text{if } x \in [0,1], \\ 1 & \text{if } x = 2. \end{cases} \end{equation*}

Both are continuous on \(E\): \(f\) is constant, and \(g\) is continuous at each point of \([0,1]\) because \(g\) vanishes on the relative neighborhood \(E \cap (-1,2) = [0,1]\) of each such point, while \(2\) is an isolated point of \(E\), so every function on \(E\) is continuous there. Moreover \(\{x \in E \mid f(x) \neq g(x)\} = \{2\}\) has measure zero, so \(f = g\) a.e. on \(E\). Yet \(f(2) \neq g(2)\).

The argument given for \([a,b]\) isolates exactly what is needed, and in fact the following characterization holds: for a measurable set \(E\), every pair of continuous functions on \(E\) that agree a.e. on \(E\) agrees on all of \(E\) if and only if

\begin{equation*} m\big(E \cap (x - r,\, x + r)\big) > 0 \quad \text{for every } x \in E \text{ and every } r > 0. \tag{\(\ast\)} \end{equation*}

Indeed, assume \((\ast)\) and suppose \(f, g\) are continuous on \(E\) with \(f = g\) a.e. If \(h = f - g\) satisfied \(h(x_0) \neq 0\) for some \(x_0 \in E\), then exactly as above there is \(\delta > 0\) with \(h \neq 0\) on \(E \cap (x_0 - \delta, x_0 + \delta)\), and by \((\ast)\) this set has positive measure; but it is contained in \(\{x \in E \mid h(x) \neq 0\}\), which as above has measure zero, a contradiction. Conversely, if \((\ast)\) fails, say \(m(E \cap (x_0 - r, x_0 + r)) = 0\) for some \(x_0 \in E\) and \(r > 0\), put \(f \equiv 0\) on \(E\) and

\begin{equation*} g(x) = \max\Big\{0,\, 1 - \frac{|x - x_0|}{r}\Big\} . \end{equation*}

Then \(g\) is continuous on all of \(\mathbb{R}\), hence on \(E\), and \(g\) vanishes off \((x_0 - r, x_0 + r)\), so \(\{x \in E \mid f(x) \neq g(x)\} \subseteq E \cap (x_0 - r, x_0 + r)\) has measure zero; but \(g(x_0) = 1 \neq 0 = f(x_0)\).

Since \((\ast)\) holds for \(E = [a,b]\) with \(a < b\), this recovers the first part. It also fails for the set \(E = [0,1] \cup \{2\}\) at the point \(x = 2\), which is why the counterexample above works.

No. One implication survives, the other does not.

The implication that does survive is the forward one: if \(f\) is continuous on \(D \cup E\), then its restrictions \(f|_D\) and \(f|_E\) are continuous. This is immediate from the definition of continuity relative to a domain. Let \(x_0 \in D\) and let \(\varepsilon > 0\). Continuity of \(f\) at \(x_0\) as a function on \(D \cup E\) gives a \(\delta > 0\) such that \(|f(x) - f(x_0)| < \varepsilon\) whenever \(x \in D \cup E\) and \(|x - x_0| < \delta\). Since \(D \subseteq D \cup E\), the same \(\delta\) works for \(f|_D\) at \(x_0\). Hence \(f|_D\) is continuous on \(D\), and likewise \(f|_E\) is continuous on \(E\).

The converse fails. Take

\begin{equation*} D = (-\infty, 0], \qquad E = (0, \infty), \qquad D \cup E = \mathbb{R}, \end{equation*}

which are measurable, and define

\begin{equation*} f(x) = \begin{cases} 0 & \text{if } x \leq 0, \\ 1 & \text{if } x > 0. \end{cases} \end{equation*}

Then \(f|_D \equiv 0\) and \(f|_E \equiv 1\) are constant, hence continuous on \(D\) and on \(E\) respectively. But \(f\) is not continuous on \(\mathbb{R}\): at \(x_0 = 0\) we have \(f(0) = 0\), while every \(\delta\)-neighborhood of \(0\) contains points \(x > 0\) with \(|f(x) - f(0)| = 1\). So \(f\) fails to be continuous at \(0\), and the analogue of Proposition 5(ii) with “continuous” in place of “measurable” is false.

The reason the analogy breaks down is structural. Measurability of \(f\) on \(D \cup E\) is decided by the sets \(\{x \in D \cup E \mid f(x) > c\}\), and each of these splits as the union

\begin{equation*} \{x \in D \mid f(x) > c\} \cup \{x \in E \mid f(x) > c\} , \end{equation*}

so the property is inherited from the pieces because the measurable sets form an algebra. Continuity, by contrast, is a local condition, and a point of \(D\) may be a limit point of \(E\); the restrictions know nothing about how \(f\) behaves as such a point is approached from within the other piece.

It is worth recording the standard positive result, which shows what extra hypothesis repairs the converse: if \(D\) and \(E\) are both closed (or both open) and \(f|_D\), \(f|_E\) are continuous, then \(f\) is continuous on \(D \cup E\). For the closed case, let \(\mathcal{C} \subseteq \mathbb{R}\) be closed; then

\begin{equation*} f^{-1}(\mathcal{C}) = (f|_D)^{-1}(\mathcal{C}) \cup (f|_E)^{-1}(\mathcal{C}), \end{equation*}

and \((f|_D)^{-1}(\mathcal{C})\) is closed in \(D\), hence (as \(D\) is closed in \(\mathbb{R}\)) closed in \(\mathbb{R}\); the same holds for \(E\), so \(f^{-1}(\mathcal{C})\) is a union of two closed sets, hence closed in \(\mathbb{R}\) and a fortiori closed in \(D \cup E\). Thus \(f\) is continuous. For the open case one argues identically with open sets, since a relatively open subset of an open set is open. In the counterexample above \(D\) is closed and \(E\) is open, and it is precisely this mismatch that is exploited.

Yes. Let \(E\) be the measurable domain of \(f\), and let

\begin{equation*} F = \{x_1, \ldots, x_n\} \subseteq E \end{equation*}

be the finite set of points of \(E\) at which \(f\) fails to be continuous; thus \(f\) is continuous at every point of \(D = E \sim F\).

First, \(F\) is measurable and \(m(F) = 0\): a single point \(\{x_k\}\) is a degenerate closed interval, so it is measurable of measure zero, and a finite union of such sets is measurable of measure zero by the countable subadditivity and the fact that the measurable sets form a \(\sigma\)-algebra (Chapter 2). Consequently \(D = E \sim F\) is measurable as well.

Now consider the two restrictions.

The restriction \(f|_D\) is continuous on \(D\). Indeed, fix \(x_0 \in D\) and \(\varepsilon > 0\). Since \(f\) is continuous at \(x_0\) as a function on \(E\), there is \(\delta > 0\) with \(|f(x) - f(x_0)| < \varepsilon\) for all \(x \in E\) with \(|x - x_0| < \delta\); as \(D \subseteq E\) this same \(\delta\) works for \(f|_D\). Observe also that \(f\) is real-valued at every point of \(D\): continuity of \(f\) at \(x_0\) in the above sense presupposes that the difference \(f(x) - f(x_0)\) is meaningful and small, which forces \(f(x_0)\) to be finite. (This is the only place where it matters that the functions of this chapter are allowed to take the values \(\pm\infty\); the exceptional points are handled separately below, and there the values are unrestricted.) So \(f|_D\) is a real-valued function that is continuous on its measurable domain \(D\), and Proposition 3 tells us that \(f|_D\) is measurable.

The restriction \(f|_F\) is measurable for the cheapest of reasons: for every real number \(c\),

\begin{equation*} \{x \in F \mid f(x) > c\} \subseteq F , \end{equation*}

and \(m(F) = 0\), so this set is a subset of a set of measure zero. Lebesgue outer measure is monotone, so any such subset has outer measure zero and hence is measurable (Chapter 2: every set of outer measure zero is measurable). Thus \(\{x \in F \mid f(x) > c\}\) is measurable for each \(c\), and by Proposition 1 the restriction \(f|_F\) is measurable on \(F\). (The same argument applies verbatim if \(f\) is allowed to take the values \(\pm\infty\) at the exceptional points.)

Finally, \(E = D \cup F\) with \(D\) and \(F\) measurable and disjoint, and both restrictions \(f|_D\) and \(f|_F\) are measurable. By Proposition 5(ii), \(f\) is measurable on \(E\). Concretely, for each \(c\),

\begin{equation*} \{x \in E \mid f(x) > c\} = \{x \in D \mid f(x) > c\} \cup \{x \in F \mid f(x) > c\}, \end{equation*}

a union of two measurable sets, hence measurable; Proposition 1 then gives the measurability of \(f\).

Two remarks. The finiteness of \(F\) is not what matters: the argument only used that \(F\) is a set of measure zero, so the conclusion holds whenever \(f\) is continuous at every point of \(E\) outside a set of measure zero, for instance outside a countable set. What one may not do is drop the measure-zero hypothesis on the exceptional set: a function continuous off an arbitrary set need not be measurable, since the values on the exceptional set are then unconstrained.

No. Measurability of every level set \(f^{-1}( c) = \{x \in \mathbb{R} \mid f(x) = c\}\) is a much weaker condition than measurability of the sets \(\{x \in \mathbb{R} \mid f(x) > c\}\), and it does not imply it. Proposition 1 records that measurability of \(f\) implies each \(f^{-1}( c)\) is measurable; the converse fails, as the following counterexample shows.

By Theorem 17 (Vitali) of Chapter 2, any set of real numbers with positive outer measure contains a subset that fails to be measurable; applying this to the interval \((0,1)\), fix a set \(A \subseteq (0,1)\) that is not measurable. Define \(f : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f(x) = \begin{cases} x & \text{if } x \in A, \\ -x & \text{if } x \in \mathbb{R} \sim A. \end{cases} \end{equation*}

Every level set of \(f\) is measurable. Fix a real number \(c\) and suppose \(f(x) = c\). Either \(x \in A\), in which case \(f(x) = x\) and so \(x = c\); or \(x \notin A\), in which case \(f(x) = -x\) and so \(x = -c\). Hence

\begin{equation*} f^{-1}( c) \subseteq \{c, -c\} , \end{equation*}

a set with at most two elements. Every finite set is measurable (it is a finite union of singletons, each a set of measure zero), and every subset of a finite set is finite, so \(f^{-1}( c)\) is measurable for each real \(c\).

Nevertheless \(f\) is not measurable. Consider \(\{x \in \mathbb{R} \mid f(x) > 0\}\) and intersect it with \((0,1)\). If \(x \in (0,1)\) and \(x \in A\), then \(f(x) = x > 0\). If \(x \in (0,1)\) and \(x \notin A\), then \(f(x) = -x < 0\). Therefore

\begin{equation*} \{x \in \mathbb{R} \mid f(x) > 0\} \cap (0,1) = A . \end{equation*}

Were \(f\) measurable, the set \(\{x \in \mathbb{R} \mid f(x) > 0\}\) would be measurable by Proposition 1, and since the measurable sets form a \(\sigma\)-algebra containing the interval \((0,1)\), the intersection \(A\) would be measurable too. This contradicts the choice of \(A\). Hence \(f\) is not measurable, even though \(f^{-1}( c)\) is measurable for every real number \(c\).

It may be worth noting that the pathology is not caused by \(f\) having repeated values: one can also arrange a one-to-one counterexample, and for a one-to-one \(f\) every set \(f^{-1}( c)\) is empty or a singleton, hence automatically measurable, so the hypothesis of the problem carries no information at all in that case. For instance, keep \(A \subseteq (0,1)\) nonmeasurable and set \(g(x) = x\) for \(x \in A\), \(g(x) = x + 2\) for \(x \in (0,1) \sim A\), \(g(x) = x - 10\) for \(x \leq 0\), and \(g(x) = x + 10\) for \(x \geq 1\). These four formulas are injective on their respective domains and have pairwise disjoint ranges contained in \((0,1)\), \((2,3)\), \((-\infty, -10]\), and \([11, \infty)\) respectively, so \(g\) is one-to-one on \(\mathbb{R}\); yet

\begin{equation*} \{x \in \mathbb{R} \mid g(x) > 1\} \cap (0,1) = (0,1) \sim A \end{equation*}

is nonmeasurable, so \(g\) is not measurable.

Yes. Testing the sets \(\{f > c\}\) at rational \(c\) already forces measurability at every real \(c\), because \(\mathbb{Q}\) is dense in \(\mathbb{R}\) and the measurable sets are closed under countable unions.

Let \(c\) be an arbitrary real number. Since the rationals are dense in \(\mathbb{R}\), we may choose a sequence \(\{c_k\}_{k=1}^{\infty}\) of rational numbers with

\begin{equation*} c < c_k \quad \text{for all } k, \qquad \lim_{k \to \infty} c_k = c ; \end{equation*}

for instance take \(c_k\) to be any rational number in the interval \((c, c + 1/k)\). I claim that

\begin{equation*} \{x \in E \mid f(x) > c\} = \bigcup_{k=1}^{\infty} \{x \in E \mid f(x) > c_k\} . \tag{\(\ast\)} \end{equation*}

The inclusion \(\supseteq\) is clear: if \(f(x) > c_k\) for some \(k\), then since \(c_k > c\) we get \(f(x) > c\).

For the inclusion \(\subseteq\), suppose \(x \in E\) and \(f(x) > c\). If \(f(x) = \infty\) then \(f(x) > c_1\) and \(x\) lies in the union. Otherwise \(f(x)\) is a real number with \(f(x) - c > 0\), so by the Archimedean property there is an index \(k\) with \(1/k < f(x) - c\), and then

\begin{equation*} c_k < c + \tfrac{1}{k} < f(x) , \end{equation*}

so \(x \in \{x \in E \mid f(x) > c_k\}\). This proves \((\ast)\).

By hypothesis each set on the right of \((\ast)\) is measurable, since each \(c_k\) is rational. The measurable sets form a \(\sigma\)-algebra (Chapter 2), so the countable union on the right is measurable. Hence \(\{x \in E \mid f(x) > c\}\) is measurable for every real number \(c\). Since the domain \(E\) is measurable by hypothesis, statement (i) of Proposition 1 holds, and therefore \(f\) is measurable by the definition of measurability given after Proposition 1.

The same argument works with \(\mathbb{Q}\) replaced by any set \(S \subseteq \mathbb{R}\) that is dense in \(\mathbb{R}\), or indeed by any countable set \(S\) with the property that for each real \(c\) there is a sequence in \(S \cap (c, \infty)\) converging to \(c\); all that is used is that the union in \((\ast)\) is countable.

Both directions are settled by computing the sets \(\{g > c\}\) and comparing them with the sets \(\{f > c\}\). Note first that the domain of \(g\) is \(\mathbb{R}\), which is measurable, and the domain of \(f\) is \(D\), measurable by hypothesis; so in each direction only the level sets need attention, and Proposition 1 applies.

Suppose first that \(f\) is measurable. Fix a real number \(c\) and split according to the sign of \(c\).

If \(c \geq 0\), then no point of \(\mathbb{R} \sim D\) belongs to \(\{x \in \mathbb{R} \mid g(x) > c\}\), since \(g\) vanishes there and \(0 \leq c\). Hence

\begin{equation*} \{x \in \mathbb{R} \mid g(x) > c\} = \{x \in D \mid g(x) > c\} = \{x \in D \mid f(x) > c\}, \end{equation*}

which is measurable because \(f\) is measurable.

If \(c < 0\), then every point of \(\mathbb{R} \sim D\) does belong to the set, since \(g = 0 > c\) there. Hence

\begin{equation*} \{x \in \mathbb{R} \mid g(x) > c\} = \{x \in D \mid f(x) > c\} \; \cup \; (\mathbb{R} \sim D). \end{equation*}

The first set is measurable since \(f\) is measurable, and \(\mathbb{R} \sim D\) is measurable because \(D\) is measurable and the measurable sets form a \(\sigma\)-algebra. Their union is therefore measurable.

In both cases \(\{x \in \mathbb{R} \mid g(x) > c\}\) is measurable, so by Proposition 1 and the definition of measurability, \(g\) is measurable on \(\mathbb{R}\).

Conversely, suppose \(g\) is measurable. Since \(g\) agrees with \(f\) on \(D\), for each real \(c\) we have

\begin{equation*} \{x \in D \mid f(x) > c\} = \{x \in D \mid g(x) > c\} = \{x \in \mathbb{R} \mid g(x) > c\} \cap D . \end{equation*}

The set \(\{x \in \mathbb{R} \mid g(x) > c\}\) is measurable because \(g\) is measurable, and \(D\) is measurable by hypothesis; since the measurable sets form a \(\sigma\)-algebra they are closed under intersection, so \(\{x \in D \mid f(x) > c\}\) is measurable. As this holds for every real \(c\) and the domain \(D\) of \(f\) is measurable, \(f\) is measurable by Proposition 1.

Alternatively, the converse direction is exactly Proposition 5(ii) applied to the function \(g\) on the measurable set \(\mathbb{R}\) and the measurable subset \(D\): measurability of \(g\) on \(\mathbb{R}\) is equivalent to measurability of its restrictions to \(D\) and to \(\mathbb{R} \sim D\), and the restriction of \(g\) to \(D\) is precisely \(f\).

The proof used nothing about the value \(0\) beyond its being a fixed real number: for any fixed \(a \in \mathbb{R}\), the extension of \(f\) by the constant \(a\) off \(D\) is measurable on \(\mathbb{R}\) if and only if \(f\) is measurable on \(D\). Only the two cases \(c \geq a\) and \(c < a\) change places.

Here \(f\) is real-valued, \(A\) ranges over the Borel subsets of \(\mathbb{R}\), and \(f^{-1}(A) = \{x \in E \mid f(x) \in A\}\). Recall that the Borel \(\sigma\)-algebra \(\mathcal{B}\) is by definition the smallest \(\sigma\)-algebra of subsets of \(\mathbb{R}\) containing all of the open sets; equivalently, \(\mathcal{B}\) is the intersection of all \(\sigma\)-algebras of subsets of \(\mathbb{R}\) that contain the open sets, so any \(\sigma\)-algebra containing every open set necessarily contains \(\mathcal{B}\).

Suppose first that \(f^{-1}(A)\) is measurable for each Borel set \(A\). For each real number \(c\) the interval \((c, \infty)\) is open, hence Borel, and

\begin{equation*} \{x \in E \mid f(x) > c\} = f^{-1}\big((c, \infty)\big) \end{equation*}

is therefore measurable. Since the domain \(E\) is measurable, statement (i) of Proposition 1 holds and \(f\) is measurable.

Conversely, suppose \(f\) is measurable. Following the hint, set

\begin{equation*} \mathcal{A} = \{A \subseteq \mathbb{R} \mid f^{-1}(A) \text{ is measurable}\} . \end{equation*}

We check that \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\), using throughout the elementary identities for inverse images and the fact that the Lebesgue measurable sets form a \(\sigma\)-algebra (Chapter 2).

First, \(\mathbb{R} \in \mathcal{A}\): since \(f\) is real-valued on \(E\), \(f^{-1}(\mathbb{R}) = E\), which is measurable by hypothesis. In particular \(\mathcal{A}\) is nonempty.

Second, \(\mathcal{A}\) is closed under complements in \(\mathbb{R}\). Let \(A \in \mathcal{A}\). A point \(x \in E\) satisfies \(f(x) \in \mathbb{R} \sim A\) if and only if \(x \notin f^{-1}(A)\), so

\begin{equation*} f^{-1}(\mathbb{R} \sim A) = E \sim f^{-1}(A) . \end{equation*}

Both \(E\) and \(f^{-1}(A)\) are measurable, and the measurable sets form a \(\sigma\)-algebra, hence an algebra, so their relative complement is measurable. Thus \(\mathbb{R} \sim A \in \mathcal{A}\).

Third, \(\mathcal{A}\) is closed under countable unions. Let \(\{A_k\}_{k=1}^{\infty}\) be a countable collection of members of \(\mathcal{A}\). A point \(x \in E\) satisfies \(f(x) \in \bigcup_{k} A_k\) if and only if \(f(x) \in A_k\) for some \(k\), so

\begin{equation*} f^{-1}\Big(\bigcup_{k=1}^{\infty} A_k\Big) = \bigcup_{k=1}^{\infty} f^{-1}(A_k) , \end{equation*}

a countable union of measurable sets, hence measurable. Thus \(\bigcup_k A_k \in \mathcal{A}\).

Therefore \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\).

Next, \(\mathcal{A}\) contains every open subset of \(\mathbb{R}\). This is exactly Proposition 2: a function defined on a measurable set is measurable if and only if the inverse image of each open set is measurable. Since \(f\) is measurable, \(f^{-1}(\mathcal{O})\) is measurable for each open \(\mathcal{O} \subseteq \mathbb{R}\), i.e. \(\mathcal{O} \in \mathcal{A}\).

So \(\mathcal{A}\) is a \(\sigma\)-algebra containing all open sets, and by the minimality in the definition of the Borel \(\sigma\)-algebra, \(\mathcal{B} \subseteq \mathcal{A}\). That is, \(f^{-1}(A)\) is measurable for every Borel set \(A\), which is what we had to prove.

A remark on the extended real-valued case. If \(f\) is permitted to take the values \(\pm\infty\), the statement as literally written is no longer correct in the direction “inverse images of Borel sets measurable \(\Rightarrow\) \(f\) measurable”: a function taking only the values \(\pm\infty\) has \(f^{-1}(A) = \emptyset\) for every \(A \subseteq \mathbb{R}\), so the hypothesis is vacuous. Concretely, with \(A_0 \subseteq (0,1)\) nonmeasurable (Theorem 17 of Chapter 2), the function on \(E = (0,1)\) equal to \(\infty\) on \(A_0\) and to \(-\infty\) on \((0,1) \sim A_0\) satisfies the hypothesis yet has \(\{x \in E \mid f(x) > 0\} = A_0\) nonmeasurable. The correct extended statement is obtained by letting \(A\) range over the Borel subsets of the extended real numbers, i.e. over the \(\sigma\)-algebra generated by the open subsets of \(\mathbb{R}\) together with the sets \((c, \infty]\) and \([-\infty, c)\). The forward direction, on the other hand, remains valid as stated for extended real-valued measurable \(f\): the argument above needs only the identity \(f^{-1}(\mathbb{R}) = \bigcup_{k=1}^{\infty} \{x \in E \mid -k < f(x) < k\}\) to see that \(f^{-1}(\mathbb{R})\) is measurable, after which one repeats the \(\sigma\)-algebra verification with \(E\) replaced by \(f^{-1}(\mathbb{R})\).

Throughout, write \(\mathcal{B}\) for the collection of Borel subsets of \(\mathbb{R}\), that is, the smallest \(\sigma\)-algebra of subsets of \(\mathbb{R}\) containing the open sets. Two features of \(\mathcal{B}\) are all that the arguments of Section 3.1 ever used about the Lebesgue measurable sets: \(\mathcal{B}\) is a \(\sigma\)-algebra, and \(\mathcal{B}\) contains every interval and every open set.

Proposition 1 for Borel sets. Let \(f\) have Borel domain \(E\). The proof of Proposition 1 proceeds as follows. The sets in (i) and (iv) are complementary in \(E\), as are those in (ii) and (iii); since \(E\) is Borel and \(\mathcal{B}\) is a \(\sigma\)-algebra, the relative complement in \(E\) of a Borel subset of \(E\) is Borel, so (i) \(\Leftrightarrow\) (iv) and (ii) \(\Leftrightarrow\) (iii). Next,

\begin{equation*} \{x \in E \mid f(x) \ge c\} = \bigcap_{k=1}^{\infty} \{x \in E \mid f(x) > c - 1/k\}, \qquad \{x \in E \mid f(x) > c\} = \bigcup_{k=1}^{\infty} \{x \in E \mid f(x) \ge c + 1/k\}, \end{equation*}

and \(\mathcal{B}\) is closed under countable intersections and countable unions, so (i) \(\Leftrightarrow\) (ii). Finally, for finite \(c\), \(f^{-1}( c) = \{f \ge c\} \cap \{f \le c\}\) is Borel, and \(f^{-1}(\infty) = \bigcap_{k=1}^{\infty}\{x \in E \mid f(x) > k\}\) is Borel, and similarly for \(-\infty\). So the four statements are equivalent and each implies that \(\{x \in E \mid f(x) = c\}\) is Borel for every extended real \(c\). Only the \(\sigma\)-algebra property was invoked, so the proposition stands verbatim with “Borel set” in place of “measurable set.”

Theorem 6 for Borel sets. Let \(f\) and \(g\) be Borel measurable real-valued functions on the Borel set \(E\). For \(\alpha \ne 0\),

\begin{equation*} \{x \in E \mid \alpha f(x) > c\} = \{x \in E \mid f(x) > c/\alpha\} \ \text{ if } \alpha > 0, \qquad \{x \in E \mid \alpha f(x) > c\} = \{x \in E \mid f(x) < c/\alpha\} \ \text{ if } \alpha < 0, \end{equation*}

and \(\alpha f \equiv 0\) is Borel measurable when \(\alpha = 0\); so \(\alpha f\) is Borel measurable. By the density of \(\mathbb{Q}\) in \(\mathbb{R}\),

\begin{equation*} \{x \in E \mid f(x) + g(x) < c\} = \bigcup_{q \in \mathbb{Q}} \big(\{x \in E \mid g(x) < c - q\} \cap \{x \in E \mid f(x) < q\}\big), \end{equation*}

a countable union of Borel sets, hence Borel; so \(f + g\), and therefore \(\alpha f + \beta g\), is Borel measurable. For products, \(fg = \frac{1}{2}[(f+g)^2 - f^2 - g^2]\), so by the linearity just proved it suffices to know that \(h^2\) is Borel measurable whenever \(h\) is, and this follows from

\begin{equation*} \{x \in E \mid h^2(x) > c\} = \{x \in E \mid h(x) > \sqrt{c}\} \cup \{x \in E \mid h(x) < -\sqrt{c}\} \ \ (c \ge 0), \qquad \{x \in E \mid h^2(x) > c\} = E \ \ (c < 0). \end{equation*}

Again every step used only that \(\mathcal{B}\) is a \(\sigma\)-algebra containing the intervals, so Theorem 6 holds with “Borel measurable” throughout.

One caveat deserves mention. Theorem 6 is stated for functions that are merely finite a.e., and the reduction to the finite case rests on Proposition 5, which asserts that a function agreeing a.e. with a measurable function is measurable. Proposition 5 is false for Borel measurability: a set of measure zero need not be Borel (by Proposition 22 of Chapter 2 there is a measurable, indeed measure-zero, subset of the Cantor set that is not Borel), and altering the identically-zero function to equal \(1\) on such a set produces a function that is not Borel measurable. So in the Borel version of Theorem 6 one must read “finite a.e.” as “finite,” or equivalently pass to the Borel set \(E_0\) on which both \(f\) and \(g\) are finite (Borel by Proposition 1) and assert Borel measurability of \(\alpha f + \beta g\) and \(fg\) on \(E_0\); the values assigned on \(E \sim E_0\) must themselves be Borel measurable if the conclusion is to hold on all of \(E\).

(i) Every Borel measurable function is Lebesgue measurable. By Theorem 9 of Chapter 2 the collection \(\mathcal{M}\) of Lebesgue measurable sets is a \(\sigma\)-algebra that contains the \(\sigma\)-algebra \(\mathcal{B}\) of Borel sets; equivalently, \(\mathcal{M}\) is a \(\sigma\)-algebra containing the open sets and \(\mathcal{B}\) is the smallest such \(\sigma\)-algebra. Hence every Borel set is Lebesgue measurable. Thus if \(f\) is Borel measurable, its domain \(E\) is a Borel set and so Lebesgue measurable, and for each \(c\) the Borel set \(\{x \in E \mid f(x) > c\}\) is Lebesgue measurable. By the definition following Proposition 1, \(f\) is Lebesgue measurable.

(ii) \(f\) Borel measurable, \(B\) Borel \(\Rightarrow\) \(f^{-1}(B)\) Borel. This is Problem 7 with \(\mathcal{B}\) in place of the Lebesgue measurable sets. There is no loss in assuming \(f\) is real-valued. Indeed, the functions of this chapter are extended real-valued, and if \(f\) takes infinite values put \(E^{\prime} = E \sim [f^{-1}(\infty) \cup f^{-1}(-\infty)]\), a Borel set by the Borel version of Proposition 1 established above; for a set \(B \subseteq \mathbb{R}\) of real numbers we have \(f^{-1}(B) = (f|_{E^{\prime}})^{-1}(B)\), and \(f|_{E^{\prime}}\) is a real-valued Borel measurable function on the Borel set \(E^{\prime}\), so the real-valued case gives the general one. Assume then that \(f : E \to \mathbb{R}\) and define

\begin{equation*} \mathcal{A} = \{A \subseteq \mathbb{R} \mid f^{-1}(A) \text{ is a Borel set}\}. \end{equation*}

Then \(\mathcal{A}\) is a \(\sigma\)-algebra of subsets of \(\mathbb{R}\): first, \(f^{-1}(\mathbb{R}) = E\) is Borel, so \(\mathbb{R} \in \mathcal{A}\); second, if \(A \in \mathcal{A}\) then \(f^{-1}(\mathbb{R} \sim A) = E \sim f^{-1}(A)\), a difference of Borel sets, hence Borel, so \(\mathbb{R} \sim A \in \mathcal{A}\); third, if \(\{A_k\}_{k=1}^{\infty} \subseteq \mathcal{A}\) then \(f^{-1}\big(\bigcup_k A_k\big) = \bigcup_k f^{-1}(A_k)\) is a countable union of Borel sets, hence Borel. Moreover \((c, \infty) \in \mathcal{A}\) for every \(c\), by the Borel measurability of \(f\). Now the \(\sigma\)-algebra generated by the rays \(\{(c,\infty) \mid c \in \mathbb{R}\}\) is all of \(\mathcal{B}\): it contains \((-\infty, c] = \mathbb{R} \sim (c,\infty)\), hence \((-\infty, c) = \bigcup_{k=1}^{\infty} (-\infty, c - 1/k]\), hence every bounded open interval \((a,b) = (a,\infty) \cap (-\infty,b)\), and hence also the unbounded open intervals \((a,\infty)\), \((-\infty,b)\) and \(\mathbb{R}\) itself. Every open subset of \(\mathbb{R}\) is the union of a countable collection of open intervals, each of them bounded or unbounded and so in every case a member of this \(\sigma\)-algebra; hence it contains the open sets and therefore contains \(\mathcal{B}\). Since \(\mathcal{A}\) is a \(\sigma\)-algebra containing all the rays, \(\mathcal{B} \subseteq \mathcal{A}\); that is, \(f^{-1}(B)\) is Borel for every Borel set \(B\).

(iii) \(f, g\) Borel measurable \(\Rightarrow\) \(f \circ g\) Borel measurable. Let \(g\) have Borel domain \(D\) and \(f\) have Borel domain \(E\), both real-valued, and let \(f \circ g\) have its natural domain \(g^{-1}(E)\), which is Borel by (ii) applied to \(g\). For each \(c\), the set \(B = \{y \in E \mid f(y) > c\}\) is Borel since \(f\) is Borel measurable, and

\begin{equation*} \{x \in g^{-1}(E) \mid f(g(x)) > c\} = g^{-1}(B), \end{equation*}

which is Borel by (ii) applied to \(g\). Hence \(f \circ g\) is Borel measurable.

(iv) \(f\) Borel measurable, \(g\) Lebesgue measurable \(\Rightarrow\) \(f \circ g\) Lebesgue measurable. Exactly the same computation, with Problem 7 supplying the transfer. By Problem 7, the collection \(\{A \subseteq \mathbb{R} \mid g^{-1}(A) \text{ is Lebesgue measurable}\}\) is a \(\sigma\)-algebra, and it contains every ray \((c,\infty)\) because \(g\) is Lebesgue measurable; by the generation argument in (ii) it therefore contains \(\mathcal{B}\). So \(g^{-1}(B)\) is Lebesgue measurable for every Borel set \(B\). In particular the domain \(g^{-1}(E)\) is measurable, and for each \(c\),

\begin{equation*} \{x \in g^{-1}(E) \mid f(g(x)) > c\} = g^{-1}\big(\{y \in E \mid f(y) > c\}\big) \end{equation*}

is the inverse image under \(g\) of a Borel set, hence Lebesgue measurable. Thus \(f \circ g\) is Lebesgue measurable.

It is worth contrasting (iv) with the Example preceding Proposition 7, where \(\chi_A \circ \psi^{-1}\) fails to be measurable even though \(\chi_A\) is Lebesgue measurable and \(\psi^{-1}\) is continuous. There \(A\) is a measurable set that is not Borel, so \(\chi_A\) is Lebesgue but not Borel measurable, and (iv) does not apply. The outer function must be Borel measurable, not merely Lebesgue measurable, for the composition argument to run.

Yes: \(E_0\) is measurable. We give a proof that uses nothing beyond Proposition 1, Proposition 5, Theorem 6, and the fact that the measurable sets are a \(\sigma\)-algebra.

The functions here are extended real-valued, so we must first say what “converges” means. Take it to mean “converges to a finite limit”; at the end we observe that the sets on which \(f_n(x) \to \infty\) and \(f_n(x) \to -\infty\) are also measurable, so the conclusion is unaffected if one prefers the wider reading.

For each pair of indices \(n, m\) put

\begin{equation*} E_{n,m} = \{x \in E \mid |f_n(x)| < \infty\} \cap \{x \in E \mid |f_m(x)| < \infty\}. \end{equation*}

By Proposition 1, \(f_n^{-1}(\infty)\) and \(f_n^{-1}(-\infty)\) are measurable, so \(\{x \in E \mid |f_n(x)| < \infty\} = E \sim [f_n^{-1}(\infty) \cup f_n^{-1}(-\infty)]\) is measurable, and hence so is \(E_{n,m}\). By Proposition 5(ii) the restrictions of \(f_n\) and \(f_m\) to \(E_{n,m}\) are measurable, and they are real-valued there, so by Theorem 6 the function \(f_n - f_m\) is measurable on \(E_{n,m}\). Consequently, for each \(k \in \mathbb{N}\) the set

\begin{equation*} A_{n,m}^{k} = \{x \in E_{n,m} \mid |f_n(x) - f_m(x)| < 1/k\} = \{x \in E_{n,m} \mid f_n(x) - f_m(x) < 1/k\} \cap \{x \in E_{n,m} \mid f_n(x) - f_m(x) > -1/k\} \end{equation*}

is measurable, being the intersection of two measurable sets.

Now define

\begin{equation*} F = \bigcap_{k=1}^{\infty} \ \bigcup_{N=1}^{\infty} \ \bigcap_{n \ge N} \ \bigcap_{m \ge N} A_{n,m}^{k}. \end{equation*}

All the unions and intersections are over countable index sets, so \(F\) is measurable because the measurable sets form a \(\sigma\)-algebra.

We claim \(F = E_0\). Suppose first that \(x \in E_0\), say \(f_n(x) \to L\) with \(L \in \mathbb{R}\). Fix \(k\). There is an \(N\) such that \(|f_n(x) - L| < 1/(2k)\) for all \(n \ge N\); in particular \(f_n(x)\) is finite for every \(n \ge N\), so \(x \in E_{n,m}\) for all \(n, m \ge N\), and by the triangle inequality

\begin{equation*} |f_n(x) - f_m(x)| \le |f_n(x) - L| + |L - f_m(x)| < 1/k \quad \text{for all } n, m \ge N. \end{equation*}

Thus \(x \in \bigcap_{n,m \ge N} A_{n,m}^{k}\). As \(k\) was arbitrary, \(x \in F\).

Conversely suppose \(x \in F\). Taking \(k = 1\) we get an \(N_1\) with \(x \in E_{n,m}\) for all \(n, m \ge N_1\), so the numbers \(f_n(x)\) are finite for \(n \ge N_1\). Given \(\epsilon > 0\), choose \(k\) with \(1/k < \epsilon\) and then \(N \ge N_1\) with \(|f_n(x) - f_m(x)| < 1/k < \epsilon\) for all \(n, m \ge N\). Hence the tail \(\{f_n(x)\}_{n \ge N_1}\) is a Cauchy sequence of real numbers, and by the completeness of \(\mathbb{R}\) it converges to some \(L \in \mathbb{R}\). Since altering finitely many terms does not affect convergence, \(\{f_n(x)\}\) converges, i.e. \(x \in E_0\).

Therefore \(E_0 = F\) is measurable.

Finally, for the wider reading of “converges,” note that

\begin{equation*} \{x \in E \mid f_n(x) \to \infty\} = \bigcap_{k=1}^{\infty} \ \bigcup_{N=1}^{\infty} \ \bigcap_{n \ge N} \{x \in E \mid f_n(x) > k\} \end{equation*}

is measurable, since each \(\{x \in E \mid f_n(x) > k\}\) is measurable by the very definition of measurability of \(f_n\) and only countable set operations are used; the set on which \(f_n(x) \to -\infty\) is measurable by the symmetric formula. Adjoining these two measurable sets to \(F\) again yields a measurable set. Either way, the answer is affirmative.

No. Proposition 7 guarantees measurability of \(f \circ g\) when the outer function is continuous and the inner one measurable; with the roles reversed the conclusion fails, and the Example preceding Proposition 7 supplies the counterexample. We spell it out.

By Proposition 21 of Chapter 2 the function

\begin{equation*} \psi(x) = \varphi(x) + x \quad \text{for } x \in [0,1], \end{equation*}

where \(\varphi\) is the Cantor-Lebesgue function, is a strictly increasing continuous function mapping \([0,1]\) onto \([0,2]\), and it maps some measurable set \(A\), a subset of the Cantor set \(C\), onto a nonmeasurable set \(\psi(A)\).

Extend \(\psi\) to all of \(\mathbb{R}\) by

\begin{equation*} \psi(x) = x \ \text{ for } x < 0, \qquad \psi(x) = x + 1 \ \text{ for } x > 1. \end{equation*}

Since \(\psi(0) = \varphi(0) + 0 = 0\) and \(\psi(1) = \varphi(1) + 1 = 2\), the extended function is continuous and strictly increasing on \(\mathbb{R}\), and it maps \(\mathbb{R}\) onto \(\mathbb{R}\). By Problem 45 of Chapter 2, a strictly increasing function defined on an interval has a continuous inverse, so

\begin{equation*} g = \psi^{-1} : \mathbb{R} \to \mathbb{R} \end{equation*}

is a continuous real-valued function defined on all of \(\mathbb{R}\).

Let \(f = \chi_A\), the characteristic function of \(A\), regarded as a function on all of \(\mathbb{R}\). Since \(A\) is measurable, \(f\) is measurable: for \(c \ge 1\) the set \(\{x \mid f(x) > c\}\) is empty, for \(0 \le c < 1\) it is \(A\), and for \(c < 0\) it is \(\mathbb{R}\), and all three are measurable.

Now consider \(h = f \circ g = \chi_A \circ \psi^{-1}\), a real-valued function on all of \(\mathbb{R}\). For \(c = 1/2\),

\begin{equation*} \{y \in \mathbb{R} \mid h(y) > 1/2\} = \{y \in \mathbb{R} \mid \psi^{-1}(y) \in A\} = \psi(A), \end{equation*}

the last equality because \(\psi\) is a bijection of \(\mathbb{R}\) onto \(\mathbb{R}\). But \(\psi(A)\) is not measurable. Hence \(h = f \circ g\) is not measurable, even though \(f\) is measurable and \(g\) is continuous.

(Equivalently, as the text puts it: if \(I\) is any open interval containing \(1\) but not \(0\), then \(h^{-1}(I) = \psi(A)\) is nonmeasurable, so by Proposition 2 the function \(h\) is not measurable.)

The obstruction is that a continuous function may carry a set of measure zero, such as the measurable set \(A \subseteq C\), onto a set of positive measure, and once measure is created in this way a measurable set can be moved onto a nonmeasurable one. Problem 11 shows that the extra hypothesis of a Lipschitz inverse rules this out and restores measurability of \(f \circ g\); Problem 8(iv) shows that requiring \(f\) to be Borel measurable rather than merely Lebesgue measurable also restores it.

Write \(h = g^{-1}\), so that \(h : \mathbb{R} \to \mathbb{R}\) is Lipschitz: there is a constant \(c \ge 0\) with

\begin{equation*} |h(u) - h(v)| \le c\,|u - v| \quad \text{for all } u, v \in \mathbb{R}. \end{equation*}

Since \(g\) is a one-to-one mapping of \(\mathbb{R}\) onto \(\mathbb{R}\) with inverse \(h\), for any set \(A \subseteq \mathbb{R}\) we have the identity

\begin{equation*} g^{-1}(A) = \{x \in \mathbb{R} \mid g(x) \in A\} = h(A). \end{equation*}

Indeed, if \(g(x) = a \in A\) then \(x = h(a) \in h(A)\); conversely if \(x = h(a)\) with \(a \in A\) then \(g(x) = a \in A\). So inverse images under \(g\) are the same thing as forward images under the Lipschitz function \(h\), and everything reduces to the following lemma, which is Problem 38 of Chapter 2 transferred from a closed bounded interval to all of \(\mathbb{R}\).

Lemma. A Lipschitz function \(h : \mathbb{R} \to \mathbb{R}\) maps measurable sets onto measurable sets.

Proof of the Lemma, in three steps.

Step 1: \(h\) maps sets of measure zero onto sets of measure zero. Let \(m^{*}(Z) = 0\) and let \(\epsilon > 0\). By the definition of outer measure there is a countable collection \(\{I_k\}_{k=1}^{\infty}\) of open, bounded intervals covering \(Z\) with \(\sum_{k} \ell(I_k) < \epsilon\). Fix \(k\). For \(u, v \in Z \cap I_k\) we have \(|h(u) - h(v)| \le c\,|u - v| \le c\,\ell(I_k)\), so the set \(h(Z \cap I_k)\) has diameter at most \(c\,\ell(I_k)\) and is therefore contained in a closed interval of length \(c\,\ell(I_k)\); by the monotonicity of outer measure and Proposition 1 of Chapter 2, which says the outer measure of an interval is its length, \(m^{*}(h(Z \cap I_k)) \le c\,\ell(I_k)\). Since \(h(Z) = \bigcup_{k} h(Z \cap I_k)\), the countable subadditivity of outer measure, Proposition 3 of Chapter 2, gives

\begin{equation*} m^{*}(h(Z)) \le \sum_{k=1}^{\infty} m^{*}(h(Z \cap I_k)) \le c \sum_{k=1}^{\infty} \ell(I_k) < c\,\epsilon . \end{equation*}

As \(\epsilon > 0\) was arbitrary, \(m^{*}(h(Z)) = 0\), and by Proposition 4 of Chapter 2 a set of outer measure zero is measurable.

Step 2: \(h\) maps \(F_{\sigma}\) sets onto \(F_{\sigma}\) sets. A Lipschitz function is continuous, and a continuous function maps compact sets onto compact sets. If \(F \subseteq \mathbb{R}\) is closed, then \(F = \bigcup_{n=1}^{\infty} (F \cap [-n, n])\) is a countable union of compact sets, so \(h(F) = \bigcup_{n=1}^{\infty} h(F \cap [-n,n])\) is a countable union of compact, hence closed, sets: \(h(F)\) is \(F_{\sigma}\). If \(S = \bigcup_{j=1}^{\infty} F_j\) is \(F_{\sigma}\), then \(h(S) = \bigcup_{j=1}^{\infty} h(F_j)\) is a countable union of \(F_{\sigma}\) sets, hence again \(F_{\sigma}\). In particular \(h(S)\) is measurable.

Step 3: Let \(A\) be measurable. By assertion (iv) of Theorem 11 of Chapter 2 there is an \(F_{\sigma}\) set \(F\) contained in \(A\) for which \(m^{*}(A \sim F) = 0\). Then

\begin{equation*} h(A) = h(F) \cup h(A \sim F), \end{equation*}

where \(h(F)\) is measurable by Step 2 and \(h(A \sim F)\) has measure zero, hence is measurable, by Step 1. As the measurable sets are closed under union, \(h(A)\) is measurable. This proves the Lemma.

Now let \(f\) be measurable with measurable domain \(E \subseteq \mathbb{R}\). The composition \(f \circ g\) has domain

\begin{equation*} \{x \in \mathbb{R} \mid g(x) \in E\} = g^{-1}(E) = h(E), \end{equation*}

which is measurable by the Lemma. Fix a real number \(c\). Since \(f\) is measurable, the set \(E_c = \{y \in E \mid f(y) > c\}\) is measurable, and

\begin{equation*} \{x \in h(E) \mid (f \circ g)(x) > c\} = \{x \mid g(x) \in E_c\} = g^{-1}(E_c) = h(E_c), \end{equation*}

which is measurable, again by the Lemma. Since this holds for every \(c\), and the domain \(h(E)\) is measurable, \(f \circ g\) is a measurable function.

Note where the hypothesis is used: the Lipschitz condition on \(g^{-1}\) is exactly what prevents the pathology of Problem 10, where the inner function was continuous with a continuous inverse but that inverse blew a set of measure zero (a subset of the Cantor set) up to a set of positive measure and thence onto a nonmeasurable set. Step 1 shows a Lipschitz map cannot do this.

Since \(f\) is a bounded measurable real-valued function on the measurable set \(E\), the Simple Approximation Lemma applies with any \(\epsilon > 0\). Apply it with \(\epsilon = 1/n\) for each natural number \(n\): there are simple functions \(\alpha_n\) and \(\beta_n\) defined on \(E\) with

\begin{equation*} \alpha_n \le f \le \beta_n \quad \text{and} \quad 0 \le \beta_n - \alpha_n < 1/n \quad \text{on } E . \end{equation*}

These need not be monotone in \(n\), so we monotonize them. Define, for each \(n\),

\begin{equation*} \varphi_n = \max\{\alpha_1, \alpha_2, \ldots, \alpha_n\}, \qquad \psi_n = \min\{\beta_1, \beta_2, \ldots, \beta_n\} \quad \text{on } E . \end{equation*}

First, \(\varphi_n\) and \(\psi_n\) are simple. Each is measurable by Proposition 8, which asserts that the maximum and the minimum of a finite family of measurable functions with common domain are measurable. Each takes only finitely many values: at every point of \(E\) the value of \(\varphi_n\) belongs to the union of the ranges of \(\alpha_1, \ldots, \alpha_n\), which is a finite set since each \(\alpha_k\) is simple, and likewise for \(\psi_n\). Being real-valued, measurable, and finitely valued, \(\varphi_n\) and \(\psi_n\) are simple.

Second, monotonicity is immediate from the definition: \(\varphi_n = \max\{\varphi_{n-1}, \alpha_n\} \ge \varphi_{n-1}\), so \(\{\varphi_n\}\) is increasing, and \(\psi_n = \min\{\psi_{n-1}, \beta_n\} \le \psi_{n-1}\), so \(\{\psi_n\}\) is decreasing.

Third, the two sequences bracket \(f\). Since \(\alpha_k \le f\) on \(E\) for every \(k\), the maximum of \(\alpha_1, \ldots, \alpha_n\) also satisfies \(\varphi_n \le f\) on \(E\); since \(\beta_k \ge f\) on \(E\) for every \(k\), we get \(\psi_n \ge f\) on \(E\). Moreover \(\alpha_n \le \varphi_n\) and \(\psi_n \le \beta_n\), by the definitions. Hence, on all of \(E\),

\begin{equation*} \alpha_n \le \varphi_n \le f \le \psi_n \le \beta_n . \end{equation*}

Finally we get uniform convergence with an explicit rate. For every \(x \in E\),

\begin{equation*} 0 \le f(x) - \varphi_n(x) \le f(x) - \alpha_n(x) \le \beta_n(x) - \alpha_n(x) < 1/n, \end{equation*}

and symmetrically

\begin{equation*} 0 \le \psi_n(x) - f(x) \le \beta_n(x) - f(x) \le \beta_n(x) - \alpha_n(x) < 1/n . \end{equation*}

Thus

\begin{equation*} \sup_{x \in E} |f(x) - \varphi_n(x)| \le 1/n \quad \text{and} \quad \sup_{x \in E} |f(x) - \psi_n(x)| \le 1/n , \end{equation*}

and since these bounds are independent of \(x\) and tend to \(0\), both \(\{\varphi_n\}\) and \(\{\psi_n\}\) converge to \(f\) uniformly on \(E\). The sequence \(\{\varphi_n\}\) is increasing and the sequence \(\{\psi_n\}\) is decreasing, as required.

Since a semisimple function is by definition real-valued, and since uniform convergence of real-valued functions to \(f\) forces \(f\) to be real-valued, we take \(f\) to be a real-valued measurable function on the measurable set \(E\). (The extended real-valued case is treated in a remark at the end.)

The point of the exercise is that, unlike simple functions, semisimple functions may use countably many values, so we need no boundedness assumption on \(f\): we may simply round \(f\) down to the nearest multiple of \(1/n\).

Fix a natural number \(n\). For each integer \(k \in \mathbb{Z}\) define

\begin{equation*} E_{n,k} = \left\{x \in E \ \middle|\ \frac{k}{n} \le f(x) < \frac{k+1}{n}\right\} = \left\{x \in E \ \middle|\ f(x) \ge \frac{k}{n}\right\} \cap \left\{x \in E \ \middle|\ f(x) < \frac{k+1}{n}\right\}. \end{equation*}

By Proposition 1 each of the two sets on the right is measurable, hence \(E_{n,k}\) is measurable. For a fixed \(n\) the sets \(\{E_{n,k}\}_{k \in \mathbb{Z}}\) are pairwise disjoint, since the intervals \([k/n, (k+1)/n)\) are pairwise disjoint, and they cover \(E\): for \(x \in E\) the number \(f(x)\) is real, so there is exactly one integer \(k\) with \(k/n \le f(x) < (k+1)/n\), namely \(k = \lfloor n f(x) \rfloor\). Thus

\begin{equation*} E = \bigcup_{k \in \mathbb{Z}} E_{n,k}, \qquad \text{the union being disjoint.} \end{equation*}

Define \(f_n\) on \(E\) by

\begin{equation*} f_n(x) = \frac{k}{n} \quad \text{for } x \in E_{n,k}, \qquad \text{that is,} \qquad f_n = \frac{1}{n}\lfloor n f \rfloor . \end{equation*}

This is well defined and real-valued on all of \(E\) by the preceding paragraph, and its range is contained in the countable set \(\{k/n \mid k \in \mathbb{Z}\}\).

The function \(f_n\) is measurable. Indeed, for each real number \(c\),

\begin{equation*} \{x \in E \mid f_n(x) > c\} = \bigcup_{\{k \in \mathbb{Z} \ \mid \ k/n > c\}} E_{n,k}, \end{equation*}

a union of a countable collection of measurable sets, hence measurable since the measurable sets are a \(\sigma\)-algebra. Being measurable, real-valued, and countably valued, each \(f_n\) is semisimple.

It remains to check uniform convergence. Let \(x \in E\) and let \(k\) be the integer with \(x \in E_{n,k}\). Then \(f_n(x) = k/n\) and \(k/n \le f(x) < (k+1)/n\), so

\begin{equation*} 0 \le f(x) - f_n(x) < \frac{1}{n}. \end{equation*}

This bound is independent of \(x\), so

\begin{equation*} \sup_{x \in E} |f(x) - f_n(x)| \le \frac{1}{n} \longrightarrow 0 \quad \text{as } n \to \infty, \end{equation*}

that is, \(\{f_n\}\) converges to \(f\) uniformly on \(E\).

Remark on extended real values. If \(f\) is permitted to take the values \(\pm\infty\), no sequence of real-valued functions can converge to \(f\) uniformly (indeed not even pointwise) at a point where \(f\) is infinite. The natural amendment is to allow the approximants to be infinite exactly where \(f\) is: define \(g_n = f_n\) on the measurable set \(\{x \in E \mid |f(x)| < \infty\}\), using the construction above applied to the restriction of \(f\) to that set (measurable by Proposition 5(ii)), and \(g_n = \infty\) on \(f^{-1}(\infty)\), \(g_n = -\infty\) on \(f^{-1}(-\infty)\); these last two sets are measurable by Proposition 1. Each \(g_n\) is measurable, takes only countably many values, and satisfies \(|g_n - f| < 1/n\) wherever \(f\) is finite while \(g_n = f\) wherever \(f\) is infinite. Thus \(\{g_n\}\) converges to \(f\) uniformly in the only sense available for extended real-valued functions.

For each natural number \(n\) define

\begin{equation*} E_n = \{x \in E \mid |f(x)| \le n\} = \{x \in E \mid f(x) \le n\} \cap \{x \in E \mid f(x) \ge -n\} . \end{equation*}

Each of the two sets on the right is measurable by Proposition 1, so each \(E_n\) is a measurable subset of \(E\). The collection \(\{E_n\}_{n=1}^{\infty}\) is ascending, since \(|f(x)| \le n\) implies \(|f(x)| \le n+1\), and

\begin{equation*} \bigcup_{n=1}^{\infty} E_n = \{x \in E \mid |f(x)| < \infty\} =: E_{\mathrm{fin}} , \end{equation*}

because a point \(x\) lies in some \(E_n\) precisely when \(|f(x)|\) is a finite number, in which case any integer \(n \ge |f(x)|\) will do.

Since \(f\) is finite a.e. on \(E\), the set

\begin{equation*} E \sim E_{\mathrm{fin}} = f^{-1}(\infty) \cup f^{-1}(-\infty) \end{equation*}

has measure zero; it is measurable by Proposition 1. Hence, by the excision property of measure (legitimate because \(m(E \sim E_{\mathrm{fin}}) = 0 < \infty\)) together with the additivity of measure over the disjoint measurable decomposition \(E = E_{\mathrm{fin}} \cup (E \sim E_{\mathrm{fin}})\),

\begin{equation*} m(E_{\mathrm{fin}}) = m(E) - m(E \sim E_{\mathrm{fin}}) = m(E) . \end{equation*}

Now apply the continuity of measure. By part (i) of Theorem 15 of Chapter 2, applied to the ascending collection \(\{E_n\}\),

\begin{equation*} \lim_{n \to \infty} m(E_n) = m\left(\bigcup_{n=1}^{\infty} E_n\right) = m(E_{\mathrm{fin}}) = m(E) . \end{equation*}

By hypothesis \(m(E) < \infty\), so this is a limit of real numbers converging to the real number \(m(E)\). Therefore, given \(\epsilon > 0\), there is a natural number \(N\) for which

\begin{equation*} m(E_N) > m(E) - \epsilon . \end{equation*}

Put \(F = E_N\). Then \(F\) is a measurable set contained in \(E\), and \(|f| \le N\) on \(F\), so \(f\) is bounded on \(F\). Moreover \(F \subseteq E\) and \(m(F) \le m(E) < \infty\), so the excision property of measure gives

\begin{equation*} m(E \sim F) = m(E) - m(F) = m(E) - m(E_N) < \epsilon . \end{equation*}

This is exactly what was to be shown.

Two remarks on the hypotheses. First, the finiteness of \(m(E)\) is essential: for \(E = \mathbb{R}\) and \(f(x) = x\), the function \(f\) is finite everywhere, yet every set on which \(f\) is bounded is contained in a bounded interval and hence has complement in \(\mathbb{R}\) of infinite measure, so no such \(F\) exists for any \(\epsilon\). Second, the finiteness a.e. of \(f\) is essential: if \(f = \infty\) on a subset of \(E\) of positive measure \(\delta\), then every set \(F\) on which \(f\) is bounded misses that subset, so \(m(E \sim F) \ge \delta\), and the conclusion fails for \(\epsilon < \delta\).

Let \(\epsilon > 0\) be given.

Step 1: a set on which \(f\) is bounded (this is Problem 14). For each natural number \(n\) put

\begin{equation*} E_n = \{x \in E \mid |f(x)| \le n\}. \end{equation*}

Here \(f\) is permitted to be extended real-valued, so we avoid appealing to the measurability of \(|f|\) and argue directly: since \(f\) is measurable, for each \(n\) both \(\{x \in E \mid f(x) \le n\}\) and \(\{x \in E \mid f(x) \ge -n\}\) are measurable by Proposition 1, and \(E_n\) is their intersection, hence measurable because the measurable sets form a \(\sigma\)-algebra. The sequence \(\{E_n\}\) is ascending and

\begin{equation*} \bigcup_{n=1}^{\infty} E_n = \{x \in E \mid |f(x)| < \infty\} = E \sim E_{\infty}, \qquad E_{\infty} = \{x \in E \mid |f(x)| = \infty\}. \end{equation*}

Because \(f\) is finite a.e. on \(E\) we have \(m(E_{\infty}) = 0\), so by the continuity of measure (Theorem 15 of Chapter 2) and the excision property,

\begin{equation*} \lim_{n \to \infty} m(E_n) = m\Big(\bigcup_{n=1}^{\infty} E_n\Big) = m(E \sim E_{\infty}) = m(E). \end{equation*}

Since \(m(E) < \infty\), every \(m(E_n)\) is finite, and by excision \(m(E \sim E_n) = m(E) - m(E_n)\). Choose \(N\) so large that \(m(E) - m(E_N) < \epsilon\) and set

\begin{equation*} F = E_N. \end{equation*}

Then \(F\) is a measurable subset of \(E\), \(m(E \sim F) < \epsilon\), and \(|f| \le N\) on \(F\); in particular \(f\) is real-valued on \(F\).

Step 2: uniform approximation on \(F\) by simple functions. By part (ii) of Proposition 5, the restriction of \(f\) to the measurable set \(F\) is measurable, and by Step 1 it is a bounded real-valued function on \(F\). Fix a natural number \(n\). The Simple Approximation Lemma, applied to \(f|_F\) with the choice \(\epsilon = 1/n\), provides simple functions \(\phi_n\) and \(\psi_n\) on \(F\) with

\begin{equation*} \phi_n \le f \le \psi_n \quad \text{and} \quad 0 \le \psi_n - \phi_n < 1/n \ \text{ on } F. \end{equation*}

Consequently

\begin{equation*} 0 \le f - \phi_n \le \psi_n - \phi_n < 1/n \quad \text{on } F. \end{equation*}

Step 3: extension to \(E\). Define \(\varphi_n\) on all of \(E\) by

\begin{equation*} \varphi_n(x) = \begin{cases} \phi_n(x) & \text{if } x \in F, \\ 0 & \text{if } x \in E \sim F. \end{cases} \end{equation*}

Each \(\varphi_n\) is real-valued and takes only finitely many values, namely the values of \(\phi_n\) together with possibly \(0\). It is measurable on \(E\): the restriction of \(\varphi_n\) to the measurable set \(F\) is \(\phi_n\), which is measurable because it is simple, and the restriction of \(\varphi_n\) to \(E \sim F\) is identically \(0\), which is measurable; so by part (ii) of Proposition 5, \(\varphi_n\) is measurable on \(E\). Being real-valued, measurable, and finitely valued, \(\varphi_n\) is a simple function on \(E\).

Step 4: conclusion. Let \(\epsilon^{\prime} > 0\) and choose \(N^{\prime}\) with \(1/N^{\prime} < \epsilon^{\prime}\). For \(n \ge N^{\prime}\) we have, by Step 2, \(|f - \varphi_n| = |f - \phi_n| < 1/n \le 1/N^{\prime} < \epsilon^{\prime}\) at every point of \(F\). Thus \(\{\varphi_n\} \to f\) uniformly on \(F\), while \(m(E \sim F) < \epsilon\). \(\blacksquare\)

Write \(I = [a, b]\). Since \(E \subseteq I\) is measurable and \(m^{*}(E) \le b - a < \infty\), Theorem 12 of Chapter 2 applies: there is a finite disjoint collection of open intervals \(\{I_k\}_{k=1}^{n}\) such that, setting \(\mathcal{O} = \bigcup_{k=1}^{n} I_k\),

\begin{equation*} m^{*}(E \,\Delta\, \mathcal{O}) < \epsilon , \end{equation*}

where, as in the text, \(E \,\Delta\, \mathcal{O} = (E \sim \mathcal{O}) \cup (\mathcal{O} \sim E)\) denotes the symmetric difference. Write \(I_k = (a_k, b_k)\) for \(1 \le k \le n\).

The step function. Define \(h\) on \(I\) by \(h = \chi_{\mathcal{O}}\) restricted to \(I\), that is,

\begin{equation*} h(x) = \begin{cases} 1 & \text{if } x \in \mathcal{O} \cap I, \\ 0 & \text{if } x \in I \sim \mathcal{O}. \end{cases} \end{equation*}

We claim \(h\) is a step function on \(I\). Let \(a = t_0 < t_1 < \cdots < t_m = b\) be the partition of \([a,b]\) whose partition points are \(a\), \(b\), and those of the finitely many numbers \(a_1, b_1, \ldots, a_n, b_n\) that lie in the open interval \((a,b)\). Fix \(j\) and consider the open interval \((t_{j-1}, t_j)\). No endpoint \(a_k\) or \(b_k\) lies in \((t_{j-1}, t_j)\). Hence for each \(k\), the set \((t_{j-1}, t_j) \cap (a_k, b_k)\) is either empty or all of \((t_{j-1}, t_j)\): indeed if some \(x \in (t_{j-1}, t_j)\) belongs to \((a_k, b_k)\) while some \(y \in (t_{j-1}, t_j)\) does not, then between \(x\) and \(y\) there would be a point of \(\{a_k, b_k\}\), and that point would lie in \((t_{j-1}, t_j)\), a contradiction. Therefore either \((t_{j-1}, t_j) \subseteq \mathcal{O}\), in which case \(h \equiv 1\) on \((t_{j-1}, t_j)\), or \((t_{j-1}, t_j) \cap \mathcal{O} = \emptyset\), in which case \(h \equiv 0\) there. So \(h\) is constant on each open interval of a finite partition of \(I\), i.e. \(h\) is a step function on \(I\).

The set \(F\). Define

\begin{equation*} F = I \sim (E \,\Delta\, \mathcal{O}) = I \sim \big[(E \sim \mathcal{O}) \cup (\mathcal{O} \sim E)\big]. \end{equation*}

Since \(E\) is measurable and \(\mathcal{O}\) is open, hence measurable, and the measurable sets form a \(\sigma\)-algebra, \(F\) is a measurable subset of \(I\).

Let \(x \in F\). Then \(x \in I\) and \(x \notin E \,\Delta\, \mathcal{O}\), so \(x \in E\) if and only if \(x \in \mathcal{O}\). Hence \(\chi_E(x) = \chi_{\mathcal{O}}(x) = h(x)\). Thus \(h = \chi_E\) on \(F\).

Finally,

\begin{equation*} I \sim F = I \cap (E \,\Delta\, \mathcal{O}) \subseteq E \,\Delta\, \mathcal{O}, \end{equation*}

so by the monotonicity of measure \(m(I \sim F) \le m(E \,\Delta\, \mathcal{O}) < \epsilon\). \(\blacksquare\)

Let \(c_1, \ldots, c_n\) be the distinct values taken by \(\psi\) and let

\begin{equation*} E_k = \{x \in I \mid \psi(x) = c_k\}, \qquad 1 \le k \le n, \end{equation*}

so that the canonical representation of \(\psi\) is

\begin{equation*} \psi = \sum_{k=1}^{n} c_k \cdot \chi_{E_k} \quad \text{on } I . \end{equation*}

Each \(E_k\) is measurable, since \(\psi\) is measurable (Proposition 1), and the \(E_k\) are disjoint with union \(I\).

Applying the preceding problem. Fix \(k\). Since \(E_k\) is a measurable subset of the closed, bounded interval \(I\), Problem 16, applied with the positive number \(\epsilon/n\), provides a step function \(h_k\) on \(I\) and a measurable subset \(F_k\) of \(I\) for which

\begin{equation*} h_k = \chi_{E_k} \text{ on } F_k \quad \text{and} \quad m(I \sim F_k) < \epsilon/n . \end{equation*}

The step function. Define

\begin{equation*} h = \sum_{k=1}^{n} c_k \cdot h_k \quad \text{on } I . \end{equation*}

A linear combination of step functions on \(I\) is again a step function on \(I\): if \(P_k\) is a partition of \(I\) on each of whose open subintervals \(h_k\) is constant, let \(P\) be the partition of \(I\) whose points are all the points of \(P_1 \cup \cdots \cup P_n\). Each open subinterval \(J\) determined by \(P\) is contained in an open subinterval determined by \(P_k\), for every \(k\), so every \(h_k\) is constant on \(J\), and hence so is \(h\). Thus \(h\) is a step function on \(I\).

The set \(F\). Define \(F = \bigcap_{k=1}^{n} F_k\), a measurable subset of \(I\). By De Morgan’s identity and the countable (here finite) subadditivity and monotonicity of measure,

\begin{equation*} m(I \sim F) = m\Big(\bigcup_{k=1}^{n} (I \sim F_k)\Big) \le \sum_{k=1}^{n} m(I \sim F_k) < n \cdot \frac{\epsilon}{n} = \epsilon . \end{equation*}

Agreement on \(F\). Let \(x \in F\). Then \(x \in F_k\) for every \(k\), so \(h_k(x) = \chi_{E_k}(x)\) for every \(k\), and therefore

\begin{equation*} h(x) = \sum_{k=1}^{n} c_k \cdot h_k(x) = \sum_{k=1}^{n} c_k \cdot \chi_{E_k}(x) = \psi(x). \end{equation*}

Hence \(h = \psi\) on \(F\) and \(m(I \sim F) < \epsilon\). \(\blacksquare\)

The function \(f\) is a bounded measurable real-valued function on the measurable set \(I\), so the Simple Approximation Lemma applies. Using it with the positive number \(\epsilon/2\), we obtain simple functions \(\phi\) and \(\eta\) on \(I\) with

\begin{equation*} \phi \le f \le \eta \quad \text{and} \quad 0 \le \eta - \phi < \epsilon/2 \ \text{ on } I . \end{equation*}

Put \(\psi = \phi\). Then \(\psi\) is a simple function on \(I\) and, at every point of \(I\),

\begin{equation*} 0 \le f - \psi \le \eta - \phi < \epsilon/2, \end{equation*}

so that

\begin{equation*} |\psi - f| < \epsilon/2 \quad \text{on all of } I . \end{equation*}

Now apply Problem 17 to the simple function \(\psi\) on the closed, bounded interval \(I\), with the same \(\epsilon > 0\): there is a step function \(h\) on \(I\) and a measurable subset \(F\) of \(I\) for which

\begin{equation*} h = \psi \text{ on } F \quad \text{and} \quad m(I \sim F) < \epsilon . \end{equation*}

Let \(x \in F\). Then \(h(x) = \psi(x)\), and therefore

\begin{equation*} |h(x) - f(x)| = |\psi(x) - f(x)| < \epsilon/2 < \epsilon . \end{equation*}

Thus \(|h - f| < \epsilon\) on \(F\) while \(m(I \sim F) < \epsilon\), as required. \(\blacksquare\)

Let \(\varphi\) and \(\psi\) be simple functions with common measurable domain \(E\) (a common domain is needed for the four operations to be defined). Let \(a_1, \ldots, a_n\) be the distinct values of \(\varphi\) and \(b_1, \ldots, b_m\) the distinct values of \(\psi\), and set

\begin{equation*} A_i = \{x \in E \mid \varphi(x) = a_i\}, \qquad B_j = \{x \in E \mid \psi(x) = b_j\}. \end{equation*}

Since \(\varphi\) and \(\psi\) are measurable, each \(A_i\) and each \(B_j\) is measurable by Proposition 1. The collection \(\{A_i\}_{i=1}^{n}\) is a disjoint cover of \(E\), as is \(\{B_j\}_{j=1}^{m}\), and hence the sets

\begin{equation*} C_{ij} = A_i \cap B_j, \qquad 1 \le i \le n, \ 1 \le j \le m, \end{equation*}

are measurable, pairwise disjoint, and their union is \(E\).

Let \(g\) denote any one of the four functions

\begin{equation*} \varphi + \psi, \qquad \varphi \cdot \psi, \qquad \max\{\varphi, \psi\}, \qquad \min\{\varphi, \psi\}, \end{equation*}

and let \(\gamma_{ij}\) denote, respectively, the number

\begin{equation*} a_i + b_j, \qquad a_i b_j, \qquad \max\{a_i, b_j\}, \qquad \min\{a_i, b_j\}. \end{equation*}

Since \(\varphi \equiv a_i\) and \(\psi \equiv b_j\) on \(C_{ij}\), in each of the four cases we have \(g \equiv \gamma_{ij}\) on \(C_{ij}\).

\(g\) takes only finitely many values. Since the \(C_{ij}\) cover \(E\), the image \(g(E)\) is contained in the finite set \(\{\gamma_{ij} \mid 1 \le i \le n, \ 1 \le j \le m\}\), which has at most \(nm\) elements. Note also that all the \(\gamma_{ij}\) are real numbers (simple functions are real-valued by definition, so no expression \(\infty - \infty\) or \(0 \cdot \infty\) can arise), so \(g\) is real-valued.

\(g\) is measurable. Let \(c\) be a real number. Because \(E\) is the disjoint union of the \(C_{ij}\) and \(g\) is constantly \(\gamma_{ij}\) on \(C_{ij}\),

\begin{equation*} \{x \in E \mid g(x) > c\} = \bigcup_{\{(i,j) \,:\, \gamma_{ij} > c\}} C_{ij}, \end{equation*}

a finite union of measurable sets, hence measurable. By Proposition 1, \(g\) is measurable on \(E\).

Thus each of \(\varphi + \psi\), \(\varphi \cdot \psi\), \(\max\{\varphi, \psi\}\) and \(\min\{\varphi, \psi\}\) is a measurable real-valued function on \(E\) taking only a finite number of values, that is, each is simple.

(Alternatively, the measurability of \(\varphi + \psi\) and \(\varphi \cdot \psi\) follows from Theorem 6 and that of \(\max\{\varphi, \psi\}\) and \(\min\{\varphi, \psi\}\) from Proposition 8; the point that remains in each case is exactly the finiteness of the set of values established above.) \(\blacksquare\)

Since a complement only has meaning relative to an ambient set, fix a set \(X\) with \(A \subseteq X\) and \(B \subseteq X\), and read all three identities as identities between functions on \(X\), with \(A^{c} = X \sim A\). (In the setting of the text \(X = \mathbb{R}\).) For a subset \(S\) of \(X\) the characteristic function \(\chi_S\) is the function on \(X\) defined by \(\chi_S(x) = 1\) if \(x \in S\) and \(\chi_S(x) = 0\) if \(x \notin S\). Two functions on \(X\) are equal precisely when they agree at every \(x \in X\), so in each case we fix \(x \in X\) and check the possible positions of \(x\) relative to \(A\) and \(B\).

First identity. Suppose \(x \in A \cap B\). Then \(\chi_{A \cap B}(x) = 1\), while \(\chi_A(x) = \chi_B(x) = 1\), so \(\chi_A(x) \cdot \chi_B(x) = 1\). Suppose \(x \notin A \cap B\). Then \(\chi_{A \cap B}(x) = 0\), and at least one of \(x \notin A\), \(x \notin B\) holds, so at least one of \(\chi_A(x)\), \(\chi_B(x)\) equals \(0\) and hence \(\chi_A(x) \cdot \chi_B(x) = 0\). In both cases the two sides agree, so \(\chi_{A \cap B} = \chi_A \cdot \chi_B\).

Second identity. Consider the four cases, writing \(u = \chi_A(x)\) and \(v = \chi_B(x)\).

• If \(x \in A\) and \(x \in B\), then \(x \in A \cup B\), so the left side is \(1\); and \(u + v - uv = 1 + 1 - 1 = 1\).
• If \(x \in A\) and \(x \notin B\), then \(x \in A \cup B\), so the left side is \(1\); and \(u + v - uv = 1 + 0 - 0 = 1\).
• If \(x \notin A\) and \(x \in B\), then \(x \in A \cup B\), so the left side is \(1\); and \(u + v - uv = 0 + 1 - 0 = 1\).
• If \(x \notin A\) and \(x \notin B\), then \(x \notin A \cup B\), so the left side is \(0\); and \(u + v - uv = 0 + 0 - 0 = 0\).

In every case the two sides agree, so \(\chi_{A \cup B} = \chi_A + \chi_B - \chi_A \cdot \chi_B\).

Third identity. If \(x \in A^c\), then \(x \notin A\), so \(\chi_{A^c}(x) = 1\) and \(1 - \chi_A(x) = 1 - 0 = 1\). If \(x \notin A^c\), then \(x \in A\), so \(\chi_{A^c}(x) = 0\) and \(1 - \chi_A(x) = 1 - 1 = 0\). Hence \(\chi_{A^c} = 1 - \chi_A\).

(Observe that the second identity can also be read off from the first two together with the third and De Morgan’s laws: \(\chi_{A \cup B} = 1 - \chi_{(A \cup B)^c} = 1 - \chi_{A^c \cap B^c} = 1 - (1 - \chi_A)(1 - \chi_B) = \chi_A + \chi_B - \chi_A \chi_B\).) \(\blacksquare\)

All functions here are extended real-valued on the measurable set \(E\), and the four functions are defined pointwise: \(\big(\sup\{f_n\}\big)(x) = \sup_n f_n(x)\), and similarly for the others. Such pointwise suprema and infima always exist in the extended real numbers.

The supremum. Put \(g = \sup\{f_n\}\). Let \(c\) be a real number. We claim

\begin{equation*} \{x \in E \mid g(x) > c\} = \bigcup_{n=1}^{\infty} \{x \in E \mid f_n(x) > c\}. \end{equation*}

Indeed, if \(g(x) > c\) then \(c\) is not an upper bound for the set \(\{f_n(x)\}_{n=1}^{\infty}\) (its supremum is \(g(x) > c\)), so there is an index \(n\) with \(f_n(x) > c\). Conversely, if \(f_n(x) > c\) for some \(n\), then \(g(x) \ge f_n(x) > c\). Each set \(\{x \in E \mid f_n(x) > c\}\) is measurable because \(f_n\) is measurable, and the measurable sets are a \(\sigma\)-algebra, so the countable union on the right is measurable. Since this holds for every real \(c\), property (i) of Proposition 1 holds for \(g\), and hence \(g = \sup\{f_n\}\) is measurable.

The infimum. Put \(h = \inf\{f_n\}\). Let \(c\) be a real number. We claim

\begin{equation*} \{x \in E \mid h(x) < c\} = \bigcup_{n=1}^{\infty} \{x \in E \mid f_n(x) < c\}. \end{equation*}

Indeed, if \(h(x) < c\) then \(c\) is not a lower bound for \(\{f_n(x)\}_{n=1}^{\infty}\), so \(f_n(x) < c\) for some \(n\); conversely if \(f_n(x) < c\) for some \(n\), then \(h(x) \le f_n(x) < c\). Again the right-hand side is a countable union of measurable sets, hence measurable, so property (iii) of Proposition 1 holds for \(h\) and \(h = \inf\{f_n\}\) is measurable.

Observe that these two arguments used nothing about the index set beyond its countability: for each natural number \(k\), the sequence \(\{f_n\}_{n \ge k}\) is again a countable family of measurable functions on \(E\), so

\begin{equation*} g_k = \inf_{n \ge k} f_n \quad \text{and} \quad G_k = \sup_{n \ge k} f_n \end{equation*}

are measurable functions on \(E\) for every \(k\).

The limit inferior. By definition,

\begin{equation*} \liminf\{f_n\} = \sup_{k \ge 1} \Big( \inf_{n \ge k} f_n \Big) = \sup_{k \ge 1} g_k . \end{equation*}

Each \(g_k\) is measurable by the preceding paragraph, so \(\{g_k\}_{k=1}^{\infty}\) is a sequence of measurable functions with common domain \(E\); by the supremum case already proved, \(\sup_{k} g_k\) is measurable. Hence \(\liminf\{f_n\}\) is measurable.

The limit superior. Similarly, by definition,

\begin{equation*} \limsup\{f_n\} = \inf_{k \ge 1} \Big( \sup_{n \ge k} f_n \Big) = \inf_{k \ge 1} G_k , \end{equation*}

each \(G_k\) is measurable, and by the infimum case already proved the pointwise infimum of the sequence \(\{G_k\}_{k=1}^{\infty}\) of measurable functions is measurable. Hence \(\limsup\{f_n\}\) is measurable. \(\blacksquare\)

Throughout, “increasing sequence” means \(f_n \le f_{n+1}\) pointwise on \([a,b]\) for every index \(n\).

We first record that \(f_n \le f\) on \([a,b]\) for every \(n\). Indeed, fix \(x \in [a,b]\) and an index \(n\). For every \(m \ge n\) we have \(f_n(x) \le f_m(x)\), and letting \(m \to \infty\) in this inequality, using \(f_m(x) \to f(x)\) and the fact that weak inequalities are preserved under limits, we get \(f_n(x) \le f(x)\). Consequently

\begin{equation*} |f(x) - f_n(x)| = f(x) - f_n(x) \quad \text{for all } x \in [a,b] \text{ and all } n, \end{equation*}

so it suffices to show that for each \(\epsilon > 0\) there is an index \(N\) with \(f - f_n < \epsilon\) on all of \([a,b]\) whenever \(n \ge N\).

Let \(\epsilon > 0\) and, following the hint, set

\begin{equation*} E_n = \{x \in [a,b] \mid f(x) - f_n(x) < \epsilon\}. \end{equation*}

Each \(E_n\) is relatively open in \([a,b]\). Indeed, \(f\) and \(f_n\) are continuous on \([a,b]\), hence so is \(g_n := f - f_n\), and \(E_n = g_n^{-1}\big((-\infty, \epsilon)\big)\) is the preimage of an open subset of \(\mathbb{R}\) under a continuous function on \([a,b]\). Thus there is an open subset \(\mathcal{O}_n\) of \(\mathbb{R}\) with \(E_n = \mathcal{O}_n \cap [a,b]\). (Concretely: if \(x_0 \in E_n\), continuity of \(g_n\) at \(x_0\) with the choice \(\eta = \epsilon - g_n(x_0) > 0\) furnishes \(\delta > 0\) such that \(g_n(x) < \epsilon\) for all \(x \in [a,b]\) with \(|x - x_0| < \delta\); taking \(\mathcal{O}_n\) to be the union of all such open intervals \((x_0 - \delta, x_0 + \delta)\) over \(x_0 \in E_n\) gives an open set with \(\mathcal{O}_n \cap [a,b] = E_n\).)

The collection \(\{E_n\}\) is ascending. Since \(f_n \le f_{n+1}\), we have \(f - f_{n+1} \le f - f_n\) on \([a,b]\), so \(f(x) - f_n(x) < \epsilon\) forces \(f(x) - f_{n+1}(x) < \epsilon\); that is, \(E_n \subseteq E_{n+1}\). The same monotonicity gives \(\mathcal{O}_n \cap [a,b] \subseteq \mathcal{O}_m \cap [a,b]\) whenever \(n \le m\).

The collection \(\{\mathcal{O}_n\}_{n=1}^{\infty}\) covers \([a,b]\). Let \(x \in [a,b]\). Since \(f_n(x) \to f(x)\), there is an index \(n\) with \(f(x) - f_n(x) < \epsilon\), that is, \(x \in E_n \subseteq \mathcal{O}_n\).

The interval \([a,b]\) is closed and bounded, so by the Heine-Borel Theorem the open cover \(\{\mathcal{O}_n\}_{n=1}^{\infty}\) of \([a,b]\) has a finite subcover, say \(\mathcal{O}_{n_1}, \dots, \mathcal{O}_{n_k}\). Put \(N = \max\{n_1, \dots, n_k\}\). For each \(i\) we have \(\mathcal{O}_{n_i} \cap [a,b] = E_{n_i} \subseteq E_N\), and therefore

\begin{equation*} [a,b] = \bigcup_{i=1}^{k} \big(\mathcal{O}_{n_i} \cap [a,b]\big) \subseteq E_N \subseteq [a,b], \end{equation*}

so \(E_N = [a,b]\).

Finally, let \(n \ge N\). Then \(E_N \subseteq E_n\), so \(E_n = [a,b]\), which says precisely that

\begin{equation*} |f(x) - f_n(x)| = f(x) - f_n(x) < \epsilon \quad \text{for every } x \in [a,b]. \end{equation*}

Since \(\epsilon > 0\) was arbitrary and \(N\) does not depend on \(x\), the convergence of \(\{f_n\}\) to \(f\) is uniform on \([a,b]\).

Two remarks on the hypotheses, each of which is essential. If \(f\) is not assumed continuous the conclusion fails: on \([0,1]\) the sequence \(f_n(x) = x^{1/n}\) is an increasing sequence of continuous functions whose pointwise limit equals \(0\) at \(x = 0\) and \(1\) on \((0,1]\), and the convergence is not uniform. If compactness of the domain is dropped the conclusion also fails: on \((0,1)\) the increasing continuous functions \(f_n(x) = \min\{nx, 1\}\) converge pointwise to the continuous function \(f \equiv 1\), but \(\sup_{0 < x < 1}|1 - f_n(x)| = 1\) for every \(n\).

Let \(E\) be measurable and let \(f\) be an extended real-valued function on \(E\). Define the positive and negative parts of \(f\) by

\begin{equation*} f^{+}(x) = \max\{f(x), 0\}, \qquad f^{-}(x) = \max\{-f(x), 0\} \qquad (x \in E). \end{equation*}

Both are nonnegative, and at each point at most one of them is nonzero, so the difference \(f^{+} - f^{-}\) never has the indeterminate form \(\infty - \infty\) and

\begin{equation*} f = f^{+} - f^{-}, \qquad |f| = f^{+} + f^{-} \qquad \text{on } E . \end{equation*}

Indeed, if \(f(x) \ge 0\) then \(f^{+}(x) = f(x)\) and \(f^{-}(x) = 0\); if \(f(x) < 0\) then \(f^{+}(x) = 0\) and \(f^{-}(x) = -f(x)\). This covers the extended values as well: if \(f(x) = \infty\) then \(f^{+}(x) = \infty\), \(f^{-}(x) = 0\); if \(f(x) = -\infty\) then \(f^{+}(x) = 0\), \(f^{-}(x) = \infty\).

If \(f\) is measurable, so are \(f^{+}\) and \(f^{-}\). For \(c \ge 0\),

\begin{equation*} \{x \in E \mid f^{+}(x) > c\} = \{x \in E \mid f(x) > c\}, \end{equation*}

which is measurable, while for \(c < 0\) the set \(\{x \in E \mid f^{+}(x) > c\}\) is all of \(E\), again measurable. Likewise, for \(c \ge 0\),

\begin{equation*} \{x \in E \mid f^{-}(x) > c\} = \{x \in E \mid f(x) < -c\}, \end{equation*}

which is measurable, and for \(c < 0\) this set is \(E\). Hence \(f^{+}\) and \(f^{-}\) are nonnegative measurable functions on \(E\) and \(f = f^{+} - f^{-}\) is the required expression.

We now prove the Simple Approximation Theorem in general, assuming it for nonnegative measurable functions.

One implication requires nothing new: every simple function is measurable, and by Proposition 9 the pointwise limit of a sequence of measurable functions on \(E\) is measurable. So if there is a sequence \(\{\varphi_n\}\) of simple functions on \(E\) converging pointwise on \(E\) to \(f\), then \(f\) is measurable.

For the converse, assume \(f\) is measurable. By the paragraphs above, \(f^{+}\) and \(f^{-}\) are nonnegative measurable functions on \(E\). Apply the nonnegative case of the theorem to each: there are sequences \(\{\varphi_n\}\) and \(\{\psi_n\}\) of simple functions on \(E\) with

\begin{equation*} 0 \le \varphi_n \le f^{+}, \quad \varphi_n \to f^{+} \ \text{pointwise on } E, \qquad 0 \le \psi_n \le f^{-}, \quad \psi_n \to f^{-} \ \text{pointwise on } E, \end{equation*}

and with each of \(\{\varphi_n\}\), \(\{\psi_n\}\) increasing. Here is why the two-sided bounds may be taken in the sharpened form \(0 \le \varphi_n \le f^{+}\) rather than merely \(|\varphi_n| \le |f^{+}|\). The nonnegative case delivers simple functions \(\widetilde{\varphi}_n\) on \(E\), increasing, with \(\widetilde{\varphi}_n \to f^{+}\) pointwise and \(|\widetilde{\varphi}_n| \le |f^{+}| = f^{+}\) on \(E\); the latter gives \(\widetilde{\varphi}_n \le f^{+}\) but not \(\widetilde{\varphi}_n \ge 0\). Replace \(\widetilde{\varphi}_n\) by \(\varphi_n = \max\{\widetilde{\varphi}_n, 0\}\). By Problem 19 the maximum of two simple functions is simple, so each \(\varphi_n\) is simple; the sequence \(\{\varphi_n\}\) is still increasing since \(t \mapsto \max\{t,0\}\) is increasing; \(0 \le \varphi_n \le \max\{f^{+},0\} = f^{+}\) because \(\widetilde{\varphi}_n \le f^{+}\) and \(0 \le f^{+}\); and at each \(x \in E\), \(\varphi_n(x) = \max\{\widetilde{\varphi}_n(x), 0\} \to \max\{f^{+}(x), 0\} = f^{+}(x)\), by continuity of \(t \mapsto \max\{t, 0\}\) on \(\mathbb{R}\) when \(f^{+}(x)\) is finite, and directly from \(\widetilde{\varphi}_n(x) \to \infty\) when \(f^{+}(x) = \infty\). The same replacement is made for \(\{\psi_n\}\).

Define \(h_n = \varphi_n - \psi_n\) on \(E\). Each \(h_n\) is simple: \(\psi_n\) simple implies \(-\psi_n = (-1) \cdot \psi_n\) is simple, and by Problem 19 the sum of two simple functions is simple. Note that \(\varphi_n\) and \(\psi_n\) are real-valued (simple functions take finitely many real values), so no indeterminate difference arises.

The bound \(|h_n| \le |f|\) holds on \(E\). Fix \(x \in E\). If \(f(x) \ge 0\) then \(f^{-}(x) = 0\), so \(0 \le \psi_n(x) \le f^{-}(x) = 0\) forces \(\psi_n(x) = 0\); hence \(h_n(x) = \varphi_n(x)\) and

\begin{equation*} |h_n(x)| = \varphi_n(x) \le f^{+}(x) = |f(x)| . \end{equation*}

If \(f(x) < 0\) then \(f^{+}(x) = 0\), so \(\varphi_n(x) = 0\), \(h_n(x) = -\psi_n(x)\), and

\begin{equation*} |h_n(x)| = \psi_n(x) \le f^{-}(x) = |f(x)| . \end{equation*}

Finally, \(h_n \to f\) pointwise on \(E\). Fix \(x \in E\). If \(f(x)\) is finite, then \(f^{+}(x)\) and \(f^{-}(x)\) are finite and

\begin{equation*} \lim_{n \to \infty} h_n(x) = \lim_{n \to \infty} \varphi_n(x) - \lim_{n \to \infty} \psi_n(x) = f^{+}(x) - f^{-}(x) = f(x). \end{equation*}

If \(f(x) = \infty\), then, as computed above, \(\psi_n(x) = 0\) for every \(n\) and \(h_n(x) = \varphi_n(x) \to f^{+}(x) = \infty = f(x)\). If \(f(x) = -\infty\), then \(\varphi_n(x) = 0\) for every \(n\) and \(h_n(x) = -\psi_n(x) \to -f^{-}(x) = -\infty = f(x)\).

Thus \(\{h_n\}\) is a sequence of simple functions on \(E\) that converges pointwise on \(E\) to \(f\) and satisfies \(|h_n| \le |f|\) on \(E\) for every \(n\), which is the general Simple Approximation Theorem.

For the last clause of the theorem, if \(f \ge 0\) on \(E\) then \(f^{-} \equiv 0\), hence \(\psi_n \equiv 0\) and \(h_n = \varphi_n\), so the approximating sequence produced is increasing, as asserted in the nonnegative case.

First note that the domain \(I\) is a measurable set, since every interval is Lebesgue measurable. So it makes sense to ask whether \(f\) is measurable, and by the criterion defining measurability it suffices to show that for each \(c \in \mathbb{R}\) the set \(\{x \in I \mid f(x) > c\}\) is measurable.

Fix a natural number \(n\) and define

\begin{equation*} g_n : I \to \mathbb{R}, \qquad g_n(x) = f(x) + \frac{x}{n}. \end{equation*}

The function \(g_n\) is strictly increasing. Indeed, let \(x, y \in I\) with \(x < y\). Since \(f\) is increasing, \(f(x) \le f(y)\), and since \(x/n < y/n\), adding gives

\begin{equation*} g_n(x) = f(x) + \frac{x}{n} < f(y) + \frac{y}{n} = g_n(y). \end{equation*}

Claim: a strictly increasing real-valued function \(g\) on an interval \(I\) is measurable. Fix \(c \in \mathbb{R}\) and put

\begin{equation*} S = \{x \in I \mid g(x) > c\}. \end{equation*}

The set \(S\) is upward closed in \(I\): if \(x \in S\) and \(y \in I\) with \(y > x\), then \(g(y) > g(x) > c\), so \(y \in S\). We show \(S\) is an interval, i.e. that it is convex. Let \(x, y \in S\) and let \(z \in \mathbb{R}\) with \(x < z < y\). Since \(x, y \in I\) and \(I\) is an interval, \(z \in I\); and since \(z > x\) and \(S\) is upward closed in \(I\), we get \(z \in S\). Thus \(S\) contains every real number lying between two of its points, so \(S\) is an interval (possibly empty, possibly a single point, possibly unbounded). Every interval is measurable, so \(S\) is measurable. As \(c \in \mathbb{R}\) was arbitrary, \(g\) is measurable, and the claim is proved.

Applying the claim to \(g_n\), we conclude that \(g_n\) is a measurable real-valued function on \(I\), for every natural number \(n\).

Now take pointwise limits. Fix \(x \in I\). Then \(x\) is a fixed real number, so \(x/n \to 0\) as \(n \to \infty\), and hence

\begin{equation*} \lim_{n \to \infty} g_n(x) = \lim_{n \to \infty}\Big(f(x) + \frac{x}{n}\Big) = f(x). \end{equation*}

Thus \(\{g_n\}\) is a sequence of measurable functions on the measurable set \(I\) that converges pointwise on \(I\) to \(f\). By Proposition 9, the pointwise limit of a sequence of measurable functions on a common measurable domain is measurable. Therefore \(f\) is measurable.

Remark. The claim above did not actually use the strictness of the increase: if \(g\) is merely increasing and \(x \in S\), \(y \in I\), \(y > x\), then \(g(y) \ge g(x) > c\), so again \(y \in S\), and the rest of the argument is unchanged. Consequently every increasing real-valued function on an interval is measurable directly, without passing to the limit. The route through the strictly increasing functions \(g_n\) is the one suggested by the problem, and it is included above in the form requested.

If \(F = \emptyset\) we may take \(g \equiv 0\), and if \(F = \mathbb{R}\) we may take \(g = f\). So assume \(F\) is a nonempty proper closed subset of \(\mathbb{R}\) and put \(\mathcal{O} = \mathbb{R} \sim F\), a nonempty open set.

Construction of the extension. By Proposition 9 of Chapter 1, \(\mathcal{O}\) is the union of a countable disjoint collection \(\{I_k\}\) of open intervals; moreover, the proof of that proposition shows that each finite endpoint of each \(I_k\) fails to belong to \(\mathcal{O}\), hence belongs to \(F\). Since \(F \neq \emptyset\), no \(I_k\) equals \(\mathbb{R}\), so each \(I_k\) is of exactly one of the three forms

\begin{equation*} (a_k, b_k) \ \text{with } a_k, b_k \in F, \qquad (-\infty, b_k) \ \text{with } b_k \in F, \qquad (a_k, \infty) \ \text{with } a_k \in F . \end{equation*}

Define \(g : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} g(x) = \begin{cases} f(x), & x \in F, \\[2pt] f(a_k) + \dfrac{f(b_k) - f(a_k)}{b_k - a_k}\,(x - a_k), & x \in I_k = (a_k, b_k) \ \text{bounded}, \\[6pt] f(b_k), & x \in I_k = (-\infty, b_k), \\[2pt] f(a_k), & x \in I_k = (a_k, \infty). \end{cases} \end{equation*}

Since the \(I_k\) are pairwise disjoint and disjoint from \(F\), \(g\) is well defined, real-valued, and \(g = f\) on \(F\); so \(g\) is an extension of \(f\). On a bounded \(I_k\) the formula is the affine interpolation between the values \(f(a_k)\) and \(f(b_k)\) at the endpoints, and it takes those values at \(a_k\) and \(b_k\), so \(g\) is continuous on the closed interval \([a_k, b_k]\) and agrees there with the endpoint values already assigned by \(f\).

Observe once and for all that for \(x\) in a bounded component \((a_k, b_k)\) the value \(g(x)\) is a convex combination of \(f(a_k)\) and \(f(b_k)\): writing \(t = (x - a_k)/(b_k - a_k) \in (0,1)\),

\begin{equation*} g(x) = (1 - t) f(a_k) + t f(b_k). \end{equation*}

Call this the convex-combination identity (C).

Continuity at points of \(\mathcal{O}\). If \(x_0 \in \mathcal{O}\), then \(x_0\) lies in some \(I_k\), which is an open interval, hence an open neighborhood of \(x_0\), and \(g\) restricted to \(I_k\) is either affine or constant, hence continuous at \(x_0\). So \(g\) is continuous at \(x_0\).

Continuity at points of \(F\). Let \(x_0 \in F\) and let \(\epsilon > 0\). Since \(f\) is continuous on \(F\) at \(x_0\), choose \(\delta > 0\) such that

\begin{equation*} |f(y) - f(x_0)| < \epsilon \quad \text{for all } y \in F \text{ with } |y - x_0| < \delta. \end{equation*}

Call this the choice-of-delta property (D). We produce \(\delta_{+} > 0\) with \(|g(x) - g(x_0)| < \epsilon\) for \(x_0 \le x < x_0 + \delta_{+}\); the argument on the left is symmetric and produces \(\delta_{-} > 0\), and then \(\delta_0 = \min\{\delta_{+}, \delta_{-}\}\) works for two-sided continuity at \(x_0\).

Case 1: \(x_0\) is the left endpoint of some component \(I_k\) of \(\mathcal{O}\). If \(I_k = (x_0, \infty)\), then \(g \equiv f(x_0) = g(x_0)\) on \([x_0, \infty)\) and any \(\delta_{+}\) works. If \(I_k = (x_0, b_k)\) is bounded, then by (C) and the fact that \(g\) is the continuous affine function on \([x_0, b_k]\) with \(g(x_0) = f(x_0)\), we have \(g(x) \to g(x_0)\) as \(x \to x_0^{+}\); choose \(\delta_{+} \le b_k - x_0\) accordingly, noting that \([x_0, x_0 + \delta_{+}) \subseteq [x_0, b_k)\).

Case 2: \(x_0\) is not the left endpoint of any component of \(\mathcal{O}\). We claim that then \(F \cap (x_0, x_0 + \eta) \neq \emptyset\) for every \(\eta > 0\). Indeed, if \(F \cap (x_0, x_0 + \eta) = \emptyset\) for some \(\eta > 0\), then \((x_0, x_0 + \eta) \subseteq \mathcal{O}\); let \(I_k = (a_k, b_k)\) be the interval of the decomposition containing the point \(x := x_0 + \eta/2\). By the construction in Proposition 9 of Chapter 1, \(I_k\) is the largest open interval that contains \(x\) and is contained in \(\mathcal{O}\); since \((x_0, x_0 + \eta)\) is an open interval containing \(x\) and contained in \(\mathcal{O}\), we get \((x_0, x_0 + \eta) \subseteq I_k\) and hence \(a_k \le x_0\). But \(a_k < x_0\) is impossible, since then \(x_0 \in (a_k, b_k) \subseteq \mathcal{O}\), contradicting \(x_0 \in F\). Hence \(a_k = x_0\) and \(x_0\) is the left endpoint of \(I_k\), contrary to the case assumption. This proves the claim.

So choose \(c \in F\) with \(x_0 < c < x_0 + \delta\), and set \(\delta_{+} = c - x_0 > 0\). Let \(x_0 \le x < x_0 + \delta_{+} = c\). If \(x \in F\), then \(|x - x_0| < \delta\) and by (D), \(|g(x) - g(x_0)| = |f(x) - f(x_0)| < \epsilon\). If \(x \in \mathcal{O}\), let \(I_k\) be its component. Since \(x_0 \in F\) and \(x_0 < x\) (as \(x_0 \notin \mathcal{O}\)), the left endpoint of \(I_k\) satisfies \(a_k \ge x_0\); in particular \(a_k\) is finite and \(a_k \in F\). Since \(c \in F\), \(c \notin I_k\), and \(x < c\), the right endpoint satisfies \(b_k \le c\); in particular \(b_k\) is finite and \(b_k \in F\). Therefore

\begin{equation*} x_0 \le a_k < b_k \le c < x_0 + \delta, \end{equation*}

so both \(a_k\) and \(b_k\) lie in \(F\) within distance \(\delta\) of \(x_0\), and (D) gives \(|f(a_k) - f(x_0)| < \epsilon\) and \(|f(b_k) - f(x_0)| < \epsilon\). By (C), with \(t \in (0,1)\),

\begin{equation*} |g(x) - g(x_0)| = |(1-t)\big(f(a_k) - f(x_0)\big) + t\big(f(b_k) - f(x_0)\big)| \le (1-t)\epsilon + t\epsilon = \epsilon . \end{equation*}

Replacing \(\epsilon\) by \(\epsilon/2\) at the outset makes this a strict inequality \(< \epsilon\), as desired.

The left-hand estimate is the mirror image, and we record it in full since it is the other half of the proof. Either \(x_0\) is the right endpoint of some component \(I_j\) of \(\mathcal{O}\), or it is not. If \(I_j = (-\infty, x_0)\), then \(g \equiv f(x_0) = g(x_0)\) on \((-\infty, x_0]\) and any \(\delta_{-}\) works. If \(I_j = (a_j, x_0)\) is bounded, then \(g\) is affine on \([a_j, x_0]\) with \(g(x_0) = f(x_0)\), hence continuous from the left at \(x_0\), and we choose \(\delta_{-} \le x_0 - a_j\) accordingly. If \(x_0\) is the right endpoint of no component, then \(F \cap (x_0 - \eta, x_0) \neq \emptyset\) for every \(\eta > 0\): otherwise \((x_0 - \eta, x_0) \subseteq \mathcal{O}\), and the component \(I_j = (a_j, b_j)\) containing \(x_0 - \eta/2\) would satisfy \((x_0 - \eta, x_0) \subseteq I_j\) by maximality, hence \(b_j \ge x_0\), while \(b_j > x_0\) would put \(x_0 \in I_j \subseteq \mathcal{O}\), contradicting \(x_0 \in F\); so \(b_j = x_0\), contrary to the case assumption. Choose then \(c^{\prime} \in F\) with \(x_0 - \delta < c^{\prime} < x_0\) and set \(\delta_{-} = x_0 - c^{\prime}\). For \(x_0 - \delta_{-} = c^{\prime} < x \le x_0\): if \(x \in F\) then \(|x - x_0| < \delta\) and (D) applies; if \(x \in \mathcal{O}\), its component \(I_j = (a_j, b_j)\) has \(b_j \le x_0\) (else \(x_0 \in I_j\)) and \(a_j \ge c^{\prime}\) (else \(c^{\prime} \in I_j\)), so \(a_j, b_j\) are finite, lie in \(F\), and satisfy \(x_0 - \delta < c^{\prime} \le a_j < b_j \le x_0\); then (D) and (C) give \(|g(x) - g(x_0)| \le \epsilon\) exactly as before. With \(\epsilon\) replaced by \(\epsilon/2\) at the outset this is \(< \epsilon\).

Hence \(g\) is continuous at every point of \(\mathbb{R}\), and \(g\) is a continuous extension of \(f\) to all of \(\mathbb{R}\).

Recall the statement of Lusin’s Theorem: \(f\) is a real-valued measurable function on \(E\), and for a given \(\epsilon > 0\) it produces a continuous function \(g\) on \(\mathbb{R}\) and a closed set \(F\) contained in \(E\) for which

\begin{equation*} f = g \ \text{ on } F \qquad \text{and} \qquad m(E \sim F) < \epsilon . \end{equation*}

The restriction of \(f\) to \(F\) is continuous. Continuity of the restriction \(f|_F\) means: for each \(x_0 \in F\) and each \(\eta > 0\) there is \(\delta > 0\) such that \(|f(x) - f(x_0)| < \eta\) for all \(x \in F\) with \(|x - x_0| < \delta\). Fix \(x_0 \in F\) and \(\eta > 0\). Since \(g\) is continuous at \(x_0\) as a function on \(\mathbb{R}\), there is \(\delta > 0\) such that

\begin{equation*} |g(x) - g(x_0)| < \eta \quad \text{for all } x \in \mathbb{R} \text{ with } |x - x_0| < \delta . \end{equation*}

Now let \(x \in F\) with \(|x - x_0| < \delta\). Since \(x, x_0 \in F\) and \(f = g\) on \(F\),

\begin{equation*} |f(x) - f(x_0)| = |g(x) - g(x_0)| < \eta . \end{equation*}

Hence \(f|_F\) is continuous at every point of \(F\), that is, \(f|_F\) is a continuous function on \(F\). (Equivalently: the restriction of a continuous function on \(\mathbb{R}\) to any subset of \(\mathbb{R}\) is continuous on that subset, and \(f|_F = g|_F\).)

Must \(f\), considered as a function on \(E\), be continuous at some point? No. The continuity supplied by Lusin’s Theorem is continuity of the restriction, that is, continuity relative to \(F\); it says nothing about the behavior of \(f\) at points of \(E\) near \(x_0\) that lie outside \(F\). There are measurable functions that are continuous at no point of their domain.

The standard example is the Dirichlet function. Let \(E = [0,1]\) and

\begin{equation*} f = \chi_{\mathbb{Q} \cap [0,1]}, \qquad \text{that is,} \quad f(x) = 1 \text{ if } x \text{ is rational, } \ f(x) = 0 \text{ if } x \text{ is irrational}. \end{equation*}

Then \(f\) is measurable, since \(\mathbb{Q} \cap [0,1]\) is countable, hence of measure zero, hence measurable, and the characteristic function of a measurable set is measurable. And \(m(E) = 1 < \infty\), so Lusin’s Theorem applies as stated in the text.

The function \(f\) is continuous at no point of \([0,1]\). Indeed, let \(x_0 \in [0,1]\) and take \(\eta = 1/2\). By the density of the rationals and of the irrationals in \(\mathbb{R}\), every interval \((x_0 - \delta, x_0 + \delta)\) with \(\delta > 0\) contains a point \(x_1 \in [0,1]\) that is rational and a point \(x_2 \in [0,1]\) that is irrational. Then \(|f(x_1) - f(x_2)| = 1\), so at least one of \(|f(x_1) - f(x_0)|\), \(|f(x_2) - f(x_0)|\) is \(\ge 1/2 = \eta\). Hence no \(\delta\) works, and \(f\) is discontinuous at \(x_0\).

It is instructive to see Lusin’s Theorem realized here. Let \(\epsilon > 0\) and enumerate \(\mathbb{Q} \cap [0,1] = \{q_k\}_{k=1}^{\infty}\). Put

\begin{equation*} \mathcal{O} = \bigcup_{k=1}^{\infty} \Big(q_k - \frac{\epsilon}{2^{k+2}},\, q_k + \frac{\epsilon}{2^{k+2}}\Big), \end{equation*}

an open set with, by countable subadditivity and monotonicity of outer measure,

\begin{equation*} m(\mathcal{O}) \le \sum_{k=1}^{\infty} \frac{\epsilon}{2^{k+1}} = \frac{\epsilon}{2} . \end{equation*}

Set \(F = [0,1] \sim \mathcal{O}\). Then \(F\) is closed (the intersection of the closed set \([0,1]\) with the closed set \(\mathbb{R} \sim \mathcal{O}\)), \(F \subseteq E\), and \(F\) contains no rational number, so \(f \equiv 0\) on \(F\) and \(f\) agrees on \(F\) with the continuous function \(g \equiv 0\) on \(\mathbb{R}\). Moreover \(E \sim F \subseteq \mathcal{O}\), so \(m(E \sim F) \le \epsilon/2 < \epsilon\). Thus the conclusion of Lusin’s Theorem holds, the restriction \(f|_F\) is continuous (it is identically \(0\)), yet \(f\) itself is continuous at no point whatsoever of \(E\) – not even at the points of \(F\).

So the answer to the question is: there need not be any point of \(E\) at which \(f\) is continuous.

Take \(E = \mathbb{R}\), which is measurable with \(m(E) = \infty\), and for each natural number \(n\) define

\begin{equation*} f_n = \chi_{[n, \infty)}, \qquad \text{that is,} \quad f_n(x) = \begin{cases} 1, & x \ge n, \\ 0, & x < n. \end{cases} \end{equation*}

Each \(f_n\) is measurable, being the characteristic function of the measurable set \([n, \infty)\). Every hypothesis of Egoroff’s Theorem except the finiteness of \(m(E)\) is satisfied; we show its conclusion fails, and in fact fails for every \(\epsilon > 0\).

The sequence converges pointwise on \(\mathbb{R}\) to the real-valued function \(f \equiv 0\). Indeed, fix \(x \in \mathbb{R}\). By the Archimedean property there is a natural number \(N > x\), and then for every \(n \ge N\) we have \(n > x\), so \(f_n(x) = 0\). Hence \(f_n(x) \to 0\).

Now let \(\epsilon > 0\) and let \(F\) be any measurable subset of \(\mathbb{R}\) – closed or not – with \(m(\mathbb{R} \sim F) < \epsilon\). We claim \(F \cap [n, \infty) \neq \emptyset\) for every natural number \(n\). Suppose to the contrary that \(F \cap [n, \infty) = \emptyset\) for some \(n\). Then \([n, \infty) \subseteq \mathbb{R} \sim F\), and by monotonicity of measure

\begin{equation*} \infty = m\big([n, \infty)\big) \le m(\mathbb{R} \sim F) < \epsilon < \infty, \end{equation*}

a contradiction. This proves the claim.

For each \(n\) choose \(x_n \in F \cap [n, \infty)\). Then \(f_n(x_n) = 1\) while \(f(x_n) = 0\), so

\begin{equation*} \sup_{x \in F} |f_n(x) - f(x)| \ge |f_n(x_n) - f(x_n)| = 1 \qquad \text{for every } n . \end{equation*}

Therefore \(\{f_n\}\) does not converge uniformly to \(f\) on \(F\): taking \(\eta = 1/2\) in the definition of uniform convergence, there is no index \(N\) for which \(|f_n - f| < 1/2\) on \(F\) for all \(n \ge N\).

Since \(F\) was an arbitrary measurable set with \(m(\mathbb{R} \sim F) < \epsilon\), and closed sets are in particular measurable, there is no closed set \(F \subseteq \mathbb{R}\) with \(m(\mathbb{R} \sim F) < \epsilon\) on which \(\{f_n\}\) converges uniformly to \(f\). Thus the conclusion of Egoroff’s Theorem fails here for every \(\epsilon > 0\), and the finiteness of \(m(E)\) cannot simply be dropped.

Remark. A variant on a set of infinite measure that fails only for small \(\epsilon\) is \(f_n = \chi_{[n, n+1)}\) on \(\mathbb{R}\), which also converges pointwise to \(0\). If \(m(\mathbb{R} \sim F) < 1\), then for each \(n\) we have \(m([n, n+1) \sim F) < 1 = m([n, n+1))\), so \([n, n+1) \cap F \neq \emptyset\), and the same argument shows \(\sup_F |f_n| \ge 1\) for all \(n\). Note the essential role of infinite measure in both examples: the sets on which \(f_n\) differs from its limit march off to infinity without shrinking in measure, and a set of finite measure cannot contain such a sequence of sets of measure bounded away from \(0\) while leaving only a small complement.

The assertion to be proved is the following. Let \(E\) have finite measure, let \(\{f_n\}\) be a sequence of measurable functions on \(E\), and let \(f\) be an extended real-valued function on \(E\) that is finite a.e. on \(E\) and such that \(\{f_n\}\) converges to \(f\) pointwise a.e. on \(E\). Then for each \(\epsilon > 0\) there is a closed set \(F\) contained in \(E\) for which

\begin{equation*} \{f_n\} \to f \ \text{ uniformly on } F \qquad \text{and} \qquad m(E \sim F) < \epsilon . \end{equation*}

Isolating the exceptional set. Let

\begin{equation*} A = \{x \in E \mid |f(x)| = \infty\}, \qquad B = \{x \in E \mid \{f_n(x)\} \text{ does not converge to } f(x)\}, \end{equation*}

and put \(E_0 = A \cup B\). By hypothesis, \(f\) is finite a.e. on \(E\), so \(A\) is contained in a set of measure zero; and \(\{f_n\} \to f\) pointwise a.e. on \(E\), so \(B\) is contained in a set of measure zero. Hence \(E_0\) is contained in the union of two sets of measure zero, which by the countable subadditivity and monotonicity of outer measure is a set of outer measure zero; therefore \(m^{*}(E_0) = 0\). Since every set of outer measure zero is measurable (the completeness of Lebesgue measure, established in Chapter 2), \(E_0\) is measurable and \(m(E_0) = 0\).

Passing to a good subdomain. Define

\begin{equation*} \widetilde{E} = E \sim E_0 . \end{equation*}

Then \(\widetilde{E}\) is measurable, being the difference of two measurable sets, and \(m(\widetilde{E}) \le m(E) < \infty\), so \(\widetilde{E}\) has finite measure. Each \(f_n\) restricted to \(\widetilde{E}\) is measurable: for every \(c \in \mathbb{R}\),

\begin{equation*} \{x \in \widetilde{E} \mid f_n(x) > c\} = \widetilde{E} \cap \{x \in E \mid f_n(x) > c\}, \end{equation*}

the intersection of two measurable sets. By the very definition of \(E_0\), on \(\widetilde{E}\) the function \(f\) is real-valued and \(\{f_n\}\) converges to \(f\) at every point.

Applying Egoroff’s Theorem. Thus \(\widetilde{E}\) has finite measure, \(\{f_n\}\) is a sequence of measurable functions on \(\widetilde{E}\), and it converges pointwise on \(\widetilde{E}\) to the real-valued function \(f\). These are exactly the hypotheses of Egoroff’s Theorem. Hence, given \(\epsilon > 0\), there is a closed set \(F\) contained in \(\widetilde{E}\) for which

\begin{equation*} \{f_n\} \to f \ \text{ uniformly on } F \qquad \text{and} \qquad m(\widetilde{E} \sim F) < \epsilon . \end{equation*}

Conclusion. The set \(F\) is closed and \(F \subseteq \widetilde{E} \subseteq E\), so \(F\) is a closed subset of \(E\), and \(\{f_n\}\) converges to \(f\) uniformly on \(F\). It remains to estimate \(m(E \sim F)\). Since \(F \subseteq \widetilde{E} = E \sim E_0\),

\begin{equation*} E \sim F = \big(E \sim \widetilde{E}\big) \cup \big(\widetilde{E} \sim F\big) = E_0 \cup \big(\widetilde{E} \sim F\big), \end{equation*}

so by the countable (here finite) subadditivity of measure,

\begin{equation*} m(E \sim F) \le m(E_0) + m(\widetilde{E} \sim F) = 0 + m(\widetilde{E} \sim F) < \epsilon . \end{equation*}

Therefore the conclusion of Egoroff’s Theorem holds verbatim under the weakened hypotheses of pointwise a.e. convergence to a function that is finite a.e. This is the assertion made in the remark immediately following the proof of Egoroff’s Theorem in the text.

We must show: if \(f\) is a real-valued measurable function on a measurable set \(E\) with \(m(E) = \infty\), then for each \(\epsilon > 0\) there is a continuous function \(g\) on \(\mathbb{R}\) and a closed set \(F\) contained in \(E\) for which \(f = g\) on \(F\) and \(m(E \sim F) < \epsilon\). The proof of Lusin’s Theorem given in the text establishes this when \(m(E) < \infty\); we bootstrap from that case.

Let \(\epsilon > 0\) be given. For each integer \(n \in \mathbb{Z}\) define \[ A_n = E \cap (n, n+1), \qquad Z = E \cap \mathbb{Z}. \] Each \(A_n\) is measurable, being the intersection of two measurable sets, and \(m(A_n) \le 1 < \infty\). The set \(Z\) is countable, so \(m(Z) = 0\). Since every real number lies either in \(\mathbb{Z}\) or in exactly one interval \((n, n+1)\), \[ E = Z \cup \bigcup_{n \in \mathbb{Z}} A_n, \] and the sets \(Z\) and \(\{A_n\}_{n \in \mathbb{Z}}\) are pairwise disjoint.

Fix a bijection \(k \mapsto n_k\) of the natural numbers onto \(\mathbb{Z}\). For each \(k\), the restriction of \(f\) to \(A_{n_k}\) is real-valued, and it is measurable on \(A_{n_k}\): for each \(c \in \mathbb{R}\), \[ \{x \in A_{n_k} \mid f(x) > c\} = \{x \in E \mid f(x) > c\} \cap A_{n_k} \] is the intersection of two measurable sets. Since \(m(A_{n_k}) < \infty\), Lusin’s Theorem in the finite measure case, which is the case proved in the text, may be applied to \(f\) on \(A_{n_k}\): there is a continuous function \(g_k\) on \(\mathbb{R}\) and a closed set \(F_{n_k}\) contained in \(A_{n_k}\) for which \[ f = g_k \text{ on } F_{n_k} \quad \text{and} \quad m(A_{n_k} \sim F_{n_k}) < \epsilon/2^{k+1}. \] Define \[ F = \bigcup_{n \in \mathbb{Z}} F_n \subseteq E. \]

We first record a separation property, to which we refer below as the localization identity. For each \(n\), \(F_n \subseteq A_n \subseteq (n, n+1)\), so the sets \(F_n\) are pairwise disjoint and, more importantly, \(F_m \cap (n, n+1) = \emptyset\) whenever \(m \neq n\); hence \[ F \cap (n, n+1) = F_n \qquad \text{for every } n \in \mathbb{Z}. \]

\(F\) is closed. Let \(x\) be a point of closure of \(F\) and choose a sequence \(\{x_j\}\) in \(F\) with \(x_j \to x\). A convergent sequence of real numbers is bounded, say \(|x_j| \le M\) for all \(j\). If \(x_j \in F_n\) then \(n < x_j < n + 1\), so \(-M - 1 \le n \le M\), and thus every \(x_j\) lies in one of the finitely many sets \(F_n\) with \(-M - 1 \le n \le M\). Therefore some single index \(n\) occurs infinitely often, that is, there is a subsequence lying entirely in \(F_n\). This subsequence also converges to \(x\), and \(F_n\) is closed, so \(x \in F_n \subseteq F\). Thus \(F\) contains all its points of closure and so is closed.

\(m(E \sim F) < \epsilon\). The localization identity gives \(A_n \sim F = A_n \sim F_n\) for each \(n\), and therefore, by the decomposition of \(E\) recorded above, \[ E \sim F \subseteq Z \cup \bigcup_{k=1}^{\infty} (A_{n_k} \sim F_{n_k}). \] Monotonicity and countable subadditivity of measure now yield \[ m(E \sim F) \le m(Z) + \sum_{k=1}^{\infty} m(A_{n_k} \sim F_{n_k}) \le 0 + \sum_{k=1}^{\infty} \frac{\epsilon}{2^{k+1}} = \frac{\epsilon}{2} < \epsilon. \]

The restriction of \(f\) to \(F\) is continuous. Since \(F\) is a subset of \(\mathbb{R}\), and hence a metric space, it suffices to verify sequential continuity. Let \(x \in F\), say \(x \in F_n\), and let \(\{x_j\}\) be a sequence in \(F\) converging to \(x\). Since \(x \in (n, n+1)\) and this interval is open, \(x_j \in (n, n+1)\) for all sufficiently large \(j\), and then \(x_j \in F \cap (n,n+1) = F_n\) by the localization identity. For such \(j\) we have \(f(x_j) = g_k(x_j)\) and \(f(x) = g_k(x)\), where \(n = n_k\); since \(g_k\) is continuous on \(\mathbb{R}\), \(f(x_j) = g_k(x_j) \to g_k(x) = f(x)\). Hence the restriction of \(f\) to \(F\) is continuous at each point of \(F\).

Finally, \(f\) restricted to \(F\) is a continuous function on the closed set \(F\), so by Problem 25 it has a continuous extension \(g\) to all of \(\mathbb{R}\). Then \(g\) is continuous on \(\mathbb{R}\), \(F\) is a closed subset of \(E\), \(f = g\) on \(F\), and \(m(E \sim F) < \epsilon\). This is exactly the conclusion of Lusin’s Theorem, now established without any assumption on \(m(E)\).

We must show: if \(f\) is an extended real-valued measurable function on a measurable set \(E\) which is finite a.e. on \(E\), then for each \(\epsilon > 0\) there is a continuous function \(g\) on \(\mathbb{R}\) and a closed set \(F\) contained in \(E\) for which \(f = g\) on \(F\) and \(m(E \sim F) < \epsilon\).

Let \(\epsilon > 0\) be given. Define \[ E_{\infty} = \{x \in E \mid f(x) = \infty\}, \qquad E_{-\infty} = \{x \in E \mid f(x) = -\infty\}, \qquad E_0 = E_{\infty} \cup E_{-\infty}. \] Since \(f\) is measurable, \[ E_{\infty} = \bigcap_{n=1}^{\infty} \{x \in E \mid f(x) > n\} \quad \text{and} \quad E_{-\infty} = \bigcap_{n=1}^{\infty} \{x \in E \mid f(x) < -n\} \] are measurable, being countable intersections of measurable sets; hence \(E_0\) is measurable. By hypothesis \(f\) is finite a.e. on \(E\), which says precisely that \(m(E_0) = 0\).

Put \(E^{\prime} = E \sim E_0\). Then \(E^{\prime}\) is measurable, and the restriction of \(f\) to \(E^{\prime}\) is a real-valued function which is measurable on \(E^{\prime}\): for each \(c \in \mathbb{R}\), \[ \{x \in E^{\prime} \mid f(x) > c\} = \{x \in E \mid f(x) > c\} \cap E^{\prime} \] is the intersection of two measurable sets.

Apply Lusin’s Theorem to the real-valued measurable function \(f\) on \(E^{\prime}\) – in the case \(m(E^{\prime}) < \infty\) this is the theorem as proved in the text, and in the case \(m(E^{\prime}) = \infty\) it is the extension established in Problem 29, so no restriction on \(m(E^{\prime})\) is needed. We obtain a continuous function \(g\) on \(\mathbb{R}\) and a closed set \(F\) contained in \(E^{\prime}\) for which \[ f = g \text{ on } F \quad \text{and} \quad m(E^{\prime} \sim F) < \epsilon. \]

This \(F\) and \(g\) do what is required for \(f\) on \(E\). Indeed, \(F \subseteq E^{\prime} \subseteq E\) and \(F\) is closed, \(g\) is continuous on all of \(\mathbb{R}\), and \(f = g\) on \(F\). Moreover, since \[ E \sim F = (E_0 \sim F) \cup (E^{\prime} \sim F) \subseteq E_0 \cup (E^{\prime} \sim F), \] the monotonicity and finite subadditivity of measure give \[ m(E \sim F) \le m(E_0) + m(E^{\prime} \sim F) = 0 + m(E^{\prime} \sim F) < \epsilon. \]

Thus Lusin’s Theorem persists when \(f\) is merely extended real-valued and finite a.e. Observe that the a.e. finiteness cannot be dispensed with in this form: on \(F\) we have \(f = g\) with \(g\) real-valued, so \(f\) is necessarily finite on \(F\), and therefore the set where \(f\) is infinite must be contained in \(E \sim F\), a set of measure less than \(\epsilon\) for every \(\epsilon > 0\); hence \(f\) must be finite a.e. for the conclusion to be possible at all.

Step 1: the case \(m(E) < \infty\). Assume first that \(E\) has finite measure. Then Egoroff’s Theorem applies to \(\{f_n\}\) on \(E\): for each integer \(k \ge 2\), taking \(\epsilon = 1/k\), there is a closed set \(A_k\) contained in \(E\) for which \[ \{f_n\} \to f \text{ uniformly on } A_k \quad \text{and} \quad m(E \sim A_k) < 1/k. \] Define \[ N = E \sim \bigcup_{k=2}^{\infty} A_k = \bigcap_{k=2}^{\infty} (E \sim A_k). \] This set is measurable, being a countable intersection of measurable sets. For each \(k \ge 2\), monotonicity of measure gives \(m(N) \le m(E \sim A_k) < 1/k\), and letting \(k \to \infty\) we conclude \(m(N) = 0\). By the very definition of \(N\), \[ E = N \cup \bigcup_{k=2}^{\infty} A_k, \] which is a decomposition of the required kind: \(N\) has measure zero and the convergence is uniform on each \(A_k\).

Step 2: the general case. Now let \(E\) be an arbitrary measurable set, of possibly infinite measure. Egoroff’s Theorem is not directly available, so we first cut \(E\) into countably many pieces of finite measure. Define \[ C_1 = E \cap [-1, 1], \qquad C_j = E \cap \{x \mid j - 1 < |x| \le j\} \ \text{ for } j \ge 2. \] Each \(C_j\) is measurable, being the intersection of \(E\) with a Borel set, and \(m(C_j) \le 2 < \infty\). Every real number \(x\) satisfies \(|x| \le 1\) or \(j-1 < |x| \le j\) for exactly one \(j \ge 2\), so \[ E = \bigcup_{j=1}^{\infty} C_j. \]

Fix \(j\). The restrictions of the \(f_n\) to \(C_j\) are measurable functions on \(C_j\), since for each \(c \in \mathbb{R}\) the set \(\{x \in C_j \mid f_n(x) > c\} = \{x \in E \mid f_n(x) > c\} \cap C_j\) is measurable; and \(\{f_n\} \to f\) pointwise on \(C_j\) with \(f\) real-valued. Since \(m(C_j) < \infty\), Step 1 applies on \(C_j\) and produces a measurable set \(N_j \subseteq C_j\) with \(m(N_j) = 0\), together with measurable sets \(A_{j,2}, A_{j,3}, \dots\) contained in \(C_j\) on each of which \(\{f_n\}\) converges uniformly to \(f\), such that \[ C_j = N_j \cup \bigcup_{k=2}^{\infty} A_{j,k}. \]

Now define \[ E_1 = \bigcup_{j=1}^{\infty} N_j. \] It is measurable as a countable union of measurable sets, and by countable subadditivity \[ m(E_1) \le \sum_{j=1}^{\infty} m(N_j) = 0, \] so \(m(E_1) = 0\).

The index set \(I = \{(j,k) \mid j \ge 1, \ k \ge 2\}\) is countably infinite, so we may fix a bijection \(i \mapsto (j(i), k(i))\) of \(\{2, 3, 4, \dots\}\) onto \(I\) and define \(E_i = A_{j(i), k(i)}\) for \(i \ge 2\). Each \(E_i\) with \(i > 1\) is measurable and \(\{f_n\}\) converges uniformly to \(f\) on it. (Some of these sets may be empty; uniform convergence on the empty set holds vacuously, so this is harmless.) Finally, \[ \bigcup_{k=1}^{\infty} E_k = \bigcup_{j=1}^{\infty} N_j \cup \bigcup_{j=1}^{\infty} \bigcup_{k=2}^{\infty} A_{j,k} = \bigcup_{j=1}^{\infty} \left( N_j \cup \bigcup_{k=2}^{\infty} A_{j,k} \right) = \bigcup_{j=1}^{\infty} C_j = E, \] and each \(E_k\) is contained in \(E\). This is the asserted decomposition.

Two remarks. First, when \(m(E) < \infty\) Step 1 alone suffices, and the sets \(E_k\) for \(k > 1\) may then be taken closed, since Egoroff’s Theorem as proved in the text delivers a closed set. In the general case each \(A_{j,k}\) is likewise closed, so the \(E_k\) with \(k > 1\) may again be taken closed. Second, no finite-measure hypothesis on \(E\) is needed for the conclusion, even though Egoroff’s Theorem itself requires one; the passage to the bounded pieces \(C_j\) is what removes that restriction. This is consistent with Problem 27, which shows that on a set of infinite measure one cannot in general find a single set \(A\) with \(m(E \sim A)\) small and uniform convergence on \(A\): the present decomposition instead uses countably many sets and pushes the exceptional behaviour into one set of measure zero.

Recall the construction. The Dirichlet function \(f\) on \([0,1]\) is defined by \(f(x) = 1\) if \(x\) is rational and \(f(x) = 0\) if \(x\) is irrational. Letting \(\{q_k\}_{k=1}^{\infty}\) be an enumeration of the rational numbers in \([0,1]\), for each natural number \(n\) the function \(f_n\) is defined on \([0,1]\) by

\begin{equation*} f_n(x) = \begin{cases} 1 & \text{if } x = q_k \text{ for some } k \text{ with } 1 \le k \le n,\\ 0 & \text{otherwise.}\end{cases} \end{equation*}

Each \(f_n\) is a step function, hence Riemann integrable: choose a partition \(P\) of \([0,1]\) whose partition points include all of \(q_1, \dots, q_n\); then \(f_n\) vanishes identically on each open subinterval determined by \(P\), which is exactly the requirement in the definition of a step function (the definition constrains the values only on the open subintervals). Moreover \(\{f_n\} \to f\) pointwise on \([0,1]\): if \(x\) is irrational then \(f_n(x) = 0 = f(x)\) for every \(n\), while if \(x\) is rational then \(x = q_K\) for exactly one index \(K\), and \(f_n(x) = 1 = f(x)\) for all \(n \ge K\).

We claim the convergence is not uniform. Indeed, we show that

\begin{equation*} \sup_{x \in [0,1]} |f(x) - f_n(x)| = 1 \quad \text{for every } n . \end{equation*}

Fix a natural number \(n\). The set \(F_n = \{q_1, \dots, q_n\}\) is finite, while the set \(\mathbb{Q} \cap [0,1]\) of rational numbers in \([0,1]\) is infinite (it contains the distinct points \(1/m\) for every natural number \(m\)). Therefore there is a rational number \(r \in [0,1]\) with \(r \notin F_n\). For this \(r\) we have \(f( r) = 1\), since \(r\) is rational, while \(f_n( r) = 0\), since \(r\) is not one of \(q_1, \dots, q_n\). Hence

\begin{equation*} |f( r) - f_n( r)| = 1 , \end{equation*}

so that \(\sup_{x \in [0,1]} |f(x) - f_n(x)| \ge 1\); since \(0 \le f_n \le f \le 1\) on \([0,1]\), the supremum equals \(1\).

Now suppose, to get a contradiction, that \(\{f_n\} \to f\) uniformly on \([0,1]\). Taking \(\epsilon = 1/2\) in the definition of uniform convergence, there would be an index \(N\) such that

\begin{equation*} |f(x) - f_n(x)| < \tfrac{1}{2} \quad \text{for all } x \in [0,1] \text{ and all } n \ge N . \end{equation*}

Taking \(n = N\) and passing to the supremum over \(x \in [0,1]\) gives \(1 \le 1/2\), a contradiction. Therefore \(\{f_n\}\) does not converge to \(f\) uniformly on \([0,1]\).

We remark that this failure is forced. As Problem 8 of this section shows, a sequence of Riemann integrable functions on a closed, bounded interval that converges uniformly has a Riemann integrable limit. Since each \(f_n\) is Riemann integrable over \([0,1]\) while \(f\) is not, as was verified in the example by exhibiting \(L(f,P) = 0\) and \(U(f,P) = 1\) for every partition \(P\) of \([0,1]\), the convergence could not possibly have been uniform. The point of the example is precisely that pointwise convergence of a uniformly bounded sequence of Riemann integrable functions is too weak for the Riemann theory, a shortcoming the Lebesgue integral repairs.

Throughout, for a partition \(P = \{x_0, x_1, \dots, x_n\}\) of \([a,b]\), with \(a = x_0 < x_1 < \cdots < x_n = b\), we write

\begin{equation*} m_i = \inf\{f(x) \mid x_{i-1} < x < x_i\}, \qquad M_i = \sup\{f(x) \mid x_{i-1} < x < x_i\}, \end{equation*}

\begin{equation*} L(f,P) = \sum_{i=1}^{n} m_i \cdot (x_i - x_{i-1}), \qquad U(f,P) = \sum_{i=1}^{n} M_i \cdot (x_i - x_{i-1}). \end{equation*}

Since \(f\) is bounded, all these infima and suprema are finite real numbers. We must show that if \(P^{\prime}\) refines \(P\), then

\begin{equation*} L(f,P) \le L(f,P^{\prime}) \quad \text{and} \quad U(f,P^{\prime}) \le U(f,P). \end{equation*}

The key monotonicity fact is elementary: if \(\emptyset \ne A \subseteq B\) and \(f\) is bounded, then

\begin{equation*} \inf_{B} f \le \inf_{A} f \quad \text{and} \quad \sup_{A} f \le \sup_{B} f , \end{equation*}

since \(\inf_B f\) is a lower bound for \(f\) on the smaller set \(A\), and \(\sup_B f\) is an upper bound for \(f\) on \(A\).

Step 1: adjoining a single point. Suppose \(P^{\prime} = P \cup \{c\}\) where \(c \notin P\); say \(x_{k-1} < c < x_k\). The partitions \(P\) and \(P^{\prime}\) have exactly the same subintervals except that the \(k\)-th subinterval \((x_{k-1}, x_k)\) of \(P\) is replaced in \(P^{\prime}\) by the two subintervals \((x_{k-1}, c)\) and \((c, x_k)\). Set

\begin{equation*} m^{\prime} = \inf\{f(x) \mid x_{k-1} < x < c\}, \qquad m^{\prime\prime} = \inf\{f(x) \mid c < x < x_k\}. \end{equation*}

All the terms of \(L(f,P^{\prime})\) and \(L(f,P)\) indexed by \(i \ne k\) agree, so

\begin{equation*} L(f,P^{\prime}) - L(f,P) = m^{\prime} \cdot (c - x_{k-1}) + m^{\prime\prime} \cdot (x_k - c) - m_k \cdot (x_k - x_{k-1}). \end{equation*}

Since \((x_{k-1}, c) \subseteq (x_{k-1}, x_k)\) and \((c, x_k) \subseteq (x_{k-1}, x_k)\), the monotonicity fact gives \(m^{\prime} \ge m_k\) and \(m^{\prime\prime} \ge m_k\). Hence

\begin{equation*} L(f,P^{\prime}) - L(f,P) \ge m_k (c - x_{k-1}) + m_k (x_k - c) - m_k (x_k - x_{k-1}) = 0 , \end{equation*}

because \((c - x_{k-1}) + (x_k - c) = x_k - x_{k-1}\). Thus \(L(f,P) \le L(f,P^{\prime})\).

Symmetrically, with \(M^{\prime} = \sup\{f(x) \mid x_{k-1} < x < c\}\) and \(M^{\prime\prime} = \sup\{f(x) \mid c < x < x_k\}\) we have \(M^{\prime} \le M_k\) and \(M^{\prime\prime} \le M_k\), so

\begin{equation*} U(f,P^{\prime}) - U(f,P) = M^{\prime} (c - x_{k-1}) + M^{\prime\prime} (x_k - c) - M_k (x_k - x_{k-1}) \le 0 , \end{equation*}

that is, \(U(f,P^{\prime}) \le U(f,P)\).

Step 2: the general case, by induction. Let \(P^{\prime}\) be any refinement of \(P\). Both are finite sets with \(P \subseteq P^{\prime}\), so \(P^{\prime} \setminus P\) is a finite set, say with \(p\) elements. We induct on \(p\). If \(p = 0\) then \(P^{\prime} = P\) and there is nothing to prove. Assume the assertion holds whenever the refinement adjoins \(p\) points, and let \(P^{\prime}\) refine \(P\) with \(P^{\prime} \setminus P\) having \(p+1\) elements. Choose \(c \in P^{\prime} \setminus P\) and put \(Q = P \cup \{c\}\). Then \(Q\) is a refinement of \(P\) adjoining a single point, and \(P^{\prime}\) is a refinement of \(Q\) with \(P^{\prime} \setminus Q\) having \(p\) elements. By Step 1 and the inductive hypothesis,

\begin{equation*} L(f,P) \le L(f,Q) \le L(f,P^{\prime}) \quad \text{and} \quad U(f,P^{\prime}) \le U(f,Q) \le U(f,P). \end{equation*}

This completes the induction, and hence the proof: under refinement lower Darboux sums increase and upper Darboux sums decrease.

Finally we note that the same argument works verbatim if, as in the footnote on page 69, the \(m_i\) and \(M_i\) are taken as infima and suprema over the closed subintervals \([x_{i-1}, x_i]\), since only the inclusion of the new subintervals in the old one was used, and this inclusion holds for the closed subintervals as well.

Let \(f\) be a bounded function on the closed, bounded interval \([a,b]\), and let \(P_1\) and \(P_2\) be any two partitions of \([a,b]\). We must show \(L(f,P_1) \le U(f,P_2)\).

First observe that for a single partition \(P = \{x_0, \dots, x_n\}\) we always have \(L(f,P) \le U(f,P)\). Indeed, for each \(i\) the numbers \(m_i = \inf\{f(x) \mid x_{i-1} < x < x_i\}\) and \(M_i = \sup\{f(x) \mid x_{i-1} < x < x_i\}\) satisfy \(m_i \le M_i\), since the interval \((x_{i-1}, x_i)\) is nonempty and any value \(f(x)\) taken there satisfies \(m_i \le f(x) \le M_i\). As each \(x_i - x_{i-1} > 0\), summing gives

\begin{equation*} L(f,P) = \sum_{i=1}^{n} m_i (x_i - x_{i-1}) \le \sum_{i=1}^{n} M_i (x_i - x_{i-1}) = U(f,P). \end{equation*}

Now let \(P = P_1 \cup P_2\). Since \(P\) is a finite subset of \([a,b]\) containing \(a\) and \(b\), it is a partition of \([a,b]\), and by construction every partition point of \(P_1\) and every partition point of \(P_2\) belongs to \(P\). Thus \(P\) is a common refinement: it refines \(P_1\) and it refines \(P_2\). By the preceding problem, lower Darboux sums increase and upper Darboux sums decrease under refinement, so

\begin{equation*} L(f, P_1) \le L(f, P) \quad \text{and} \quad U(f, P) \le U(f, P_2). \end{equation*}

Combining these with \(L(f,P) \le U(f,P)\) yields

\begin{equation*} L(f, P_1) \le L(f, P) \le U(f, P) \le U(f, P_2). \end{equation*}

Hence each lower Darboux sum is no greater than each upper Darboux sum.

Now we conclude the comparison of the integrals. Fix a partition \(P_2\) of \([a,b]\). By what we just proved, the number \(U(f,P_2)\) is an upper bound for the set

\begin{equation*} \{L(f,P_1) \mid P_1 \text{ a partition of } [a,b]\}. \end{equation*}

This set is nonempty and bounded above, and its supremum is by definition the lower Riemann integral. Since the supremum is the least upper bound, we get

\begin{equation*} ( R)\underline{\int_a^b} f = \sup\{L(f,P_1) \mid P_1 \text{ a partition of } [a,b]\} \le U(f, P_2). \end{equation*}

As \(P_2\) was an arbitrary partition of \([a,b]\), the number \(( R)\underline{\int_a^b} f\) is a lower bound for the set

\begin{equation*} \{U(f,P_2) \mid P_2 \text{ a partition of } [a,b]\}, \end{equation*}

and since the infimum is the greatest lower bound,

\begin{equation*} ( R)\underline{\int_a^b} f \le \inf\{U(f,P_2) \mid P_2 \text{ a partition of } [a,b]\} = ( R)\overline{\int_a^b} f . \end{equation*}

Thus the lower Riemann integral is no greater than the upper Riemann integral. This justifies the assertion made in the text that the upper integral is always at least as large as the lower integral, so that Riemann integrability is exactly the assertion that the two coincide.

Since \(f\) is Riemann integrable over \([a,b]\), the lower and upper Riemann integrals are equal; call their common value

\begin{equation*} I = ( R)\underline{\int_a^b} f = ( R)\overline{\int_a^b} f = ( R)\int_a^b f . \end{equation*}

Since \(f\) is bounded, \(I\) is a real number.

Fix a natural number \(n\). By definition,

\begin{equation*} I = \sup\{L(f,P) \mid P \text{ a partition of } [a,b]\}, \end{equation*}

so \(I - 1/(2n)\) is not an upper bound for the set of lower Darboux sums; hence there is a partition \(P_n^{\prime}\) of \([a,b]\) with

\begin{equation*} L(f, P_n^{\prime}) > I - \frac{1}{2n}. \end{equation*}

Likewise,

\begin{equation*} I = \inf\{U(f,P) \mid P \text{ a partition of } [a,b]\}, \end{equation*}

so \(I + 1/(2n)\) is not a lower bound for the set of upper Darboux sums; hence there is a partition \(P_n^{\prime\prime}\) of \([a,b]\) with

\begin{equation*} U(f, P_n^{\prime\prime}) < I + \frac{1}{2n}. \end{equation*}

Define \(P_n = P_n^{\prime} \cup P_n^{\prime\prime}\). This is a finite subset of \([a,b]\) containing \(a\) and \(b\), hence a partition of \([a,b]\), and it is a refinement of both \(P_n^{\prime}\) and \(P_n^{\prime\prime}\). By Problem 2, refinement increases lower Darboux sums and decreases upper Darboux sums, so

\begin{equation*} L(f, P_n) \ge L(f, P_n^{\prime}) > I - \frac{1}{2n}, \qquad U(f, P_n) \le U(f, P_n^{\prime\prime}) < I + \frac{1}{2n}. \end{equation*}

Therefore

\begin{equation*} 0 \le U(f, P_n) - L(f, P_n) < \left( I + \frac{1}{2n} \right) - \left( I - \frac{1}{2n} \right) = \frac{1}{n}, \end{equation*}

where the first inequality is the observation, proved in Problem 3, that \(L(f,P) \le U(f,P)\) for every partition \(P\).

Since \(1/n \to 0\), the squeeze principle gives

\begin{equation*} \lim_{n \to \infty} \left[ U(f, P_n) - L(f, P_n) \right] = 0 , \end{equation*}

so \(\{P_n\}\) is a sequence of partitions of the required kind.

We add one remark that will be useful in the following problem: the sequence just constructed also satisfies \(L(f,P_n) \to I\) and \(U(f,P_n) \to I\), since both lie in the interval \(\left( I - 1/(2n), I + 1/(2n) \right)\). Thus for a Riemann integrable function the Riemann integral is the limit of the Darboux sums along this sequence of partitions.

Since \(f\) is bounded on the closed, bounded interval \([a,b]\), both the lower Riemann integral

\begin{equation*} ( R)\underline{\int_a^b} f = \sup\{L(f,P) \mid P \text{ a partition of } [a,b]\} \end{equation*}

and the upper Riemann integral

\begin{equation*} ( R)\overline{\int_a^b} f = \inf\{U(f,P) \mid P \text{ a partition of } [a,b]\} \end{equation*}

are real numbers: if \(|f| \le M\) on \([a,b]\), then every Darboux sum lies in \([-M(b-a), M(b-a)]\), so the relevant sets are nonempty and bounded. By Problem 3,

\begin{equation*} ( R)\underline{\int_a^b} f \le ( R)\overline{\int_a^b} f . \end{equation*}

Fix a natural number \(n\). Since \(( R)\underline{\int_a^b} f\) is the supremum of all lower Darboux sums, it dominates the particular lower sum \(L(f, P_n)\); since \(( R)\overline{\int_a^b} f\) is the infimum of all upper Darboux sums, it is dominated by the particular upper sum \(U(f, P_n)\). Combining these with the inequality above gives the chain

\begin{equation*} L(f, P_n) \le ( R)\underline{\int_a^b} f \le ( R)\overline{\int_a^b} f \le U(f, P_n). \end{equation*}

Consequently

\begin{equation*} 0 \le ( R)\overline{\int_a^b} f - ( R)\underline{\int_a^b} f \le U(f, P_n) - L(f, P_n) \quad \text{for every } n . \end{equation*}

The left-hand quantity does not depend on \(n\), and by hypothesis the right-hand quantity tends to \(0\) as \(n \to \infty\). Therefore, letting \(n \to \infty\),

\begin{equation*} 0 \le ( R)\overline{\int_a^b} f - ( R)\underline{\int_a^b} f \le 0 , \end{equation*}

so that

\begin{equation*} ( R)\underline{\int_a^b} f = ( R)\overline{\int_a^b} f . \end{equation*}

Since \(f\) is bounded on \([a,b]\) and its lower and upper Riemann integrals coincide, \(f\) is by definition Riemann integrable over \([a,b]\), and its Riemann integral is this common value. Indeed, the displayed chain of inequalities shows moreover that

\begin{equation*} ( R)\int_a^b f = \lim_{n \to \infty} L(f, P_n) = \lim_{n \to \infty} U(f, P_n), \end{equation*}

since both \(L(f,P_n)\) and \(U(f,P_n)\) are squeezed toward the common value by the same estimate.

Together with the preceding problem this gives the Cauchy criterion for Riemann integrability: a bounded function \(f\) on \([a,b]\) is Riemann integrable over \([a,b]\) if and only if there is a sequence \(\{P_n\}\) of partitions with \(U(f,P_n) - L(f,P_n) \to 0\); equivalently, if and only if for each \(\epsilon > 0\) there is a partition \(P\) with \(U(f,P) - L(f,P) < \epsilon\).

Let \(f\) be continuous on the closed, bounded interval \([a,b]\). We may assume \(a < b\), the case \(a = b\) being trivial.

First, \(f\) is bounded on \([a,b]\). Indeed, \([a,b]\) is a nonempty closed, bounded set of real numbers, so by the Extreme Value Theorem of Chapter 1 the continuous function \(f\) takes a minimum and a maximum value on \([a,b]\); in particular \(f\) is bounded there. Thus the lower and upper Darboux sums and the lower and upper Riemann integrals of \(f\) over \([a,b]\) are defined, and the hypothesis of boundedness required in Problem 5 is met.

Second, \(f\) is uniformly continuous on \([a,b]\): by Theorem 23 of Chapter 1, a continuous real-valued function on a closed, bounded set of real numbers is uniformly continuous. Written out, for each \(\epsilon > 0\) there is a \(\delta > 0\) such that

\begin{equation*} |f(u) - f(v)| < \epsilon \quad \text{whenever } u, v \in [a,b] \text{ and } |u - v| < \delta . \end{equation*}

Now we construct the required sequence of partitions. Fix a natural number \(n\). Apply uniform continuity with \(\epsilon = 1/n\) to obtain \(\delta_n > 0\) such that

\begin{equation*} |f(u) - f(v)| < \frac{1}{n} \quad \text{whenever } u, v \in [a,b] \text{ and } |u - v| < \delta_n . \end{equation*}

Choose a natural number \(k_n\) so large that \((b-a)/k_n < \delta_n\), which is possible by the Archimedean property of \(\mathbb{R}\), and let \(P_n\) be the regular partition of \([a,b]\) into \(k_n\) subintervals of equal length \((b-a)/k_n\), that is,

\begin{equation*} P_n = \left\{ x_i = a + i \cdot \frac{b-a}{k_n} \ \middle|\ 0 \le i \le k_n \right\}. \end{equation*}

Fix \(i\) with \(1 \le i \le k_n\) and set, as usual,

\begin{equation*} m_i = \inf\{f(x) \mid x_{i-1} < x < x_i\}, \qquad M_i = \sup\{f(x) \mid x_{i-1} < x < x_i\}. \end{equation*}

For any two points \(u, v\) in \((x_{i-1}, x_i)\) we have \(|u - v| < x_i - x_{i-1} = (b-a)/k_n < \delta_n\), hence \(f(u) - f(v) < 1/n\). Taking the supremum over \(u\) and then the infimum over \(v\) in this inequality gives

\begin{equation*} M_i - m_i \le \frac{1}{n} . \end{equation*}

(In detail: for fixed \(v\), \(f(u) < f(v) + 1/n\) for all \(u \in (x_{i-1},x_i)\), so \(M_i \le f(v) + 1/n\); thus \(M_i - 1/n\) is a lower bound for \(f\) on \((x_{i-1},x_i)\), so \(M_i - 1/n \le m_i\).)

Therefore

\begin{equation*} U(f, P_n) - L(f, P_n) = \sum_{i=1}^{k_n} (M_i - m_i)(x_i - x_{i-1}) \le \frac{1}{n} \sum_{i=1}^{k_n} (x_i - x_{i-1}) = \frac{b-a}{n}. \end{equation*}

Since \((b-a)/n \to 0\) as \(n \to \infty\), and \(U(f,P_n) - L(f,P_n) \ge 0\) for every \(n\) by Problem 3, we conclude

\begin{equation*} \lim_{n \to \infty} \left[ U(f, P_n) - L(f, P_n) \right] = 0 . \end{equation*}

As \(f\) is bounded on \([a,b]\) and \(\{P_n\}\) is a sequence of partitions of \([a,b]\) with this property, Problem 5 tells us that \(f\) is Riemann integrable over \([a,b]\).

Let \(f\) be increasing on \([0,1]\), meaning \(f(u) \le f(v)\) whenever \(0 \le u \le v \le 1\).

First, \(f\) is bounded on \([0,1]\): for every \(x \in [0,1]\) we have \(0 \le x \le 1\), hence

\begin{equation*} f(0) \le f(x) \le f(1). \end{equation*}

So the Darboux sums and the lower and upper Riemann integrals of \(f\) over \([0,1]\) are defined. (Note that no continuity is assumed; an increasing function may have infinitely many jump discontinuities.)

Fix a natural number \(n\) and let \(P_n = \{x_0, x_1, \dots, x_n\}\) with \(x_i = i/n\), so that each subinterval has length \(x_i - x_{i-1} = 1/n\). Set

\begin{equation*} m_i = \inf\{f(x) \mid x_{i-1} < x < x_i\}, \qquad M_i = \sup\{f(x) \mid x_{i-1} < x < x_i\}. \end{equation*}

We claim that

\begin{equation*} f(x_{i-1}) \le m_i \le M_i \le f(x_i) \qquad \text{for } 1 \le i \le n . \end{equation*}

Indeed, if \(x_{i-1} < x < x_i\), then since \(f\) is increasing, \(f(x_{i-1}) \le f(x) \le f(x_i)\). Thus \(f(x_{i-1})\) is a lower bound for \(f\) on \((x_{i-1}, x_i)\) and \(f(x_i)\) is an upper bound, so \(f(x_{i-1}) \le m_i\) and \(M_i \le f(x_i)\); and \(m_i \le M_i\) because the interval \((x_{i-1},x_i)\) is nonempty. In particular

\begin{equation*} 0 \le M_i - m_i \le f(x_i) - f(x_{i-1}). \end{equation*}

Therefore

\begin{equation*} \begin{aligned} U(f, P_n) - L(f, P_n) &= \sum_{i=1}^{n} (M_i - m_i)(x_i - x_{i-1}) \\ &= \frac{1}{n} \sum_{i=1}^{n} (M_i - m_i) \\ &\le \frac{1}{n} \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right] \\ &= \frac{1}{n} \left[ f(x_n) - f(x_0) \right] \\ &= \frac{1}{n} \left[ f(1) - f(0) \right], \end{aligned} \end{equation*}

the fourth equality holding because the sum telescopes: all the intermediate values \(f(x_1), \dots, f(x_{n-1})\) cancel in pairs. This is the asserted estimate.

Finally, the number \(f(1) - f(0)\) is a fixed nonnegative real number independent of \(n\), so

\begin{equation*} 0 \le U(f, P_n) - L(f, P_n) \le \frac{f(1) - f(0)}{n} \longrightarrow 0 \quad \text{as } n \to \infty , \end{equation*}

the left-hand inequality being the fact, established in Problem 3, that \(L(f,P) \le U(f,P)\) for every partition \(P\). By the squeeze principle,

\begin{equation*} \lim_{n \to \infty} \left[ U(f, P_n) - L(f, P_n) \right] = 0 . \end{equation*}

Since \(f\) is a bounded function on \([0,1]\) and \(\{P_n\}\) is a sequence of partitions of \([0,1]\) with this property, Problem 5 shows that \(f\) is Riemann integrable over \([0,1]\).

The same argument applies on any closed, bounded interval \([a,b]\), with \(P_n\) the regular partition into \(n\) subintervals of length \((b-a)/n\), yielding \(U(f,P_n) - L(f,P_n) \le \frac{b-a}{n}\left[ f(b) - f(a) \right]\); hence every monotone function on a closed, bounded interval is Riemann integrable there.

Throughout, for a bounded function \(g\) on \([a,b]\) and a partition \(P=\{a=x_0<x_1<\cdots<x_k=b\}\) we write

\begin{equation*} U(g,P)=\sum_{i=1}^k M_i(g)\,(x_i-x_{i-1}),\qquad L(g,P)=\sum_{i=1}^k m_i(g)\,(x_i-x_{i-1}), \end{equation*}

where \(M_i(g)=\sup\{g(x): x_{i-1}\le x\le x_i\}\) and \(m_i(g)=\inf\{g(x): x_{i-1}\le x\le x_i\}\).

Step 1: \(f\) is bounded. Put \(\eta_n=\sup_{x\in[a,b]}|f_n(x)-f(x)|\), a number in \([0,\infty]\). Uniform convergence of \(\{f_n\}\) to \(f\) on \([a,b]\) says precisely that for each \(\epsilon>0\) there is an index \(N\) with \(|f_n-f|<\epsilon\) on all of \([a,b]\) for \(n\ge N\), that is, \(\eta_n\le\epsilon\) for \(n\ge N\); so \(\eta_n\) is finite for all large \(n\) and \(\eta_n\to 0\). In particular there is an index \(N\) with \(\eta_N\le 1\). Since \(f_N\) is bounded, say \(|f_N|\le M\) on \([a,b]\), we get \(|f|\le |f_N|+|f-f_N|\le M+1\) on \([a,b]\). So \(f\) is a bounded function on \([a,b]\) and its Darboux sums are defined.

Step 2: a comparison of Darboux sums. Let \(g\) and \(h\) be bounded on \([a,b]\) and suppose \(|g-h|\le \eta\) on \([a,b]\). Then \(g\le h+\eta\) on \([a,b]\), so for every subinterval \([x_{i-1},x_i]\) of a partition \(P\) we have \(M_i(g)\le M_i(h)+\eta\), and likewise \(m_i(g)\ge m_i(h)-\eta\). Multiplying by the lengths \(x_i-x_{i-1}\) and summing, and using \(\sum_{i=1}^k (x_i-x_{i-1})=b-a\),

\begin{equation*} U(g,P)\le U(h,P)+\eta\,(b-a),\qquad L(g,P)\ge L(h,P)-\eta\,(b-a). \tag{\(\ast\)} \end{equation*}

Step 3: \(f\) is Riemann integrable. We may assume \(a<b\), since for \(a=b\) every bounded function is trivially Riemann integrable with integral \(0\). Let \(k\) be a natural number. Since \(\eta_n\to 0\), choose an index \(n_k\) for which

\begin{equation*} \eta_{n_k}=\sup_{x\in[a,b]}|f_{n_k}(x)-f(x)|<\frac{1}{4k(b-a)}. \end{equation*}

The function \(f_{n_k}\) is Riemann integrable over \([a,b]\), so by Problem 4 of this chapter there is a sequence of partitions of \([a,b]\) along which the difference of the upper and lower Darboux sums of \(f_{n_k}\) tends to \(0\); choose from it a partition \(P_k\) of \([a,b]\) for which

\begin{equation*} U(f_{n_k},P_k)-L(f_{n_k},P_k)<\frac{1}{2k}. \end{equation*}

By \((\ast)\), applied with \(g=f\), \(h=f_{n_k}\) and \(\eta=1/[4k(b-a)]\),

\begin{equation*} U(f,P_k)-L(f,P_k)\le \bigl[U(f_{n_k},P_k)-L(f_{n_k},P_k)\bigr]+2\cdot\frac{1}{4k(b-a)}\cdot (b-a)<\frac{1}{2k}+\frac{1}{2k}=\frac1k . \end{equation*}

Thus \(\{P_k\}\) is a sequence of partitions of \([a,b]\) with \(\lim_{k\to\infty}\bigl[U(f,P_k)-L(f,P_k)\bigr]=0\), and therefore \(f\) is Riemann integrable over \([a,b]\) by Problem 5.

Step 4: the integrals converge. Yes, it is true that \(\lim_{n\to\infty}\int_a^b f_n=\int_a^b f\). Indeed, fix an index \(n\) large enough that \(\eta_n<\infty\) (by Step 1 all but finitely many indices are such) and apply \((\ast)\) with \(g=f\), \(h=f_n\), \(\eta=\eta_n\): for every partition \(P\),

\begin{equation*} U(f,P)\le U(f_n,P)+\eta_n (b-a). \end{equation*}

Taking the infimum over all partitions \(P\) of \([a,b]\) on both sides gives the corresponding inequality for the upper Riemann integrals,

\begin{equation*} ( R)\overline{\int_a^b} f\le ( R)\overline{\int_a^b} f_n+\eta_n(b-a). \end{equation*}

Both \(f\) and \(f_n\) are Riemann integrable, so their upper Riemann integrals equal their integrals, and hence

\begin{equation*} \int_a^b f\le \int_a^b f_n+\eta_n (b-a). \end{equation*}

Interchanging the roles of \(f\) and \(f_n\) (the hypothesis \(|f-f_n|\le\eta_n\) is symmetric) gives \(\int_a^b f_n\le \int_a^b f+\eta_n(b-a)\). Together,

\begin{equation*} \left|\int_a^b f_n-\int_a^b f\right|\le \eta_n\,(b-a). \end{equation*}

Since \(\eta_n\to 0\) by uniform convergence, the right-hand side tends to \(0\), and therefore \(\lim_{n\to\infty}\int_a^b f_n=\int_a^b f\).

Note that uniform convergence is essential for both conclusions: the Dirichlet function example of this section exhibits a uniformly bounded, increasing sequence of Riemann integrable functions on \([0,1]\) converging pointwise to a function that is not Riemann integrable.

\(f\) is measurable. First note that \(E\) itself is measurable: it has outer measure zero, so Proposition 4 of Chapter 2 applies. Hence it makes sense to ask whether a function on \(E\) is measurable and to form \(\int_E\). Now let \(c\) be a real number and set

\begin{equation*} A_c=\{x\in E \mid f(x)>c\}. \end{equation*}

Then \(A_c\subseteq E\), so by the monotonicity of Lebesgue outer measure, \(m^*(A_c)\le m^*(E)=m(E)=0\), that is, \(A_c\) has outer measure zero. By Proposition 4 of Chapter 2, a set of outer measure zero is measurable. Hence \(A_c\) is measurable for every \(c\), and therefore \(f\) is a measurable function on \(E\). (Note that no hypothesis beyond \(m(E)=0\) is needed here: every real-valued function on a set of measure zero is measurable. Boundedness is only used below, to make the integral of a bounded function over a set of finite measure meaningful.)

\(\int_E f=0\). Since \(m(E)=0<\infty\), \(E\) is a set of finite measure, and \(f\) is a bounded measurable function on \(E\); so by Theorem 4, \(f\) is integrable over \(E\) and \(\int_E f\) is the common value of the upper and lower Lebesgue integrals of \(f\) over \(E\).

We claim that every simple function \(\varphi\) on \(E\) has \(\int_E\varphi=0\). Indeed, let \(\varphi=\sum_{i=1}^n a_i\cdot\chi_{E_i}\) be the canonical representation of \(\varphi\) on \(E\), so that the \(E_i\) are measurable subsets of \(E\). By monotonicity of measure, \(m(E_i)\le m(E)=0\) for each \(i\), so \(m(E_i)=0\), and by the definition of the integral of a simple function,

\begin{equation*} \int_E \varphi=\sum_{i=1}^n a_i\cdot m(E_i)=\sum_{i=1}^n a_i\cdot 0=0 . \end{equation*}

Now, \(f\) is bounded, say \(|f|\le M\) on \(E\). The constant function \(\varphi\equiv -M\) is simple with \(\varphi\le f\) on \(E\), and the constant function \(\psi\equiv M\) is simple with \(f\le\psi\) on \(E\); thus the families of simple functions occurring in the definitions of the lower and upper Lebesgue integrals are nonempty. By the claim, every simple \(\varphi\le f\) has \(\int_E\varphi=0\) and every simple \(\psi\ge f\) has \(\int_E\psi=0\). Therefore the lower Lebesgue integral of \(f\) over \(E\) equals

\begin{equation*} \sup\left\{\int_E\varphi \;\Big|\; \varphi \text{ simple},\ \varphi\le f \text{ on } E\right\}=\sup\{0\}=0, \end{equation*}

and likewise the upper Lebesgue integral equals \(\inf\{0\}=0\). The two agree, confirming integrability, and

\begin{equation*} \int_E f=0 . \end{equation*}

First observe that both integrals are defined. The restriction of \(f\) to \(A\) is bounded and measurable on \(A\), and \(m(A)\le m(E)<\infty\) by monotonicity of measure, so \(\int_A f\) is defined by Theorem 4. Likewise \(f\cdot\chi_A\) is bounded on \(E\) (by the same bound as \(f\)) and measurable on \(E\), being the product of two measurable functions, so \(\int_E f\cdot\chi_A\) is defined by Theorem 4. In particular the upper and lower Lebesgue integrals of \(f\cdot\chi_A\) over \(E\) both equal \(\int_E f\cdot\chi_A\).

If \(A\) is empty both sides are zero, since \(f\cdot\chi_A\) is then the zero function on \(E\), whose integral over \(E\) is \(0\) by Lemma 1, and \(\int_\emptyset f=0\); so assume \(A\) is nonempty.

A device for transporting simple functions. Let \(\sigma\) be a simple function on \(A\), with canonical representation \(\sigma=\sum_{i=1}^n a_i\cdot\chi_{A_i}\) on \(A\), where the \(A_i=\sigma^{-1}(a_i)\) are disjoint measurable subsets of \(A\) whose union is \(A\); they are measurable subsets of \(E\) as well. Define \(\tilde\sigma\) on \(E\) by

\begin{equation*} \tilde\sigma=\sigma \ \text{ on } A,\qquad \tilde\sigma=0 \ \text{ on } E\sim A . \end{equation*}

Then \(\tilde\sigma\) takes only finitely many values and is measurable on \(E\), so it is simple on \(E\), and

\begin{equation*} \tilde\sigma=\sum_{i=1}^n a_i\cdot\chi_{A_i}+0\cdot\chi_{E\sim A}\quad\text{on }E, \end{equation*}

a representation over the finite disjoint collection \(\{A_1,\dots,A_n,E\sim A\}\) of measurable subsets of \(E\). By Lemma 1,

\begin{equation*} \int_E\tilde\sigma=\sum_{i=1}^n a_i\cdot m(A_i)+0\cdot m(E\sim A)=\sum_{i=1}^n a_i\cdot m(A_i)=\int_A\sigma . \tag{\(\dagger\)} \end{equation*}

The inequality \(\int_E f\cdot\chi_A\le\int_A f\). Let \(\psi\) be a simple function on \(A\) with \(f\le\psi\) on \(A\). On \(A\) we have \(f\cdot\chi_A=f\le\psi=\tilde\psi\), while on \(E\sim A\) we have \(f\cdot\chi_A=0=\tilde\psi\). Hence \(f\cdot\chi_A\le\tilde\psi\) on \(E\), and since \(\int_E f\cdot\chi_A\) equals the upper Lebesgue integral of \(f\cdot\chi_A\) over \(E\), which is an infimum over all such competitors, \((\dagger)\) gives

\begin{equation*} \int_E f\cdot\chi_A\le\int_E\tilde\psi=\int_A\psi . \end{equation*}

Taking the infimum over all simple \(\psi\) on \(A\) with \(\psi\ge f\), and using that this infimum is the upper Lebesgue integral of \(f\) over \(A\), which equals \(\int_A f\), we obtain \(\int_E f\cdot\chi_A\le\int_A f\).

The inequality \(\int_E f\cdot\chi_A\ge\int_A f\). Let \(\varphi\) be a simple function on \(A\) with \(\varphi\le f\) on \(A\). On \(A\) we have \(\tilde\varphi=\varphi\le f=f\cdot\chi_A\), and on \(E\sim A\) we have \(\tilde\varphi=0=f\cdot\chi_A\). Hence \(\tilde\varphi\le f\cdot\chi_A\) on \(E\), so, \(\int_E f\cdot\chi_A\) being also the lower Lebesgue integral of \(f\cdot\chi_A\) over \(E\),

\begin{equation*} \int_A\varphi=\int_E\tilde\varphi\le\int_E f\cdot\chi_A . \end{equation*}

Taking the supremum over all simple \(\varphi\le f\) on \(A\) gives \(\int_A f\le \int_E f\cdot\chi_A\).

Combining the two inequalities,

\begin{equation*} \int_A f=\int_E f\cdot\chi_A . \end{equation*}

No. The theorem fails for the Riemann integral, and it fails in the strongest possible way: the hypotheses can hold while the Riemann integral of the limit function does not even exist, so the asserted conclusion is meaningless.

The counterexample. This is exactly the Dirichlet function example of Section 4.1. The set of rational numbers in \([0,1]\) is countable; let \(\{q_k\}_{k=1}^\infty\) be an enumeration of it. For each natural number \(n\) define \(f_n\) on \([0,1]\) by

\begin{equation*} f_n(x)=\begin{cases} 1 & \text{if } x=q_k \text{ for some } k \text{ with } 1\le k\le n,\\[2pt] 0 & \text{otherwise.}\end{cases} \end{equation*}

Each \(f_n\) is a step function, hence Riemann integrable over \([0,1]\); indeed \(f_n\) differs from the zero function at only finitely many points, so for every \(\epsilon>0\) one can cover those \(n\) points by finitely many intervals of total length less than \(\epsilon\) and thereby produce a partition \(P\) with \(U(f_n,P)-L(f_n,P)<\epsilon\). Consequently

\begin{equation*} ( R)\int_0^1 f_n=0\quad\text{for every }n . \end{equation*}

Moreover the sequence is uniformly bounded, \(|f_n|\le 1\) on \([0,1]\) for all \(n\), and it is increasing, so it converges pointwise on \([0,1]\) to the Dirichlet function

\begin{equation*} f(x)=\begin{cases} 1 & \text{if } x \text{ is rational},\\ 0 & \text{if } x \text{ is irrational.}\end{cases} \end{equation*}

Thus every hypothesis of the Bounded Convergence Theorem is satisfied, with \(E=[0,1]\), \(M=1\), and each \(f_n\) Riemann integrable. But, as shown in Section 4.1, by the density of the rationals and of the irrationals every partition \(P\) of \([0,1]\) has \(L(f,P)=0\) and \(U(f,P)=1\), so

\begin{equation*} ( R)\underline{\int_0^1} f=0<1=( R)\overline{\int_0^1} f , \end{equation*}

and \(f\) is not Riemann integrable over \([0,1]\). Hence \(\lim_{n\to\infty}( R)\int_0^1 f_n\) exists (it is \(0\)) but there is no number \(( R)\int_0^1 f\) for it to equal. The Bounded Convergence Theorem therefore does not hold for the Riemann integral.

A remark on what does survive. If one adds to the hypotheses the assumption that the limit function \(f\) is itself Riemann integrable, then the conclusion is true; but this is a theorem about the Lebesgue integral in disguise. Suppose each \(f_n\), and also \(f\), is Riemann integrable over the closed, bounded interval \([a,b]\), that \(|f_n|\le M\) on \([a,b]\) for all \(n\), and that \(\{f_n\}\to f\) pointwise on \([a,b]\).

To invoke the Bounded Convergence Theorem we need the \(f_n\) to be measurable. That a bounded Riemann integrable function is measurable is the converse of Theorem 4, which the text states but defers to Theorem 7 of Chapter 5; for a closed, bounded interval it can be proved with the tools at hand, as follows. Let \(g\) be Riemann integrable over \([a,b]\). By the reformulation of the upper and lower Riemann integrals in terms of step functions used in the proof of Theorem 3, for each natural number \(n\) there are step functions \(\varphi_n\le g\le\psi_n\) on \([a,b]\) with \(( R)\int_a^b\varphi_n>( R)\int_a^b g-1/(2n)\) and \(( R)\int_a^b\psi_n<( R)\int_a^b g+1/(2n)\), so that \(( R)\int_a^b(\psi_n-\varphi_n)<1/n\). Replacing \(\varphi_n\) by \(\max\{\varphi_1,\dots,\varphi_n\}\) and \(\psi_n\) by \(\min\{\psi_1,\dots,\psi_n\}\) leaves step functions (a maximum or a minimum of finitely many step functions is again a step function, as one sees by passing to a common refinement of their partitions) that still satisfy \(\varphi_n\le g\le\psi_n\) and, since the new \(\psi_n-\varphi_n\) is no larger than the old one, still satisfy the estimate; so we may assume \(\{\varphi_n\}\) is increasing and \(\{\psi_n\}\) is decreasing. Then \(\{\varphi_n\}\) is increasing and bounded above by the bounded function \(g\), and \(\{\psi_n\}\) is decreasing and bounded below by \(g\), so both converge pointwise on \([a,b]\), say to \(\varphi_*\) and \(\psi_*\). Step functions are simple, hence measurable, so \(\varphi_*\) and \(\psi_*\) are measurable by Proposition 9 of Chapter 3, and \(\varphi_*\le g\le\psi_*\) on \([a,b]\). Put \(h=\psi_*-\varphi_*\), a nonnegative bounded measurable function with \(h\le\psi_n-\varphi_n\) for every \(n\). Since the Lebesgue and Riemann integrals of a step function agree, the monotonicity and linearity of integration, Theorem 5, give

\begin{equation*} 0\le\int_{[a,b]}h\le\int_{[a,b]}(\psi_n-\varphi_n)=( R)\int_a^b(\psi_n-\varphi_n)<\frac1n\quad\text{for every }n, \end{equation*}

so \(\int_{[a,b]}h=0\). For a natural number \(k\) let \(E_k=\{x\in[a,b]\mid h(x)>1/k\}\), a measurable set. Since \((1/k)\cdot\chi_{E_k}\le h\) on \([a,b]\), Lemma 1 and monotonicity give \((1/k)\cdot m(E_k)\le\int_{[a,b]}h=0\), so \(m(E_k)=0\); by countable subadditivity of measure, \(\{x\mid h(x)>0\}=\bigcup_{k=1}^\infty E_k\) has measure zero. Hence \(\psi_*=\varphi_*\) a.e. on \([a,b]\), and since \(\varphi_*\le g\le\psi_*\) we get \(g=\varphi_*\) a.e. on \([a,b]\). By Proposition 5 of Chapter 3, \(g\) is measurable.

Applying this to each \(f_n\), the sequence \(\{f_n\}\) consists of measurable functions on \([a,b]\), a set of finite measure \(b-a\), is uniformly bounded by \(M\), and converges pointwise to \(f\). The Bounded Convergence Theorem therefore gives \(\lim_{n\to\infty}\int_{[a,b]}f_n=\int_{[a,b]}f\), and by Theorem 3 each of these Lebesgue integrals equals the corresponding Riemann integral, so \(\lim_{n\to\infty}( R)\int_a^b f_n=( R)\int_a^b f\). So the failure exhibited above is precisely a failure of the class of Riemann integrable functions to be closed under uniformly bounded pointwise limits; this closure defect is one of the principal shortcomings of the Riemann integral that the Lebesgue integral removes.

By definition, \(f=g\) a.e. on \(E\) means there is a subset \(E_0\) of \(E\) with \(m(E_0)=0\) for which

\begin{equation*} f=g \quad\text{on } E\sim E_0 . \end{equation*}

(The set \(\{x\in E\mid f(x)\ne g(x)\}\) is contained in \(E_0\); being a subset of a set of measure zero it has outer measure zero and so is itself measurable, by Proposition 4 of Chapter 2. We may therefore assume \(E_0\) is measurable, replacing it by that set if necessary. Also \(E\sim E_0\) is measurable, being the difference of two measurable sets.)

Step 1: \(g\) is measurable on \(E\). This is immediate from Proposition 5 of Chapter 3, according to which a function that agrees almost everywhere on \(E\) with a measurable function is itself measurable. For completeness we give the verification. Let \(c\) be a real number. Then

\begin{equation*} \{x\in E\mid g(x)>c\}=\bigl(\{x\in E\mid f(x)>c\}\cap (E\sim E_0)\bigr)\;\cup\;\{x\in E_0\mid g(x)>c\} , \end{equation*}

since \(g=f\) off \(E_0\). The first set is the intersection of two measurable sets, hence measurable. The second set is a subset of \(E_0\), hence has outer measure zero by monotonicity of outer measure, hence is measurable by Proposition 4 of Chapter 2. A union of two measurable sets is measurable, so \(\{x\in E\mid g>c\}\) is measurable for every \(c\), and therefore \(g\) is measurable on \(E\).

Step 2: both integrals exist. \(f\) and \(g\) are bounded measurable functions on the set \(E\) of finite measure, so by Theorem 4 both are integrable over \(E\), and \(\int_E f\) and \(\int_E g\) are defined.

Step 3: the difference has integral zero. Put \(h=g-f\) on \(E\). Then \(h\) is bounded and measurable on \(E\), and \(h=0\) on \(E\sim E_0\). Choose \(M\ge 0\) with \(|h|\le M\) on \(E\) (possible since \(f\) and \(g\) are bounded). Then

\begin{equation*} 0\le |h|\le M\cdot\chi_{E_0}\quad\text{on }E, \end{equation*}

because on \(E_0\) we have \(|h|\le M=M\cdot\chi_{E_0}\), while on \(E\sim E_0\) both sides are \(0\). The function \(M\cdot\chi_{E_0}=M\cdot\chi_{E_0}+0\cdot\chi_{E\sim E_0}\) is simple on \(E\), expressed over the finite disjoint collection \(\{E_0,\,E\sim E_0\}\) of measurable subsets of \(E\), so by Lemma 1

\begin{equation*} \int_E M\cdot\chi_{E_0}=M\cdot m(E_0)+0\cdot m(E\sim E_0)=M\cdot 0=0 . \end{equation*}

The functions \(|h|\) and \(M\cdot\chi_{E_0}\) are bounded and measurable on \(E\), so by the monotonicity of integration, Theorem 5, applied to the inequalities \(0\le|h|\le M\cdot\chi_{E_0}\),

\begin{equation*} 0=\int_E 0\le \int_E |h|\le\int_E M\cdot\chi_{E_0}=0 , \end{equation*}

whence \(\int_E|h|=0\).

Step 4: conclusion. By Corollary 7 and then Step 3,

\begin{equation*} \left|\int_E g-\int_E f\right| = \left|\int_E (g-f)\right| = \left|\int_E h\right|\le\int_E|h|=0 , \end{equation*}

where the first equality uses the linearity of integration, Theorem 5. Therefore

\begin{equation*} \int_E f=\int_E g . \end{equation*}

Thus the Lebesgue integral of a bounded measurable function over a set of finite measure is unchanged by altering the function on a set of measure zero. Note that the statement of the problem does not assume \(g\) to be measurable, and it does not need to: measurability of \(g\) is a consequence of the measurability of \(f\) and the equality \(f=g\) a.e. on \(E\), as Step 1 shows. It is, however, essential that \(g\) be assumed bounded, since otherwise \(\int_E g\) need not be defined by the theory of this section.

No. Uniform boundedness cannot be dropped, even for a set of finite measure and even if each individual \(f_n\) is bounded and the convergence is pointwise everywhere.

A counterexample. Take \(E=[0,1]\), so \(m(E)=1<\infty\). For each natural number \(n\) define

\begin{equation*} f_n=n\cdot\chi_{(0,\,1/n]}\quad\text{on }[0,1], \end{equation*}

that is, \(f_n(x)=n\) if \(0<x\le 1/n\) and \(f_n(x)=0\) otherwise.

Each \(f_n\) is a simple function on \([0,1]\): it takes the two values \(n\) and \(0\), on the measurable sets \((0,1/n]\) and \([0,1]\sim(0,1/n]\) respectively. In particular each \(f_n\) is a bounded measurable function on \(E\), so \(\int_E f_n\) is defined, and by Lemma 1,

\begin{equation*} \int_{[0,1]} f_n=n\cdot m\bigl((0,1/n]\bigr)+0\cdot m\bigl([0,1]\sim(0,1/n]\bigr)=n\cdot\frac1n=1 . \end{equation*}

The sequence converges pointwise on \([0,1]\) to \(f\equiv 0\). Indeed, \(f_n(0)=0\) for every \(n\); and if \(0<x\le 1\), choose by the Archimedean property an index \(N\) with \(1/N<x\), so that for all \(n\ge N\) we have \(1/n\le 1/N<x\) and hence \(f_n(x)=0\). Thus \(\{f_n\}\to f\) pointwise on \([0,1]\), where \(f\equiv 0\) is bounded and measurable with

\begin{equation*} \int_{[0,1]} f=0 . \end{equation*}

Every hypothesis of the Bounded Convergence Theorem holds here except uniform boundedness: the \(f_n\) are measurable on a set of finite measure and converge pointwise on \(E\). But

\begin{equation*} \lim_{n\to\infty}\int_{[0,1]} f_n=1\ne 0=\int_{[0,1]} f , \end{equation*}

so the conclusion fails. Of course \(\{|f_n|\}\) is not uniformly bounded: \(\sup_{[0,1]}|f_n|=n\to\infty\), so there is no single \(M\ge 0\) with \(|f_n|\le M\) on \(E\) for all \(n\).

Remark. This is essentially the example given in this section just before the statement of the Bounded Convergence Theorem, where \(f_n\) is taken to be the continuous “tent” of height \(n\) that vanishes for \(x\ge 2/n\), satisfies \(f_n(1/n)=n\) and \(f_n(0)=0\), and is linear on \([0,1/n]\) and on \([1/n,2/n]\); that \(f_n\) also has \(\int_0^1 f_n=1\) for every \(n\) while \(\{f_n\}\to 0\) pointwise on \([0,1]\). The mechanism is the same in both cases: a fixed amount of “mass,” namely \(1\), is carried on sets of measure tending to \(0\) by functions of height tending to \(\infty\), so in the limit the mass escapes rather than being distributed by the limit function. A uniform bound \(M\) prevents this, since it forces the integral carried on a set \(A\) to be at most \(M\cdot m(A)\), which is small when \(m(A)\) is small.

Note also that the failure is not caused by any pathology of the limit: here \(f\equiv 0\) is as well behaved as possible, and the convergence is pointwise at every point of \(E\), not merely almost everywhere. Nor can it be repaired by shrinking \(E\): the example lives on a set of measure \(1\). Uniform boundedness is genuinely needed, and it is exactly what the proof of the Bounded Convergence Theorem uses, through the term \(2M\cdot m(E\sim A)\) in inequality (7), to control the integral over the small set on which Egoroff’s Theorem does not deliver uniform convergence.

Proposition 8 asserts: if \(\{f_n\}\) is a sequence of bounded measurable functions on a set of finite measure \(E\) and \(\{f_n\}\to f\) uniformly on \(E\), then \(\lim_{n\to\infty}\int_E f_n=\int_E f\). The Bounded Convergence Theorem asserts the same conclusion under the hypotheses that the \(f_n\) are measurable on the set \(E\) of finite measure, that the sequence is uniformly pointwise bounded (there is an \(M\ge 0\) with \(|f_n|\le M\) on \(E\) for all \(n\)), and that \(\{f_n\}\to f\) pointwise on \(E\).

So it suffices to show that the hypotheses of Proposition 8 imply those of the Bounded Convergence Theorem. Two of them are immediate: the functions \(f_n\) are measurable on \(E\), and \(m(E)<\infty\), by assumption. It remains to verify pointwise convergence and uniform boundedness.

Uniform convergence implies pointwise convergence. Let \(x\in E\) and let \(\epsilon>0\). By uniform convergence there is an index \(N\), depending on \(\epsilon\) alone, such that \(|f_n-f|<\epsilon\) on all of \(E\) for \(n\ge N\); in particular \(|f_n(x)-f(x)|<\epsilon\) for all \(n\ge N\). Hence \(\{f_n(x)\}\to f(x)\). As \(x\in E\) was arbitrary, \(\{f_n\}\to f\) pointwise on \(E\).

Uniform convergence, together with the boundedness of each \(f_n\), implies uniform boundedness. Apply the definition of uniform convergence with \(\epsilon=1\): there is an index \(N\) for which

\begin{equation*} |f_n-f|<1 \quad\text{on }E \text{ for all } n\ge N . \end{equation*}

Since \(f_N\) is bounded, choose \(M_N\ge 0\) with \(|f_N|\le M_N\) on \(E\). Then for every \(x\in E\),

\begin{equation*} |f(x)|\le |f(x)-f_N(x)|+|f_N(x)|< 1+M_N , \end{equation*}

so \(f\) is bounded on \(E\). Consequently, for every \(n\ge N\) and every \(x\in E\),

\begin{equation*} |f_n(x)|\le |f_n(x)-f(x)|+|f(x)|< 1+(1+M_N)=M_N+2 . \end{equation*}

Each of the finitely many functions \(f_1,\dots,f_{N-1}\) is bounded, say \(|f_j|\le M_j\) on \(E\) for \(1\le j\le N-1\). Set

\begin{equation*} M=\max\{M_1,\dots,M_{N-1},\,M_N+2\}, \end{equation*}

a maximum of finitely many real numbers, hence a real number. Then \(|f_n|\le M\) on \(E\) for every natural number \(n\): for \(n<N\) by the choice of \(M_n\), and for \(n\ge N\) by the displayed estimate. So \(\{f_n\}\) is uniformly pointwise bounded on \(E\).

Conclusion. All the hypotheses of the Bounded Convergence Theorem hold for \(\{f_n\}\), \(f\), \(E\) and \(M\). Therefore

\begin{equation*} \lim_{n\to\infty}\int_E f_n=\int_E f , \end{equation*}

which is precisely the conclusion of Proposition 8. (For this statement to be meaningful one needs \(\int_E f\) to be defined: \(f\) is measurable, being the pointwise limit on \(E\) of the sequence of measurable functions \(\{f_n\}\), by Proposition 9 of Chapter 3; it is bounded, as shown above; and \(m(E)<\infty\); so \(f\) is integrable over \(E\) by Theorem 4. The same remark occurs in the proof of Proposition 8 itself.)

Hence Proposition 8 is indeed the special case of the Bounded Convergence Theorem in which the pointwise convergence is strengthened to uniform convergence. The two hypotheses of Proposition 8 that look weaker than those of the Bounded Convergence Theorem, namely that only each individual \(f_n\) is assumed bounded rather than the family being uniformly bounded, are not really weaker: uniform convergence manufactures the uniform bound, as the argument above shows. Note that this is a genuine implication and not a circularity: no step of the proof of the Bounded Convergence Theorem given in the text invokes Proposition 8. That proof rests on Egoroff’s Theorem, Corollary 7, and the linearity, monotonicity and additivity over domains of the integral; the mention of “the easy proof of the preceding proposition” in its opening paragraph is expository, motivating the appeal to Egoroff’s Theorem rather than serving as a step in the argument.

There are three assertions to check. The first two are an audit of everything in Section 4.2 that precedes the Bounded Convergence Theorem: we must confirm that no step invokes the countable additivity of \(m\), and that exactly one step invokes its finite additivity. The third is the assertion that continuity of measure is a consequence of countable additivity, together with the observation that the proof of the Bounded Convergence Theorem does rest on it, through Egoroff’s Theorem.

Throughout, additivity of \(m\) means additivity of Lebesgue measure over disjoint measurable sets: finite additivity is the special case \(n < \infty\) of Proposition 6 of Chapter 2, and countable additivity is Proposition 13 of Chapter 2. It is important to keep this separate from the countable subadditivity of outer measure \(m^*\) and from the Caratheodory criterion; those two belong to the construction of the \(\sigma\)-algebra of measurable sets and are what make the measurability machinery of Chapter 3 (in particular, that the measurable sets are closed under countable unions, and hence that a pointwise limit of measurable functions is measurable) work. The Remark is about the additivity of \(m\) itself, and it is that property whose uses we now trace.

Assertion 1 and 2: the audit of Section 4.2.

(a) The definition of the integral of a simple function. For a simple function \(\psi\) on a set \(E\) of finite measure with canonical representation \(\psi = \sum_{i=1}^{n} a_i \cdot \chi_{E_i}\), where the \(a_i\) are the distinct values of \(\psi\) and \(E_i = \{x \in E \mid \psi(x) = a_i\}\), one defines

\begin{equation*} \int_E \psi = \sum_{i=1}^{n} a_i \cdot m(E_i). \end{equation*}

This is a definition. It requires only that each \(E_i\) be measurable (it is, since \(\psi\) is measurable) and of finite measure (it is, being a subset of \(E\), by the monotonicity of measure). No additivity of any kind is invoked.

(b) Lemma 1. Here \(\{E_i\}_{i=1}^{n}\) is a finite disjoint collection of measurable subsets of \(E\), the \(a_i\) are real numbers, \(\varphi = \sum_{i=1}^{n} a_i \cdot \chi_{E_i}\), and the conclusion is \(\int_E \varphi = \sum_{i=1}^{n} a_i \cdot m(E_i)\). The proof groups the indices: with \(\{\lambda_1, \dots, \lambda_m\}\) the distinct values of \(\varphi\), \(A_j = \{x \in E \mid \varphi(x) = \lambda_j\}\), and \(I_j = \{i \mid a_i = \lambda_j\}\), one has that \(A_j\) is the disjoint union of the finitely many sets \(E_i\) with \(i \in I_j\), and the proof asserts

\begin{equation*} m(A_j) = \sum_{i \in I_j} m(E_i) \quad \text{for } 1 \le j \le m. \end{equation*}

This is exactly the finite additivity of Lebesgue measure. This is the single appearance of additivity of \(m\) in Section 4.2 prior to the Bounded Convergence Theorem, as we now confirm by inspecting each remaining step.

(c) Proposition 2 (linearity and monotonicity for simple functions). Given simple \(\varphi, \psi\) on \(E\), the proof chooses a finite disjoint collection \(\{E_i\}_{i=1}^{n}\) of measurable sets whose union is \(E\) and on each of which both \(\varphi\) and \(\psi\) are constant. Such a collection is produced by pure set algebra: take the nonempty sets among the intersections of a level set of \(\varphi\) with a level set of \(\psi\). These are measurable because the measurable sets are closed under finite intersection, which is a structural property of the \(\sigma\)-algebra, not an additivity statement. Lemma 1 is then applied to \(\varphi\), to \(\psi\), and to \(\alpha\varphi + \beta\psi\), and the identity \(\sum_i (\alpha a_i + \beta b_i) m(E_i) = \alpha \sum_i a_i m(E_i) + \beta \sum_i b_i m(E_i)\) is arithmetic. For monotonicity one needs only that a nonnegative simple function \(\eta\) has \(\int_E \eta \ge 0\), which is immediate from the definition in (a) and the nonnegativity of \(m\). So Proposition 2 uses additivity only through Lemma 1.

The text immediately after Proposition 2 also observes that, by linearity, the disjointness hypothesis on \(\{E_i\}\) in Lemma 1 is unnecessary. That deduction uses only Proposition 2, hence no additivity beyond the single use in (b).

(d) Step functions. The text next observes that a step function is simple (a singleton and an interval are measurable, and a step function takes finitely many values), and that for a step function on a closed, bounded interval the Riemann and Lebesgue integrals agree. This uses that the measure of an interval is its length and that the measure of a singleton is zero. The first is Proposition 1 of Chapter 2, a statement about outer measure proved by a Heine-Borel compactness argument; the second is immediate from the definition of outer measure. Neither is an additivity statement about \(m\). The agreement of the two integrals is then the linearity of Proposition 2 applied to a finite linear combination of characteristic functions of intervals.

(e) The upper and lower Lebesgue integrals, and the definition of integrability. These are definitions. That both are finite and that the lower integral does not exceed the upper follows from the monotonicity half of Proposition 2. No additivity beyond (b).

(f) Theorem 3 (a Riemann integrable function is Lebesgue integrable, with the same integral). The proof compares the supremum and infimum over step functions with the supremum and infimum over simple functions, using only that every step function is simple and the fact recorded in (d). No additivity beyond (b).

(g) Theorem 4 (a bounded measurable function on a set of finite measure is integrable). The proof invokes the Simple Approximation Lemma of Chapter 3 with \(\epsilon = 1/n\) to get simple \(\varphi_n \le f \le \psi_n\) with \(0 \le \psi_n - \varphi_n \le 1/n\) on \(E\), and then applies the linearity and monotonicity of Proposition 2 to conclude \(0 \le \int_E \psi_n - \int_E \varphi_n \le (1/n) \cdot m(E)\), together with \(m(E) < \infty\). The Simple Approximation Lemma itself partitions the bounded range of \(f\) into finitely many intervals and uses the measurability of the preimages; it invokes no property of \(m\) whatsoever. So Theorem 4 uses additivity only through Lemma 1, and uses the finiteness of \(m(E)\).

(h) Theorem 5 (linearity and monotonicity of integration for bounded measurable functions). The proof passes between the upper and lower Lebesgue integrals using Theorem 4 and Proposition 2 only: for \(\alpha \ne 0\) the correspondence \(\psi \leftrightarrow \psi/\alpha\) between simple functions dominating \(\alpha f\) and those dominating (or dominated by) \(f\) is a bijection, and for \(\alpha = \beta = 1\) one compares \(\int_E(f+g)\) with sums \(\int_E \psi_1 + \int_E \psi_2\) and \(\int_E \varphi_1 + \int_E \varphi_2\). Monotonicity uses only that \(\int_E \eta \ge 0\) for the simple function \(\eta \equiv 0\) compared with a nonnegative \(h\). No new use of additivity.

(h’) The Example of the Dirichlet function. Between Theorem 3 and Theorem 4 the text computes \(\int_{[0,1]} \chi_E = 1 \cdot m(E) = 0\) for \(E\) the rationals in \([0,1]\). This uses only the definition of the integral of a simple function together with the fact that a countable set has measure zero — and the latter is a statement about outer measure, proved in Chapter 2 from the countable subadditivity of \(m^*\) and the fact that a singleton has outer measure zero. Countable subadditivity of \(m^*\) is not the countable additivity of \(m\); it is available before any additivity of \(m\) is proved. So this Example adds no use of additivity either.

(i) Corollary 6 (additivity over domains of integration). For disjoint measurable \(A, B \subseteq E\) the proof uses the pointwise set identity \(\chi_{A \cup B} = \chi_A + \chi_B\) — which holds because \(A\) and \(B\) are disjoint, and is a statement about functions, not about \(m\) — together with Problem 10 (\(\int_{E_1} f = \int_E f \cdot \chi_{E_1}\), itself a consequence of the definitions and of Theorem 5) and the linearity of Theorem 5. Note in particular that the name of this corollary is misleading for our purposes: the additivity being proved here is a property of the integral, deduced from linearity, and it is not an application of the additivity of \(m\).

(j) Corollary 7 (\(|\int_E f| \le \int_E |f|\)). Immediate from \(-|f| \le f \le |f|\) and Theorem 5.

(k) Proposition 8 (passage to the limit for uniformly convergent sequences). The proof uses that a uniform limit of measurable functions is measurable, then, for \(n \ge N\) with \(|f - f_n| < \epsilon/m(E)\) on \(E\), applies Theorem 5 and Corollary 7 to get \(|\int_E f - \int_E f_n| \le [\epsilon/m(E)] \cdot m(E) = \epsilon\). Again \(m(E) < \infty\) is used, but no additivity beyond Lemma 1.

This exhausts everything in Section 4.2 before the Bounded Convergence Theorem. The countable additivity of \(m\) is never invoked, and its finite additivity is invoked exactly once, in the proof of Lemma 1. That verifies the first two assertions.

Assertion 3: the Bounded Convergence Theorem uses Egoroff’s Theorem, and Egoroff’s Theorem uses continuity of measure, which follows from countable additivity.

First, the Bounded Convergence Theorem. Its proof fixes \(\epsilon > 0\) and, since \(m(E) < \infty\) and \(f\) is real-valued, applies Egoroff’s Theorem to produce a measurable \(A \subseteq E\) on which \(\{f_n\} \to f\) uniformly and with \(m(E \sim A) < \epsilon/4M\); the estimate

\begin{equation*} \left| \int_E f_n - \int_E f \right| \le \int_A |f_n - f| + 2M \cdot m(E \sim A) \end{equation*}

is then finished by uniform convergence on \(A\). So Egoroff’s Theorem is genuinely a hypothesis of the argument.

Second, Egoroff’s Theorem uses the continuity of measure. Its proof rests on Lemma 10 of Chapter 3, whose proof sets, for \(\eta > 0\),

\begin{equation*} E_n = \{x \in E \mid |f(x) - f_k(x)| < \eta \text{ for all } k \ge n\}, \end{equation*}

observes that \(\{E_n\}_{n=1}^{\infty}\) is an ascending collection of measurable sets with \(\bigcup_{n=1}^{\infty} E_n = E\) because \(\{f_n\} \to f\) pointwise on \(E\), and then concludes

\begin{equation*} m(E) = \lim_{n \to \infty} m(E_n) \end{equation*}

from the continuity of measure, Theorem 15(i) of Chapter 2. Since \(m(E) < \infty\), an index \(N\) can then be chosen with \(m(E_N) > m(E) - \delta\), and the excision property gives \(m(E \sim E_N) < \delta\). (The proof of Egoroff’s Theorem proper additionally uses the countable subadditivity of measure to bound \(m(E \sim \bigcap_{n} A_n)\), and Theorem 11 of Chapter 2 to shrink \(A\) to a closed set — but the essential point for the Remark is the appeal to continuity of measure.)

Third, continuity of measure is a consequence of countable additivity. Let \(\{A_k\}_{k=1}^{\infty}\) be an ascending collection of measurable sets and put \(A_0 = \emptyset\). The sets \(A_k \sim A_{k-1}\), \(k \ge 1\), are measurable (the measurable sets are a \(\sigma\)-algebra) and pairwise disjoint: if \(j < k\) then \(A_j \subseteq A_{k-1}\), so \(A_j \sim A_{j-1}\) is disjoint from \(A_k \sim A_{k-1}\). Moreover, since the collection is ascending,

\begin{equation*} \bigcup_{k=1}^{\infty} A_k = \bigcup_{k=1}^{\infty} \left[ A_k \sim A_{k-1} \right] \quad \text{and} \quad A_n = \bigcup_{k=1}^{n} \left[ A_k \sim A_{k-1} \right] \text{ for each } n. \end{equation*}

Indeed, if \(x\) belongs to \(\bigcup_k A_k\), there is a least index \(k\) with \(x \in A_k\), and then \(x \in A_k \sim A_{k-1}\). By the countable additivity of Lebesgue measure, Proposition 13 of Chapter 2, applied to the left-hand union, and by finite additivity applied to the right-hand one,

\begin{equation*} m\left( \bigcup_{k=1}^{\infty} A_k \right) = \sum_{k=1}^{\infty} m(A_k \sim A_{k-1}) = \lim_{n \to \infty} \sum_{k=1}^{n} m(A_k \sim A_{k-1}) = \lim_{n \to \infty} m(A_n), \end{equation*}

which is Theorem 15(i). (Part (ii), for descending collections with \(m(B_1) < \infty\), follows by applying (i) to the ascending collection \(\{B_1 \sim B_k\}\) and using excision, which is legitimate because \(m(B_1) < \infty\).)

Assembling the three points: the development of Section 4.2 up to, but not including, the Bounded Convergence Theorem rests on Lebesgue measure only through its monotonicity, the value it assigns to intervals, and one application of finite additivity in Lemma 1; whereas the Bounded Convergence Theorem itself rests, via Egoroff’s Theorem and the continuity of measure, on the countable additivity of Lebesgue measure. All the assertions of the Remark are verified.

One caveat, for honesty: the Remark, and hence this exercise, is a bookkeeping claim about which properties of \(m\) the given proofs happen to use. It is not a claim that the Bounded Convergence Theorem cannot be proved without countable additivity, and nothing above should be read as establishing such a claim.

Since \(f\) is bounded and measurable on \(E\) and \(m(E) < \infty\), \(f\) is integrable over \(E\) by Theorem 4, so the hypothesis \(\int_E f = 0\) makes sense.

For each natural number \(n\) define

\begin{equation*} E_n = \left\{ x \in E \;\middle|\; f(x) \ge \tfrac{1}{n} \right\}. \end{equation*}

Each \(E_n\) is measurable, since \(f\) is a measurable function, and \(E_n \subseteq E\), so \(m(E_n) \le m(E) < \infty\) by the monotonicity of measure.

Fix \(n\). The function \(\frac{1}{n} \cdot \chi_{E_n}\) is a simple, hence bounded and measurable, function on \(E\), and its integral over \(E\) is computed from the definition of the integral of a simple function on a set of finite measure:

\begin{equation*} \int_E \tfrac{1}{n} \cdot \chi_{E_n} = \tfrac{1}{n} \cdot m(E_n). \end{equation*}

(If \(E_n = E\) the canonical representation of \(\frac1n \chi_{E_n}\) on \(E\) has the single value \(1/n\), taken on \(E\); if \(E_n = \emptyset\) it has the single value \(0\), and then both sides are \(0\); otherwise it has the two distinct values \(1/n\) and \(0\), taken on \(E_n\) and on \(E \sim E_n\), and the term \(0 \cdot m(E \sim E_n)\) contributes nothing. In every case the displayed formula holds.)

Now \(\frac{1}{n} \cdot \chi_{E_n} \le f\) on \(E\): at a point \(x \in E_n\) the left side equals \(1/n \le f(x)\) by the definition of \(E_n\), and at a point \(x \in E \sim E_n\) the left side equals \(0 \le f(x)\) because \(f\) is nonnegative. Both functions are bounded and measurable on the set \(E\) of finite measure, so the monotonicity of integration, Theorem 5, applies and gives

\begin{equation*} \tfrac{1}{n} \cdot m(E_n) = \int_E \tfrac{1}{n} \cdot \chi_{E_n} \le \int_E f = 0. \end{equation*}

Since \(m(E_n) \ge 0\), we conclude \(m(E_n) = 0\) for every \(n\).

Finally, because \(f \ge 0\) on \(E\),

\begin{equation*} \{ x \in E \mid f(x) \ne 0 \} = \{ x \in E \mid f(x) > 0 \} = \bigcup_{n=1}^{\infty} E_n, \end{equation*}

the second equality because \(f(x) > 0\) holds if and only if \(f(x) \ge 1/n\) for some natural number \(n\) (choose \(n\) with \(1/n \le f(x)\), possible by the Archimedean property). By the countable subadditivity of Lebesgue measure,

\begin{equation*} m\left( \{ x \in E \mid f(x) \ne 0 \} \right) \le \sum_{n=1}^{\infty} m(E_n) = 0. \end{equation*}

Hence \(\{x \in E \mid f(x) \ne 0\}\) has measure zero, that is, \(f = 0\) a.e. on \(E\).

The function \(f \equiv \infty\) is a nonnegative extended real-valued measurable function on \(E\): for every \(c\) the set \(\{x \in E \mid f(x) > c\}\) equals \(E\), which is measurable. So its integral is defined by (8),

\begin{equation*} \int_E f = \sup \left\{ \int_E h \;\middle|\; h \text{ bounded, measurable, of finite support and } 0 \le h \le f \text{ on } E \right\}. \end{equation*}

We show every number in the set on the right equals \(0\), and that the set is nonempty; the supremum is then \(0\).

The set is nonempty: the function \(h \equiv 0\) on \(E\) is bounded, measurable, vanishes outside the set \(E_0 = \emptyset\) of finite measure (so it has finite support), and satisfies \(0 \le h \le f\) since \(f \equiv \infty\).

Now let \(h\) be any competitor, that is, \(h\) is bounded and measurable on \(E\), \(0 \le h \le f\) on \(E\), and \(h\) has finite support. Since \(m(E) = 0 < \infty\), the set \(E\) itself is an admissible choice of the set outside of which \(h\) vanishes (vacuously, \(h = 0\) on \(E \sim E = \emptyset\)), so the integral of \(h\) over \(E\) is simply the integral of the bounded measurable function \(h\) over the set \(E\) of finite measure, in the sense of Section 4.2. No appeal to the independence of the choice of \(E_0\) is needed here.

We claim \(\int_E h = 0\). Let \(\psi\) be any simple function on \(E\) with \(h \le \psi\) on \(E\); such \(\psi\) exist, since \(\psi \equiv M\) works for any \(M\) bounding \(h\). Write the canonical representation \(\psi = \sum_{i=1}^{n} a_i \cdot \chi_{A_i}\), where the \(A_i\) are the level sets of \(\psi\) in \(E\). Each \(A_i \subseteq E\), so \(m(A_i) \le m(E) = 0\) by the monotonicity of measure, whence

\begin{equation*} \int_E \psi = \sum_{i=1}^{n} a_i \cdot m(A_i) = 0. \end{equation*}

Thus every number in the defining set for the upper Lebesgue integral of \(h\) over \(E\) equals \(0\), so that upper integral is \(0\). Since \(h\) is bounded and measurable on the set \(E\) of finite measure, Theorem 4 says \(h\) is integrable over \(E\) and \(\int_E h\) is the common value of the upper and lower integrals; hence \(\int_E h = 0\). (This is precisely the content of Problem 9: a bounded function on a set of measure zero is measurable with integral zero. If \(E = \emptyset\) the assertion is trivial.)

(The same conclusion holds if one prefers to compute \(\int_E h\) as \(\int_{E_0} h\) for a set \(E_0 \subseteq E\) of finite measure outside which \(h\) vanishes: then \(m(E_0) \le m(E) = 0\) and the identical argument on \(E_0\) gives \(\int_{E_0} h = 0\).)

Therefore \(\int_E h = 0\) for every competitor \(h\), and consequently

\begin{equation*} \int_E f = \sup\{0\} = 0. \end{equation*}

Note that the same argument shows more generally that the integral of any nonnegative measurable function over a set of measure zero is zero; the extreme case \(f \equiv \infty\) makes the point that the convention \(\infty \cdot 0 = 0\) built into definition (8) is consistent.

Recall the situation. A measurable function \(f\) on a measurable set \(E\) is said to vanish outside a set of finite measure, or to have finite support, provided there is a measurable subset \(E_0\) of \(E\) with \(m(E_0) < \infty\) and \(f = 0\) on \(E \sim E_0\). If \(f\) is in addition bounded and measurable on \(E\), one defines

\begin{equation*} \int_E f = \int_{E_0} f . \end{equation*}

Two things must be checked for this to be a legitimate definition: that the right-hand side makes sense, and that it does not depend on which set \(E_0\) is used.

The right-hand side is defined. Let \(E_0\) be as above. The restriction of \(f\) to \(E_0\) is a bounded measurable function on a set of finite measure, so by Theorem 4 it is integrable over \(E_0\) and \(\int_{E_0} f\) is a well-defined real number.

Independence of the choice of \(E_0\). Let \(E_0\) and \(E_1\) be measurable subsets of \(E\), each of finite measure, with

\begin{equation*} f = 0 \text{ on } E \sim E_0 \quad \text{and} \quad f = 0 \text{ on } E \sim E_1 . \end{equation*}

We must show \(\int_{E_0} f = \int_{E_1} f\).

Set \(A = E_0 \cap E_1\) and \(B = E_0 \sim E_1\). Then \(A\) and \(B\) are disjoint measurable subsets of \(E_0\) with \(A \cup B = E_0\), and \(m(E_0) < \infty\), so the restriction of \(f\) to \(E_0\) is a bounded measurable function on a set of finite measure. Corollary 6, additivity over domains of integration, applies and yields

\begin{equation*} \int_{E_0} f = \int_{A} f + \int_{B} f . \end{equation*}

Now \(B = E_0 \sim E_1 \subseteq E \sim E_1\), and \(f = 0\) on \(E \sim E_1\). Hence \(f\) vanishes identically on \(B\), and \(m(B) \le m(E_0) < \infty\), so \(\int_B f = \int_B 0 = 0\): indeed the zero function on \(B\) is simple with canonical representation \(0 \cdot \chi_B\), whose integral is \(0 \cdot m(B) = 0\). (If \(B = \emptyset\) the integral is \(0\) trivially.) Therefore

\begin{equation*} \int_{E_0} f = \int_{E_0 \cap E_1} f . \end{equation*}

The argument is symmetric in \(E_0\) and \(E_1\): taking \(A = E_1 \cap E_0\) and \(B^{\prime} = E_1 \sim E_0 \subseteq E \sim E_0\), on which \(f\) vanishes, Corollary 6 applied on \(E_1\) gives

\begin{equation*} \int_{E_1} f = \int_{E_0 \cap E_1} f + \int_{E_1 \sim E_0} f = \int_{E_0 \cap E_1} f . \end{equation*}

Comparing the two displays, \(\int_{E_0} f = \int_{E_1} f\). So the value assigned to \(\int_E f\) is the same for every admissible choice of \(E_0\), and the definition is proper.

Remark. The set \(E_0\) may always be taken to be the support \(\{x \in E \mid f(x) \ne 0\}\) itself: this set is measurable since \(f\) is measurable, it is contained in any admissible \(E_0\) and hence has finite measure, and \(f\) vanishes on its complement in \(E\). So the definition could equivalently have been made with this canonical choice; the computation above shows the more flexible formulation agrees with it. Note also that when \(m(E) < \infty\) we may take \(E_0 = E\), so the new definition is consistent with the integral of Section 4.2.

The answer is

\begin{equation*} \int_0^1 f = \begin{cases} \dfrac{1}{\alpha + 1} & \text{if } \alpha > -1, \\[2mm] \infty & \text{if } \alpha \le -1. \end{cases} \end{equation*}

Setting up. The function \(f\) is real-valued and nonnegative on \([0,1]\): for \(x \in (0,1]\) we have \(x^{\alpha} = e^{\alpha \ln x} \in (0, \infty)\), and \(f(0) = 0\). It is measurable. Indeed, \(f\) is continuous on \((0,1]\), so for each real \(c\) the set \(\{x \in (0,1] \mid f(x) > c\}\) is relatively open in \((0,1]\), hence the intersection of an open set with \((0,1]\), hence measurable; and

\begin{equation*} \{x \in [0,1] \mid f(x) > c\} = \{x \in (0,1] \mid f(x) > c\} \cup S_c, \quad S_c = \begin{cases} \{0\} & \text{if } c < 0, \\ \emptyset & \text{if } c \ge 0, \end{cases} \end{equation*}

a union of two measurable sets. So \(f\) is a nonnegative measurable function on \([0,1]\) and \(\int_{[0,1]} f\) is defined by (8). Note that the value assigned at the single point \(0\) is irrelevant, since \(\{0\}\) has measure zero and the excision formula (15) of Theorem 11 gives \(\int_{[0,1]} f = \int_{(0,1]} f\).

A consistency remark. Below we integrate functions that are simultaneously nonnegative measurable (so that (8) applies) and bounded measurable of finite support (so that the integral of Section 4.2 applies). The two integrals agree: if \(g\) is nonnegative, bounded and measurable on \(E\) with finite support, then \(g\) is itself a competitor in (8), so the supremum in (8) is at least the Section 4.2 integral of \(g\); and every competitor \(h\) in (8) satisfies \(0 \le h \le g\), so, choosing a set \(E_0\) of finite measure outside which both \(g\) and \(h\) vanish, the monotonicity of Theorem 5 gives \(\int_{E_0} h \le \int_{E_0} g\), that is, the Section 4.2 integral of \(g\) is an upper bound for the competitors. Hence the two values coincide, and we may write \(\int_E g\) without ambiguity.

Exhausting by an increasing sequence. For each natural number \(n \ge 2\) define

\begin{equation*} f_n = f \cdot \chi_{[1/n,\, 1]} \quad \text{on } [0,1]. \end{equation*}

Each \(f_n\) is nonnegative and measurable. The sequence is increasing: the sets \([1/n, 1]\) ascend with \(n\) and \(f \ge 0\), so \(f_n \le f_{n+1}\) pointwise on \([0,1]\). It converges to \(f\) pointwise on \([0,1]\): if \(x \in (0,1]\), then \(1/n < x\) for all large \(n\), so \(f_n(x) = f(x)\) eventually; and \(f_n(0) = 0 = f(0)\) for every \(n\). By the Monotone Convergence Theorem,

\begin{equation*} \int_{[0,1]} f = \lim_{n \to \infty} \int_{[0,1]} f_n . \end{equation*}

Computing \(\int_{[0,1]} f_n\). On the closed, bounded interval \([1/n, 1]\) the function \(x \mapsto x^{\alpha}\) is continuous, hence bounded there and, by Problem 6 of Section 4.1, Riemann integrable there. Consequently \(f_n\) is a bounded measurable function on \([0,1]\), and it has finite support since \(m([0,1]) = 1 < \infty\). Since \([0,1]\) is the union of the disjoint measurable sets \([0, 1/n)\) and \([1/n, 1]\), additivity over domains for nonnegative measurable functions (Theorem 11), together with the fact that \(f_n\) vanishes identically on \([0, 1/n)\) and so has integral zero there by Proposition 9, gives

\begin{equation*} \int_{[0,1]} f_n = \int_{[1/n,\,1]} x^{\alpha} \, dx . \end{equation*}

By Theorem 3, the Lebesgue integral of a Riemann integrable function on a closed, bounded interval equals its Riemann integral, so this is the elementary integral \(\int_{1/n}^{1} x^{\alpha}\, dx\), which the Fundamental Theorem of Calculus evaluates as

\begin{equation*} \int_{1/n}^{1} x^{\alpha}\, dx = \begin{cases} \dfrac{1 - (1/n)^{\alpha+1}}{\alpha+1} & \text{if } \alpha \ne -1, \\[2mm] \ln n & \text{if } \alpha = -1. \end{cases} \end{equation*}

Passing to the limit. Three cases.

If \(\alpha > -1\), then \(\alpha + 1 > 0\), so \((1/n)^{\alpha+1} \to 0\) and

\begin{equation*} \int_{[0,1]} f = \lim_{n \to \infty} \frac{1 - (1/n)^{\alpha+1}}{\alpha+1} = \frac{1}{\alpha+1}. \end{equation*}

If \(\alpha = -1\), then \(\int_{[0,1]} f = \lim_{n\to\infty} \ln n = \infty\).

If \(\alpha < -1\), then \(\alpha + 1 < 0\), so \((1/n)^{\alpha+1} = n^{-(\alpha+1)} \to \infty\); the numerator \(1 - (1/n)^{\alpha+1}\) tends to \(-\infty\) while the denominator \(\alpha+1\) is a fixed negative number, so the quotient tends to \(+\infty\) and \(\int_{[0,1]} f = \infty\).

Summary and remarks. Thus \(f\) is integrable over \([0,1]\) exactly when \(\alpha > -1\), in which case \(\int_0^1 f = 1/(\alpha+1)\); for \(\alpha \le -1\) the integral is \(\infty\) and \(f\) is not integrable. Two consistency checks. For \(\alpha > 0\) the function \(f\) is continuous on all of \([0,1]\), since \(x^{\alpha} \to 0 = f(0)\) as \(x \to 0^{+}\); hence it is Riemann integrable over \([0,1]\) by Problem 6 of Section 4.1, and Theorem 3 gives directly \(\int_0^1 x^{\alpha}\, dx = 1/(\alpha+1)\), in agreement with the formula above. For \(\alpha = 0\) the function \(f\) is the step function equal to \(1\) on \((0,1]\) and \(0\) at the origin, whose integral is \(1 \cdot m((0,1]) = 1 = 1/(0+1)\), again in agreement. And for \(\alpha \le -1\) one sees the divergence without computation from Chebychev’s Inequality applied to the sets \(\{f \ge \lambda\}\), or simply by noting \(x^{\alpha} \ge x^{-1}\) on \((0,1]\) for \(\alpha \le -1\) and using monotonicity of integration, Theorem 10, together with the case \(\alpha = -1\).

Call the asserted property \((P)\):

\((P)\): whenever \(\{f_n\}\) is a sequence of nonnegative measurable functions on \(E\) converging to \(f\) pointwise on \(E\), and \(M \ge 0\) satisfies \(\int_E f_n \le M\) for all \(n\), then \(\int_E f \le M\).

We also record the a.e. variant, in which pointwise convergence on \(E\) is weakened to pointwise convergence a.e. on \(E\); call it \((P^{\prime})\). The two are equivalent, and the equivalence is proved in the first of the two remarks at the end. In the arguments below we use whichever form is convenient, and we work with \((P^{\prime})\) where the a.e. hypothesis of Fatou’s Lemma as printed in the text makes it natural.

Fatou’s Lemma implies \((P)\) and \((P^{\prime})\). Let \(\{f_n\}\), \(f\) and \(M\) be as in the hypothesis of \((P^{\prime})\) (which includes that of \((P)\)). The functions \(f_n\) are nonnegative and measurable and converge pointwise a.e. on \(E\) to \(f\), so Fatou’s Lemma applies and gives

\begin{equation*} \int_E f \le \liminf_{n \to \infty} \int_E f_n . \end{equation*}

Since \(\int_E f_n \le M\) for every \(n\), the sequence of real numbers \(\left\{ \int_E f_n \right\}\) is bounded above by \(M\), hence so is its lower limit: \(\liminf_n \int_E f_n \le M\). Combining,

\begin{equation*} \int_E f \le M , \end{equation*}

which is what was to be shown. This proves the first assertion of the problem.

\((P)\) implies Fatou’s Lemma. Assume \((P)\) holds for every sequence as described; by the first remark below, \((P^{\prime})\) then holds too, and it is \((P^{\prime})\) we use. Let \(\{g_n\}\) be a sequence of nonnegative measurable functions on \(E\) with \(\{g_n\} \to g\) pointwise a.e. on \(E\); we must show

\begin{equation*} \int_E g \le \liminf_{n \to \infty} \int_E g_n . \end{equation*}

Write \(L = \liminf_{n \to \infty} \int_E g_n\), an element of \([0, \infty]\) since each \(\int_E g_n \ge 0\).

If \(L = \infty\) the inequality is trivially true, so assume \(L < \infty\).

Let \(\epsilon > 0\). Since \(L = \lim_{k \to \infty} \left( \inf_{n \ge k} \int_E g_n \right)\) is finite, the lower limit is attained along a subsequence: there is a subsequence \(\{g_{n_k}\}_{k=1}^{\infty}\), with \(n_1 < n_2 < \cdots\), for which

\begin{equation*} \lim_{k \to \infty} \int_E g_{n_k} = L . \end{equation*}

Hence there is an index \(K\) such that \(\int_E g_{n_k} \le L + \epsilon\) for all \(k \ge K\).

Consider the sequence \(\{h_k\}_{k=1}^{\infty}\) defined by \(h_k = g_{n_{K + k - 1}}\), that is, the tail of the chosen subsequence beginning at index \(K\). Each \(h_k\) is a nonnegative measurable function on \(E\); the sequence \(\{h_k\}\) converges pointwise a.e. on \(E\) to \(g\), because it is a subsequence of a sequence that does so, and a subsequence of a convergent sequence of real (or extended real) numbers converges to the same limit; and

\begin{equation*} \int_E h_k \le L + \epsilon \quad \text{for every } k . \end{equation*}

So \((P^{\prime})\) applies with the bound \(M = L + \epsilon \ge 0\) and yields

\begin{equation*} \int_E g \le L + \epsilon . \end{equation*}

Since \(\epsilon > 0\) was arbitrary, \(\int_E g \le L = \liminf_n \int_E g_n\), which is Fatou’s Lemma.

Two remarks on the bookkeeping. First, the promised equivalence of \((P)\) and \((P^{\prime})\). Trivially \((P^{\prime})\) implies \((P)\). Conversely, assume \((P)\) and let \(\{f_n\}\) be nonnegative measurable on \(E\) with \(\{f_n\} \to f\) pointwise on \(E \sim E_0\), where \(E_0 \subseteq E\) is measurable with \(m(E_0) = 0\), and with \(\int_E f_n \le M\) for all \(n\). The function \(f\) is measurable on \(E\). Indeed, define \(g\) on \(E\) by \(g = f\) on \(E \sim E_0\) and \(g = 0\) on \(E_0\). On \(E \sim E_0\) the function \(g\) is the pointwise limit of the measurable functions \(f_n\) restricted there, hence measurable on \(E \sim E_0\) by Proposition 9 of Chapter 3; \(g\) is measurable on \(E_0\) as well, being constant; and a function whose restrictions to the two measurable pieces \(E_0\) and \(E \sim E_0\) of \(E\) are measurable is measurable on \(E\), since \(\{g > c\} = \{x \in E_0 \mid g(x) > c\} \cup \{x \in E \sim E_0 \mid g(x) > c\}\). Finally \(f = g\) on \(E \sim E_0\) and \(m(E_0) = 0\), so \(f = g\) a.e. on \(E\) and Proposition 5(i) of Chapter 3 gives the measurability of \(f\) on \(E\). The restrictions of the \(f_n\) to \(E \sim E_0\) are nonnegative and measurable and converge to \(f\) pointwise on all of \(E \sim E_0\), and by the excision formula (15) of Theorem 11,

\begin{equation*} \int_{E \sim E_0} f_n = \int_E f_n \le M \quad \text{for every } n . \end{equation*}

So \((P)\), applied on the set \(E \sim E_0\), gives \(\int_{E \sim E_0} f \le M\), and one more application of (15) gives \(\int_E f = \int_{E \sim E_0} f \le M\). This is \((P^{\prime})\). (This is exactly the reduction performed at the start of the proof of Fatou’s Lemma in the text.)

Second, the two implications are genuinely a proof of equivalence and not a circularity: the second implication derives Fatou’s Lemma from \((P)\) using nothing but the definition of \(\liminf\) and the excision property, so \((P)\) could be taken as the official statement of Fatou’s Lemma and the version printed in the text recovered from it. Note incidentally that \((P)\) is strictly a statement about an upper bound: it does not assert \(\int_E f \le \liminf \int_E f_n\) directly, but the freedom to pass to a subsequence realising the lower limit converts one into the other.

Part 1: approximation by a simple function of finite support.

Since \(f\) is integrable over \(E\), \(\int_E f < \infty\), and by Proposition 13 the set \(Z = \{x \in E \mid f(x) = \infty\}\) has measure zero.

Step 1: choose a bounded competitor close to the supremum. By the definition (8) of \(\int_E f\) as a supremum, and since that supremum is finite, there is a bounded measurable function \(h_1\) on \(E\) of finite support with \(0 \le h_1 \le f\) on \(E\) and

\begin{equation*} \int_E h_1 > \int_E f - \frac{\epsilon}{2}. \end{equation*}

Step 2: replace \(h_1\) by a simple function below it. Let \(E_0\) be a measurable subset of \(E\) with \(m(E_0) < \infty\) and \(h_1 = 0\) on \(E \sim E_0\); by Problem 18, \(\int_E h_1 = \int_{E_0} h_1\) regardless of which such \(E_0\) is taken. The restriction of \(h_1\) to \(E_0\) is a bounded, real-valued measurable function, so the Simple Approximation Lemma of Chapter 3, applied with

\begin{equation*} \epsilon^{\prime} = \frac{\epsilon}{2\,(m(E_0) + 1)} > 0, \end{equation*}

provides simple functions \(\varphi\) and \(\psi\) on \(E_0\) with

\begin{equation*} \varphi \le h_1 \le \psi \quad \text{and} \quad 0 \le \psi - \varphi < \epsilon^{\prime} \quad \text{on } E_0 . \end{equation*}

Define \(\eta_0 = \max\{\varphi, 0\}\) on \(E_0\). It is simple. Indeed, if \(\varphi\) has canonical representation \(\varphi = \sum_{i=1}^{p} d_i \chi_{D_i}\) on \(E_0\), then \(\eta_0\) takes only the finitely many values in \(\{ \max\{d_i, 0\} : 1 \le i \le p \}\), and for each such value \(a\) the level set \(\{x \in E_0 \mid \eta_0(x) = a\}\) is the union of those \(D_i\) with \(\max\{d_i,0\} = a\), a finite union of measurable sets. So \(\eta_0\) is a measurable real-valued function taking finitely many values, that is, a simple function. Since \(h_1 \ge 0\) and \(\varphi \le h_1\), we have \(0 \le \eta_0 \le h_1\) on \(E_0\), and

\begin{equation*} 0 \le h_1 - \eta_0 \le h_1 - \varphi \le \psi - \varphi < \epsilon^{\prime} \quad \text{on } E_0 . \end{equation*}

Now define \(\eta\) on all of \(E\) by \(\eta = \eta_0\) on \(E_0\) and \(\eta = 0\) on \(E \sim E_0\). Then \(\eta\) is a simple function on \(E\), it vanishes outside the set \(E_0\) of finite measure, so it has finite support, and

\begin{equation*} 0 \le \eta \le h_1 \le f \quad \text{on } E . \end{equation*}

Step 3: the estimate. All three functions \(h_1\), \(\eta\) and \(h_1 - \eta\) are bounded and measurable on \(E\) and vanish on \(E \sim E_0\), where \(m(E_0) < \infty\). So each of their integrals over \(E\) is by definition the corresponding integral over the single set \(E_0\) of finite measure, and Theorem 5, the linearity and monotonicity of integration for bounded measurable functions on a set of finite measure, applies to them on \(E_0\). Since \(0 \le h_1 - \eta \le \epsilon^{\prime}\) on \(E_0\), monotonicity against the constant function \(\epsilon^{\prime}\) gives

\begin{equation*} \int_E (h_1 - \eta) = \int_{E_0} (h_1 - \eta) \le \epsilon^{\prime} \cdot m(E_0) < \frac{\epsilon}{2}, \end{equation*}

and linearity on \(E_0\) gives \(\int_{E_0} \eta = \int_{E_0} h_1 - \int_{E_0}(h_1 - \eta)\), that is,

\begin{equation*} \int_E \eta = \int_E h_1 - \int_E (h_1 - \eta) > \left( \int_E f - \frac{\epsilon}{2} \right) - \frac{\epsilon}{2} = \int_E f - \epsilon . \end{equation*}

(Here \(\int_E \eta\) in the sense of the supremum definition (8) agrees with \(\int_{E_0} \eta\) in the sense of Section 4.2: \(\eta\) is itself a competitor in (8), so the supremum is at least \(\int_{E_0}\eta\), while every competitor \(h\) in (8) satisfies \(0 \le h \le \eta\) and hence \(\int h \le \int_{E_0} \eta\) by the monotonicity of Theorem 5 on a set of finite measure containing the supports of both. The same remark applies to \(h_1\).)

Step 4: conclusion of Part 1. Since \(0 \le \eta \le f\) on \(E\), we have \(|f - \eta| = f - \eta\) on \(E\) (with the convention \(\infty - c = \infty\) on the null set \(Z\)), and \(f - \eta\) is a nonnegative measurable function with \((f - \eta) + \eta = f\) pointwise on \(E\). Its measurability is immediate and does not require \(f\) to be finite: writing the canonical representation \(\eta = \sum_{i} c_i \chi_{F_i}\) with \(\{F_i\}\) a finite disjoint measurable partition of \(E\), for each real \(t\)

\begin{equation*} \{ x \in E \mid f(x) - \eta(x) > t \} = \bigcup_{i} \left( F_i \cap \{ x \in E \mid f(x) > t + c_i \} \right), \end{equation*}

a finite union of measurable sets. By the linearity of integration for nonnegative measurable functions, Theorem 10,

\begin{equation*} \int_E f = \int_E (f - \eta) + \int_E \eta . \end{equation*}

Both \(\int_E f\) and \(\int_E \eta\) are finite (the latter because \(\eta \le f\) and monotonicity, (12) of Theorem 10), so we may subtract and conclude

\begin{equation*} \int_E |f - \eta| = \int_E (f - \eta) = \int_E f - \int_E \eta < \epsilon . \end{equation*}

This is the required simple function.

Part 2: approximation by a step function when \(E = [a,b]\).

Let \(E = [a,b]\) be a closed, bounded interval. Note that every function on \(E\) has finite support, since \(m(E) = b - a < \infty\); so the substance of the claim is the integral estimate. Note also that \(h\) is not required to satisfy \(0 \le h \le f\).

By Part 1 there is a simple function \(\eta\) on \(E\) with finite support, \(0 \le \eta \le f\) on \(E\), and

\begin{equation*} \int_E |f - \eta| < \frac{\epsilon}{2}. \end{equation*}

It therefore suffices to produce a step function \(h\) on \(E\) with \(\int_E |\eta - h| < \epsilon/2\), for then the triangle inequality finishes the proof, as follows. Let \(Z = \{x \in E \mid f(x) = \infty\}\), a set of measure zero by Proposition 13. On \(E \sim Z\) all three functions \(f, \eta, h\) are finite, so the pointwise inequality \(|f - h| \le |f - \eta| + |\eta - h|\) holds there; by the excision formula (15) of Theorem 11 and then the monotonicity and linearity of integration for nonnegative measurable functions, Theorem 10,

\begin{equation*} \int_E |f - h| = \int_{E \sim Z} |f - h| \le \int_{E \sim Z} |f - \eta| + \int_{E \sim Z} |\eta - h| = \int_E |f - \eta| + \int_E |\eta - h| < \epsilon . \end{equation*}

Construction of the step function. Let

\begin{equation*} \eta = \sum_{k=1}^{n} a_k \cdot \chi_{E_k} \end{equation*}

be the canonical representation of \(\eta\) on \(E\), where \(a_1, \dots, a_n\) are the distinct nonzero values taken by \(\eta\) and \(E_k = \{x \in E \mid \eta(x) = a_k\}\) (we discard the value \(0\), which contributes nothing). The \(E_k\) are disjoint measurable subsets of \([a,b]\), so each has finite measure. Put

\begin{equation*} \delta = \frac{\epsilon}{2 \left( 1 + \sum_{k=1}^{n} |a_k| \right)} > 0 . \end{equation*}

Each \(E_k\) is measurable of finite outer measure, so Theorem 12 of Chapter 2 applies: there is a finite disjoint collection of open intervals \(\{I_{k,j}\}_{j=1}^{m_k}\) such that, setting \(O_k = \bigcup_{j=1}^{m_k} I_{k,j}\),

\begin{equation*} m^{*}\left( E_k \, \Delta \, O_k \right) < \delta , \end{equation*}

where \(E_k \, \Delta \, O_k = [E_k \sim O_k] \cup [O_k \sim E_k]\) denotes the symmetric difference. Since \(E_k\) and \(O_k\) are both measurable, so is \(E_k \, \Delta \, O_k\), and the displayed inequality reads \(m\left( E_k \, \Delta \, O_k \right) < \delta\). Define

\begin{equation*} h = \sum_{k=1}^{n} a_k \cdot \chi_{O_k \cap [a,b]} \quad \text{on } [a,b] . \end{equation*}

Each set \(O_k \cap [a,b]\) is a finite disjoint union of intervals, so each \(\chi_{O_k \cap [a,b]}\) is a step function on \([a,b]\); a finite linear combination of step functions is a step function, so \(h\) is a step function on \([a,b]\), and it has finite support because \(m([a,b]) < \infty\).

The estimate. For every \(x \in [a,b]\),

\begin{equation*} |\eta(x) - h(x)| = \left| \sum_{k=1}^{n} a_k \left[ \chi_{E_k}(x) - \chi_{O_k \cap [a,b]}(x) \right] \right| \le \sum_{k=1}^{n} |a_k| \cdot \chi_{E_k \, \Delta \, (O_k \cap [a,b])}(x), \end{equation*}

since \(|\chi_A - \chi_B| = \chi_{A \Delta B}\) for any sets \(A, B\). Because \(E_k \subseteq [a,b]\), we have \(E_k \sim (O_k \cap [a,b]) = E_k \sim O_k\) and \((O_k \cap [a,b]) \sim E_k \subseteq O_k \sim E_k\), so

\begin{equation*} E_k \, \Delta \, (O_k \cap [a,b]) \subseteq E_k \, \Delta \, O_k . \end{equation*}

All the functions here are bounded and measurable on the set \([a,b]\) of finite measure, so the monotonicity and linearity of integration, Theorem 5, apply. Since for each \(k\)

\begin{equation*} \int_{[a,b]} \chi_{E_k \, \Delta \, (O_k \cap [a,b])} = m\left( E_k \, \Delta \, (O_k \cap [a,b]) \right) \le m\left( E_k \, \Delta \, O_k \right) < \delta , \end{equation*}

we obtain

\begin{equation*} \int_{[a,b]} |\eta - h| \le \sum_{k=1}^{n} |a_k| \cdot m\left( E_k \, \Delta \, O_k \right) \le \delta \sum_{k=1}^{n} |a_k| = \frac{\epsilon}{2} \cdot \frac{\sum_{k=1}^{n} |a_k|}{1 + \sum_{k=1}^{n} |a_k|} < \frac{\epsilon}{2}. \end{equation*}

(If \(n = 0\), that is if \(\eta \equiv 0\) on \(E\), take \(h \equiv 0\); the estimate is then trivial.)

Combining with the triangle inequality established above, \(\int_E |f - h| < \epsilon\), as required.

Remark. The step function \(h\) produced in Part 2 need not satisfy \(0 \le h \le f\), and in general no step function can: \(f\) may be unbounded, and more to the point the sets \(O_k\) may spill outside the \(E_k\). What the two parts together express is the density, in the \(L^1\) sense, of the simple functions of finite support in the nonnegative integrable functions on an arbitrary measurable set, and of the step functions in the nonnegative integrable functions on a closed, bounded interval.

Let \(E\) be a measurable subset of \(\mathbb{R}\) and write \(E^c = \mathbb{R} \setminus E\), so that \(E\) and \(E^c\) are disjoint measurable subsets of \(\mathbb{R}\) whose union is \(\mathbb{R}\).

Each \(f_n\) is nonnegative and measurable, and \(f\), being the pointwise limit of measurable functions, is nonnegative and measurable as well. Put

\begin{equation*} L = \int_{\mathbb{R}} f = \lim_{n \to \infty} \int_{\mathbb{R}} f_n . \end{equation*}

Since \(f\) is integrable over \(\mathbb{R}\) we have \(L < \infty\). By Theorem 11 of Chapter 4 (Additivity Over Domains of Integration), applied to the nonnegative measurable function \(f\) and the disjoint sets \(E\), \(E^c\),

\begin{equation*} \int_E f + \int_{E^c} f = \int_{\mathbb{R}} f = L < \infty , \end{equation*}

so both \(\int_E f\) and \(\int_{E^c} f\) are finite. Likewise, for every index \(n\),

\begin{equation*} \int_E f_n + \int_{E^c} f_n = \int_{\mathbb{R}} f_n . \tag{\(\ast\)} \end{equation*}

Since \(\int_{\mathbb{R}} f_n \to L < \infty\), there is an index \(N\) for which \(\int_{\mathbb{R}} f_n \le L + 1 < \infty\) whenever \(n \ge N\); by \((\ast)\) and the nonnegativity of the integrands, both \(\int_E f_n\) and \(\int_{E^c} f_n\) are then finite for \(n \ge N\). Thus for \(n \ge N\) we may rewrite \((\ast)\) as

\begin{equation*} \int_E f_n = \int_{\mathbb{R}} f_n - \int_{E^c} f_n , \end{equation*}

an honest difference of real numbers.

The restriction of \(\{f_n\}\) to \(E\) is a sequence of nonnegative measurable functions converging pointwise on \(E\) to \(f\), so Fatou’s Lemma gives

\begin{equation*} \int_E f \le \liminf_{n \to \infty} \int_E f_n . \tag{1} \end{equation*}

The same reasoning on \(E^c\) gives

\begin{equation*} \int_{E^c} f \le \liminf_{n \to \infty} \int_{E^c} f_n . \tag{2} \end{equation*}

Now we bound the upper limit on \(E\). Since the sequence of real numbers \(\left\{\int_{\mathbb{R}} f_n\right\}\) converges to the finite number \(L\), for any sequence \(\{b_n\}\) of real numbers we have \(\limsup_{n} \left( \int_{\mathbb{R}} f_n - b_n \right) = L - \liminf_n b_n\). Taking \(b_n = \int_{E^c} f_n\) (which is a real number for \(n \ge N\)) and using \((2)\),

\begin{equation*} \limsup_{n \to \infty} \int_E f_n = L - \liminf_{n \to \infty} \int_{E^c} f_n \le L - \int_{E^c} f = \int_{\mathbb{R}} f - \int_{E^c} f = \int_E f , \end{equation*}

where the inequality follows from \((2)\) upon multiplying by \(-1\), and the last equality is the additivity relation for \(f\) recorded above, a legitimate subtraction because \(\int_{E^c} f\) is finite.

Combining this with \((1)\),

\begin{equation*} \int_E f \le \liminf_{n \to \infty} \int_E f_n \le \limsup_{n \to \infty} \int_E f_n \le \int_E f . \end{equation*}

Hence all three quantities agree, the limit \(\lim_{n \to \infty} \int_E f_n\) exists, and

\begin{equation*} \lim_{n \to \infty} \int_E f_n = \int_E f . \end{equation*}

Since \(E\) was an arbitrary measurable set, the proof is complete.

For each natural number \(n\) put \(I_n = [n, n+1)\) and

\begin{equation*} u_n = a_n \cdot \chi_{I_n} . \end{equation*}

The intervals \(I_1, I_2, \dots\) are pairwise disjoint and their union is \(E = [1, \infty)\). Consequently, for each \(x \in E\) there is exactly one index \(n\) with \(x \in I_n\), and for that index \(f(x) = a_n = u_n(x)\), while \(u_k(x) = 0\) for every \(k \ne n\). Therefore

\begin{equation*} f = \sum_{n=1}^{\infty} u_n \quad \text{pointwise on } E , \end{equation*}

the series having at most one nonzero term at each point.

Each \(u_n\) is a nonnegative measurable function on \(E\): indeed \(I_n\) is measurable, so \(\chi_{I_n}\) is measurable, and a constant multiple of a measurable function is measurable. In particular \(f\), being a pointwise sum of measurable functions (equivalently: \(f\) is constant on each of the countably many measurable pieces \(I_n\), so \(\{x \in E \mid f(x) > c\}\) is a countable union of intervals for every \(c\)), is a nonnegative measurable function on \(E\).

Each \(u_n\) is a nonnegative simple function of finite support: it takes only the two values \(a_n\) and \(0\), and \(\{x \in E \mid u_n(x) \ne 0\} \subseteq I_n\) has measure at most \(1\). Hence, by the definition of the integral of a simple function,

\begin{equation*} \int_E u_n = a_n \cdot m(I_n) = a_n \cdot 1 = a_n . \end{equation*}

(If \(a_n = 0\) this reads \(0 = 0\).)

Now apply Corollary 12 of Chapter 4, which asserts that if \(f = \sum_{n=1}^{\infty} u_n\) pointwise a.e. on \(E\) with each \(u_n\) nonnegative and measurable, then \(\int_E f = \sum_{n=1}^{\infty} \int_E u_n\). We conclude

\begin{equation*} \int_E f = \sum_{n=1}^{\infty} \int_E u_n = \sum_{n=1}^{\infty} a_n , \end{equation*}

the equality holding in the extended sense: both sides are simultaneously finite with the same value, or both equal \(\infty\).

For completeness, here is the argument behind Corollary 12 in this concrete case. Let \(s_N = \sum_{n=1}^{N} u_n\). Since the \(u_n\) are nonnegative, \(\{s_N\}\) is an increasing sequence of nonnegative measurable functions on \(E\), and \(s_N \to f\) pointwise on \(E\) (for \(x \in I_n\) we have \(s_N(x) = f(x)\) as soon as \(N \ge n\)). The Monotone Convergence Theorem and the linearity of integration for nonnegative measurable functions give

\begin{equation*} \int_E f = \lim_{N \to \infty} \int_E s_N = \lim_{N \to \infty} \sum_{n=1}^{N} \int_E u_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n = \sum_{n=1}^{\infty} a_n . \end{equation*}

Throughout, a function is said to be of finite support when the set on which it fails to vanish has finite measure, and \(f\) is permitted to take the value \(\infty\).

Part (i). For each natural number \(n\) define \(g_n\) on \(E\) by

\begin{equation*} g_n(x) = \min \left\{ n, \; \frac{\lfloor 2^n f(x) \rfloor}{2^n} \right\}, \end{equation*}

with the convention \(g_n(x) = n\) where \(f(x) = \infty\), and set

\begin{equation*} E_n = E \cap [-n, n], \qquad \varphi_n = g_n \cdot \chi_{E_n} . \end{equation*}

Each \(\varphi_n\) is a nonnegative simple function of finite support. Indeed, \(g_n\) takes values only in the finite set \(\{k/2^n \mid k = 0, 1, \dots, n 2^n\}\), and for each such value \(c\) the set \(\{x \in E \mid g_n(x) = c\}\) is measurable because it is described by inequalities on \(f\), which is measurable; hence \(g_n\) is simple. Multiplying by \(\chi_{E_n}\) keeps the range inside that same finite set, and \(E_n\) is measurable, so \(\varphi_n\) is simple. Moreover \(\{x \in E \mid \varphi_n(x) \ne 0\} \subseteq E_n \subseteq [-n, n]\), a set of measure at most \(2n < \infty\); thus \(\varphi_n\) has finite support. Nonnegativity is clear, and since \(\lfloor 2^n t \rfloor / 2^n \le t\) for \(t \ge 0\) we also have \(0 \le \varphi_n \le g_n \le f\) on \(E\).

The sequence is increasing. First, \(\{g_n\}\) is increasing: for a real number \(t \ge 0\) we have \(2^{n+1} t \ge 2 \lfloor 2^n t \rfloor\), and \(2 \lfloor 2^n t \rfloor\) is an integer, so by the defining property of the floor function \(\lfloor 2^{n+1} t \rfloor \ge 2 \lfloor 2^n t \rfloor\); dividing by \(2^{n+1}\) gives

\begin{equation*} \frac{\lfloor 2^{n+1} t \rfloor}{2^{n+1}} \ge \frac{\lfloor 2^n t \rfloor}{2^n}, \end{equation*}

and since also \(n+1 \ge n\), the minimum of the two increasing quantities increases: \(g_{n+1} \ge g_n\) (this also holds trivially where \(f = \infty\), where \(g_n = n\)). Second, \(E_n \subseteq E_{n+1}\), so \(\chi_{E_n} \le \chi_{E_{n+1}}\). As all quantities are nonnegative,

\begin{equation*} \varphi_n = g_n \chi_{E_n} \le g_{n+1} \chi_{E_{n+1}} = \varphi_{n+1} \quad \text{on } E . \end{equation*}

The sequence converges pointwise to \(f\). Fix \(x \in E\). Since \(x\) is a real number, \(x \in E_n\) for all \(n \ge |x|\), and for such \(n\), \(\varphi_n(x) = g_n(x)\). If \(f(x) < \infty\), then for every \(n > f(x)\) we have \(\lfloor 2^n f(x)\rfloor / 2^n \le f(x) < n\), so \(g_n(x) = \lfloor 2^n f(x)\rfloor / 2^n\) and

\begin{equation*} 0 \le f(x) - g_n(x) < \frac{1}{2^n} , \end{equation*}

whence \(\varphi_n(x) \to f(x)\). If \(f(x) = \infty\), then \(\varphi_n(x) = n \to \infty = f(x)\). In both cases \(\varphi_n(x) \to f(x)\), which proves (i).

Part (ii). Write

\begin{equation*} S = \sup \left\{ \int_E \varphi \;\middle|\; \varphi \text{ simple, of finite support, } 0 \le \varphi \le f \text{ on } E \right\}. \end{equation*}

First, \(S \le \int_E f\). A simple function \(\varphi\) of finite support with \(0 \le \varphi \le f\) takes finitely many real values, so it is a bounded measurable function of finite support with \(0 \le \varphi \le f\) on \(E\). By the very definition of the integral of the nonnegative measurable function \(f\) over \(E\),

\begin{equation*} \int_E f = \sup \left\{ \int_E h \;\middle|\; h \text{ bounded, measurable, of finite support, } 0 \le h \le f \text{ on } E \right\}, \end{equation*}

each such \(\int_E \varphi\) is one of the numbers over which this supremum is taken. Hence \(S \le \int_E f\).

Second, \(\int_E f \le S\). Let \(\{\varphi_n\}\) be the sequence produced in part (i). Each \(\varphi_n\) is simple, of finite support, and satisfies \(0 \le \varphi_n \le f\) on \(E\), so \(\int_E \varphi_n \le S\) for every \(n\). Since \(\{\varphi_n\}\) is an increasing sequence of nonnegative measurable functions on \(E\) converging pointwise on \(E\) to \(f\), the Monotone Convergence Theorem gives

\begin{equation*} \int_E f = \lim_{n \to \infty} \int_E \varphi_n \le S . \end{equation*}

The two inequalities together give \(\int_E f = S\), which is (ii).

The limit function \(f\) is nonnegative and measurable on \(E\), being the pointwise limit on \(E\) of the sequence \(\{f_n\}\) of nonnegative measurable functions. Thus \(\int_E f\) is defined as an element of \([0, \infty]\), as is each \(\int_E f_n\).

Upper bound. Fix an index \(n\). By hypothesis \(0 \le f_n \le f\) on \(E\), so the monotonicity of integration for nonnegative measurable functions gives

\begin{equation*} \int_E f_n \le \int_E f . \end{equation*}

Since this holds for every \(n\), the upper limit obeys

\begin{equation*} \limsup_{n \to \infty} \int_E f_n \le \int_E f . \tag{1} \end{equation*}

Lower bound. The sequence \(\{f_n\}\) consists of nonnegative measurable functions on \(E\) and converges pointwise on \(E\) to \(f\), so Fatou’s Lemma applies and gives

\begin{equation*} \int_E f \le \liminf_{n \to \infty} \int_E f_n . \tag{2} \end{equation*}

Conclusion. Combining \((2)\) and \((1)\),

\begin{equation*} \int_E f \le \liminf_{n \to \infty} \int_E f_n \le \limsup_{n \to \infty} \int_E f_n \le \int_E f . \end{equation*}

Hence the lower and upper limits of the sequence of extended real numbers \(\left\{\int_E f_n\right\}\) coincide and equal \(\int_E f\). Therefore the limit exists in \([0, \infty]\) and

\begin{equation*} \lim_{n \to \infty} \int_E f_n = \int_E f . \end{equation*}

Observe that no integrability hypothesis was needed: if \(\int_E f = \infty\) the chain of inequalities forces \(\int_E f_n \to \infty\) as well. The hypothesis \(f_n \le f\) is what rules out the loss of mass exhibited by the examples following Fatou’s Lemma (for instance \(f_n = n \cdot \chi_{(0, 1/n)}\) on \(E = (0,1]\), where no such dominating limit function is available since \(f \equiv 0\) there).

The Monotone Convergence Theorem asserts that if \(\{f_n\}\) is an increasing sequence of nonnegative measurable functions on \(E\) converging pointwise a.e. on \(E\) to \(f\), then \(\int_E f_n \to \int_E f\). We exhibit a decreasing sequence of nonnegative measurable functions converging pointwise everywhere for which the conclusion fails.

An example on a set of infinite measure. Let \(E = \mathbb{R}\) and for each natural number \(n\) define

\begin{equation*} f_n = \chi_{[n, \infty)} . \end{equation*}

Each \(f_n\) is nonnegative and measurable, being the characteristic function of a measurable set. Since \([n+1, \infty) \subseteq [n, \infty)\), we have \(f_{n+1} \le f_n\) on \(\mathbb{R}\), so \(\{f_n\}\) is decreasing. It converges pointwise on all of \(\mathbb{R}\) to \(f \equiv 0\): given \(x \in \mathbb{R}\), choose a natural number \(n_0 > x\); then \(x \notin [n, \infty)\) and so \(f_n(x) = 0\) for all \(n \ge n_0\).

Now compute the integrals. For each \(n\) and each natural number \(k\), the bounded measurable function \(\chi_{[n, n+k]}\) has finite support and satisfies \(0 \le \chi_{[n, n+k]} \le f_n\) on \(\mathbb{R}\), and \(\int_{\mathbb{R}} \chi_{[n, n+k]} = m([n, n+k]) = k\). Since \(\int_{\mathbb{R}} f_n\) is the supremum of the integrals of such functions, \(\int_{\mathbb{R}} f_n \ge k\) for every \(k\), that is,

\begin{equation*} \int_{\mathbb{R}} f_n = \infty \quad \text{for every } n . \end{equation*}

On the other hand \(\int_{\mathbb{R}} f = \int_{\mathbb{R}} 0 = 0\). Therefore

\begin{equation*} \lim_{n \to \infty} \int_{\mathbb{R}} f_n = \infty \ne 0 = \int_{\mathbb{R}} f , \end{equation*}

so the conclusion of the Monotone Convergence Theorem fails for this decreasing sequence.

An example on a set of finite measure. Infinite measure is not the essential feature. Let \(E = (0, 1)\) and define

\begin{equation*} f_n(x) = \frac{1}{x} \cdot \chi_{(0, 1/n)}(x) \quad \text{for } x \in (0,1). \end{equation*}

Each \(f_n\) is nonnegative and measurable, and since \((0, 1/(n+1)) \subseteq (0, 1/n)\) the sequence is decreasing. It converges pointwise on \((0,1)\) to \(f \equiv 0\): for fixed \(x \in (0,1)\) we have \(f_n(x) = 0\) once \(1/n \le x\). Yet \(\int_{(0,1)} f_n = \infty\) for every \(n\). To see this, fix \(n\) and choose a natural number \(N\) with \(2^{-N} < 1/n\). For \(j \ge N\) set \(A_j = \left(2^{-(j+1)}, 2^{-j}\right)\); these are pairwise disjoint subsets of \((0, 1/n)\), on each of which \(1/x > 2^{j}\), and \(m(A_j) = 2^{-(j+1)}\). For a natural number \(M \ge N\) the function

\begin{equation*} h_M = \sum_{j=N}^{M} 2^{j} \chi_{A_j} \end{equation*}

is bounded, measurable, of finite support, and satisfies \(0 \le h_M \le f_n\) on \((0,1)\), while

\begin{equation*} \int_{(0,1)} h_M = \sum_{j=N}^{M} 2^{j} \cdot 2^{-(j+1)} = \frac{M - N + 1}{2} . \end{equation*}

Since \(\int_{(0,1)} f_n\) is the supremum of the integrals of such functions and \(M\) is arbitrary, \(\int_{(0,1)} f_n = \infty\). Meanwhile \(\int_{(0,1)} f = 0\), so again \(\lim_n \int_E f_n \ne \int_E f\).

Remark. The failure in both examples is caused by non-integrability of the first term, and this is no accident. Suppose \(\{f_n\}\) is a decreasing sequence of nonnegative measurable functions on \(E\) converging pointwise on \(E\) to \(f\), and suppose \(f_1\) is integrable over \(E\). By Proposition 13 of Chapter 4, \(f_1\) is finite a.e. on \(E\), so after excising a set of measure zero (permissible by \((15)\)) we may assume \(f_1\) is finite on \(E\). Then \(\{f_1 - f_n\}\) is an increasing sequence of nonnegative measurable functions converging pointwise on \(E\) to \(f_1 - f\), so the Monotone Convergence Theorem gives \(\int_E (f_1 - f_n) \to \int_E (f_1 - f)\). Now \(f_1 - f_n\) and \(f_n\) are nonnegative measurable functions on \(E\) whose sum is \(f_1\), so the linearity of integration for nonnegative measurable functions, Theorem 10 of Chapter 4, gives \(\int_E (f_1 - f_n) + \int_E f_n = \int_E f_1\); since \(0 \le f_n \le f_1\) and \(\int_E f_1 < \infty\), monotonicity makes both summands finite and we may write \(\int_E (f_1 - f_n) = \int_E f_1 - \int_E f_n\). The same argument applied to \(f_1 - f\) and \(f\) gives \(\int_E (f_1 - f) = \int_E f_1 - \int_E f\). Cancelling the finite number \(\int_E f_1\) yields \(\int_E f_n \to \int_E f\). Hence any counterexample must have \(\int_E f_n = \infty\) for every \(n\), as above.

No convergence of \(\{f_n\}\) is assumed; the point is that the pointwise lower limit always exists as an extended real-valued function and obeys the Fatou inequality.

For each natural number \(n\) define the function \(g_n\) on \(E\) by

\begin{equation*} g_n(x) = \inf_{k \ge n} f_k(x), \qquad x \in E , \end{equation*}

and set \(g = \liminf_{n \to \infty} f_n\), that is, \(g(x) = \lim_{n \to \infty} g_n(x)\) for \(x \in E\).

Step 1: each \(g_n\) is a nonnegative measurable function on \(E\). Nonnegativity is immediate, since each \(f_k \ge 0\). For measurability, note that it suffices to show that \(\{x \in E \mid g_n(x) \ge c\}\) is measurable for every real number \(c\), since the complements of these sets are the sets \(\{g_n < c\}\) and measurability of a function is equivalent to measurability of all the sets \(\{g_n < c\}\). Now for every real \(c\),

\begin{equation*} \{x \in E \mid g_n(x) \ge c\} = \bigcap_{k \ge n} \{x \in E \mid f_k(x) \ge c\} , \end{equation*}

because \(\inf_{k \ge n} f_k(x) \ge c\) precisely when \(f_k(x) \ge c\) for every \(k \ge n\). Each set on the right is measurable since \(f_k\) is measurable, and the countable intersection of measurable sets is measurable. As the sets \(\{g_n \ge c\}\) are measurable for every real \(c\), the function \(g_n\) is measurable. (This is the standard fact that the pointwise infimum of a countable family of measurable functions is measurable.)

Step 2: \(\{g_n\}\) is increasing and converges pointwise on \(E\) to \(g\). Fix \(x \in E\). The family \(\{f_k(x) \mid k \ge n+1\}\) is a subfamily of \(\{f_k(x) \mid k \ge n\}\), so its infimum is at least as large:

\begin{equation*} g_n(x) \le g_{n+1}(x) . \end{equation*}

Thus \(\{g_n(x)\}\) is an increasing sequence in \([0, \infty]\), hence convergent in \([0,\infty]\), and its limit is by definition \(\liminf_{n} f_n(x) = g(x)\). In particular \(g\) is a nonnegative measurable function on \(E\), being a pointwise limit of nonnegative measurable functions, so \(\int_E g\) is defined.

Step 3: comparison of integrals. Fix \(n\) and let \(k \ge n\). By construction \(0 \le g_n \le f_k\) on \(E\), so the monotonicity of integration for nonnegative measurable functions gives

\begin{equation*} \int_E g_n \le \int_E f_k \qquad \text{for every } k \ge n . \end{equation*}

Taking the infimum over \(k \ge n\) on the right,

\begin{equation*} \int_E g_n \le \inf_{k \ge n} \int_E f_k . \tag{\(\ast\)} \end{equation*}

Step 4: pass to the limit. The sequence \(\{g_n\}\) is an increasing sequence of nonnegative measurable functions on \(E\) converging pointwise on \(E\) to \(g\), so the Monotone Convergence Theorem yields

\begin{equation*} \int_E g = \lim_{n \to \infty} \int_E g_n . \end{equation*}

The right-hand sides of \((\ast)\) form an increasing sequence in \([0,\infty]\) whose limit is, by definition, \(\liminf_{n \to \infty} \int_E f_n\). Letting \(n \to \infty\) in \((\ast)\) therefore gives

\begin{equation*} \int_E \liminf_{n \to \infty} f_n = \int_E g = \lim_{n \to \infty} \int_E g_n \le \lim_{n \to \infty} \left( \inf_{k \ge n} \int_E f_k \right) = \liminf_{n \to \infty} \int_E f_n , \end{equation*}

which is the asserted inequality.

This contains Fatou’s Lemma as stated in the text: if \(\{f_n\} \to f\) pointwise a.e. on \(E\), then \(\liminf_n f_n = f\) a.e. on \(E\), and integrals over \(E\) are unchanged by altering the integrand on a set of measure zero, by \((15)\).

Here \(\chi_C\) denotes the characteristic function of \(C\), regarded as a function on \(E\); since \(C\) is measurable, \(\chi_C\) is measurable on \(E\), and therefore so is the product \(f \cdot \chi_C\).

Step 1: the identity for nonnegative measurable functions. Let \(h\) be a nonnegative measurable function on \(E\) and let \(C \subseteq E\) be measurable. We claim

\begin{equation*} \int_C h = \int_E h \cdot \chi_C . \tag{\(\ast\)} \end{equation*}

Indeed, \(h \cdot \chi_C\) is a nonnegative measurable function on \(E\), and \(C\) and \(E \setminus C\) are disjoint measurable subsets of \(E\) with union \(E\). By Theorem 11 of Chapter 4 (Additivity Over Domains of Integration),

\begin{equation*} \int_E h \cdot \chi_C = \int_C h \cdot \chi_C + \int_{E \setminus C} h \cdot \chi_C . \end{equation*}

On \(C\) we have \(\chi_C \equiv 1\), so \(h \cdot \chi_C = h\) there and the first integral equals \(\int_C h\); the integral over a set depends only on the restriction of the integrand to that set. On \(E \setminus C\) we have \(\chi_C \equiv 0\), so \(h \cdot \chi_C\) is the zero function there and the second integral is \(0\) (the only bounded measurable function \(\psi\) of finite support with \(0 \le \psi \le 0\) is \(\psi \equiv 0\), whose integral is \(0\); hence the defining supremum is \(0\)). This proves \((\ast)\).

Step 2: integrability of the two sides. Since \(f\) is integrable over \(E\), \(|f|\) is integrable over \(E\). Because \(|f \cdot \chi_C| \le |f|\) on \(E\), Proposition 16 of Chapter 4 (the Integral Comparison Test) shows that \(f \cdot \chi_C\) is integrable over \(E\), so the right-hand side of the desired identity is a well-defined real number. Moreover, applying \((\ast)\) with \(h = |f|\),

\begin{equation*} \int_C |f| = \int_E |f| \cdot \chi_C \le \int_E |f| < \infty , \end{equation*}

where the inequality is the monotonicity of integration for nonnegative measurable functions applied to \(|f| \cdot \chi_C \le |f|\). Hence \(|f|\) is integrable over \(C\), i.e. \(f\) is integrable over \(C\), and the left-hand side is a well-defined real number too.

Step 3: reduction to positive and negative parts. Write \(f = f^{+} - f^{-}\), where \(f^{+} = \max\{f, 0\}\) and \(f^{-} = \max\{-f, 0\}\) are nonnegative measurable functions on \(E\), each integrable over \(E\) by Proposition 14 of Chapter 4. At each point of \(E\), at most one of \(f^{+}, f^{-}\) is nonzero, and multiplying by the factor \(\chi_C\), which takes only the values \(0\) and \(1\), preserves the description of the positive and negative parts: for \(x \in E\),

\begin{equation*} (f \cdot \chi_C)^{+}(x) = \max\{ f(x) \chi_C(x), 0\} = \chi_C(x) \max\{f(x), 0\} = f^{+}(x) \chi_C(x) , \end{equation*}

using \(\chi_C(x) \in \{0, 1\}\), and likewise \((f \cdot \chi_C)^{-} = f^{-} \cdot \chi_C\) on \(E\).

Therefore, by the definition of the general Lebesgue integral applied on \(C\) and then on \(E\), together with \((\ast)\) applied to the nonnegative measurable functions \(f^{+}\) and \(f^{-}\),

\begin{equation*} \begin{aligned} \int_C f &= \int_C f^{+} - \int_C f^{-} \\ &= \int_E f^{+} \cdot \chi_C - \int_E f^{-} \cdot \chi_C \\ &= \int_E (f \cdot \chi_C)^{+} - \int_E (f \cdot \chi_C)^{-} \\ &= \int_E f \cdot \chi_C . \end{aligned} \end{equation*}

All four integrals of nonnegative functions appearing here are finite, by Step 2 and monotonicity, so the subtractions are legitimate. This is the asserted identity.

Note that the argument uses only Theorem 11, which concerns nonnegative measurable functions; it does not use Corollary 18 (Additivity Over Domains of Integration for integrable functions), whose proof in the text invokes the present result.

The answer to both questions is no: in each case one implication is true and the other fails.

First observe that each \(a_n\) is a well-defined real number. Indeed, \(f\) is measurable and bounded on the bounded set \([n, n+1]\), which has finite measure, so \(|f| \le M_n\) there for some \(M_n \ge 0\) and hence, by the monotonicity of integration for nonnegative functions, \(\int_{[n,n+1]} |f| \le M_n \cdot 1 < \infty\). Thus \(f\) is integrable over \([n, n+1]\) and \(a_n = \int_n^{n+1} f\) is finite.

The implication that is true. Suppose \(f\) is integrable over \([1, \infty)\). For \(n \ge 1\) put \(u_n = |f| \cdot \chi_{[n, n+1)}\), a nonnegative measurable function on \([1, \infty)\). Since the intervals \([n, n+1)\) are disjoint with union \([1, \infty)\), we have \(|f| = \sum_{n=1}^{\infty} u_n\) pointwise on \([1, \infty)\), and therefore, by Corollary 12 of Chapter 4,

\begin{equation*} \int_{[1,\infty)} |f| \;=\; \sum_{n=1}^{\infty} \int_{[1,\infty)} u_n \;=\; \sum_{n=1}^{\infty} \int_{[n,n+1)} |f| , \end{equation*}

where the last equality is the additivity over domains of integration for nonnegative functions (Theorem 11 of Chapter 4), which gives \(\int_{[1,\infty)} u_n = \int_{[n,n+1)} u_n + \int_{[1,\infty) \sim [n,n+1)} u_n = \int_{[n,n+1)} |f| + 0\). By the Integral Comparison Test (Proposition 16 of Chapter 4),

\begin{equation*} |a_n| \;=\; \left| \int_{[n,n+1)} f \right| \;\le\; \int_{[n,n+1)} |f| , \end{equation*}

where we have used that the value of the integral of an integrable function is unchanged by excising the single point \(n+1\), a set of measure zero, so that \(a_n = \int_{[n,n+1]} f = \int_{[n,n+1)} f\); this is (18) of Proposition 15 of Chapter 4. Hence

\begin{equation*} \sum_{n=1}^{\infty} |a_n| \;\le\; \sum_{n=1}^{\infty} \int_{[n,n+1)} |f| \;=\; \int_{[1,\infty)} |f| \;<\; \infty . \end{equation*}

So integrability of \(f\) implies that \(\sum_{n=1}^{\infty} a_n\) converges absolutely, and in particular that it converges. This is the true half of both assertions.

The first converse fails. Define \(f\) on \([1, \infty)\) by

\begin{equation*} f(x) \;=\; \frac{(-1)^n}{n} \quad \text{for } n \le x < n+1,\ n = 1, 2, \dots \end{equation*}

Then \(f\) is measurable (it is constant on each interval of a countable disjoint measurable decomposition of \([1,\infty)\)) and \(|f| \le 1\), so \(f\) is certainly bounded on bounded sets. Here \(a_n = (-1)^n/n\), and \(\sum_{n=1}^{\infty} (-1)^n/n\) converges by the alternating series test. But by the computation above,

\begin{equation*} \int_{[1,\infty)} |f| \;=\; \sum_{n=1}^{\infty} \int_{[n,n+1)} |f| \;=\; \sum_{n=1}^{\infty} \frac{1}{n} \;=\; \infty , \end{equation*}

so \(|f|\) is not integrable, that is, \(f\) is not integrable over \([1, \infty)\). Thus convergence of \(\sum_{n=1}^{\infty} a_n\) does not imply integrability, and the answer to the first question is no.

The second converse fails. Define \(f\) on \([1, \infty)\) by setting, for each natural number \(n\), \(f(x) = 1\) for \(n \le x < n + 1/2\) and \(f(x) = -1\) for \(n + 1/2 \le x < n+1\). Again \(f\) is measurable and \(|f| \equiv 1\), so \(f\) is bounded on bounded sets. Fix \(n\). Excising the single point \(n+1\), which changes no integral by (18), and then applying additivity over domains for integrable functions (Corollary 18 of Chapter 4) to the disjoint decomposition \([n, n+1) = [n,\, n+1/2) \cup [n+1/2,\, n+1)\),

\begin{equation*} a_n \;=\; \int_{[n,\,n+1)} f \;=\; \int_{[n,\, n+1/2)} 1 + \int_{[n+1/2,\, n+1)} (-1) \;=\; \tfrac{1}{2} - \tfrac{1}{2} \;=\; 0 . \end{equation*}

Hence \(\sum_{n=1}^{\infty} |a_n| = 0 < \infty\), so the series converges absolutely. Yet \(\int_{[1,\infty)} |f| = \sum_{n=1}^{\infty} \int_{[n,n+1)} 1 = \sum_{n=1}^{\infty} 1 = \infty\), so \(f\) is not integrable over \([1, \infty)\). Thus the answer to the second question is also no.

Remark. What is true in general is that \(f\) is integrable over \([1, \infty)\) if and only if \(\sum_{n=1}^{\infty} b_n < \infty\), where \(b_n = \int_n^{n+1} |f|\); this is exactly the identity \(\int_{[1,\infty)} |f| = \sum_{n=1}^{\infty} b_n\) established above. In particular, if \(f\) is nonnegative, then \(a_n = b_n\) and both assertions of the problem do hold: \(f\) is integrable over \([1, \infty)\) if and only if \(\sum_{n=1}^{\infty} a_n\) converges, and for a series of nonnegative terms convergence and absolute convergence coincide.

We first record the generalized form of Fatou’s Lemma that is the content of Problem 27: if \(\{h_n\}\) is a sequence of nonnegative measurable functions on \(E\), then

\begin{equation*} \int_E \liminf h_n \;\le\; \liminf \int_E h_n . \end{equation*}

Call this the generalized Fatou inequality. Indeed, put \(u_k = \inf_{n \ge k} h_n\). Each \(u_k\) is nonnegative and measurable, \(\{u_k\}\) is increasing, and \(u_k \to \liminf h_n\) pointwise on \(E\). By the Monotone Convergence Theorem, \(\lim_{k \to \infty} \int_E u_k = \int_E \liminf h_n\). On the other hand \(u_k \le h_n\) on \(E\) for every \(n \ge k\), so by the monotonicity of integration for nonnegative functions \(\int_E u_k \le \inf_{n \ge k} \int_E h_n\). Letting \(k \to \infty\) gives the generalized Fatou inequality.

Now we set up the excision. For each \(n\) let \(E_n\) be the set of measure zero outside of which \(|f_n| \le g\), and let \(E_0\) be the set of measure zero outside of which \(g\) is finite; such a set exists because \(g\) is a nonnegative integrable function, by Proposition 13 of Chapter 4. The set \(Z = E_0 \cup \bigcup_{n=1}^{\infty} E_n\) has measure zero by the countable subadditivity of Lebesgue measure. By the excision property (15) for nonnegative measurable functions, and by (18) of Proposition 15 of Chapter 4 for the integrable ones, the integral over \(E\) of any of the functions appearing in the asserted inequalities equals its integral over \(E \sim Z\), and the functions \(\liminf f_n\), \(\limsup f_n\) are unchanged off \(Z\). Hence, replacing \(E\) by \(E \sim Z\), we may and do assume that

\begin{equation*} |f_n| \le g \ \text{ on } E \ \text{ for every } n, \qquad \text{and } g \text{ is finite on } E . \end{equation*}

Three preliminary integrability remarks. First, each \(f_n\) is measurable and dominated in absolute value by the integrable function \(g\), so by the Integral Comparison Test (Proposition 16 of Chapter 4) each \(f_n\) is integrable over \(E\) and \(\int_E f_n\) is a real number; thus the two middle quantities in the assertion are the lower and upper limits of a bounded sequence of real numbers, and the middle inequality \(\liminf \int_E f_n \le \limsup \int_E f_n\) is immediate. Second, since \(|f_n| \le g\) for all \(n\) we also have \(|\liminf f_n| \le g\) and \(|\limsup f_n| \le g\) on \(E\), so both \(\liminf f_n\) and \(\limsup f_n\) are integrable over \(E\) by the same Integral Comparison Test; in particular all four quantities in the assertion are finite. Third, \(g + f_n\) and \(g - f_n\) are properly defined (because \(g\) is finite), measurable, integrable over \(E\) by the Integral Comparison Test since \(|g \pm f_n| \le 2g\), and nonnegative on \(E\) because \(|f_n| \le g\).

The left-hand inequality. Apply the generalized Fatou inequality to the nonnegative measurable sequence \(\{g + f_n\}\). Since \(g\) is finite, \(\liminf (g + f_n) = g + \liminf f_n\) pointwise on \(E\), so

\begin{equation*} \int_E g + \int_E \liminf f_n \;=\; \int_E \left( g + \liminf f_n \right) \;\le\; \liminf \int_E (g + f_n) \;=\; \int_E g + \liminf \int_E f_n , \end{equation*}

where the first and last equalities use the linearity of integration (Theorem 17 of Chapter 4), legitimate because \(g\), \(\liminf f_n\) and each \(f_n\) are integrable over \(E\), and the last equality also uses that \(\int_E g\) is a fixed real number, so that \(\liminf ( \int_E g + \int_E f_n ) = \int_E g + \liminf \int_E f_n\). Since \(\int_E g\) is finite, it may be cancelled from both ends, giving

\begin{equation*} \int_E \liminf f_n \;\le\; \liminf \int_E f_n . \end{equation*}

The right-hand inequality. Apply the generalized Fatou inequality to the nonnegative measurable sequence \(\{g - f_n\}\). Since \(g\) is finite, \(\liminf (g - f_n) = g - \limsup f_n\) pointwise on \(E\), so, again by Theorem 17,

\begin{equation*} \int_E g - \int_E \limsup f_n \;\le\; \liminf \int_E (g - f_n) \;=\; \int_E g + \liminf \left( - \int_E f_n \right) \;=\; \int_E g - \limsup \int_E f_n . \end{equation*}

Cancelling the finite number \(\int_E g\) and multiplying by \(-1\) reverses the inequality and yields

\begin{equation*} \limsup \int_E f_n \;\le\; \int_E \limsup f_n . \end{equation*}

Combining the three inequalities gives the asserted chain.

First note that the proposed definition makes sense for any single such decomposition. If \(f = g + h\) with \(g\) finite, measurable and integrable over \(E\) and \(h \ge 0\) on \(E\), then \(h = f - g\) is a properly defined extended real-valued function (no expression \(\infty - \infty\) arises, since \(g\) is finite) and it is measurable, being the difference of the measurable functions \(f\) and \(g\). Therefore \(\int_E h\) is defined as an element of \([0, \infty]\), while \(\int_E g\) is a real number, and the sum \(\int_E g + \int_E h\) is an unambiguous element of \((-\infty, \infty]\).

Now suppose we have two such decompositions,

\begin{equation*} f \;=\; g_1 + h_1 \;=\; g_2 + h_2 \quad \text{on } E, \end{equation*}

where \(g_1, g_2\) are finite and integrable over \(E\) and \(h_1, h_2\) are nonnegative measurable functions on \(E\). We must show

\begin{equation*} \int_E g_1 + \int_E h_1 \;=\; \int_E g_2 + \int_E h_2 . \end{equation*}

The difficulty is that \(h_1\) and \(h_2\) may take the value \(\infty\), so we may not simply subtract. We therefore rearrange the identity \(g_1 + h_1 = g_2 + h_2\) into an identity between nonnegative functions. Write \(g_i = g_i^{+} - g_i^{-}\), where \(g_i^{+}\) and \(g_i^{-}\) are the nonnegative and finite positive and negative parts of \(g_i\); each of \(g_1^{+}, g_1^{-}, g_2^{+}, g_2^{-}\) is a nonnegative measurable function on \(E\) which is integrable over \(E\), by Proposition 14 of Chapter 4. Adding the finite nonnegative function \(g_1^{-} + g_2^{-}\) to both sides of \(g_1 + h_1 = g_2 + h_2\) gives

\begin{equation*} h_1 + g_1^{+} + g_2^{-} \;=\; h_2 + g_2^{+} + g_1^{-} \quad \text{on } E . \end{equation*}

Let us verify this identity carefully, since extended values occur. At a point \(x\) where \(h_1(x)\) and \(h_2(x)\) are both finite, every term is a real number and the identity is the elementary rearrangement

\begin{equation*} h_1 + (g_1 + g_1^{-}) + g_2^{-} \;=\; (g_1 + h_1) + g_1^{-} + g_2^{-} \;=\; (g_2 + h_2) + g_2^{-} + g_1^{-} \;=\; h_2 + (g_2 + g_2^{-}) + g_1^{-}, \end{equation*}

using \(g_i + g_i^{-} = g_i^{+}\). At a point \(x\) where \(h_1(x) = \infty\), we get \(f(x) = g_1(x) + \infty = \infty\) because \(g_1(x)\) is finite; since \(g_2(x)\) is finite as well and \(f(x) = g_2(x) + h_2(x)\), necessarily \(h_2(x) = \infty\). Thus \(h_1(x) = \infty\) if and only if \(h_2(x) = \infty\), and at such a point both sides of the displayed identity equal \(\infty\) (each is the sum of \(\infty\) with finite nonnegative numbers). So the identity holds at every point of \(E\), and every function appearing in it is nonnegative and measurable.

Integrate. By the linearity of integration for nonnegative measurable functions,

\begin{equation*} \int_E h_1 + \int_E g_1^{+} + \int_E g_2^{-} \;=\; \int_E h_2 + \int_E g_2^{+} + \int_E g_1^{-}, \end{equation*}

an equality of elements of \([0, \infty]\). The four numbers \(\int_E g_1^{+}, \int_E g_1^{-}, \int_E g_2^{+}, \int_E g_2^{-}\) are finite because \(g_1\) and \(g_2\) are integrable over \(E\). Hence we may subtract the finite number \(\int_E g_1^{-} + \int_E g_2^{-}\) from both sides, which is a legitimate operation in \((-\infty, \infty]\), to obtain

\begin{equation*} \int_E h_1 + \left( \int_E g_1^{+} - \int_E g_1^{-} \right) \;=\; \int_E h_2 + \left( \int_E g_2^{+} - \int_E g_2^{-} \right), \end{equation*}

that is, by the definition of the integral of an integrable function,

\begin{equation*} \int_E h_1 + \int_E g_1 \;=\; \int_E h_2 + \int_E g_2 . \end{equation*}

This is precisely the required independence: the value \(\int_E g + \int_E h\) does not depend on the decomposition chosen. (In particular both sides are simultaneously \(+\infty\), or simultaneously the same real number.)

Finally, the new definition is consistent with the old one. If \(f\) is itself integrable over \(E\), then \(f\) is finite a.e. on \(E\) by Proposition 15 of Chapter 4; excising a set of measure zero, which changes no integral by (18) of that same proposition, we may take \(f\) finite on \(E\) and use the decomposition \(g = f\), \(h \equiv 0\), for which \(\int_E g + \int_E h = \int_E f\), the integral of \(f\) in the sense of Section 4.4. By the independence just proved, every admissible decomposition gives this same value.

We must prove Theorem 19 of Chapter 4: let \(\{f_n\}\) be a sequence of measurable functions on \(E\) that converges pointwise a.e. on \(E\) to \(f\), and suppose there is a sequence \(\{g_n\}\) of nonnegative measurable functions on \(E\) that converges pointwise a.e. on \(E\) to \(g\) and dominates \(\{f_n\}\) in the sense that \(|f_n| \le g_n\) on \(E\) for all \(n\); if \(\lim_{n \to \infty} \int_E g_n = \int_E g < \infty\), then \(\lim_{n \to \infty} \int_E f_n = \int_E f\).

Preliminaries. Since \(\lim_{n \to \infty} \int_E g_n = \int_E g < \infty\), there is an index \(N\) such that \(\int_E g_n < \infty\) for all \(n \ge N\); that is, \(g_n\) is integrable over \(E\) for all \(n \ge N\). As the conclusion concerns only the behavior of \(\int_E f_n\) for large \(n\), we may discard the first \(N-1\) terms and assume that every \(g_n\) is integrable over \(E\). Then, for each \(n\), \(f_n\) is measurable and \(|f_n| \le g_n\) with \(g_n\) nonnegative and integrable, so by the Integral Comparison Test (Proposition 16 of Chapter 4) each \(f_n\) is integrable over \(E\) and \(\int_E f_n\) is a real number.

Next, \(g\) is nonnegative and measurable, being an a.e. pointwise limit of such functions, and \(\int_E g < \infty\), so \(g\) is integrable over \(E\) and hence finite a.e. on \(E\) by Proposition 13 of Chapter 4. Also each \(g_n\) is finite a.e. on \(E\), for the same reason. Passing to the limit in \(|f_n| \le g_n\) at each point where both \(f_n \to f\) and \(g_n \to g\), we get \(|f| \le g\) a.e. on \(E\); since \(f\) is measurable as an a.e. pointwise limit of measurable functions, the Integral Comparison Test shows \(f\) is integrable over \(E\).

Let \(Z\) be the union of the following sets of measure zero: the set where \(\{f_n\}\) fails to converge to \(f\), the set where \(\{g_n\}\) fails to converge to \(g\), the set where \(g\) is infinite, the set where \(|f| > g\), and, for each \(n\), the set where \(g_n\) is infinite. By the countable subadditivity of Lebesgue measure, \(m(Z) = 0\); by the excision property (15) applied to positive and negative parts, no integral occurring below is changed if \(E\) is replaced by \(E \sim Z\). So we may and do assume that on \(E\) we have: \(f_n \to f\) pointwise, \(g_n \to g\) pointwise, every \(g_n\) and \(g\) are finite, \(|f_n| \le g_n\) for all \(n\), and \(|f| \le g\).

The two Fatou arguments. For each \(n\) the functions \(g_n - f_n\) and \(g_n + f_n\) are properly defined (all the functions involved are finite on \(E\)), measurable, and nonnegative on \(E\), since \(|f_n| \le g_n\). Moreover \(g_n - f_n \to g - f\) and \(g_n + f_n \to g + f\) pointwise on \(E\), and \(g - f\), \(g + f\) are nonnegative because \(|f| \le g\).

Apply Fatou’s Lemma to \(\{g_n - f_n\}\):

\begin{equation*} \int_E (g - f) \;\le\; \liminf \int_E (g_n - f_n) . \end{equation*}

Both \(g_n\) and \(f_n\) are integrable over \(E\), so by the linearity of integration (Theorem 17 of Chapter 4), \(\int_E (g_n - f_n) = \int_E g_n - \int_E f_n\); likewise \(\int_E (g - f) = \int_E g - \int_E f\), since \(g\) and \(f\) are integrable over \(E\). Because \(\lim_{n \to \infty} \int_E g_n\) exists and is the finite number \(\int_E g\), we may split the lower limit:

\begin{equation*} \liminf \left( \int_E g_n - \int_E f_n \right) \;=\; \int_E g + \liminf \left( - \int_E f_n \right) \;=\; \int_E g - \limsup \int_E f_n . \end{equation*}

Hence

\begin{equation*} \int_E g - \int_E f \;\le\; \int_E g - \limsup \int_E f_n , \end{equation*}

and cancelling the finite number \(\int_E g\) gives

\begin{equation*} \limsup \int_E f_n \;\le\; \int_E f . \end{equation*}

Now apply Fatou’s Lemma to \(\{g_n + f_n\}\):

\begin{equation*} \int_E (g + f) \;\le\; \liminf \int_E (g_n + f_n) . \end{equation*}

Exactly as before, by Theorem 17 and the finiteness of \(\lim_{n \to \infty} \int_E g_n = \int_E g\),

\begin{equation*} \int_E g + \int_E f \;\le\; \int_E g + \liminf \int_E f_n , \end{equation*}

so that

\begin{equation*} \int_E f \;\le\; \liminf \int_E f_n . \end{equation*}

Conclusion. Combining the two displayed inequalities,

\begin{equation*} \int_E f \;\le\; \liminf \int_E f_n \;\le\; \limsup \int_E f_n \;\le\; \int_E f , \end{equation*}

so all four quantities are equal and \(\lim_{n \to \infty} \int_E f_n\) exists and equals \(\int_E f\). This is the assertion of Theorem 19.

Note that the hypothesis \(\lim_{n \to \infty} \int_E g_n = \int_E g < \infty\) was used exactly where the ordinary Dominated Convergence Theorem uses the constancy of the dominating function: it is what permits the terms \(\int_E g_n\) to be pulled out of the lower limit as the finite number \(\int_E g\) and then cancelled.

Throughout, \(f\) and each \(f_n\) are integrable over \(E\), so \(|f|\) and each \(|f_n|\) are nonnegative integrable functions on \(E\) and all the integrals below are finite real numbers. Also, \(f - f_n\) is measurable and \(|f - f_n| \le |f| + |f_n|\), which is integrable, so by the Integral Comparison Test (Proposition 16 of Chapter 4) \(f - f_n\) is integrable over \(E\) and \(\int_E |f - f_n| < \infty\).

Necessity. Suppose \(\int_E |f - f_n| \to 0\). For each \(n\), the reverse triangle inequality for real numbers gives

\begin{equation*} \big| \, |f_n(x)| - |f(x)| \, \big| \;\le\; |f(x) - f_n(x)| \quad \text{for a.e. } x \in E \end{equation*}

(at every point where \(f\) and \(f_n\) are finite, hence a.e. on \(E\) by Proposition 15 of Chapter 4). By the linearity of integration (Theorem 17 of Chapter 4), the Integral Comparison Test, and the monotonicity of integration,

\begin{equation*} \left| \int_E |f_n| - \int_E |f| \right| \;=\; \left| \int_E \big( |f_n| - |f| \big) \right| \;\le\; \int_E \big| \, |f_n| - |f| \, \big| \;\le\; \int_E |f - f_n| . \end{equation*}

Since the right-hand side tends to \(0\), we conclude \(\lim_{n \to \infty} \int_E |f_n| = \int_E |f|\). Observe that this half did not use the pointwise convergence of \(\{f_n\}\).

Sufficiency. Suppose \(\lim_{n \to \infty} \int_E |f_n| = \int_E |f|\). Define, for each \(n\),

\begin{equation*} h_n \;=\; |f - f_n| \quad \text{and} \quad g_n \;=\; |f| + |f_n| \quad \text{on } E . \end{equation*}

We check the hypotheses of the General Lebesgue Dominated Convergence Theorem (Theorem 19 of Chapter 4) for the sequence \(\{h_n\}\) dominated by \(\{g_n\}\).

Each \(h_n\) is measurable, and \(h_n \to 0\) a.e. on \(E\): indeed at a.e. point \(x\) we have \(f_n(x) \to f(x)\) with \(f(x)\) finite (Proposition 15), so \(|f(x) - f_n(x)| \to 0\). Each \(g_n\) is nonnegative and measurable, and \(g_n \to 2|f|\) a.e. on \(E\), since \(|f_n| \to |f|\) a.e. by the continuity of the absolute value. The domination \(|h_n| = |f - f_n| \le |f| + |f_n| = g_n\) holds a.e. on \(E\) (at every point where both \(f\) and \(f_n\) are finite); excising a set of measure zero, which by (15) changes none of the integrals involved, we may assume it holds on all of \(E\).

Finally, the crucial hypothesis on the integrals of the dominating sequence holds: by the linearity of integration for nonnegative functions and the assumed convergence,

\begin{equation*} \lim_{n \to \infty} \int_E g_n \;=\; \lim_{n \to \infty} \left( \int_E |f| + \int_E |f_n| \right) \;=\; \int_E |f| + \int_E |f| \;=\; \int_E 2|f| \;<\; \infty , \end{equation*}

the finiteness because \(f\) is integrable over \(E\), and \(\int_E 2|f|\) is precisely the integral of the a.e. pointwise limit of \(\{g_n\}\).

Therefore Theorem 19 applies to \(\{h_n\}\) with dominating sequence \(\{g_n\}\) and limit function \(0\), and yields

\begin{equation*} \lim_{n \to \infty} \int_E |f - f_n| \;=\; \lim_{n \to \infty} \int_E h_n \;=\; \int_E 0 \;=\; 0 . \end{equation*}

This completes the proof of both implications.

Remark. The ordinary Lebesgue Dominated Convergence Theorem is not enough here: there is in general no single integrable function dominating all the \(|f - f_n|\), and this is exactly the situation the generalized theorem was designed for.

Here \(\int_{-n}^{n} f\) denotes \(\int_{[-n,n]} f\), and since \(f\) is nonnegative and measurable all the integrals occurring are defined as elements of \([0, \infty]\); the assertion is an equality in \([0, \infty]\).

For each natural number \(n\) define

\begin{equation*} f_n \;=\; f \cdot \chi_{[-n,n]} \quad \text{on } \mathbb{R}. \end{equation*}

Each \(f_n\) is nonnegative and measurable, being the product of the nonnegative measurable function \(f\) with the characteristic function of a measurable set.

The sequence \(\{f_n\}\) is increasing on \(\mathbb{R}\). Indeed, \([-n, n] \subseteq [-(n+1), n+1]\), so \(\chi_{[-n,n]} \le \chi_{[-(n+1), n+1]}\) and, \(f\) being nonnegative, \(f_n \le f_{n+1}\) on \(\mathbb{R}\).

The sequence \(\{f_n\}\) converges pointwise on \(\mathbb{R}\) to \(f\). Let \(x \in \mathbb{R}\) and choose a natural number \(N > |x|\), which is possible by the Archimedean property. Then \(x \in [-n, n]\) for all \(n \ge N\), so \(f_n(x) = f(x)\) for all \(n \ge N\); in particular \(f_n(x) \to f(x)\).

The Monotone Convergence Theorem therefore applies and gives

\begin{equation*} \lim_{n \to \infty} \int_{\mathbb{R}} f_n \;=\; \int_{\mathbb{R}} f , \end{equation*}

this limit existing in \([0, \infty]\) since \(\{ \int_{\mathbb{R}} f_n \}\) is increasing by the monotonicity of integration for nonnegative functions.

It remains to identify \(\int_{\mathbb{R}} f_n\) with \(\int_{-n}^{n} f\). The sets \([-n, n]\) and \(\mathbb{R} \sim [-n, n]\) are disjoint measurable sets whose union is \(\mathbb{R}\), so by the additivity over domains of integration for nonnegative measurable functions (Theorem 11 of Chapter 4),

\begin{equation*} \int_{\mathbb{R}} f_n \;=\; \int_{[-n,n]} f_n \;+\; \int_{\mathbb{R} \sim [-n,n]} f_n . \end{equation*}

On \([-n, n]\) we have \(f_n = f\), so \(\int_{[-n,n]} f_n = \int_{[-n,n]} f = \int_{-n}^{n} f\). On \(\mathbb{R} \sim [-n, n]\) we have \(f_n \equiv 0\), so \(\int_{\mathbb{R} \sim [-n,n]} f_n = 0\) by the monotonicity of integration for nonnegative functions (the integral of the zero function over any measurable set is \(0\), being the supremum of the integrals of bounded measurable functions of finite support lying below it, all of which are \(0\)). Hence

\begin{equation*} \int_{\mathbb{R}} f_n \;=\; \int_{-n}^{n} f \quad \text{for every } n . \end{equation*}

Substituting this into the conclusion of the Monotone Convergence Theorem gives

\begin{equation*} \lim_{n \to \infty} \int_{-n}^{n} f \;=\; \int_{\mathbb{R}} f , \end{equation*}

as required. Note that no integrability assumption was needed: if \(\int_{\mathbb{R}} f = \infty\), the assertion says that the increasing sequence \(\{ \int_{-n}^{n} f \}\) diverges to \(\infty\), which is exactly what the Monotone Convergence Theorem delivers in the extended sense.

Throughout, \(E = [0,1]\) and, for \(y \in [0,1]\), we write \(f_y\) for the function \(x \mapsto f(x, y)\) on \(E\). By hypothesis each \(f_y\) is measurable on \(E\) and \(|f_y| \le g\) on \(E\), and \(g\) is integrable over \(E\); note that \(g\) is automatically nonnegative, since \(g(x) \ge |f(x, 0)| \ge 0\) for every \(x \in E\). Hence, by the Integral Comparison Test (Proposition 16 of Chapter 4), each \(f_y\) is integrable over \(E\), so

\begin{equation*} H(y) \;=\; \int_0^1 f(x, y)\, dx \end{equation*}

is a well-defined real number for every \(y \in [0,1]\).

The limit function \(f\) is measurable and integrable. Fix any sequence \(\{y_n\}\) in \((0,1]\) with \(y_n \to 0\). Then for each \(x \in E\) we have \(f_{y_n}(x) = f(x, y_n) \to f(x)\) by the hypothesis \(\lim_{y \to 0} f(x, y) = f(x)\). Thus \(f\) is the pointwise limit on \(E\) of the sequence of measurable functions \(\{f_{y_n}\}\) and is therefore measurable on \(E\). Moreover \(|f(x)| = \lim_n |f(x, y_n)| \le g(x)\) for all \(x \in E\), so \(f\) is integrable over \(E\) by the Integral Comparison Test, and \(\int_0^1 f(x)\, dx\) is a real number.

The first assertion. Let \(\{y_n\}\) be an arbitrary sequence in \((0, 1]\) with \(y_n \to 0\). The sequence of measurable functions \(\{f_{y_n}\}\) converges pointwise on \(E\) to \(f\), and it is dominated on \(E\) by the single integrable function \(g\), since \(|f_{y_n}| \le g\) for every \(n\). By the Lebesgue Dominated Convergence Theorem,

\begin{equation*} \lim_{n \to \infty} H(y_n) \;=\; \lim_{n \to \infty} \int_0^1 f(x, y_n)\, dx \;=\; \int_0^1 f(x)\, dx . \end{equation*}

We now pass from sequences to the limit as \(y \to 0\). Let \(L = \int_0^1 f(x)\, dx\) and suppose, by way of contradiction, that \(H(y)\) does not converge to \(L\) as \(y \to 0\). Then there is an \(\epsilon_0 > 0\) such that for every \(\delta > 0\) there is a \(y \in (0, 1]\) with \(y < \delta\) and \(|H(y) - L| \ge \epsilon_0\). Taking \(\delta = 1/n\) for each natural number \(n\) produces a sequence \(\{y_n\}\) in \((0,1]\) with \(y_n < 1/n\), hence \(y_n \to 0\), and \(|H(y_n) - L| \ge \epsilon_0\) for all \(n\). This contradicts the convergence \(H(y_n) \to L\) established above for every such sequence. Therefore

\begin{equation*} \lim_{y \to 0} \int_0^1 f(x, y)\, dx \;=\; \int_0^1 f(x)\, dx , \end{equation*}

which is the first assertion.

The second assertion. Assume now in addition that for each fixed \(x \in [0,1]\) the function \(y \mapsto f(x, y)\) is continuous on \([0, 1]\), and let \(h(y) = \int_0^1 f(x, y)\, dx\), which is the function \(H\) above and is real-valued by the first paragraph. Fix \(y_0 \in [0, 1]\); we show \(h\) is continuous at \(y_0\).

Let \(\{y_n\}\) be any sequence in \([0,1]\) with \(y_n \to y_0\). For each fixed \(x \in [0,1]\), continuity of \(f(x, \cdot)\) at \(y_0\) gives \(f(x, y_n) \to f(x, y_0)\). Thus \(\{f_{y_n}\}\) is a sequence of measurable functions on \(E\) converging pointwise on \(E\) to the measurable function \(f_{y_0}\), and \(|f_{y_n}| \le g\) on \(E\) for every \(n\), with \(g\) integrable over \(E\). By the Lebesgue Dominated Convergence Theorem,

\begin{equation*} \lim_{n \to \infty} h(y_n) \;=\; \lim_{n \to \infty} \int_0^1 f(x, y_n)\, dx \;=\; \int_0^1 f(x, y_0)\, dx \;=\; h(y_0) . \end{equation*}

Since \(\{y_n\}\) was an arbitrary sequence in \([0,1]\) converging to \(y_0\), the sequential criterion for continuity gives that \(h\) is continuous at \(y_0\). Explicitly, were \(h\) discontinuous at \(y_0\), there would be an \(\epsilon_0 > 0\) and, for each natural number \(n\), a point \(y_n \in [0,1]\) with \(|y_n - y_0| < 1/n\) and \(|h(y_n) - h(y_0)| \ge \epsilon_0\); this sequence converges to \(y_0\) yet \(h(y_n) \not\to h(y_0)\), contradicting what was just proved. As \(y_0 \in [0,1]\) was arbitrary, \(h\) is a continuous function of \(y\) on \([0, 1]\).

For the assertion to have content the inner integral must be defined, so we assume, as the statement implicitly does, that \(x \mapsto f(x, y_0)\) is integrable over \([0,1]\) for at least one \(y_0 \in [0,1]\). This much cannot be dispensed with: the function \(f(x, y) = 1/x\) for \(0 < x \le 1\), \(f(0, y) = 0\), is real-valued on \(Q\), is measurable in \(x\) for each \(y\), and has \(\partial f/\partial y \equiv 0\), so it meets every stated hypothesis with \(g \equiv 0\), yet \(\int_0^1 f(x, y)\, dx = \infty\) for every \(y\). We first check that integrability at a single \(y_0\) forces integrability for every \(y\).

Step 1: \(f(\cdot, y)\) is integrable over \([0,1]\) for each \(y \in [0,1]\).

Fix \(x \in [0,1]\). By hypothesis the function \(t \mapsto f(x, t)\) is differentiable at every point of \([0,1]\), hence continuous there. Therefore, given \(y \in [0,1]\) with \(y \ne y_0\), the Mean Value Theorem applied to \(t \mapsto f(x,t)\) on the closed interval with endpoints \(y_0\) and \(y\) provides a point \(\xi\) strictly between them with

\begin{equation*} f(x, y) - f(x, y_0) = \frac{\partial f}{\partial y}(x, \xi)\,(y - y_0). \end{equation*}

Since \(|y - y_0| \le 1\) and \(|\partial f/\partial y| \le g\) on \(Q\), we conclude

\begin{equation*} |f(x, y)| \le |f(x, y_0)| + g(x) \quad \text{for all } x \in [0,1]. \end{equation*}

The right-hand side is integrable over \([0,1]\), and \(f(\cdot, y)\) is measurable by hypothesis; by the Integral Comparison Test (Proposition 16 of Chapter 4), \(f(\cdot, y)\) is integrable over \([0,1]\). Thus

\begin{equation*} F(y) := \int_0^1 f(x, y)\, dx \end{equation*}

is a well-defined real number for every \(y \in [0,1]\).

Step 2: the difference quotients are dominated by \(g\).

Fix \(y \in [0,1]\) and let \(\{y_n\}\) be any sequence in \([0,1]\) with \(y_n \ne y\) for all \(n\) and \(y_n \to y\). Define, for \(x \in [0,1]\),

\begin{equation*} h_n(x) = \frac{f(x, y_n) - f(x, y)}{y_n - y}. \end{equation*}

Each \(h_n\) is measurable on \([0,1]\), being a linear combination of the measurable functions \(f(\cdot, y_n)\) and \(f(\cdot, y)\). Exactly as in Step 1, the Mean Value Theorem applied to \(t \mapsto f(x,t)\) on the interval with endpoints \(y\) and \(y_n\) yields a point \(\xi_{n,x}\) between \(y\) and \(y_n\) (so \(\xi_{n,x} \in [0,1]\)) with

\begin{equation*} h_n(x) = \frac{\partial f}{\partial y}(x, \xi_{n,x}), \quad \text{hence} \quad |h_n(x)| \le g(x) \ \text{ for all } x \in [0,1] \text{ and all } n. \end{equation*}

Step 3: pointwise convergence and passage to the limit.

For each fixed \(x \in [0,1]\), the very definition of the partial derivative \(\partial f/\partial y (x, y)\), which exists by hypothesis, gives

\begin{equation*} \lim_{n \to \infty} h_n(x) = \frac{\partial f}{\partial y}(x, y). \end{equation*}

(At \(y = 0\) and \(y = 1\) this derivative is the appropriate one-sided derivative, and the sequences \(\{y_n\}\) necessarily approach \(y\) from one side.) Hence \(\partial f/\partial y(\cdot, y)\) is measurable on \([0,1]\), being a pointwise limit of measurable functions, and it satisfies \(|\partial f/\partial y(\cdot, y)| \le g\); by the Integral Comparison Test it is integrable over \([0,1]\).

The sequence \(\{h_n\}\) consists of measurable functions on \([0,1]\), converges pointwise to \(\partial f/\partial y(\cdot, y)\), and is dominated on \([0,1]\) by the integrable function \(g\). By the Lebesgue Dominated Convergence Theorem,

\begin{equation*} \lim_{n \to \infty} \int_0^1 h_n(x)\, dx = \int_0^1 \frac{\partial f}{\partial y}(x, y)\, dx . \end{equation*}

On the other hand, by the linearity of integration (Theorem 17 of Chapter 4; all three functions involved are integrable by Step 1),

\begin{equation*} \int_0^1 h_n(x)\, dx = \frac{1}{y_n - y}\left[ \int_0^1 f(x, y_n)\, dx - \int_0^1 f(x, y)\, dx \right] = \frac{F(y_n) - F(y)}{y_n - y}. \end{equation*}

Therefore

\begin{equation*} \lim_{n \to \infty} \frac{F(y_n) - F(y)}{y_n - y} = \int_0^1 \frac{\partial f}{\partial y}(x, y)\, dx . \end{equation*}

Step 4: conclusion.

The sequence \(\{y_n\}\) in \([0,1] \setminus \{y\}\) converging to \(y\) was arbitrary, and the limit obtained does not depend on it. By the sequential characterization of limits of a function of a real variable, the limit

\begin{equation*} \lim_{t \to y,\ t \in [0,1],\ t \ne y} \frac{F(t) - F(y)}{t - y} \end{equation*}

exists and equals \(\int_0^1 (\partial f/\partial y)(x, y)\, dx\). That is, \(F\) is differentiable at \(y\) (one-sidedly at the endpoints \(y = 0\) and \(y = 1\)) and

\begin{equation*} \frac{d}{dy}\left[ \int_0^1 f(x, y)\, dx \right] = \int_0^1 \frac{\partial f}{\partial y}(x, y)\, dx \quad \text{for all } y \in [0,1], \end{equation*}

which is what we were to prove.

Here \(E\) is a measurable set of real numbers and \(f\) is integrable over \(E\).

First, each \(E_n = E \cap \{x \in \mathbb{R} : |x| \ge n\} = E \cap \big( (-\infty, -n] \cup [n, \infty) \big)\) is measurable, being the intersection of the measurable set \(E\) with a closed set. Moreover \(f\) is integrable over \(E_n\), since \(E_n\) is a measurable subset of \(E\) and hence, by the monotonicity of integration for nonnegative functions,

\begin{equation*} \int_{E_n} |f| \le \int_E |f| < \infty . \end{equation*}

Thus \(\int_{E_n} f\) is a well-defined real number for every \(n\).

Observe that \(\{E_n\}_{n=1}^\infty\) is a descending collection of measurable sets, since \(|x| \ge n+1\) implies \(|x| \ge n\), and that

\begin{equation*} \bigcap_{n=1}^\infty E_n = \varnothing , \end{equation*}

because each \(x \in E\) is a real number and therefore fails to satisfy \(|x| \ge n\) as soon as \(n > |x|\) (by the Archimedean property of \(\mathbb{R}\)).

We give a direct argument. Define \(g_n = |f| \cdot \chi_{E_n}\) on \(E\). Each \(g_n\) is measurable, and for every \(x \in E\) we have \(g_n(x) \to 0\): indeed, once \(n > |x|\) we have \(x \notin E_n\) and so \(g_n(x) = 0\). Furthermore \(0 \le g_n \le |f|\) on \(E\), and \(|f|\) is integrable over \(E\). By the Lebesgue Dominated Convergence Theorem,

\begin{equation*} \lim_{n \to \infty} \int_{E_n} |f| = \lim_{n \to \infty} \int_E g_n = \int_E 0 = 0 . \end{equation*}

By the Integral Comparison Test (Proposition 16 of Chapter 4), which supplies the triangle inequality for integrals,

\begin{equation*} \left| \int_{E_n} f \right| \le \int_{E_n} |f| \quad \text{for every } n . \end{equation*}

Hence, given \(\epsilon > 0\), choose \(N\) so that \(\int_{E_n} |f| < \epsilon\) for all \(n \ge N\); then for all \(n \ge N\),

\begin{equation*} \left| \int_{E_n} f \right| \le \int_{E_n} |f| < \epsilon . \end{equation*}

Alternatively, the conclusion is immediate from the Continuity of Integration (Theorem 21 of Chapter 4): since \(f\) is integrable over \(E\) and \(\{E_n\}\) is a descending countable collection of measurable subsets of \(E\) with \(\bigcap_{n=1}^\infty E_n = \varnothing\), formula (25) gives

\begin{equation*} \lim_{n \to \infty} \int_{E_n} f = \int_{\bigcap_{n=1}^\infty E_n} f = \int_\varnothing f = 0 , \end{equation*}

and the definition of this limit supplies, for each \(\epsilon > 0\), a natural number \(N\) such that \(\left| \int_{E_n} f \right| < \epsilon\) whenever \(n \ge N\).

Throughout, recall that in Royden a measurable function \(f\) is said to be integrable over \(E\) provided \(\int_E |f| < \infty\). So to show \(f\) is not integrable over \([1,\infty)\) we show \(\int_{[1,\infty)} |f| = \infty\).

(i) The step function \(f(x) = (-1)^n/n\) for \(n \le x < n+1\).

The function \(f\) is measurable on \([1, \infty)\): it is constant on each of the intervals \([n, n+1)\), \(n \ge 1\), so for any \(c\) the set \(\{x : f(x) > c\}\) is a countable union of such intervals, hence measurable.

Fix a natural number \(N \ge 2\). The interval \([1, N)\) is the disjoint union of \([n, n+1)\) for \(n = 1, \dots, N-1\), so by the additivity over domains of integration and the fact that \(\int_{[n,n+1)} c = c \cdot m([n,n+1)) = c\),

\begin{equation*} \int_1^N f = \sum_{n=1}^{N-1} \frac{(-1)^n}{n}\, m\big([n, n+1)\big) = \sum_{n=1}^{N-1} \frac{(-1)^n}{n} . \end{equation*}

The alternating series \(\sum_{n=1}^\infty (-1)^n/n\) converges (its terms alternate in sign and \(1/n\) decreases monotonically to \(0\); this is the alternating series test), with sum \(-\ln 2\). Therefore

\begin{equation*} \lim_{N \to \infty} \int_1^N f = \sum_{n=1}^\infty \frac{(-1)^n}{n} = -\ln 2 , \end{equation*}

so the limit exists and is finite.

On the other hand, \(|f| = 1/n\) on \([n, n+1)\), so on \([1, \infty)\) we have the pointwise identity

\begin{equation*} |f| = \sum_{n=1}^\infty u_n, \qquad u_n := \frac{1}{n}\, \chi_{[n, n+1)} , \end{equation*}

with each \(u_n\) nonnegative and measurable. Note that Theorem 20 of Chapter 4 is not available here, since it presupposes that \(f\) is integrable over \([1, \infty)\), which is exactly what is in question; but Corollary 12 of Chapter 4, which is the countable additivity of integration for nonnegative measurable functions and assumes no integrability, applies and gives

\begin{equation*} \int_{[1, \infty)} |f| = \sum_{n=1}^\infty \int_{[1,\infty)} u_n = \sum_{n=1}^\infty \frac{1}{n}\, m\big([n, n+1)\big) = \sum_{n=1}^\infty \frac{1}{n} = \infty , \end{equation*}

the harmonic series being divergent. Hence \(f\) is not integrable over \([1, \infty)\).

(ii) \(f(x) = (\sin x)/x\) on \([1, \infty)\).

This \(f\) is continuous on \([1, \infty)\), hence measurable. On each bounded interval \([1, b]\) it is continuous and therefore Riemann integrable, and by Theorem 3 of Chapter 4 its Riemann and Lebesgue integrals over \([1,b]\) agree. Integrating by parts (a legitimate Riemann-integral manipulation, both integrands being continuous),

\begin{equation*} \int_1^b \frac{\sin x}{x}\, dx = \left[ \frac{-\cos x}{x} \right]_1^b - \int_1^b \frac{\cos x}{x^2}\, dx = \cos 1 - \frac{\cos b}{b} - \int_1^b \frac{\cos x}{x^2}\, dx . \end{equation*}

Now \(|\cos b / b| \le 1/b \to 0\) as \(b \to \infty\). Also, the function \(x \mapsto (\cos x)/x^2\) satisfies \(|\cos x / x^2| \le 1/x^2\) on \([1, \infty)\), and \(1/x^2\) is integrable over \([1, \infty)\). Indeed, the functions \(x^{-2}\chi_{[1,n]}\) form an increasing sequence of nonnegative measurable functions converging pointwise to \(x^{-2}\) on \([1, \infty)\), so the Monotone Convergence Theorem, together with Theorem 3 of Chapter 4 for the integrals over \([1,n]\), gives

\begin{equation*} \int_{[1,\infty)} \frac{dx}{x^2} = \lim_{n \to \infty} \int_1^n \frac{dx}{x^2} = \lim_{n\to\infty}\left( 1 - \frac{1}{n} \right) = 1 < \infty . \end{equation*}

Hence \((\cos x)/x^2\) is integrable over \([1, \infty)\), and applying the Lebesgue Dominated Convergence Theorem to \((\cos x)/x^2 \cdot \chi_{[1, n]}\), which converges pointwise to \((\cos x)/x^2\) and is dominated by \(1/x^2\), we get

\begin{equation*} \lim_{n \to \infty} \int_1^n \frac{\cos x}{x^2}\, dx = \int_{[1, \infty)} \frac{\cos x}{x^2}\, dx . \end{equation*}

Combining, the limit exists:

\begin{equation*} \lim_{n \to \infty} \int_1^n \frac{\sin x}{x}\, dx = \cos 1 - \int_{[1,\infty)} \frac{\cos x}{x^2}\, dx \in \mathbb{R} . \end{equation*}

To see that \(f\) is not integrable over \([1, \infty)\), note that for each integer \(k \ge 1\) the interval \([k\pi, (k+1)\pi)\) is contained in \([1, \infty)\), and on it \(x \le (k+1)\pi\), so

\begin{equation*} \int_{[k\pi, (k+1)\pi)} \frac{|\sin x|}{x}\, dx \ge \frac{1}{(k+1)\pi} \int_{k\pi}^{(k+1)\pi} |\sin x|\, dx = \frac{2}{(k+1)\pi} , \end{equation*}

since \(\int_{k\pi}^{(k+1)\pi} |\sin x|\, dx = 2\) for every \(k\). These intervals are pairwise disjoint measurable subsets of \([1, \infty)\). Writing \(v_k = |f| \cdot \chi_{[k\pi, (k+1)\pi)}\), the \(v_k\) are nonnegative measurable functions with \(\sum_{k=1}^\infty v_k \le |f|\) pointwise on \([1,\infty)\), so Corollary 12 of Chapter 4 together with the monotonicity of integration for nonnegative functions gives

\begin{equation*} \int_{[1, \infty)} \frac{|\sin x|}{x}\, dx \ge \sum_{k=1}^\infty \int_{[1,\infty)} v_k \ge \sum_{k=1}^\infty \frac{2}{(k+1)\pi} = \frac{2}{\pi} \sum_{k=2}^\infty \frac{1}{k} = \infty . \end{equation*}

Hence \(f\) is not integrable over \([1, \infty)\).

Does this contradict the continuity of integration?

No. Theorem 21 of Chapter 4 (the Continuity of Integration) begins with the hypothesis “Let \(f\) be integrable over \(E\).” Here one would want to take \(E = [1, \infty)\) and \(E_n = [1, n]\), an ascending countable collection of measurable subsets of \(E\) with \(\bigcup_{n=1}^\infty E_n = [1, \infty)\), and conclude from (24) that

\begin{equation*} \int_{[1, \infty)} f = \lim_{n \to \infty} \int_{[1, n]} f . \end{equation*}

But in both examples \(f\) fails to be integrable over \(E = [1, \infty)\), so the hypothesis of the theorem is not met and the left-hand side is not even defined: \(\int_{[1,\infty)} f^+ = \int_{[1,\infty)} f^- = \infty\) in each case, so the difference \(\int f^+ - \int f^-\) is meaningless.

These examples therefore show only that the converse of the theorem fails: the existence of \(\lim_{n\to\infty} \int_{E_n} f\) for an ascending sequence \(\{E_n\}\) exhausting \(E\) does not imply that \(f\) is integrable over \(E\). The limit in question is an improper (conditionally convergent) integral, and the Lebesgue integral is by design an absolutely convergent theory.

Throughout, note first that if \(A\) is any measurable subset of \(E\), then \(f\) is integrable over \(A\): \(f\) is measurable on \(A\), and by the monotonicity of integration for nonnegative measurable functions,

\begin{equation*} \int_A |f| \le \int_E |f| < \infty . \end{equation*}

So every integral appearing below is a well-defined real number, and finite. We also use freely the Additivity Over Domains of Integration (Corollary 18 of Chapter 4): if \(A\) and \(B\) are disjoint measurable subsets of \(E\), then \(\int_{A \cup B} f = \int_A f + \int_B f\).

(i) The ascending case.

Let \(\{E_n\}_{n=1}^\infty\) be ascending, that is \(E_n \subseteq E_{n+1}\) for all \(n\), and set \(E_\infty = \bigcup_{n=1}^\infty E_n\). Define \(E_0 = \varnothing\) and, for each \(k \ge 1\),

\begin{equation*} F_k = E_k \sim E_{k-1} . \end{equation*}

Each \(F_k\) is a measurable subset of \(E\). The collection \(\{F_k\}_{k=1}^\infty\) is disjoint: if \(j < k\) then \(F_j \subseteq E_j \subseteq E_{k-1}\) while \(F_k \cap E_{k-1} = \varnothing\). Moreover, for each \(n \ge 1\),

\begin{equation*} \bigcup_{k=1}^n F_k = E_n , \end{equation*}

which follows by induction: it is trivial for \(n = 1\) since \(F_1 = E_1\), and if it holds for \(n\) then \(\bigcup_{k=1}^{n+1} F_k = E_n \cup (E_{n+1} \sim E_n) = E_{n+1}\), using \(E_n \subseteq E_{n+1}\). Taking the union over all \(n\),

\begin{equation*} \bigcup_{k=1}^\infty F_k = \bigcup_{n=1}^\infty E_n = E_\infty . \end{equation*}

Thus \(\{F_k\}_{k=1}^\infty\) is a disjoint countable collection of measurable subsets of \(E_\infty\) whose union is \(E_\infty\), and \(f\) is integrable over \(E_\infty\). By the Countable Additivity of Integration (Theorem 20 of Chapter 4), applied with \(E_\infty\) in the role of \(E\),

\begin{equation*} \int_{E_\infty} f = \sum_{k=1}^\infty \int_{F_k} f = \lim_{n \to \infty} \sum_{k=1}^n \int_{F_k} f , \end{equation*}

the last equality being the definition of the sum of a series. But for each fixed \(n\), \(E_n\) is the disjoint union of the finitely many measurable sets \(F_1, \dots, F_n\), so a finite induction on the additivity over domains of integration gives

\begin{equation*} \sum_{k=1}^n \int_{F_k} f = \int_{E_n} f . \end{equation*}

Combining the last two displays,

\begin{equation*} \int_{\bigcup_{n=1}^\infty E_n} f = \int_{E_\infty} f = \lim_{n \to \infty} \int_{E_n} f , \end{equation*}

which is (24).

(ii) The descending case.

Let \(\{E_n\}_{n=1}^\infty\) be descending, that is \(E_{n+1} \subseteq E_n\) for all \(n\), and set \(E_\infty = \bigcap_{n=1}^\infty E_n\). Then \(E_\infty \subseteq E_n \subseteq E_1\) for every \(n\).

Define \(G_n = E_1 \sim E_n\) for \(n \ge 1\). Since \(\{E_n\}\) is descending, \(\{G_n\}\) is an ascending collection of measurable subsets of \(E\). Moreover, by De Morgan’s identities,

\begin{equation*} \bigcup_{n=1}^\infty G_n = \bigcup_{n=1}^\infty (E_1 \sim E_n) = E_1 \sim \bigcap_{n=1}^\infty E_n = E_1 \sim E_\infty . \end{equation*}

Applying part (i) to the ascending collection \(\{G_n\}\),

\begin{equation*} \int_{E_1 \sim E_\infty} f = \lim_{n \to \infty} \int_{E_1 \sim E_n} f . \end{equation*}

Now \(E_1\) is the disjoint union of \(E_\infty\) and \(E_1 \sim E_\infty\), and also, for each \(n\), the disjoint union of \(E_n\) and \(E_1 \sim E_n\). By additivity over domains,

\begin{equation*} \int_{E_1} f = \int_{E_\infty} f + \int_{E_1 \sim E_\infty} f , \qquad \int_{E_1} f = \int_{E_n} f + \int_{E_1 \sim E_n} f . \end{equation*}

Every term here is a finite real number, so we may solve and subtract. From the second identity, \(\int_{E_n} f = \int_{E_1} f - \int_{E_1 \sim E_n} f\); letting \(n \to \infty\) and using the limit established above,

\begin{equation*} \lim_{n \to \infty} \int_{E_n} f = \int_{E_1} f - \int_{E_1 \sim E_\infty} f = \int_{E_\infty} f , \end{equation*}

the last equality being the first identity. That is,

\begin{equation*} \int_{\bigcap_{n=1}^\infty E_n} f = \lim_{n \to \infty} \int_{E_n} f , \end{equation*}

which is (25). This completes the proof.

Remark. The finiteness of \(\int_{E_1} f\), which is where the integrability of \(f\) over \(E\) is used in part (ii), is essential; it is what licenses the cancellation. This is exactly parallel to the proof of the continuity of measure, where the descending case requires \(m(E_1) < \infty\).

\(F\) is properly defined.

Fix \(x \in \mathbb{R}\). The set \((-\infty, x]\) is closed, hence measurable, and \(f\) is measurable on it. Since \(|f| \ge 0\) and \((-\infty, x] \subseteq \mathbb{R}\), monotonicity of integration gives

\begin{equation*} \int_{(-\infty, x]} |f| \le \int_{\mathbb{R}} |f| < \infty , \end{equation*}

so \(f\) is integrable over \((-\infty, x]\) and \(F(x) = \int_{(-\infty, x]} f\) is a well-defined real number.

\(F\) is (uniformly) continuous.

Let \(\epsilon > 0\). Since \(f\) is integrable over \(\mathbb{R}\), Proposition 23 of Chapter 4 provides a \(\delta > 0\) such that

\begin{equation*} \text{if } A \subseteq \mathbb{R} \text{ is measurable and } m(A) < \delta, \text{ then } \int_A |f| < \epsilon . \end{equation*}

Let \(x, y \in \mathbb{R}\) with \(|x - y| < \delta\); without loss of generality \(x < y\). The set \((-\infty, y]\) is the disjoint union of the measurable sets \((-\infty, x]\) and \((x, y]\), so by the Additivity Over Domains of Integration (Corollary 18 of Chapter 4),

\begin{equation*} F(y) = \int_{(-\infty, y]} f = \int_{(-\infty, x]} f + \int_{(x, y]} f = F(x) + \int_{(x,y]} f , \end{equation*}

all terms being finite. Hence, by the Integral Comparison Test (Proposition 16 of Chapter 4),

\begin{equation*} |F(y) - F(x)| = \left| \int_{(x, y]} f \right| \le \int_{(x, y]} |f| < \epsilon , \end{equation*}

since \(m((x, y]) = y - x = |x - y| < \delta\). As \(\delta\) depends only on \(\epsilon\), \(F\) is uniformly continuous on \(\mathbb{R}\), and in particular continuous.

\(F\) need not be Lipschitz.

The identity \(F(y) - F(x) = \int_{(x,y]} f\) shows that if \(|f| \le M\) a.e. on \(\mathbb{R}\) then \(|F(y) - F(x)| \le M|y - x|\), so boundedness of \(f\) would give the Lipschitz property. Without boundedness it can fail. Define

\begin{equation*} f(x) = \begin{cases} \dfrac{1}{\sqrt{x}}, & 0 < x \le 1, \\[4pt] 0, & \text{otherwise}. \end{cases} \end{equation*}

Then \(f\) is nonnegative and measurable on \(\mathbb{R}\) (it is continuous off the two points \(0\) and \(1\)). The functions \(f \cdot \chi_{[1/n, 1]}\) are nonnegative, measurable, and increase pointwise to \(f\), so by the Monotone Convergence Theorem, together with Theorem 3 of Chapter 4 (which identifies the Lebesgue and Riemann integrals of a continuous function on a closed bounded interval),

\begin{equation*} \int_{\mathbb{R}} f = \lim_{n \to \infty} \int_{1/n}^1 \frac{dt}{\sqrt{t}} = \lim_{n \to \infty} \left( 2 - \frac{2}{\sqrt{n}} \right) = 2 < \infty . \end{equation*}

Thus \(f\) is integrable over \(\mathbb{R}\). The same computation applied on \((0, x]\) for \(0 < x \le 1\) gives

\begin{equation*} F(x) = \int_{(-\infty, x]} f = \int_{(0, x]} \frac{dt}{\sqrt{t}} = 2\sqrt{x} , \qquad 0 \le x \le 1 , \end{equation*}

and \(F(x) = 0\) for \(x \le 0\).

If \(F\) were Lipschitz on \(\mathbb{R}\) with constant \(C\), then for every \(x \in (0, 1]\) we would have

\begin{equation*} 2\sqrt{x} = |F(x) - F(0)| \le C\,|x - 0| = Cx, \quad \text{i.e.} \quad \frac{2}{\sqrt{x}} \le C . \end{equation*}

Letting \(x \to 0^+\) makes the left-hand side unbounded, a contradiction. Hence \(F\) is not Lipschitz.

So \(F\) is always uniformly continuous, but it is Lipschitz only under an extra hypothesis such as essential boundedness of \(f\).

We exhibit a sequence of functions on \(E = \mathbb{R}\) which is uniformly integrable over \(\mathbb{R}\) and converges pointwise on all of \(\mathbb{R}\) to a function that is not integrable over \(\mathbb{R}\). This shows that the conclusion of Proposition 25 fails when the hypothesis of finite measure is dropped, so the hypothesis \(m(E) < \infty\) cannot be removed.

Define, for every natural number \(n\),

\begin{equation*} f_n = \chi_{[-n, n]} \quad \text{on } \mathbb{R} . \end{equation*}

Each \(f_n\) is measurable, being the characteristic function of a closed set, and each \(f_n\) is in fact integrable over \(\mathbb{R}\), with \(\int_{\mathbb{R}} |f_n| = m([-n, n]) = 2n < \infty\); so no objection can be raised that the members of the family are themselves badly behaved.

\(\{f_n\} \to f\) pointwise on all of \(\mathbb{R}\), where \(f \equiv 1\). Indeed, fix \(x \in \mathbb{R}\) and use the Archimedean property to choose \(N > |x|\); then \(x \in [-n, n]\) for all \(n \ge N\), so \(f_n(x) = 1\) for all \(n \ge N\). Hence \(f_n(x) \to 1\). In particular the convergence holds pointwise a.e. on \(\mathbb{R}\).

The family \(\{f_n\}_{n=1}^\infty\) is uniformly integrable over \(\mathbb{R}\).

Let \(\epsilon > 0\) and take \(\delta = \epsilon\). If \(A \subseteq \mathbb{R}\) is measurable with \(m(A) < \delta\), then for every \(n\), since \(0 \le f_n \le 1\), monotonicity of integration for nonnegative functions gives

\begin{equation*} \int_A |f_n| = \int_A \chi_{[-n,n]} \le \int_A 1 = m(A) < \delta = \epsilon . \end{equation*}

Thus this single \(\delta\) responds to the \(\epsilon\) challenge simultaneously for every member of the family, which is precisely criterion (27) for uniform integrability.

The limit function is not integrable over \(\mathbb{R}\).

The limit is \(f \equiv 1\), and

\begin{equation*} \int_{\mathbb{R}} |f| = \int_{\mathbb{R}} 1 = m(\mathbb{R}) = \infty . \end{equation*}

Indeed, for every natural number \(n\) the function \(\chi_{[-n,n]}\) is a nonnegative measurable function with \(\chi_{[-n,n]} \le 1\) on \(\mathbb{R}\), so by monotonicity of integration \(\int_{\mathbb{R}} 1 \ge \int_{\mathbb{R}} \chi_{[-n,n]} = 2n\) for every \(n\), and hence \(\int_{\mathbb{R}} 1 = \infty\). Therefore \(f\) is not integrable over \(\mathbb{R}\).

So all the hypotheses of Proposition 25 except the finiteness of \(m(E)\) are satisfied, yet the conclusion fails. Hence Proposition 25 is false if \(E = \mathbb{R}\).

Remark. An even simpler family works, if one is willing to admit members that are not themselves integrable: the constant sequence \(f_n \equiv 1\) is uniformly integrable over \(\mathbb{R}\) by the identical \(\delta = \epsilon\) computation, and converges pointwise to the non-integrable \(f \equiv 1\). Royden’s definition of uniform integrability imposes no integrability on the members of the family, so this is a legitimate counterexample too; the family \(\{\chi_{[-n,n]}\}\) was chosen above only to forestall the objection.

What goes wrong on a set of infinite measure is that uniform integrability controls only the mass carried on sets of small measure. On a set of finite measure this is enough: Lemma 22 lets one split \(E\) into finitely many, say \(N\), pieces each of measure less than the \(\delta\) responding to \(\epsilon = 1\), whence \(\int_E |f_n| < N\) for every \(n\), and Fatou’s Lemma then bounds \(\int_E |f|\) by \(N\). When \(m(E) = \infty\) that finite splitting is unavailable: \(\mathbb{R}\) is not the union of finitely many sets of measure less than \(\delta\), and indeed in the example above \(\int_{\mathbb{R}} |f_n| = 2n\) is unbounded in \(n\).

Only one of the two implications in Theorem 26 uses the nonnegativity of the \(h_n\). The implication

\begin{equation*} \{h_n\} \text{ uniformly integrable} \ \Longrightarrow \ \lim_{n \to \infty} \int_E h_n = 0 \end{equation*}

is just the Vitali Convergence Theorem applied to the pointwise limit \(h \equiv 0\), and the Vitali Convergence Theorem makes no sign assumption; so that implication survives. It is the converse,

\begin{equation*} \lim_{n \to \infty} \int_E h_n = 0 \ \Longrightarrow \ \{h_n\} \text{ uniformly integrable} , \end{equation*}

that fails when the \(h_n\) are allowed to change sign. We construct a counterexample.

Take \(E = [0, 1]\), which has finite measure \(m(E) = 1\). For each natural number \(n\) define

\begin{equation*} h_n = n \cdot \chi_{\left( 0,\, \frac{1}{2n} \right)} \; - \; n \cdot \chi_{\left[ \frac{1}{2n},\, \frac{1}{n} \right)} . \end{equation*}

Explicitly,

\begin{equation*} h_n(x) = \begin{cases} n, & 0 < x < \dfrac{1}{2n}, \\[6pt] -n, & \dfrac{1}{2n} \le x < \dfrac{1}{n}, \\[6pt] 0, & \text{otherwise on } [0,1] . \end{cases} \end{equation*}

Each \(h_n\) is integrable over \(E\). It is a simple function taking finitely many values on measurable sets of finite measure, and

\begin{equation*} \int_E |h_n| = n \cdot \frac{1}{2n} + n \cdot \frac{1}{2n} = 1 < \infty . \end{equation*}

\(\{h_n\} \to 0\) pointwise on \(E\). We have \(h_n(0) = 0\) for every \(n\). If \(0 < x \le 1\), choose \(N\) with \(1/N < x\) (Archimedean property); then for all \(n \ge N\) we have \(x \ge 1/N > 1/n\), hence \(h_n(x) = 0\). So \(h_n(x) \to 0\) for every \(x \in [0,1]\); in particular the convergence holds a.e. on \(E\), and \(h \equiv 0\).

\(\lim_{n \to \infty} \int_E h_n = 0\). By the linearity of integration, for every \(n\),

\begin{equation*} \int_E h_n = n \cdot m\!\left( \left(0, \tfrac{1}{2n}\right) \right) - n \cdot m\!\left( \left[ \tfrac{1}{2n}, \tfrac{1}{n} \right) \right) = n \cdot \frac{1}{2n} - n \cdot \frac{1}{2n} = 0 . \end{equation*}

Hence \(\int_E h_n = 0\) for every \(n\) and the limit is \(0\). The positive and negative parts cancel exactly; this is what the sign restriction in Theorem 26 was preventing.

\(\{h_n\}\) is not uniformly integrable over \(E\). Take the \(\epsilon\) challenge \(\epsilon = 1/2\) and let \(\delta > 0\) be arbitrary. Choose \(n\) so large that \(1/(2n) < \delta\), and set

\begin{equation*} A_n = \left( 0, \frac{1}{2n} \right) \subseteq E , \qquad m(A_n) = \frac{1}{2n} < \delta . \end{equation*}

Then

\begin{equation*} \int_{A_n} |h_n| = n \cdot \frac{1}{2n} = \frac{1}{2} \not< \epsilon . \end{equation*}

So no \(\delta > 0\) responds to the challenge \(\epsilon = 1/2\) in criterion (27), and \(\{h_n\}\) fails to be uniformly integrable over \(E\).

Thus \(E\) has finite measure, each \(h_n\) is integrable over \(E\), \(\{h_n\} \to 0\) pointwise on \(E\), and \(\lim_{n\to\infty} \int_E h_n = 0\), yet \(\{h_n\}\) is not uniformly integrable. Theorem 26 is therefore false without the assumption that the \(h_n\) are nonnegative.

Remark on where the proof breaks. In Royden’s proof of the converse direction, from \(\lim_{n\to\infty}\int_E h_n = 0\) one chooses \(N\) with \(\int_E h_n < \epsilon\) for \(n \ge N\) and then asserts inequality (30): since \(h_n \ge 0\) on \(E\), for any measurable \(A \subseteq E\) one has \(\int_A h_n \le \int_E h_n < \epsilon\). It is exactly this monotonicity step that requires \(h_n \ge 0\); for the \(h_n\) above, \(\int_E h_n = 0\) while \(\int_{A_n} |h_n| = 1/2\).

Recall the definition: a family \(\mathcal{F}\) of measurable functions on \(E\) is uniformly integrable over \(E\) provided for each \(\epsilon > 0\) there is a \(\delta > 0\) such that for each \(f \in \mathcal{F}\),

\begin{equation*} \text{if } A \subseteq E \text{ is measurable and } m(A) < \delta, \text{ then } \int_A |f| < \epsilon . \end{equation*}

The definition is stated for families of measurable, possibly extended real-valued, functions, so before anything else we must know that the pointwise sum \(\alpha f_n + \beta g_n\) is defined a.e. on \(E\). We claim that every member \(h\) of a uniformly integrable family on \(E\) is finite a.e. on \(E\). Suppose not, so that \(S = \{ x \in E \mid |h(x)| = \infty \}\) has \(m(S) > 0\). Let \(\delta > 0\) respond to the \(\epsilon = 1\) challenge for the family. The sets \(S \cap [-n, n]\) ascend to \(S\), so by the continuity of measure (Theorem 15 of Chapter 2) there is an \(n\) with \(m(S \cap [-n,n]) > 0\). The function \(t \mapsto m(S \cap [-n, t])\) is nondecreasing on \([-n,n]\) and increases by at most \(t^{\prime} - t\) over \([t, t^{\prime}]\), hence is continuous; it equals \(0\) at \(t = -n\) and \(m(S \cap [-n,n]) > 0\) at \(t = n\). By the Intermediate Value Theorem there is a \(t\) for which \(A = S \cap [-n,t]\) satisfies \(0 < m(A) < \delta\). But \(|h| \equiv \infty\) on \(A\) and \(m(A) > 0\), so \(\int_A |h| = \infty\), contradicting \(\int_A |h| < 1\). Hence \(m(S) = 0\).

Consequently, for each \(n\) there is a measurable \(A_n \subseteq E\) with \(m(E \sim A_n) = 0\) on which both \(f_n\) and \(g_n\) are finite and on which \(\alpha f_n + \beta g_n\) is therefore defined. As in the convention adopted in Section 4.4, the integral of \(\alpha f_n + \beta g_n\) over a measurable \(A \subseteq E\) means its integral over \(A \cap A_n\); this is unambiguous, since the integral of a nonnegative measurable function over a set of measure zero is \(0\) (immediately from the definition of that integral as a supremum) and so, by the additivity of integration over domains (Theorem 11), excising the null set \(E \sim A_n\) changes nothing. Wherever it is defined we have the pointwise triangle inequality

\begin{equation*} |\alpha f_n + \beta g_n| \le |\alpha|\,|f_n| + |\beta|\,|g_n| . \end{equation*}

Now set

\begin{equation*} M = |\alpha| + |\beta| + 1 > 0 . \end{equation*}

Let \(\epsilon > 0\). Since \(\{f_n\}\) is uniformly integrable over \(E\), there is a \(\delta_1 > 0\) such that

\begin{equation*} \text{if } A \subseteq E \text{ is measurable and } m(A) < \delta_1, \text{ then } \int_A |f_n| < \frac{\epsilon}{2M} \quad \text{for all } n . \end{equation*}

Since \(\{g_n\}\) is uniformly integrable over \(E\), there is a \(\delta_2 > 0\) such that

\begin{equation*} \text{if } A \subseteq E \text{ is measurable and } m(A) < \delta_2, \text{ then } \int_A |g_n| < \frac{\epsilon}{2M} \quad \text{for all } n . \end{equation*}

Define \(\delta = \min\{\delta_1, \delta_2\} > 0\).

Let \(A \subseteq E\) be measurable with \(m(A) < \delta\) and let \(n\) be a natural number. Since \(m(A) < \delta_1\) and \(m(A) < \delta_2\), the monotonicity and linearity of integration for nonnegative measurable functions (Theorem 10) together with the pointwise inequality above give

\begin{equation*} \int_A |\alpha f_n + \beta g_n| \le \int_A \bigl( |\alpha|\,|f_n| + |\beta|\,|g_n| \bigr) = |\alpha| \int_A |f_n| + |\beta| \int_A |g_n| \le M \int_A |f_n| + M \int_A |g_n| < M \cdot \frac{\epsilon}{2M} + M \cdot \frac{\epsilon}{2M} = \epsilon . \end{equation*}

Here we used \(|\alpha| \le M\) and \(|\beta| \le M\), and the final inequality is strict because \(M > 0\) and both of the last two integrals are strictly less than \(\epsilon/(2M)\).

Since \(\delta\) was chosen independently of \(n\) and of \(A\), this \(\delta\) responds to the \(\epsilon\) challenge for the sequence \(\{\alpha f_n + \beta g_n\}\). Therefore \(\{\alpha f_n + \beta g_n\}\) is uniformly integrable over \(E\).

Throughout, “finite support” means vanishing outside a set of finite measure.

(i) First assume \(f \ge 0\) on \(\mathbb{R}\). By the definition of the integral of a nonnegative measurable function,

\begin{equation*} \int_{\mathbb{R}} f = \sup \left\{ \int_{\mathbb{R}} h \;\Big|\; h \text{ bounded, measurable, of finite support, } 0 \le h \le f \text{ on } \mathbb{R} \right\}, \end{equation*}

and this supremum is finite since \(f\) is integrable. Hence there is a bounded measurable function \(h\) of finite support with \(0 \le h \le f\) on \(\mathbb{R}\) and

\begin{equation*} \int_{\mathbb{R}} f - \frac{\epsilon}{2} < \int_{\mathbb{R}} h \le \int_{\mathbb{R}} f . \end{equation*}

The function \(f - h\) is nonnegative and measurable, and by the linearity of integration for nonnegative measurable functions (Theorem 10), \(\int_{\mathbb{R}} h + \int_{\mathbb{R}} (f-h) = \int_{\mathbb{R}} f\). Since \(\int_{\mathbb{R}} h\) is finite,

\begin{equation*} \int_{\mathbb{R}} |f - h| = \int_{\mathbb{R}} (f-h) = \int_{\mathbb{R}} f - \int_{\mathbb{R}} h < \frac{\epsilon}{2} . \end{equation*}

Let \(E_0\) be a measurable set of finite measure outside of which \(h\) vanishes, and choose \(M > 0\) with \(0 \le h \le M\) on \(\mathbb{R}\). If \(m(E_0) = 0\) then \(h = 0\) a.e. and we may take \(\eta = 0\); so assume \(m(E_0) > 0\). Apply the Simple Approximation Lemma to the bounded measurable function \(h\) restricted to \(E_0\), with the challenge \(\epsilon^{\prime} = \epsilon / \bigl( 2\, m(E_0) + 1 \bigr)\): there are simple functions \(\varphi \le h \le \psi\) on \(E_0\) with \(0 \le \psi - \varphi < \epsilon^{\prime}\) on \(E_0\); in particular \(|h - \varphi| < \epsilon^{\prime}\) on \(E_0\). Replacing \(\varphi\) by \(\max\{\varphi, 0\}\) (still simple, and still within \(\epsilon^{\prime}\) of \(h\) since \(h \ge 0\)), we may assume \(\varphi \ge 0\). Define \(\eta\) on \(\mathbb{R}\) by \(\eta = \varphi\) on \(E_0\) and \(\eta = 0\) on \(\mathbb{R} \sim E_0\). Then \(\eta\) is a simple function on \(\mathbb{R}\) vanishing outside the set \(E_0\) of finite measure, so \(\eta\) has finite support, and by the monotonicity of integration together with additivity over domains,

\begin{equation*} \int_{\mathbb{R}} |h - \eta| = \int_{E_0} |h - \varphi| \le \epsilon^{\prime} \cdot m(E_0) < \frac{\epsilon}{2} . \end{equation*}

Therefore, by the triangle inequality \(|f - \eta| \le |f - h| + |h - \eta|\) and the monotonicity and linearity of integration,

\begin{equation*} \int_{\mathbb{R}} |f - \eta| \le \int_{\mathbb{R}} |f - h| + \int_{\mathbb{R}} |h - \eta| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

Now the general case. Write \(f = f^+ - f^-\). By Proposition 14 both \(f^+\) and \(f^-\) are nonnegative and integrable over \(\mathbb{R}\). By the case just settled, there are simple functions \(\eta_1, \eta_2\) on \(\mathbb{R}\) of finite support with

\begin{equation*} \int_{\mathbb{R}} |f^+ - \eta_1| < \frac{\epsilon}{2}, \qquad \int_{\mathbb{R}} |f^- - \eta_2| < \frac{\epsilon}{2} . \end{equation*}

Set \(\eta = \eta_1 - \eta_2\). A difference of simple functions is simple (Problem 19 of Chapter 3), and \(\eta\) vanishes outside the union of two sets of finite measure, so \(\eta\) has finite support. By Proposition 15, \(f\) is finite a.e. on \(\mathbb{R}\); at every point where \(f\) is finite,

\begin{equation*} |f - \eta| = |(f^+ - \eta_1) - (f^- - \eta_2)| \le |f^+ - \eta_1| + |f^- - \eta_2| . \end{equation*}

Since a set of measure zero does not affect an integral (Proposition 15), monotonicity and linearity give

\begin{equation*} \int_{\mathbb{R}} |f - \eta| \le \int_{\mathbb{R}} |f^+ - \eta_1| + \int_{\mathbb{R}} |f^- - \eta_2| < \epsilon . \end{equation*}

(ii) By part (i) choose a simple function \(\eta\) on \(\mathbb{R}\) of finite support with \(\int_{\mathbb{R}} |f - \eta| < \epsilon/2\). Write \(\eta = \sum_{k=1}^{p} a_k \chi_{E_k}\) with the \(E_k\) disjoint measurable sets and \(a_k \ne 0\); since \(\eta\) has finite support, \(m(E_k) < \infty\) for each \(k\). For each \(k\), the sets \(E_k \sim [-N, N]\) descend to \(\bigcap_{N} (E_k \sim [-N,N]) = \emptyset\) as \(N \to \infty\) and \(m(E_k) < \infty\), so by the continuity of measure \(m(E_k \sim [-N,N]) \to 0\). Choose a natural number \(N\) so large that

\begin{equation*} \sum_{k=1}^{p} |a_k| \cdot m\bigl( E_k \sim [-N,N] \bigr) < \frac{\epsilon}{4} . \end{equation*}

Put \(I = [-N, N]\) and \(\eta^{\prime} = \eta \cdot \chi_I\). Then \(\eta^{\prime}\) is simple, vanishes outside \(I\), and

\begin{equation*} \int_{\mathbb{R}} |\eta - \eta^{\prime}| = \int_{\mathbb{R} \sim I} |\eta| \le \sum_{k=1}^{p} |a_k| \cdot m\bigl(E_k \sim I\bigr) < \frac{\epsilon}{4} . \end{equation*}

Let \(M = \max_k |a_k|\), so \(|\eta^{\prime}| \le M\) on \(\mathbb{R}\). Let \(\epsilon^{\prime} > 0\) be a number to be fixed presently. The restriction of \(\eta^{\prime}\) to the closed, bounded interval \(I\) is a bounded measurable function, so by Problem 18 of Chapter 3 there is a step function \(h\) on \(I\) and a measurable subset \(F \subseteq I\) for which

\begin{equation*} |h - \eta^{\prime}| < \epsilon^{\prime} \text{ on } F \quad \text{and} \quad m(I \sim F) < \epsilon^{\prime} . \end{equation*}

Replace \(h\) by its truncation \(\tilde h = \min\{\max\{h, -M\}, M\}\). Truncating a step function at the levels \(\pm M\) again yields a step function on \(I\), and since \(|\eta^{\prime}| \le M\), truncation does not increase the distance to \(\eta^{\prime}\): \(|\tilde h - \eta^{\prime}| \le |h - \eta^{\prime}|\) pointwise on \(I\). Thus \(|\tilde h - \eta^{\prime}| < \epsilon^{\prime}\) on \(F\) and \(|\tilde h - \eta^{\prime}| \le 2M\) on all of \(I\). Consequently, by additivity over domains and monotonicity,

\begin{equation*} \int_I |\eta^{\prime} - \tilde h| = \int_F |\eta^{\prime} - \tilde h| + \int_{I \sim F} |\eta^{\prime} - \tilde h| \le \epsilon^{\prime} \cdot m(I) + 2M \cdot \epsilon^{\prime} = \epsilon^{\prime} \,(2N + 2M) . \end{equation*}

Choose \(\epsilon^{\prime} > 0\) small enough that \(\epsilon^{\prime}(2N + 2M) < \epsilon/4\).

Define \(s\) on \(\mathbb{R}\) by \(s = \tilde h\) on \(I\) and \(s = 0\) on \(\mathbb{R} \sim I\). Then \(s\) is a step function on \(\mathbb{R}\) (adjoin the two unbounded complementary intervals to the partition) which vanishes outside the closed, bounded interval \(I\), and since \(\eta^{\prime}\) also vanishes outside \(I\),

\begin{equation*} \int_{\mathbb{R}} |\eta^{\prime} - s| = \int_I |\eta^{\prime} - \tilde h| < \frac{\epsilon}{4} . \end{equation*}

Adding the three estimates,

\begin{equation*} \int_{\mathbb{R}} |f - s| \le \int_{\mathbb{R}} |f - \eta| + \int_{\mathbb{R}} |\eta - \eta^{\prime}| + \int_{\mathbb{R}} |\eta^{\prime} - s| < \frac{\epsilon}{2} + \frac{\epsilon}{4} + \frac{\epsilon}{4} = \epsilon . \end{equation*}

(iii) By part (ii) choose a step function \(s\) vanishing outside a closed, bounded interval \([a, b]\) with \(\int_{\mathbb{R}} |f - s| < \epsilon/2\). Let

\begin{equation*} a = x_0 < x_1 < \cdots < x_n = b \end{equation*}

be a partition of \([a,b]\) such that \(s\) is constant, say \(s \equiv c_i\), on the open interval \((x_{i-1}, x_i)\) for \(1 \le i \le n\). Let \(M = \max_i |c_i|\) (if \(M = 0\) then \(s = 0\) a.e. and \(g \equiv 0\) works, so assume \(M > 0\)).

Let \(\sigma > 0\) satisfy

\begin{equation*} \sigma < \min_{1 \le i \le n} (x_i - x_{i-1}) \quad \text{and} \quad 2M (n+1) \sigma < \frac{\epsilon}{2} . \end{equation*}

Define \(g : \mathbb{R} \to \mathbb{R}\) as follows. Let \(J_i = (x_i - \sigma/2,\; x_i + \sigma/2)\) for \(0 \le i \le n\); by the choice of \(\sigma\) these \(n+1\) intervals are pairwise disjoint. Outside \(\bigcup_{i=0}^n J_i\) set \(g\) equal to the value \(c_i\) on each \((x_{i-1}, x_i) \sim (J_{i-1} \cup J_i)\) and equal to \(0\) outside \([a, b]\). On each \(J_i\) define \(g\) to be the affine function joining the value of \(g\) at the left endpoint \(x_i - \sigma/2\) to its value at the right endpoint \(x_i + \sigma/2\) (these two one-sided values are the constants just prescribed; for \(i = 0\) the left value is \(0\) and for \(i = n\) the right value is \(0\)). Then \(g\) is continuous on \(\mathbb{R}\): it is locally constant off the closed intervals \(\overline{J_i}\), affine on each \(J_i\), and the definitions agree at the endpoints of each \(J_i\). Moreover \(g\) vanishes outside the bounded set \([a - \sigma, b + \sigma]\), and \(|g| \le M\) everywhere, since on each \(J_i\) it is a convex combination of two values of modulus at most \(M\).

By construction \(g = s\) except possibly on \(\bigcup_{i=0}^{n} \overline{J_i}\), a set of measure \((n+1)\sigma\), and \(|g - s| \le 2M\) everywhere. Hence

\begin{equation*} \int_{\mathbb{R}} |s - g| \le 2M \cdot (n+1)\sigma < \frac{\epsilon}{2} . \end{equation*}

Therefore

\begin{equation*} \int_{\mathbb{R}} |f - g| \le \int_{\mathbb{R}} |f - s| + \int_{\mathbb{R}} |s - g| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon , \end{equation*}

and \(g\) is continuous on \(\mathbb{R}\) and vanishes outside a bounded set.

Measurability of \(\hat f\). For \(c \ge 0\),

\begin{equation*} \{ x \in \mathbb{R} \mid \hat f(x) > c \} = \{ x \in E \mid f(x) > c \}, \end{equation*}

while for \(c < 0\),

\begin{equation*} \{ x \in \mathbb{R} \mid \hat f(x) > c \} = \{ x \in E \mid f(x) > c \} \cup (\mathbb{R} \sim E) . \end{equation*}

In both cases the set is measurable, since \(f\) is measurable on the measurable set \(E\) and \(\mathbb{R} \sim E\) is measurable. Hence \(\hat f\) is measurable on \(\mathbb{R}\).

Next suppose first that \(f \ge 0\) on \(E\), so that \(\hat f \ge 0\) on \(\mathbb{R}\). We claim

\begin{equation*} \int_{\mathbb{R}} \hat f = \int_E f . \end{equation*}

Let \(h\) be bounded, measurable, of finite support on \(\mathbb{R}\) with \(0 \le h \le \hat f\) on \(\mathbb{R}\). Since \(\hat f = 0\) on \(\mathbb{R} \sim E\) and \(h \ge 0\), we get \(h = 0\) on \(\mathbb{R} \sim E\). Since \(h\) is nonnegative and measurable, the additivity of integration over domains (Theorem 11) applies and gives

\begin{equation*} \int_{\mathbb{R}} h = \int_E h + \int_{\mathbb{R} \sim E} h = \int_E h . \end{equation*}

The restriction of \(h\) to \(E\) is bounded, measurable, of finite support, and satisfies \(0 \le h \le f\) on \(E\); therefore, by the definition of \(\int_E f\) as a supremum, \(\int_{\mathbb{R}} h = \int_E h \le \int_E f\). Taking the supremum over all such \(h\) gives \(\int_{\mathbb{R}} \hat f \le \int_E f\).

Conversely, let \(h\) be bounded, measurable, of finite support on \(E\) with \(0 \le h \le f\) on \(E\), and let \(\hat h\) be its extension by \(0\) to \(\mathbb{R}\). Then \(\hat h\) is bounded, measurable (by the same argument as for \(\hat f\)), of finite support, and \(0 \le \hat h \le \hat f\) on \(\mathbb{R}\). Again by additivity over domains, \(\int_E h = \int_{\mathbb{R}} \hat h \le \int_{\mathbb{R}} \hat f\). Taking the supremum over all such \(h\) gives \(\int_E f \le \int_{\mathbb{R}} \hat f\). The two inequalities give the claimed equality for nonnegative \(f\).

For general integrable \(f\) over \(E\), observe that

\begin{equation*} (\hat f)^+ = \widehat{f^+}, \qquad (\hat f)^- = \widehat{f^-}, \qquad |\hat f| = \widehat{|f|} . \end{equation*}

Applying the nonnegative case to \(|f|\) gives \(\int_{\mathbb{R}} |\hat f| = \int_E |f| < \infty\), so \(\hat f\) is integrable over \(\mathbb{R}\). Applying it to \(f^+\) and to \(f^-\) and using the definition of the general integral,

\begin{equation*} \int_{\mathbb{R}} \hat f = \int_{\mathbb{R}} (\hat f)^+ - \int_{\mathbb{R}} (\hat f)^- = \int_E f^+ - \int_E f^- = \int_E f . \end{equation*}

Now let \(\epsilon > 0\). Since \(\hat f\) is integrable over \(\mathbb{R}\), part (i) of the preceding problem provides a simple function \(\tilde\eta\) on \(\mathbb{R}\) of finite support with

\begin{equation*} \int_{\mathbb{R}} |\hat f - \tilde\eta| < \epsilon . \end{equation*}

Let \(\eta\) be the restriction of \(\tilde\eta\) to \(E\). The restriction of a simple function to a measurable subset is a simple function on that subset, so \(\eta\) is simple on \(E\). Since \(|\hat f - \tilde\eta|\) is nonnegative and \(\hat f = f\) on \(E\), the monotonicity together with the additivity over domains of integration for nonnegative functions gives

\begin{equation*} \int_E |f - \eta| = \int_E |\hat f - \tilde\eta| \le \int_E |\hat f - \tilde\eta| + \int_{\mathbb{R} \sim E} |\hat f - \tilde\eta| = \int_{\mathbb{R}} |\hat f - \tilde\eta| < \epsilon . \end{equation*}

Similarly, part (iii) of the preceding problem provides a function \(\tilde g\) that is continuous on \(\mathbb{R}\) and vanishes outside a bounded set, with \(\int_{\mathbb{R}} |\hat f - \tilde g| < \epsilon\). Let \(g\) be the restriction of \(\tilde g\) to \(E\); a restriction of a continuous function is continuous on its domain, so \(g\) is continuous on \(E\), and exactly as above

\begin{equation*} \int_E |f - g| = \int_E |\hat f - \tilde g| \le \int_{\mathbb{R}} |\hat f - \tilde g| < \epsilon . \end{equation*}

This is what was to be shown.

First note that for each natural number \(n\) the function \(x \mapsto f(x) \cos nx\) is measurable on \(\mathbb{R}\). Indeed, \(f\) is integrable, hence finite a.e. on \(\mathbb{R}\) by Proposition 15, while \(x \mapsto \cos nx\) is continuous and therefore measurable (Proposition 3 of Chapter 3); by Theorem 6 of Chapter 3 the product of two measurable functions that are finite a.e. is measurable off the null set where \(f\) fails to be finite, and hence, by Proposition 5 of Chapter 3 together with the completeness of Lebesgue measure, it is measurable on all of \(\mathbb{R}\) however it is defined on that null set. Moreover

\begin{equation*} |f(x) \cos nx| \le |f(x)| \quad \text{for all } x . \end{equation*}

Since \(|f|\) is integrable over \(\mathbb{R}\), the Integral Comparison Test (Proposition 16) shows that \(f(x) \cos nx\) is integrable over \(\mathbb{R}\) and that all the integrals below are finite.

Step 1: the case of a step function. Let \(s\) be a step function on \(\mathbb{R}\) that vanishes outside a closed, bounded interval \([a, b]\), and let

\begin{equation*} a = x_0 < x_1 < \cdots < x_p = b \end{equation*}

be a partition of \([a,b]\) with \(s \equiv c_i\) on \((x_{i-1}, x_i)\) for \(1 \le i \le p\). The finite set of partition points has measure zero, so it does not affect any integral (Proposition 15); hence for each \(n\), by the additivity over domains and the linearity of integration,

\begin{equation*} \int_{-\infty}^{\infty} s(x) \cos nx \, dx = \sum_{i=1}^{p} c_i \int_{x_{i-1}}^{x_i} \cos nx \, dx . \end{equation*}

On each bounded interval the integrand is continuous, so the Lebesgue integral agrees with the Riemann integral (Theorem 3 of this chapter), and for \(n \ge 1\),

\begin{equation*} \int_{x_{i-1}}^{x_i} \cos nx \, dx = \frac{\sin n x_i - \sin n x_{i-1}}{n}, \qquad \left| \int_{x_{i-1}}^{x_i} \cos nx \, dx \right| \le \frac{2}{n} . \end{equation*}

Therefore

\begin{equation*} \left| \int_{-\infty}^{\infty} s(x) \cos nx \, dx \right| \le \frac{2}{n} \sum_{i=1}^{p} |c_i| \longrightarrow 0 \quad \text{as } n \to \infty . \end{equation*}

Thus the assertion holds for such step functions.

Step 2: the general case. Let \(\epsilon > 0\). Since \(f\) is integrable over \(\mathbb{R}\), by the approximation property (ii) of Problem 44 there is a step function \(s\) on \(\mathbb{R}\) which vanishes outside a closed, bounded interval and satisfies

\begin{equation*} \int_{\mathbb{R}} |f - s| < \frac{\epsilon}{2} . \end{equation*}

(The function \(s\) is bounded and vanishes outside a set of finite measure, hence is integrable over \(\mathbb{R}\); so \(f - s\) is integrable and the following manipulations are legitimate by Theorem 17.) For each \(n\), by the linearity of integration and the Integral Comparison Test,

\begin{equation*} \left| \int_{\mathbb{R}} f(x) \cos nx \, dx - \int_{\mathbb{R}} s(x) \cos nx \, dx \right| = \left| \int_{\mathbb{R}} \bigl( f(x) - s(x) \bigr) \cos nx \, dx \right| \le \int_{\mathbb{R}} |f - s| \cdot |\cos nx| \le \int_{\mathbb{R}} |f - s| < \frac{\epsilon}{2} , \end{equation*}

where we used \(|\cos nx| \le 1\) and the monotonicity of integration for nonnegative functions.

By Step 1 there is a natural number \(N\) such that

\begin{equation*} \left| \int_{\mathbb{R}} s(x) \cos nx \, dx \right| < \frac{\epsilon}{2} \quad \text{for all } n \ge N . \end{equation*}

Hence for \(n \ge N\),

\begin{equation*} \left| \int_{-\infty}^{\infty} f(x) \cos nx \, dx \right| \le \left| \int_{\mathbb{R}} \bigl( f - s \bigr)(x) \cos nx \, dx \right| + \left| \int_{\mathbb{R}} s(x) \cos nx \, dx \right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

Since \(\epsilon > 0\) was arbitrary,

\begin{equation*} \lim_{n \to \infty} \int_{-\infty}^{\infty} f(x) \cos nx \, dx = 0 . \end{equation*}

For a function \(h\) on \(\mathbb{R}\) and a number \(t\), write \(h_t(x) = h(x+t)\).

(i) First, \(f_t\) is measurable: for each \(c\),

\begin{equation*} \{ x \mid f_t(x) > c \} = \{ x \mid f(x+t) > c \} = \{ y \mid f(y) > c \} - t , \end{equation*}

which is measurable by Proposition 10 of Chapter 2 (the translate of a measurable set is measurable).

Now suppose first that \(f \ge 0\) on \(\mathbb{R}\). We claim

\begin{equation*} \int_{\mathbb{R}} f = \int_{\mathbb{R}} f_t . \end{equation*}

Consider the map \(h \mapsto h_t\). If \(h\) is bounded, measurable, of finite support, and \(0 \le h \le f\) on \(\mathbb{R}\), then \(h_t\) is bounded, measurable (same argument as above), and vanishes outside \(E_0 - t\) where \(h\) vanishes outside the finite-measure set \(E_0\); since Lebesgue measure is translation invariant (Theorem 14 of Chapter 2), \(m(E_0 - t) = m(E_0) < \infty\), so \(h_t\) has finite support. Also \(0 \le h_t \le f_t\) on \(\mathbb{R}\). Moreover \(\int_{\mathbb{R}} h = \int_{\mathbb{R}} h_t\). Indeed, for a simple function \(\varphi = \sum_{k=1}^{p} a_k \chi_{E_k}\) with the \(E_k\) of finite measure we have \(\varphi_t = \sum_{k=1}^{p} a_k \chi_{E_k - t}\), so by the translation invariance of measure \(\int_{\mathbb{R}} \varphi = \sum_k a_k m(E_k) = \sum_k a_k m(E_k - t) = \int_{\mathbb{R}} \varphi_t\); and the integral of a bounded measurable function of finite support over a set of finite measure is, by definition, the common value of the infimum of integrals of simple functions above it and the supremum of integrals of simple functions below it, and \(\varphi \le h\) on \(\mathbb{R}\) if and only if \(\varphi_t \le h_t\) on \(\mathbb{R}\). Hence \(\int_{\mathbb{R}} h = \int_{\mathbb{R}} h_t\).

The map \(h \mapsto h_t\) is a bijection of the family of bounded measurable functions of finite support lying between \(0\) and \(f\) onto the corresponding family for \(f_t\) (its inverse is \(k \mapsto k_{-t}\)). Since it preserves integrals, the two suprema in the definition of \(\int_{\mathbb{R}} f\) and \(\int_{\mathbb{R}} f_t\) coincide:

\begin{equation*} \int_{\mathbb{R}} f = \int_{\mathbb{R}} f_t \qquad (f \ge 0) . \end{equation*}

For general integrable \(f\), note \((f_t)^+ = (f^+)_t\) and \((f_t)^- = (f^-)_t\), and likewise \(|f_t| = |f|_t\). Applying the nonnegative case to \(|f|\) shows \(f_t\) is integrable over \(\mathbb{R}\), and applying it to \(f^+\) and \(f^-\) gives

\begin{equation*} \int_{-\infty}^{\infty} f(x+t)\,dx = \int_{\mathbb{R}} (f^+)_t - \int_{\mathbb{R}} (f^-)_t = \int_{\mathbb{R}} f^+ - \int_{\mathbb{R}} f^- = \int_{-\infty}^{\infty} f(x)\, dx . \end{equation*}

(ii) Let \(M > 0\) be such that \(|g| \le M\) on \(\mathbb{R}\). By part (i), \(f_t\) is integrable over \(\mathbb{R}\) for every \(t\), hence \(f - f_t\) is integrable (Theorem 17), and since \(g\) is bounded and measurable, \(g \cdot (f - f_t)\) is integrable over \(\mathbb{R}\) by Problem 48. Moreover, by the Integral Comparison Test,

\begin{equation*} \left| \int_{\mathbb{R}} g \cdot [f - f_t] \right| \le M \int_{\mathbb{R}} |f - f_t| . \end{equation*}

Call this estimate \((\ast)\); the same reasoning bounds \(\bigl| \int_{\mathbb{R}} g \cdot u \bigr|\) by \(M \int_{\mathbb{R}} |u|\) for any function \(u\) integrable over \(\mathbb{R}\).

Case 1: \(f\) continuous on \(\mathbb{R}\) and vanishing outside a bounded set. Say \(f = 0\) outside \([-N, N]\). Then \(f\) is continuous on the compact interval \([-N-1, N+1]\) and vanishes on a neighbourhood of each of its endpoints, from which it follows that \(f\) is uniformly continuous on all of \(\mathbb{R}\). (Explicitly: \(f\) is uniformly continuous on the compact set \([-N-1, N+1]\); \(f\) is identically \(0\) off \([-N,N]\); combining the two, given \(\epsilon^{\prime} > 0\) pick \(\delta^{\prime} \in (0,1)\) that works for uniform continuity on \([-N-1, N+1]\), and note that if \(|x - y| < \delta^{\prime}\) and not both \(x, y\) lie in \([-N-1, N+1]\), then both lie outside \([-N, N]\), so \(f(x) = f(y) = 0\).)

Let \(\epsilon > 0\). Choose \(\delta \in (0, 1)\) such that

\begin{equation*} |f(x) - f(y)| < \frac{\epsilon}{M (2N + 2) + 1} \quad \text{whenever } |x - y| < \delta . \end{equation*}

Let \(|t| < \delta\). If \(|x| > N + 1\) then \(|x + t| > N\) as well, so \(f(x) - f(x+t) = 0\). Hence \(f - f_t\) vanishes outside \([-N-1, N+1]\), a set of measure \(2N+2\), and \(|f - f_t| < \epsilon / (M(2N+2)+1)\) everywhere. By monotonicity of integration and \((\ast)\),

\begin{equation*} \left| \int_{\mathbb{R}} g \cdot [f - f_t] \right| \le M \int_{[-N-1,N+1]} |f - f_t| \le M \cdot (2N+2) \cdot \frac{\epsilon}{M(2N+2)+1} < \epsilon . \end{equation*}

Thus the limit is \(0\) in this case.

Case 2: general integrable \(f\). Let \(\epsilon > 0\). By the approximation property (iii) of Problem 44 there is a function \(h\), continuous on \(\mathbb{R}\) and vanishing outside a bounded set, with

\begin{equation*} \int_{\mathbb{R}} |f - h| < \frac{\epsilon}{3(M+1)} . \end{equation*}

For any \(t\), decompose (all four functions being integrable, by part (i), Theorem 17 and Problem 48)

\begin{equation*} \int_{\mathbb{R}} g\cdot[f - f_t] = \int_{\mathbb{R}} g \cdot [f - h] + \int_{\mathbb{R}} g \cdot [h - h_t] + \int_{\mathbb{R}} g \cdot [h_t - f_t] . \end{equation*}

The first term is bounded, by \((\ast)\)-type estimation, by \(M \int_{\mathbb{R}} |f - h| < \epsilon/3\). For the third term, note \(|h_t - f_t| = |h - f|_t\), so by part (i) applied to the integrable nonnegative function \(|h - f|\),

\begin{equation*} \left| \int_{\mathbb{R}} g \cdot [h_t - f_t] \right| \le M \int_{\mathbb{R}} |h - f|_t = M \int_{\mathbb{R}} |h - f| < \frac{\epsilon}{3} . \end{equation*}

By Case 1 applied to \(h\), there is a \(\delta > 0\) such that

\begin{equation*} \left| \int_{\mathbb{R}} g \cdot [h - h_t] \right| < \frac{\epsilon}{3} \quad \text{whenever } |t| < \delta . \end{equation*}

Therefore, for \(0 < |t| < \delta\),

\begin{equation*} \left| \int_{-\infty}^{\infty} g(x) \cdot [f(x) - f(x+t)] \, dx \right| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon . \end{equation*}

Since \(\epsilon > 0\) was arbitrary, the limit is \(0\).

Choose \(M \ge 0\) with \(|g| \le M\) on \(E\).

First we check that \(f \cdot g\) is measurable on \(E\). Since \(f\) is integrable over \(E\), Proposition 15 tells us that \(f\) is finite a.e. on \(E\); let

\begin{equation*} E_0 = \{ x \in E \mid |f(x)| = \infty \}, \end{equation*}

so \(m(E_0) = 0\). On \(E \sim E_0\) both \(f\) and \(g\) are real-valued measurable functions, so their product \(f \cdot g\) is measurable on \(E \sim E_0\) by Theorem 6 of Chapter 3. Since Lebesgue measure is complete, a function on \(E\) that agrees with a measurable function outside a set of measure zero is measurable on \(E\) (Proposition 5 of Chapter 3); as \(f \cdot g\) restricted to \(E \sim E_0\) is measurable and \(m(E_0) = 0\), \(f \cdot g\) is measurable on \(E\) whatever convention is used for the product at points of \(E_0\).

Next, at every point of \(E\),

\begin{equation*} |f \cdot g| = |f| \cdot |g| \le M \cdot |f| . \end{equation*}

The function \(M|f|\) is nonnegative and, by the linearity of integration (Theorem 17, or Theorem 10 for nonnegative functions), integrable over \(E\) with \(\int_E M|f| = M \int_E |f| < \infty\). Thus \(M|f|\) is a nonnegative integrable function on \(E\) that dominates \(f \cdot g\) in the sense of the Integral Comparison Test.

By the Integral Comparison Test (Proposition 16), \(f \cdot g\) is integrable over \(E\) and

\begin{equation*} \left| \int_E f \cdot g \right| \le \int_E |f \cdot g| \le M \int_E |f| . \end{equation*}

We prove (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (iv) \(\Rightarrow\) (i).

We use repeatedly the following consequence of Proposition 15: if \(h\) is integrable over a measurable set \(B\) and \(h = 0\) a.e. on \(B\), then \(\int_B h = 0\). Indeed, letting \(B_0 = \{ x \in B \mid h(x) \ne 0\}\), we have \(m(B_0) = 0\), so by Proposition 15, \(\int_B h = \int_{B \sim B_0} h = 0\), the last integral vanishing because its integrand is identically \(0\).

(i) \(\Rightarrow\) (ii). Let \(g\) be a bounded measurable function on \(\mathbb{R}\). By Problem 48, \(f g\) is integrable over \(\mathbb{R}\). Since \(f = 0\) a.e. on \(\mathbb{R}\), also \(f g = 0\) a.e. on \(\mathbb{R}\), and therefore \(\int_{\mathbb{R}} f g = 0\) by the remark above.

(ii) \(\Rightarrow\) (iii). Let \(A\) be a measurable set. The characteristic function \(\chi_A\) is measurable and bounded (by \(1\)) on \(\mathbb{R}\), so by hypothesis \(\int_{\mathbb{R}} f \chi_A = 0\). On the other hand \(f \chi_A\) is integrable over \(\mathbb{R}\) (Problem 48), \(f \chi_A = f\) on \(A\) and \(f\chi_A = 0\) on \(\mathbb{R} \sim A\); hence, by the additivity of integration over domains (Corollary 18),

\begin{equation*} \int_{\mathbb{R}} f \chi_A = \int_A f \chi_A + \int_{\mathbb{R} \sim A} f \chi_A = \int_A f + 0 = \int_A f . \end{equation*}

Therefore \(\int_A f = 0\).

(iii) \(\Rightarrow\) (iv). Every open subset \(\mathcal{O}\) of \(\mathbb{R}\) is measurable, so this is immediate.

(iv) \(\Rightarrow\) (i). Assume \(\int_{\mathcal{O}} f = 0\) for every open set \(\mathcal{O}\).

Step 1: \(\int_A f = 0\) for every measurable set \(A\) of finite measure. Let \(A\) be such a set and let \(\epsilon > 0\). Since \(f\) is integrable over \(\mathbb{R}\), Proposition 23 provides a \(\delta > 0\) such that

\begin{equation*} \text{if } B \subseteq \mathbb{R} \text{ is measurable and } m(B) < \delta, \text{ then } \int_B |f| < \epsilon . \end{equation*}

By assertion (i) of Theorem 11 of Chapter 2 (outer approximation by open sets), there is an open set \(\mathcal{O}\) containing \(A\) with \(m(\mathcal{O} \sim A) < \delta\). Note that \(f\) is integrable over every measurable subset \(B\) of \(\mathbb{R}\): by the additivity of integration over domains for nonnegative measurable functions (Theorem 11), \(\int_{\mathbb{R}} |f| = \int_B |f| + \int_{\mathbb{R} \sim B} |f| \ge \int_B |f|\), so \(\int_B |f| < \infty\) and the Integral Comparison Test applies. Hence, applying the additivity of integration over domains (Corollary 18) to the disjoint decomposition \(\mathcal{O} = A \cup (\mathcal{O} \sim A)\),

\begin{equation*} \int_A f = \int_{\mathcal{O}} f - \int_{\mathcal{O} \sim A} f = 0 - \int_{\mathcal{O} \sim A} f . \end{equation*}

By the Integral Comparison Test and the choice of \(\delta\),

\begin{equation*} \left| \int_A f \right| = \left| \int_{\mathcal{O} \sim A} f \right| \le \int_{\mathcal{O} \sim A} |f| < \epsilon . \end{equation*}

As \(\epsilon > 0\) was arbitrary, \(\int_A f = 0\).

Step 2: \(f = 0\) a.e. Let

\begin{equation*} A^+ = \{ x \in \mathbb{R} \mid f(x) > 0 \}, \qquad A^- = \{x \in \mathbb{R} \mid f(x) < 0 \} . \end{equation*}

For natural numbers \(n\) put

\begin{equation*} A_n^+ = \left\{ x \in (-n, n) \;\Big|\; f(x) > \frac{1}{n} \right\} . \end{equation*}

Each \(A_n^+\) is measurable and \(m(A_n^+) \le 2n < \infty\), so by Step 1, \(\int_{A_n^+} f = 0\). But \(f > 1/n\) on \(A_n^+\), so by the monotonicity of integration for nonnegative functions,

\begin{equation*} 0 = \int_{A_n^+} f \ge \int_{A_n^+} \frac{1}{n} = \frac{1}{n} \, m(A_n^+) \ge 0 . \end{equation*}

Hence \(m(A_n^+) = 0\) for every \(n\). Since

\begin{equation*} A^+ = \bigcup_{n=1}^{\infty} A_n^+ \end{equation*}

(if \(f(x) > 0\), choose \(n\) with \(|x| < n\) and \(f(x) > 1/n\), which is possible whether \(f(x)\) is finite and positive or \(f(x) = \infty\)), the countable subadditivity of measure gives \(m(A^+) = 0\).

Applying the same argument to \(-f\), which also satisfies the hypothesis of (iv) since \(\int_{\mathcal{O}} (-f) = -\int_{\mathcal{O}} f = 0\) for every open \(\mathcal{O}\), we obtain \(m(A^-) = 0\).

Therefore \(m(\{x \in \mathbb{R} \mid f(x) \ne 0\}) = m(A^+) + m(A^-) = 0\), that is, \(f = 0\) a.e. on \(\mathbb{R}\).

This closes the cycle, so the four assertions are equivalent.

Recall the definition in force: \(\mathcal{F}\) is uniformly integrable over \(E\) provided for each \(\epsilon>0\) there is a \(\delta>0\) such that for each \(f\in\mathcal{F}\), if \(A\subseteq E\) is measurable and \(m(A)<\delta\), then \(\int_A |f| < \epsilon\). So the content of the problem is that the absolute value may be moved outside the integral sign in this criterion.

Observe first that each \(f\in\mathcal{F}\) is integrable over \(E\), hence integrable over every measurable subset \(A\) of \(E\): indeed \(|f|\cdot\chi_A \le |f|\) on \(E\), so \(|f|\) is integrable over \(A\) by the monotonicity of integration for nonnegative functions, and therefore \(f\) is integrable over \(A\) by the definition of the general Lebesgue integral. In particular all the integrals written below are finite.

The criterion is necessary. Assume \(\mathcal{F}\) is uniformly integrable over \(E\). Let \(\epsilon>0\) and let \(\delta>0\) respond to the \(\epsilon\) challenge in the definition of uniform integrability. Let \(f\in\mathcal{F}\) and let \(A\subseteq E\) be measurable with \(m(A)<\delta\). Since \(-|f| \le f \le |f|\) on \(A\), the monotonicity of integration (Theorem 17 of Chapter 4) gives

\begin{equation*} -\int_A |f| \;\le\; \int_A f \;\le\; \int_A |f| , \end{equation*}

that is, \(\left| \int_A f \right| \le \int_A |f| < \epsilon\). Thus the same \(\delta\) responds to the \(\epsilon\) challenge in the stated criterion.

The criterion is sufficient. Assume that for each \(\epsilon>0\) there is a \(\delta>0\) such that for every \(f\in\mathcal{F}\) and every measurable \(A\subseteq E\) with \(m(A)<\delta\) we have \(\left|\int_A f\right| < \epsilon\). Let \(\epsilon>0\) be given and let \(\delta>0\) respond to the challenge \(\epsilon/2\) in this criterion. We claim this same \(\delta\) responds to the \(\epsilon\) challenge in the definition of uniform integrability.

Indeed, let \(f\in\mathcal{F}\) and let \(A\subseteq E\) be measurable with \(m(A)<\delta\). Since \(f\) is measurable, the sets

\begin{equation*} A^{+} = \{x \in A \mid f(x) \ge 0\}, \qquad A^{-} = \{x \in A \mid f(x) < 0\} \end{equation*}

are measurable. They are disjoint, their union is \(A\), and each is a measurable subset of \(E\) with

\begin{equation*} m(A^{+}) \le m(A) < \delta, \qquad m(A^{-}) \le m(A) < \delta \end{equation*}

by the monotonicity of Lebesgue measure. Hence, by the choice of \(\delta\),

\begin{equation*} \left| \int_{A^{+}} f \right| < \frac{\epsilon}{2}, \qquad \left| \int_{A^{-}} f \right| < \frac{\epsilon}{2} . \end{equation*}

Now \(|f| = f\) on \(A^{+}\) and \(|f| = -f\) on \(A^{-}\). Since \(|f|\) is integrable over \(E\) and \(A = A^{+} \cup A^{-}\) is a disjoint union of measurable subsets of \(E\), the additivity over domains of integration (Corollary 18 of Chapter 4), applied to \(|f|\) and then to \(f\) over each piece, yields

\begin{equation*} \int_A |f| \;=\; \int_{A^{+}} |f| + \int_{A^{-}} |f| \;=\; \int_{A^{+}} f \;-\; \int_{A^{-}} f . \end{equation*}

But \(f \ge 0\) on \(A^{+}\) forces \(\int_{A^{+}} f \ge 0\), so \(\int_{A^{+}} f = \left|\int_{A^{+}} f\right|\); and \(f \le 0\) on \(A^{-}\) forces \(\int_{A^{-}} f \le 0\), so \(-\int_{A^{-}} f = \left|\int_{A^{-}} f\right|\). Therefore

\begin{equation*} \int_A |f| \;=\; \left| \int_{A^{+}} f \right| + \left| \int_{A^{-}} f \right| \;<\; \frac{\epsilon}{2} + \frac{\epsilon}{2} \;=\; \epsilon . \end{equation*}

Since \(f\in\mathcal{F}\) and the measurable set \(A\subseteq E\) with \(m(A)<\delta\) were arbitrary, \(\mathcal{F}\) is uniformly integrable over \(E\).

Throughout, \(\mathcal{F}\) is uniformly integrable over \(E\) means: for each \(\epsilon>0\) there is a \(\delta>0\) such that for each \(f\in\mathcal{F}\), if \(A\subseteq E\) is measurable and \(m(A)<\delta\), then \(\int_A |f| < \epsilon\). As in the preceding problem, each \(f\in\mathcal{F}\) is integrable over every measurable subset of \(E\), so all integrals below are finite. We also use repeatedly the following consequence of the additivity over domains of integration (Corollary 18 of Chapter 4) for the nonnegative integrable function \(|f|\): if \(A \subseteq B\) are measurable subsets of \(E\), then, writing \(B = A \cup (B \sim A)\) as a disjoint union,

\begin{equation*} \int_B |f| = \int_A |f| + \int_{B \sim A} |f| \;\ge\; \int_A |f| , \end{equation*}

since \(\int_{B\sim A} |f| \ge 0\) by the monotonicity of integration for nonnegative functions.

The criterion is necessary. Assume \(\mathcal{F}\) is uniformly integrable over \(E\), let \(\epsilon>0\), and let \(\delta>0\) respond to the \(\epsilon\) challenge in the definition. If \(\mathcal{U}\) is open and \(m(E\cap\mathcal{U})<\delta\), then \(A = E\cap\mathcal{U}\) is a measurable subset of \(E\) (open sets are measurable, and the intersection of two measurable sets is measurable) of measure less than \(\delta\). Hence \(\int_{E\cap\mathcal{U}} |f| < \epsilon\) for every \(f\in\mathcal{F}\). So the same \(\delta\) responds to the \(\epsilon\) challenge in the stated criterion.

The criterion is sufficient. Assume that for each \(\epsilon>0\) there is a \(\delta>0\) such that for all \(f\in\mathcal{F}\), if \(\mathcal{U}\) is open and \(m(E\cap\mathcal{U})<\delta\), then \(\int_{E\cap\mathcal{U}} |f| < \epsilon\).

Let \(\epsilon>0\) be given. Choose \(\delta_0>0\) responding to this \(\epsilon\) challenge in the assumed criterion, and define

\begin{equation*} \delta = \frac{\delta_0}{2} > 0 . \end{equation*}

We claim \(\delta\) responds to the \(\epsilon\) challenge in the definition of uniform integrability. Let \(f\in\mathcal{F}\) and let \(A\subseteq E\) be measurable with \(m(A) < \delta\).

Since \(A\) is measurable, by the outer approximation property of measurable sets, Theorem 11(i) of Chapter 2, there is an open set \(\mathcal{U}\) containing \(A\) for which

\begin{equation*} m^{*}(\mathcal{U} \sim A) < \frac{\delta_0}{2} . \end{equation*}

Now \(E\cap\mathcal{U} \subseteq A \cup (\mathcal{U}\sim A)\), because a point of \(E\cap\mathcal{U}\) that fails to lie in \(A\) lies in \(\mathcal{U}\sim A\). Hence, by the monotonicity and finite subadditivity of outer measure,

\begin{equation*} m(E\cap\mathcal{U}) \;\le\; m(A) + m^{*}(\mathcal{U}\sim A) \;<\; \frac{\delta_0}{2} + \frac{\delta_0}{2} \;=\; \delta_0 . \end{equation*}

Therefore, by the choice of \(\delta_0\),

\begin{equation*} \int_{E\cap\mathcal{U}} |f| < \epsilon . \end{equation*}

Finally, \(A \subseteq E \cap \mathcal{U}\), since \(A\subseteq E\) and \(A\subseteq\mathcal{U}\). By the monotonicity over domains noted at the outset, applied to the nonnegative function \(|f|\),

\begin{equation*} \int_A |f| \;\le\; \int_{E\cap\mathcal{U}} |f| \;<\; \epsilon . \end{equation*}

Since \(f\in\mathcal{F}\) and the measurable set \(A\subseteq E\) with \(m(A)<\delta\) were arbitrary, \(\mathcal{F}\) is uniformly integrable over \(E\).

(a) No. A uniform bound on the integrals does not prevent the mass of \(|f|\) from concentrating on sets of arbitrarily small measure.

For each natural number \(n\) define

\begin{equation*} f_n = n \cdot \chi_{(0,\,1/n]} \quad \text{on } [0,1]. \end{equation*}

Each \(f_n\) is a nonnegative simple function, hence bounded and measurable on the set \([0,1]\) of finite measure, and so integrable over \([0,1]\), with

\begin{equation*} \int_0^1 |f_n| = n \cdot m\big((0,1/n]\big) = n \cdot \frac{1}{n} = 1 \le 1 . \end{equation*}

Thus \(f_n \in \mathcal{F}\) for every \(n\).

Now take the challenge \(\epsilon = 1/2\) and let \(\delta>0\) be arbitrary. By the Archimedean property of \(\mathbb{R}\) choose a natural number \(n\) with \(1/n < \delta\), and set \(A_n = (0,1/n]\). Then \(A_n\) is a measurable subset of \([0,1]\) with \(m(A_n) = 1/n < \delta\), yet

\begin{equation*} \int_{A_n} |f_n| = n \cdot m(A_n) = 1 > \frac{1}{2} = \epsilon . \end{equation*}

So no \(\delta>0\) responds to the challenge \(\epsilon = 1/2\), and \(\mathcal{F}\) is not uniformly integrable over \([0,1]\).

(b) Yes. Here the whole family is dominated by a single integrable function, namely the constant function \(1\).

Let \(f\in\mathcal{F}\). Being continuous on \([0,1]\), \(f\) is measurable (Proposition 3 of Chapter 3), and it is bounded, since \(|f|\le 1\); as \(m([0,1]) = 1 < \infty\), \(f\) is integrable over \([0,1]\) by Theorem 4 of Chapter 4. Let \(\epsilon>0\) and put \(\delta = \epsilon\). If \(A\subseteq[0,1]\) is measurable with \(m(A)<\delta\), then \(|f| \le 1\) on \(A\), so by the monotonicity of integration for nonnegative functions (Theorem 10 of Chapter 4),

\begin{equation*} \int_A |f| \;\le\; \int_A 1 \;=\; m(A) \;<\; \delta \;=\; \epsilon . \end{equation*}

Since \(\delta\) was chosen independently of \(f\in\mathcal{F}\), the family \(\mathcal{F}\) is uniformly integrable over \([0,1]\). This is in fact an instance of the Example following the definition of uniform integrability in Section 4.6, taken with \(g \equiv 1\): continuity plays no role, only the uniform bound \(|f|\le 1\) matters.

(c) Yes. The hypothesis controls \(\int_A |f|\) only when \(A\) is a closed, bounded interval; but by Proposition 9 of Chapter 1 every open set is assembled from intervals, and by the preceding problem it suffices to test the uniform integrability criterion on open sets.

Step 1: the estimate on open sets. We claim that for each \(f\in\mathcal{F}\) and each open set \(\mathcal{U}\subseteq\mathbb{R}\),

\begin{equation*} \int_{[0,1]\cap\,\mathcal{U}} |f| \;\le\; m\big([0,1]\cap\mathcal{U}\big) . \end{equation*}

Put \(\mathcal{V} = (0,1)\cap\mathcal{U}\), an open subset of \(\mathbb{R}\). The set \(\big([0,1]\cap\mathcal{U}\big) \sim \mathcal{V}\) is contained in \(\{0,1\}\) and so has measure zero; hence \(m([0,1]\cap\mathcal{U}) = m(\mathcal{V})\), and since \(|f|\) is a nonnegative measurable function on \([0,1]\), Theorem 11 of Chapter 4 (additivity over domains of integration, together with the vanishing of the integral over a subset of measure zero) gives

\begin{equation*} \int_{[0,1]\cap\,\mathcal{U}} |f| \;=\; \int_{\mathcal{V}} |f| . \end{equation*}

If \(\mathcal{V} = \emptyset\), both sides of the claimed inequality are \(0\) and there is nothing to prove. Otherwise, by Proposition 9 of Chapter 1, \(\mathcal{V}\) is the union of a countable disjoint collection \(\{I_k\}\) of open intervals. Since \(\mathcal{V}\subseteq(0,1)\), each \(I_k\) is bounded, say \(I_k = (a_k, b_k)\) with \(0 \le a_k < b_k \le 1\), so that \([a_k,b_k]\subseteq[0,1]\). The set \([a_k,b_k] \sim (a_k,b_k) = \{a_k,b_k\}\) has measure zero, so again by Theorem 11 of Chapter 4 and the defining hypothesis on \(f\),

\begin{equation*} \int_{I_k} |f| \;=\; \int_{[a_k,\,b_k]} |f| \;=\; \int_{a_k}^{b_k} |f| \;\le\; b_k - a_k \;=\; m(I_k) . \end{equation*}

Since \(f\) is integrable over \([0,1]\) it is integrable over the measurable subset \(\mathcal{V}\) (indeed \(|f|\cdot\chi_{\mathcal{V}} \le |f|\) on \([0,1]\), so \(\int_{\mathcal{V}} |f| \le \int_0^1 |f| < \infty\) by the monotonicity of integration for nonnegative functions, Theorem 10 of Chapter 4), and \(\mathcal{V}\) is the disjoint union of the countable collection \(\{I_k\}\) of measurable sets. Therefore, by the Countable Additivity of Integration (Theorem 20 of Chapter 4) applied to the integrable function \(|f|\) over \(\mathcal{V}\), and then by the countable additivity of Lebesgue measure,

\begin{equation*} \int_{\mathcal{V}} |f| \;=\; \sum_{k} \int_{I_k} |f| \;\le\; \sum_{k} m(I_k) \;=\; m(\mathcal{V}) . \end{equation*}

If the collection \(\{I_k\}\) is finite, the same conclusion follows from Corollary 18 of Chapter 4 and the finite additivity of measure. Combining the three displays proves the claim.

Step 2: conclusion. Let \(\epsilon>0\) and put \(\delta = \epsilon\). If \(\mathcal{U}\) is open and \(m([0,1]\cap\mathcal{U}) < \delta\), then by Step 1, for every \(f\in\mathcal{F}\),

\begin{equation*} \int_{[0,1]\cap\,\mathcal{U}} |f| \;\le\; m\big([0,1]\cap\mathcal{U}\big) \;<\; \delta \;=\; \epsilon . \end{equation*}

Thus \(\delta\) responds, uniformly over all of \(\mathcal{F}\), to the \(\epsilon\) challenge in the criterion of the preceding problem (Problem 51). Therefore \(\mathcal{F}\) is uniformly integrable over \([0,1]\).

One can avoid invoking Problem 51 and argue directly: given a measurable \(A\subseteq[0,1]\) with \(m(A) < \epsilon/2\), use Theorem 11(i) of Chapter 2 to choose an open set \(\mathcal{U}\) containing \(A\) with \(m^{*}(\mathcal{U}\sim A) < \epsilon/2\); then \(m([0,1]\cap\mathcal{U}) \le m(A) + m^{*}(\mathcal{U}\sim A) < \epsilon\), and by Step 1 together with the monotonicity over domains of the integral of the nonnegative function \(|f|\),

\begin{equation*} \int_A |f| \;\le\; \int_{[0,1]\cap\,\mathcal{U}} |f| \;\le\; m\big([0,1]\cap\mathcal{U}\big) \;<\; \epsilon . \end{equation*}

We prove the two implications separately.

Assume first that \(\{h_n\}\) is uniformly integrable and tight over \(E\). Since \(\{h_n\} \to 0\) pointwise almost everywhere on \(E\), the Vitali Convergence Theorem of this section (the version for sets of possibly infinite measure) applies with \(f \equiv 0\) and yields

\begin{equation*} \lim_{n \to \infty} \int_E h_n = \int_E 0 = 0 . \end{equation*}

Conversely, assume \(\lim_{n \to \infty} \int_E h_n = 0\). Let \(\epsilon > 0\). Choose a natural number \(N\) for which

\begin{equation*} \int_E h_n < \epsilon \quad \text{for all } n \ge N . \end{equation*}

Since each \(h_n\) is nonnegative, monotonicity of integration together with additivity over domains gives, for every measurable subset \(A\) of \(E\) and every \(n \ge N\),

\begin{equation*} \int_A |h_n| = \int_A h_n \le \int_E h_n < \epsilon . \tag{1} \end{equation*}

Thus the tail of the sequence satisfies the uniform integrability and the tightness criteria for this \(\epsilon\) with no restriction at all on \(A\), and with the empty set serving as the set of finite measure.

It remains to handle the finitely many initial indices. If \(N = 1\) there are none, and we may take \(\delta = 1\) and \(E_0 = \emptyset\). Otherwise consider the finite collection \(\{h_n\}_{n=1}^{N-1}\), each member of which is integrable over \(E\). By Proposition 23 of Chapter 4, for each \(n\) with \(1 \le n \le N - 1\) there is a \(\delta_n > 0\) such that

\begin{equation*} \text{if } A \subseteq E \text{ is measurable and } m(A) < \delta_n, \text{ then } \int_A |h_n| < \epsilon . \end{equation*}

Define \(\delta = \min\{\delta_1, \dots, \delta_{N-1}\} > 0\). (This is exactly Proposition 24 of Chapter 4: a finite collection of integrable functions is uniformly integrable.) By (1), the same \(\delta\) responds to the \(\epsilon\) challenge for every index \(n\), so \(\{h_n\}_{n=1}^{\infty}\) is uniformly integrable over \(E\).

For tightness, apply Proposition 1 of this section to each of the integrable functions \(h_1, \dots, h_{N-1}\): there are sets \(E_1, \dots, E_{N-1}\) of finite measure for which

\begin{equation*} \int_{E \sim E_n} h_n < \epsilon \quad \text{for } 1 \le n \le N-1 . \end{equation*}

Put \(E_0 = \bigcup_{n=1}^{N-1} E_n\). Being a finite union of sets of finite measure, \(E_0\) has finite measure. For \(1 \le n \le N - 1\) we have \(E \sim E_0 \subseteq E \sim E_n\), so by monotonicity

\begin{equation*} \int_{E \sim E_0} h_n \le \int_{E \sim E_n} h_n < \epsilon , \end{equation*}

while for \(n \ge N\) inequality (1), applied with \(A = E \sim E_0\), gives \(\int_{E \sim E_0} h_n < \epsilon\). Hence \(E_0\) responds to the \(\epsilon\) challenge for the whole sequence, and \(\{h_n\}\) is tight over \(E\).

Since \(\epsilon > 0\) was arbitrary, \(\{h_n\}\) is uniformly integrable and tight over \(E\). The proof is complete.

Let \(\epsilon > 0\).

Uniform integrability. For each index \(k\) with \(1 \le k \le n\), the function \(f_k\) is integrable over \(E\), so by Proposition 23 of Chapter 4 there is a \(\delta_k > 0\) for which

\begin{equation*} \text{if } A \subseteq E \text{ is measurable and } m(A) < \delta_k, \text{ then } \int_A |f_k| < \epsilon . \end{equation*}

Define \(\delta = \min\{\delta_1, \dots, \delta_n\}\). Since this is a minimum of finitely many positive numbers, \(\delta > 0\). If \(A \subseteq E\) is measurable with \(m(A) < \delta\), then \(m(A) < \delta_k\) for every \(k\), and hence \(\int_A |f_k| < \epsilon\) for every \(k\). Thus \(\delta\) responds to the \(\epsilon\) challenge regarding the criterion for the family \(\{f_k\}_{k=1}^{n}\) to be uniformly integrable over \(E\). This is Proposition 24 of Chapter 4.

Tightness. For each index \(k\) with \(1 \le k \le n\), the function \(f_k\) is integrable over \(E\), so by Proposition 1 of this section there is a set \(E_k \subseteq E\) of finite measure for which

\begin{equation*} \int_{E \sim E_k} |f_k| < \epsilon . \end{equation*}

Define

\begin{equation*} E_0 = \bigcup_{k=1}^{n} E_k . \end{equation*}

Then \(E_0\) is a measurable subset of \(E\), and by the finite subadditivity of measure \(m(E_0) \le \sum_{k=1}^{n} m(E_k) < \infty\), so \(E_0\) has finite measure. Moreover, for each \(k\) we have \(E_0 \supseteq E_k\) and therefore \(E \sim E_0 \subseteq E \sim E_k\). By the monotonicity of integration over domains (the integrand \(|f_k|\) being nonnegative),

\begin{equation*} \int_{E \sim E_0} |f_k| \le \int_{E \sim E_k} |f_k| < \epsilon \quad \text{for } 1 \le k \le n . \end{equation*}

Hence \(E_0\) responds to the \(\epsilon\) challenge regarding the criterion for tightness.

Since \(\epsilon > 0\) was arbitrary, the finite family \(\{f_k\}_{k=1}^{n}\) is both uniformly integrable and tight over \(E\).

Set \(M = |\alpha| + |\beta| + 1\), so that \(M > 0\).

We first check that the functions \(\alpha f_n + \beta g_n\) are properly defined. As the book observes just after the definition of tightness, a family that is uniformly integrable and tight over \(E\) consists of functions that are integrable over \(E\); hence each \(f_n\) and each \(g_n\) is integrable over \(E\) and therefore, by Proposition 15 of Chapter 4, finite a.e. on \(E\). Let \(Z\) be the union over all \(n\) of the sets where \(f_n\) or \(g_n\) fails to be finite. Being a countable union of sets of measure zero, \(Z\) has measure zero by the countable subadditivity of measure. Off \(Z\) every combination \(\alpha f_n + \beta g_n\) is defined and finite, and it is measurable there as a linear combination of measurable functions; since altering a function on a set of measure zero changes neither its measurability nor any of its integrals, we may and do work on \(E \sim Z\), and all integrals below are unaffected.

Pointwise on \(E \sim Z\) we have

\begin{equation*} |\alpha f_n + \beta g_n| \le |\alpha| \, |f_n| + |\beta| \, |g_n| . \end{equation*}

Consequently, by the monotonicity and linearity of integration for nonnegative measurable functions, for every measurable subset \(A\) of \(E\),

\begin{equation*} \int_A |\alpha f_n + \beta g_n| \le |\alpha| \int_A |f_n| + |\beta| \int_A |g_n| . \tag{2} \end{equation*}

Uniform integrability. Let \(\epsilon > 0\). Since \(\{f_n\}\) is uniformly integrable over \(E\), there is a \(\delta_1 > 0\) such that for every index \(n\),

\begin{equation*} \text{if } A \subseteq E \text{ is measurable and } m(A) < \delta_1, \text{ then } \int_A |f_n| < \epsilon / M . \end{equation*}

Since \(\{g_n\}\) is uniformly integrable over \(E\), there is a \(\delta_2 > 0\) that does the same for the \(g_n\)’s with the same bound \(\epsilon / M\). Define \(\delta = \min\{\delta_1, \delta_2\} > 0\). If \(A \subseteq E\) is measurable with \(m(A) < \delta\), then by (2), for every index \(n\),

\begin{equation*} \int_A |\alpha f_n + \beta g_n| \le |\alpha| \cdot \frac{\epsilon}{M} + |\beta| \cdot \frac{\epsilon}{M} = \frac{|\alpha| + |\beta|}{M} \, \epsilon < \epsilon , \end{equation*}

since \(|\alpha| + |\beta| < M\). Thus \(\delta\) responds to the \(\epsilon\) challenge, and \(\{\alpha f_n + \beta g_n\}\) is uniformly integrable over \(E\).

Tightness. Let \(\epsilon > 0\). By the tightness of \(\{f_n\}\) there is a set \(E_1 \subseteq E\) of finite measure for which \(\int_{E \sim E_1} |f_n| < \epsilon / M\) for all \(n\), and by the tightness of \(\{g_n\}\) there is a set \(E_2 \subseteq E\) of finite measure for which \(\int_{E \sim E_2} |g_n| < \epsilon / M\) for all \(n\). Define \(E_0 = E_1 \cup E_2\). Then \(m(E_0) \le m(E_1) + m(E_2) < \infty\), so \(E_0\) has finite measure, and

\begin{equation*} E \sim E_0 \subseteq E \sim E_1, \qquad E \sim E_0 \subseteq E \sim E_2 . \end{equation*}

Hence, by monotonicity of integration over domains and by (2) applied with \(A = E \sim E_0\), for every index \(n\),

\begin{equation*} \int_{E \sim E_0} |\alpha f_n + \beta g_n| \le |\alpha| \int_{E \sim E_1} |f_n| + |\beta| \int_{E \sim E_2} |g_n| < \frac{|\alpha| + |\beta|}{M} \, \epsilon < \epsilon . \end{equation*}

Thus \(E_0\) responds to the \(\epsilon\) challenge for tightness, and \(\{\alpha f_n + \beta g_n\}\) is tight over \(E\).

Suppose first that \(\{f_n\}\) is uniformly integrable and tight over \(E\), and let \(\epsilon > 0\). By tightness there is a measurable subset \(E_0\) of \(E\) of finite measure for which

\begin{equation*} \int_{E \sim E_0} |f_n| < \epsilon / 2 \quad \text{for all } n . \tag{3} \end{equation*}

By uniform integrability there is a \(\delta > 0\) such that

\begin{equation*} \text{if } B \subseteq E \text{ is measurable and } m(B) < \delta, \text{ then } \int_B |f_n| < \epsilon / 2 \text{ for all } n . \tag{4} \end{equation*}

We claim this \(E_0\) and this \(\delta\) respond to the \(\epsilon\) challenge. Indeed, let \(A\) be a measurable subset of \(E\) with \(m(A \cap E_0) < \delta\), and let \(n\) be any index. Decompose \(A\) as the disjoint union of the measurable sets \(A \cap E_0\) and \(A \sim E_0\). By the additivity over domains of integration of the nonnegative function \(|f_n|\),

\begin{equation*} \int_A |f_n| = \int_{A \cap E_0} |f_n| + \int_{A \sim E_0} |f_n| . \end{equation*}

The first term is less than \(\epsilon / 2\) by (4), since \(m(A \cap E_0) < \delta\). For the second, \(A \sim E_0 \subseteq E \sim E_0\), so by monotonicity and (3) it is at most \(\int_{E \sim E_0} |f_n| < \epsilon / 2\). Adding,

\begin{equation*} \int_A |f_n| < \epsilon / 2 + \epsilon / 2 = \epsilon , \end{equation*}

which is what was to be shown.

Conversely, suppose that for each \(\epsilon > 0\) there is a measurable subset \(E_0\) of \(E\) of finite measure and a \(\delta > 0\) with the stated property. Let \(\epsilon > 0\) and let \(E_0\) and \(\delta\) be so chosen.

For uniform integrability: if \(A \subseteq E\) is measurable and \(m(A) < \delta\), then by monotonicity of measure \(m(A \cap E_0) \le m(A) < \delta\), and therefore \(\int_A |f_n| < \epsilon\) for every index \(n\). Thus this same \(\delta\) responds to the \(\epsilon\) challenge in the definition of uniform integrability.

For tightness: take \(A = E \sim E_0\). Then \(A \cap E_0 = \emptyset\), so \(m(A \cap E_0) = 0 < \delta\), and the hypothesis gives

\begin{equation*} \int_{E \sim E_0} |f_n| < \epsilon \quad \text{for all } n . \end{equation*}

Since \(E_0\) has finite measure, \(E_0\) responds to the \(\epsilon\) challenge in the definition of tightness.

As \(\epsilon > 0\) was arbitrary, \(\{f_n\}\) is uniformly integrable and tight over \(E\).

Suppose first that \(\{f_n\}\) is uniformly integrable and tight over \(\mathbb{R}\), and let \(\epsilon > 0\).

By uniform integrability there is a \(\delta > 0\) such that

\begin{equation*} \text{if } A \subseteq \mathbb{R} \text{ is measurable and } m(A) < \delta, \text{ then } \int_A |f_n| < \epsilon / 3 \text{ for all } n . \tag{5} \end{equation*}

By tightness there is a measurable set \(E_0 \subseteq \mathbb{R}\) of finite measure for which

\begin{equation*} \int_{\mathbb{R} \sim E_0} |f_n| < \epsilon / 3 \quad \text{for all } n . \tag{6} \end{equation*}

Since \(m(E_0) < \infty\), and since the sets \(E_0 \sim (-k, k)\), for \(k\) a natural number, form a descending sequence of measurable sets of finite measure whose intersection is empty, Theorem 15 of Chapter 2 (the Continuity of Measure) tells us that \(\lim_{k \to \infty} m(E_0 \sim (-k, k)) = 0\). Choose \(r > 0\) (a natural number will do) for which

\begin{equation*} m(E_0 \sim (-r, r)) < \delta . \end{equation*}

Now

\begin{equation*} \mathbb{R} \sim (-r, r) \subseteq (\mathbb{R} \sim E_0) \cup (E_0 \sim (-r, r)) , \end{equation*}

so by the monotonicity and the (finite) subadditivity of integration of the nonnegative function \(|f_n|\) over domains, together with (5) and (6),

\begin{equation*} \int_{\mathbb{R} \sim (-r, r)} |f_n| \le \int_{\mathbb{R} \sim E_0} |f_n| + \int_{E_0 \sim (-r, r)} |f_n| < \epsilon / 3 + \epsilon / 3 = 2 \epsilon / 3 \quad \text{for all } n . \tag{7} \end{equation*}

We claim this \(r\) and this \(\delta\) respond to the \(\epsilon\) challenge. Indeed, let \(\mathcal{O}\) be an open subset of \(\mathbb{R}\) with \(m(\mathcal{O} \cap (-r, r)) < \delta\) and let \(n\) be any index. Splitting \(\mathcal{O}\) into the disjoint measurable pieces \(\mathcal{O} \cap (-r, r)\) and \(\mathcal{O} \sim (-r, r)\) and using additivity over domains, (5) and (7),

\begin{equation*} \int_{\mathcal{O}} |f_n| = \int_{\mathcal{O} \cap (-r, r)} |f_n| + \int_{\mathcal{O} \sim (-r, r)} |f_n| < \frac{\epsilon}{3} + \int_{\mathbb{R} \sim (-r, r)} |f_n| < \frac{\epsilon}{3} + \frac{2 \epsilon}{3} = \epsilon . \end{equation*}

(Openness of \(\mathcal{O}\) was not needed here; the estimate holds for every measurable set.)

Conversely, suppose that for each \(\epsilon > 0\) there are positive numbers \(r\) and \(\delta\) with the stated property. Let \(\epsilon > 0\), and let \(r > 0\) and \(\delta > 0\) respond to this \(\epsilon\).

Tightness. Take \(E_0 = (-r, r)\), a set of finite measure \(2r\). The set \(\mathcal{O} = \mathbb{R} \sim [-r, r]\) is open and \(\mathcal{O} \cap (-r, r) = \emptyset\), so \(m(\mathcal{O} \cap (-r, r)) = 0 < \delta\) and therefore \(\int_{\mathcal{O}} |f_n| < \epsilon\) for all \(n\). Since \(\mathbb{R} \sim (-r, r)\) differs from \(\mathcal{O}\) only by the two-point set \(\{-r, r\}\), which has measure zero, the integrals over these two domains agree, and hence

\begin{equation*} \int_{\mathbb{R} \sim E_0} |f_n| = \int_{\mathcal{O}} |f_n| < \epsilon \quad \text{for all } n . \end{equation*}

Thus \(E_0\) responds to the \(\epsilon\) challenge for tightness.

Uniform integrability. We claim \(\delta / 2\) responds to the \(\epsilon\) challenge. Let \(A \subseteq \mathbb{R}\) be measurable with \(m(A) < \delta / 2\). Since \(A\) is measurable, assertion (i) of Theorem 11 of Chapter 2 provides an open set \(\mathcal{O}\) containing \(A\) for which \(m(\mathcal{O} \sim A) < \delta / 2\). By the additivity of measure,

\begin{equation*} m(\mathcal{O}) = m(A) + m(\mathcal{O} \sim A) < \delta / 2 + \delta / 2 = \delta , \end{equation*}

and hence \(m(\mathcal{O} \cap (-r, r)) \le m(\mathcal{O}) < \delta\). By hypothesis \(\int_{\mathcal{O}} |f_n| < \epsilon\) for every index \(n\), and since \(A \subseteq \mathcal{O}\), monotonicity of integration over domains gives

\begin{equation*} \int_A |f_n| \le \int_{\mathcal{O}} |f_n| < \epsilon \quad \text{for all } n . \end{equation*}

Since \(\epsilon > 0\) was arbitrary, \(\{f_n\}\) is uniformly integrable and tight over \(\mathbb{R}\).

Recall that, by the definition of convergence in measure, the hypothesis \(\{f_n\} \to f\) in measure on \(E\) carries with it the assertion that \(f\) and each \(f_n\) is measurable and finite a.e. on \(E\); and \(g\) is measurable and finite a.e. on \(E\) by hypothesis. Let \(Z\) be the set of points of \(E\) at which \(f\) or \(g\) fails to be finite. Being the union of two sets of measure zero, \(Z\) has measure zero.

Suppose first that \(f = g\) a.e. on \(E\), and let \(Z^{\prime} = \{x \in E \mid f(x) \ne g(x)\}\), a set of measure zero. Write \(W = Z \cup Z^{\prime}\), so \(m(W) = 0\) by the countable subadditivity of measure. Fix \(\eta > 0\) and an index \(n\). If \(x \in E \sim W\) then \(f(x) = g(x)\) and this common value is a real number, so the two differences \(f_n(x) - g(x)\) and \(f_n(x) - f(x)\) are the same extended real number and

\begin{equation*} |f_n(x) - g(x)| > \eta \iff |f_n(x) - f(x)| > \eta \quad \text{for } x \in E \sim W . \end{equation*}

Consequently

\begin{equation*} \{x \in E \mid |f_n - g| > \eta\} \subseteq W \cup \{x \in E \mid |f_n - f| > \eta\} , \end{equation*}

and so, by the monotonicity and countable subadditivity of measure together with \(m(W) = 0\),

\begin{equation*} m\{x \in E \mid |f_n - g| > \eta\} \le m\{x \in E \mid |f_n - f| > \eta\} . \end{equation*}

Since \(\{f_n\} \to f\) in measure, the right-hand side tends to \(0\) as \(n \to \infty\), and hence so does the left-hand side. As \(\eta > 0\) was arbitrary, and as \(g\) is by hypothesis measurable and finite a.e. on \(E\), we conclude \(\{f_n\} \to g\) in measure on \(E\).

Conversely, suppose \(\{f_n\} \to g\) in measure on \(E\) as well. Fix \(\eta > 0\). Let \(x \in E \sim Z\) be a point at which \(|f(x) - g(x)| > \eta\), and let \(n\) be any index. Since \(f(x)\) and \(g(x)\) are finite, the triangle inequality gives

\begin{equation*} \eta < |f(x) - g(x)| \le |f(x) - f_n(x)| + |f_n(x) - g(x)| , \end{equation*}

and hence at least one of the two summands exceeds \(\eta / 2\) (this remains true, trivially, if \(f_n(x)\) is infinite). Therefore, for every index \(n\),

\begin{equation*} \{x \in E \mid |f(x) - g(x)| > \eta\} \subseteq Z \cup \{x \in E \mid |f_n(x) - f(x)| > \eta/2\} \cup \{x \in E \mid |f_n(x) - g(x)| > \eta/2\} . \end{equation*}

By the monotonicity and countable subadditivity of measure, together with \(m(Z) = 0\),

\begin{equation*} m\{x \in E \mid |f(x) - g(x)| > \eta\} \le m\{x \in E \mid |f_n - f| > \eta/2\} + m\{x \in E \mid |f_n - g| > \eta/2\} . \end{equation*}

The left-hand side does not depend on \(n\), while by hypothesis both terms on the right tend to \(0\) as \(n \to \infty\). Letting \(n \to \infty\) we conclude

\begin{equation*} m\{x \in E \mid |f(x) - g(x)| > \eta\} = 0 \quad \text{for every } \eta > 0 . \end{equation*}

Finally, on \(E \sim Z\) the functions \(f\) and \(g\) are finite, so

\begin{equation*} \{x \in E \sim Z \mid f(x) \ne g(x)\} = \bigcup_{k=1}^{\infty} \{x \in E \sim Z \mid |f(x) - g(x)| > 1/k\} , \end{equation*}

a countable union of sets of measure zero, which therefore has measure zero by the countable subadditivity of measure. Adjoining the null set \(Z\), we conclude that \(\{x \in E \mid f(x) \ne g(x)\}\) has measure zero, that is, \(f = g\) a.e. on \(E\).

Throughout, recall that convergence in measure presupposes that the limit function is measurable and finite a.e. on \(E\). All products below are defined a.e. on \(E\) and are measurable, and the exceptional null sets may be discarded without affecting any of the measures computed.

We first record an elementary lemma.

Lemma. If \(\{u_n\} \to u\) and \(\{v_n\} \to v\) in measure on \(E\), and \(\alpha\), \(\beta\) are real numbers, then \(\{\alpha u_n + \beta v_n\} \to \alpha u + \beta v\) in measure on \(E\).

Proof of the Lemma. Set \(M = |\alpha| + |\beta| + 1\), so \(M > 0\) and \(|\alpha| + |\beta| < M\). By the definition of convergence in measure, \(u\), \(v\) and every \(u_n\), \(v_n\) is measurable and finite a.e. on \(E\); let \(Z\) be the union of the countably many sets of measure zero on which one of these functions fails to be finite. By the countable subadditivity of measure \(m(Z) = 0\), and off \(Z\) every function in sight is real-valued, so all the algebraic manipulations below are legitimate there and no integral or measure is affected by discarding \(Z\). Let \(\eta > 0\). For \(x \in E \sim Z\),

\begin{equation*} |(\alpha u_n + \beta v_n)(x) - (\alpha u + \beta v)(x)| \le |\alpha| \, |u_n(x) - u(x)| + |\beta| \, |v_n(x) - v(x)| , \end{equation*}

so if the left side exceeds \(\eta\) then \(|u_n(x) - u(x)| > \eta / (2M)\) or \(|v_n(x) - v(x)| > \eta / (2M)\) (otherwise the right side would be at most \((|\alpha| + |\beta|) \eta / (2M) < \eta\)). Hence

\begin{equation*} \{|\alpha u_n + \beta v_n - (\alpha u + \beta v)| > \eta\} \subseteq Z \cup \{|u_n - u| > \eta/(2M)\} \cup \{|v_n - v| > \eta/(2M)\} , \end{equation*}

and by the monotonicity and countable subadditivity of measure, together with \(m(Z) = 0\), the measure of the left-hand set is at most the sum of the measures of the last two sets, each of which tends to \(0\) as \(n \to \infty\). Since \(\eta > 0\) was arbitrary, and since \(\alpha u + \beta v\) is measurable and finite a.e. on \(E\) because \(u\) and \(v\) are, the Lemma is proved.

Part 1: \(\{f_n \cdot g\} \to f \cdot g\) in measure on \(E\).

Let \(\eta > 0\) and \(\epsilon > 0\). For each natural number \(k\) define

\begin{equation*} E_k = \{x \in E \mid |g(x)| > k\} . \end{equation*}

These are measurable sets that descend, and \(\bigcap_{k=1}^{\infty} E_k = \{x \in E \mid |g(x)| = \infty\}\), a set of measure zero since \(g\) is finite a.e. on \(E\). Since \(m(E_1) \le m(E) < \infty\), Theorem 15 of Chapter 2 (the Continuity of Measure) gives \(\lim_{k \to \infty} m(E_k) = 0\). (This is exactly where the finiteness of \(m(E)\) is used.) Choose \(k\) with \(m(E_k) < \epsilon\).

Let \(Z\) be the union of the countably many sets of measure zero on which one of \(f\), \(g\), \(f_1\), \(f_2, \dots\) fails to be finite; by the countable subadditivity of measure \(m(Z) = 0\). Suppose \(x \in E\) satisfies \(|f_n(x) g(x) - f(x) g(x)| > \eta\) and \(x \notin Z \cup E_k\). Then \(f(x)\), \(g(x)\), \(f_n(x)\) are all real numbers and \(|g(x)| \le k\), so the factorization \(f_n(x) g(x) - f(x) g(x) = g(x) [f_n(x) - f(x)]\) is legitimate and

\begin{equation*} \eta < |g(x)| \, |f_n(x) - f(x)| \le k \, |f_n(x) - f(x)| , \end{equation*}

so \(|f_n(x) - f(x)| > \eta / k\). Therefore

\begin{equation*} \{x \in E \mid |f_n g - f g| > \eta\} \subseteq Z \cup E_k \cup \{x \in E \mid |f_n - f| > \eta / k\} , \end{equation*}

and hence, by the monotonicity and countable subadditivity of measure, together with \(m(Z) = 0\) and \(m(E_k) < \epsilon\),

\begin{equation*} m\{x \in E \mid |f_n g - f g| > \eta\} \le \epsilon + m\{x \in E \mid |f_n - f| > \eta / k\} . \end{equation*}

Since \(\{f_n\} \to f\) in measure, letting \(n \to \infty\) yields

\begin{equation*} \limsup_{n \to \infty} m\{x \in E \mid |f_n g - f g| > \eta\} \le \epsilon . \end{equation*}

As \(\epsilon > 0\) was arbitrary, this limit superior is \(0\), that is,

\begin{equation*} \lim_{n \to \infty} m\{x \in E \mid |f_n(x) g(x) - f(x) g(x)| > \eta\} = 0 . \end{equation*}

Since \(\eta > 0\) was arbitrary and \(f \cdot g\) is measurable and finite a.e. on \(E\), we conclude \(\{f_n \cdot g\} \to f \cdot g\) in measure on \(E\).

Part 2: \(\{f_n^2\} \to f^2\) in measure on \(E\).

Let \(Z\) again denote the null set off which \(f\) and all the \(f_n\) are finite; every set equality below is asserted on \(E \sim Z\), and adjoining or deleting \(Z\) changes no measure.

First, \(\{f_n - f\} \to 0\) in measure on \(E\) by the Lemma (take \(u_n = f_n\), \(v_n\) the constant sequence \(f\), \(\alpha = 1\), \(\beta = -1\); the constant sequence trivially converges to \(f\) in measure). Next, for \(\eta > 0\) and \(x \in E \sim Z\) the numbers \(f_n(x)\) and \(f(x)\) are real, so \((f_n(x) - f(x))^2 > \eta\) holds precisely when \(|f_n(x) - f(x)| > \sqrt{\eta}\). Hence

\begin{equation*} \{x \in E \mid (f_n(x) - f(x))^2 > \eta\} \subseteq Z \cup \{x \in E \mid |f_n(x) - f(x)| > \sqrt{\eta}\} , \end{equation*}

and the measure of the right-hand side tends to \(0\) as \(n \to \infty\). Since \((f_n - f)^2\) is measurable and finite a.e. on \(E\), this says \(\{(f_n - f)^2\} \to 0\) in measure on \(E\).

Second, \(f\) itself is a measurable function on \(E\) that is finite a.e. on \(E\), so Part 1 applied with \(g = f\) gives \(\{f_n \cdot f\} \to f^2\) in measure on \(E\).

Now put \(h_n = (f_n - f)^2 + [2 f f_n - f^2]\). The sequence \(\{(f_n - f)^2\}\) converges in measure to \(0\), and the sequence \(\{2 f f_n - f^2\}\) converges in measure to \(2 f^2 - f^2 = f^2\) (apply the Lemma with \(u_n = f f_n \to f^2\), \(\alpha = 2\), and the constant sequence \(v_n = f^2 \to f^2\), \(\beta = -1\)). Hence the Lemma, applied once more with \(\alpha = \beta = 1\), gives \(\{h_n\} \to 0 + f^2 = f^2\) in measure on \(E\).

Finally, at each point of \(E \sim Z\) all the quantities involved are real numbers and the algebraic identity

\begin{equation*} (f_n - f)^2 + 2 f f_n - f^2 = f_n^2 - 2 f f_n + f^2 + 2 f f_n - f^2 = f_n^2 \end{equation*}

holds, so \(h_n = f_n^2\) on \(E \sim Z\). Therefore, for each \(\eta > 0\),

\begin{equation*} \{x \in E \mid |f_n^2 - f^2| > \eta\} \subseteq Z \cup \{x \in E \mid |h_n - f^2| > \eta\} , \end{equation*}

and since \(m(Z) = 0\) the measure of the left-hand set tends to \(0\) as \(n \to \infty\). As \(f^2\) is measurable and finite a.e. on \(E\), we conclude

\begin{equation*} \{f_n^2\} \to f^2 \quad \text{in measure on } E . \end{equation*}

Part 3: if also \(\{g_n\} \to g\) in measure on \(E\), then \(\{f_n \cdot g_n\} \to f \cdot g\) in measure on \(E\).

Here \(g\) is measurable and finite a.e. on \(E\), as required by the definition of convergence in measure. By the Lemma, \(\{f_n + g_n\} \to f + g\) in measure on \(E\). The argument of Part 2 used nothing about the sequence \(\{f_n\}\) beyond the facts that \(m(E) < \infty\) and that \(\{f_n\}\) converges in measure on \(E\) to a function that is measurable and finite a.e.; so it applies verbatim to each of the three sequences \(\{f_n + g_n\}\), \(\{f_n\}\) and \(\{g_n\}\), whose limits \(f + g\), \(f\) and \(g\) are all measurable and finite a.e. on \(E\). Thus

\begin{equation*} \{(f_n + g_n)^2\} \to (f + g)^2, \quad \{f_n^2\} \to f^2, \quad \{g_n^2\} \to g^2 \quad \text{in measure on } E . \end{equation*}

Applying the Lemma twice to the linear combination

\begin{equation*} k_n = \tfrac{1}{2} (f_n + g_n)^2 - \tfrac{1}{2} f_n^2 - \tfrac{1}{2} g_n^2 \end{equation*}

gives

\begin{equation*} \{k_n\} \to \tfrac{1}{2} \left[ (f + g)^2 - f^2 - g^2 \right] \quad \text{in measure on } E . \end{equation*}

Let \(W\) be the null set off which \(f\), \(g\) and every \(f_n\), \(g_n\) is finite. At each point of \(E \sim W\) all these values are real numbers, so the identity \((a + b)^2 - a^2 - b^2 = 2ab\) gives \(k_n = f_n g_n\) there, and likewise \(\tfrac{1}{2}[(f + g)^2 - f^2 - g^2] = f \cdot g\) there. Exactly as in Part 2, for each \(\eta > 0\),

\begin{equation*} \{x \in E \mid |f_n g_n - f g| > \eta\} \subseteq W \cup \{x \in E \mid |k_n - \tfrac{1}{2}[(f+g)^2 - f^2 - g^2]| > \eta\} , \end{equation*}

whose measure tends to \(0\). Since \(f \cdot g\) is measurable and finite a.e. on \(E\), we conclude that \(\{f_n \cdot g_n\} \to f \cdot g\) in measure on \(E\). This completes the proof.

Throughout, recall the standing convention that when we write \(\{f_n\} \to f\) in measure on \(E\) we are assuming \(f\) and each \(f_n\) is measurable and finite a.e. on \(E\). We isolate two elementary observations and then treat the four theorems in turn.

Observation A (subsequences inherit convergence in measure). If \(\{f_n\} \to f\) in measure on \(E\) and \(\{f_{n_k}\}\) is any subsequence, then \(\{f_{n_k}\} \to f\) in measure on \(E\). Indeed, given \(\eta > 0\) and \(\epsilon > 0\) choose \(N\) with \(m\{x \in E \mid |f_n(x) - f(x)| > \eta\} < \epsilon\) for all \(n \ge N\); since \(n_k \ge k\), the same inequality holds for all \(k \ge N\).

Observation B (the subsequence criterion for numerical sequences). Let \(\{a_n\}\) be a sequence in \(\mathbb{R}\) and \(a \in \mathbb{R}\). If every subsequence of \(\{a_n\}\) possesses a further subsequence converging to \(a\), then \(a_n \to a\). Indeed, if \(a_n \not\to a\) there are \(\epsilon_0 > 0\) and indices \(n_1 < n_2 < \cdots\) with \(|a_{n_k} - a| \ge \epsilon_0\) for all \(k\); no subsequence of \(\{a_{n_k}\}\) can converge to \(a\), a contradiction.

The engine driving everything is the Riesz Theorem (Theorem 4 of this chapter): if \(\{f_n\} \to f\) in measure on \(E\), then some subsequence of \(\{f_n\}\) converges to \(f\) pointwise a.e. on \(E\). Combining it with Observation A: every subsequence of \(\{f_n\}\) has in turn a further subsequence converging to \(f\) pointwise a.e. on \(E\). Call this property \((\ast)\).

1. Fatou’s Lemma. Claim: if \(\{f_n\}\) is a sequence of nonnegative measurable functions on \(E\) and \(\{f_n\} \to f\) in measure on \(E\), then \(f \ge 0\) a.e. on \(E\) (indeed, by the Riesz Theorem some subsequence converges to \(f\) pointwise a.e., and a pointwise limit of nonnegative numbers is nonnegative), so that \(\int_E f\) is defined, and

\begin{equation*} \int_E f \le \liminf_{n \to \infty} \int_E f_n . \end{equation*}

Set \(L = \liminf_{n \to \infty} \int_E f_n \in [0, \infty]\). Choose a subsequence \(\{f_{n_k}\}\) with \(\lim_{k \to \infty} \int_E f_{n_k} = L\). By \((\ast)\) there is a further subsequence \(\{f_{n_{k_j}}\}\) converging to \(f\) pointwise a.e. on \(E\). Fatou’s Lemma in its original form (Chapter 4) applies to this further subsequence and gives

\begin{equation*} \int_E f \le \liminf_{j \to \infty} \int_E f_{n_{k_j}} = \lim_{k \to \infty} \int_E f_{n_k} = L, \end{equation*}

the middle equality because a subsequence of a convergent sequence of numbers has the same limit. This is the assertion.

1. The Monotone Convergence Theorem. Claim: if \(\{f_n\}\) is an increasing sequence of nonnegative measurable functions on \(E\) and \(\{f_n\} \to f\) in measure on \(E\), then \(\lim_{n \to \infty} \int_E f_n = \int_E f\).

Since \(f_n \le f_{n+1}\) on \(E\), the numerical sequence \(\left\{\int_E f_n\right\}\) is increasing, so it converges in \([0, \infty]\); call the limit \(L\). Part 1 already gives \(\int_E f \le L\). For the reverse inequality, fix an index \(p\). By the Riesz Theorem select a subsequence \(\{f_{n_k}\}\) converging to \(f\) pointwise a.e. on \(E\), say on \(E \sim E_0\) where \(E_0\) is a measurable subset of \(E\) with \(m(E_0) = 0\). For \(x \in E \sim E_0\) and any \(k\) with \(n_k \ge p\) we have \(f_p(x) \le f_{n_k}(x)\) by monotonicity, and letting \(k \to \infty\) gives \(f_p(x) \le f(x)\). Hence \(f_p \le f\) a.e. on \(E\), and by the monotonicity of integration for nonnegative measurable functions \(\int_E f_p \le \int_E f\). Letting \(p \to \infty\) yields \(L \le \int_E f\). Therefore \(L = \int_E f\).

1. The Lebesgue Dominated Convergence Theorem. Claim: if \(\{f_n\}\) is a sequence of measurable functions on \(E\), \(g\) is integrable over \(E\) with \(|f_n| \le g\) on \(E\) for all \(n\), and \(\{f_n\} \to f\) in measure on \(E\), then \(f\) is integrable over \(E\) and \(\lim_{n \to \infty} \int_E f_n = \int_E f\).

First, \(f\) is integrable. By the Riesz Theorem some subsequence \(\{f_{n_k}\}\) converges to \(f\) pointwise a.e. on \(E\); passing to the limit in \(|f_{n_k}| \le g\) gives \(|f| \le g\) a.e. on \(E\), and since \(f\) is measurable and \(g\) is integrable, \(f\) is integrable over \(E\) by the Integral Comparison Test (Proposition 16 of Chapter 4). In particular each \(\int_E f_n\) and \(\int_E f\) is a real number.

Now let \(\{\int_E f_{n_k}\}\) be an arbitrary subsequence of the numerical sequence \(\{\int_E f_n\}\). By \((\ast)\) there is a further subsequence \(\{f_{n_{k_j}}\}\) converging to \(f\) pointwise a.e. on \(E\); it is dominated by \(g\), so the Dominated Convergence Theorem in its original form gives \(\int_E f_{n_{k_j}} \to \int_E f\). Thus every subsequence of \(\left\{\int_E f_n\right\}\) has a further subsequence converging to \(\int_E f\), and Observation B gives \(\lim_{n \to \infty} \int_E f_n = \int_E f\).

1. The Vitali Convergence Theorem. Claim: let \(\{f_n\}\) be a sequence of measurable functions on \(E\) that is uniformly integrable and tight over \(E\) (if \(m(E) < \infty\), tightness is automatic, since one may take \(E_0 = E\) in its definition), and suppose \(\{f_n\} \to f\) in measure on \(E\). Then \(f\) is integrable over \(E\) and \(\lim_{n \to \infty} \int_E f_n = \int_E f\).

The argument is verbatim that of part 3, with domination replaced by uniform integrability and tightness. First note that each \(f_n\) is integrable over \(E\): this is exactly what uniform integrability together with tightness gives, by the remark preceding the statement of the general Vitali Convergence Theorem in Section 5.1 (which appeals to Proposition 23 of Chapter 4). So \(\left\{\int_E f_n\right\}\) is a sequence of real numbers. Let \(\left\{\int_E f_{n_k}\right\}\) be any subsequence. By \((\ast)\) choose a further subsequence \(\{f_{n_{k_j}}\}\) converging to \(f\) pointwise a.e. on \(E\). This further subsequence is itself uniformly integrable and tight over \(E\), since both conditions are inherited by subsequences (the \(\delta\) and the set of finite measure produced for the whole family serve for any subfamily). Hence the Vitali Convergence Theorem in its original form (the general version stated in Section 5.1 of this chapter; for \(m(E) < \infty\) the version of Section 4.6 suffices) tells us that \(f\) is integrable over \(E\) and \(\int_E f_{n_{k_j}} \to \int_E f\). The integrability of \(f\) is thereby established once and for all (from a single such subsequence), so \(\int_E f\) is a real number; and Observation B applied to \(\left\{\int_E f_n\right\}\) gives \(\lim_{n \to \infty} \int_E f_n = \int_E f\).

In each of the four cases the hypothesis of a.e. pointwise convergence was used only along a subsequence, which the Riesz Theorem supplies from convergence in measure; and in each case convergence in measure is a genuinely weaker hypothesis, since by Proposition 3 pointwise a.e. convergence on a set of finite measure implies convergence in measure while the example following Proposition 3 shows the converse fails.

Proposition 3 asserts: if \(m(E) < \infty\), \(\{f_n\}\) is a sequence of measurable functions on \(E\) converging pointwise a.e. on \(E\) to \(f\), and \(f\) is finite a.e. on \(E\), then \(\{f_n\} \to f\) in measure on \(E\). We exhibit a set of infinite measure on which the conclusion fails.

Let \(E = \mathbb{R}\), which has infinite measure, and for each natural number \(n\) define

\begin{equation*} f_n = \chi_{[n,\, n+1)} . \end{equation*}

Each \(f_n\) is measurable, being the characteristic function of a measurable set, and each is real-valued. Let \(f \equiv 0\) on \(\mathbb{R}\); then \(f\) is measurable and finite everywhere.

The sequence converges pointwise on all of \(\mathbb{R}\) to \(f\). Indeed, fix \(x \in \mathbb{R}\) and choose a natural number \(N > x\). For every \(n \ge N\) we have \(n > x\), so \(x \notin [n, n+1)\) and hence \(f_n(x) = 0\). Thus \(f_n(x) \to 0 = f(x)\). So the hypotheses of Proposition 3 hold except for the finiteness of \(m(E)\).

The sequence does not converge to \(f\) in measure. Take \(\eta = 1/2\). For each \(n\),

\begin{equation*} \{x \in \mathbb{R} \mid |f_n(x) - f(x)| > 1/2\} = \{x \in \mathbb{R} \mid \chi_{[n, n+1)}(x) > 1/2\} = [n,\, n+1), \end{equation*}

a set of measure \(1\). Hence for \(\epsilon = 1/2\) there is no index \(N\) for which

\begin{equation*} m\{x \in \mathbb{R} \mid |f_n(x) - f(x)| > 1/2\} < \epsilon \quad \text{for all } n \ge N, \end{equation*}

since the left-hand side equals \(1\) for every \(n\). Therefore \(\{f_n\}\) fails to converge to \(f\) in measure on \(\mathbb{R}\), and Proposition 3 is false for sets of infinite measure.

The mechanism is exactly the one Egoroff’s Theorem, on which the proof of Proposition 3 rests, cannot control on a set of infinite measure: the “bad set” on which \(f_n\) is far from its limit does not shrink, it merely escapes to infinity. One can of course produce the same failure on a set of infinite measure that is not all of \(\mathbb{R}\), for instance on \(E = [0, \infty)\) with the same functions.

It is worth noting that a related but distinct phenomenon occurs even on sets of finite measure with the roles reversed: as the Example following Proposition 3 shows, on \(E = [0,1]\) there is a sequence converging in measure to \(f \equiv 0\) that converges pointwise at no point of \([0,1]\). Thus on a set of finite measure the implication of Proposition 3 is strict, while on a set of infinite measure it is not even valid.

Let \(E\) be measurable with \(m(E) < \infty\). We are to show: if \(\{f_n\} \to f\) in measure on \(E\) and \(\{g_n\} \to g\) in measure on \(E\), and \(\alpha, \beta\) are real numbers, then

\begin{equation*} \{\alpha f_n + \beta g_n\} \to \alpha f + \beta g \quad \text{in measure on } E . \end{equation*}

By induction this gives the assertion for a linear combination of any finite number of sequences.

First a word on well-definedness. By our standing convention \(f\), \(g\) and each \(f_n\), \(g_n\) are measurable and finite a.e. on \(E\). Let

\begin{equation*} E_0 = \{x \in E \mid \text{at least one of } f(x), g(x), f_1(x), g_1(x), f_2(x), g_2(x), \ldots \text{ is infinite}\} . \end{equation*}

This is a countable union of sets of measure zero, so \(m(E_0) = 0\). On \(E \sim E_0\) every one of these functions is real-valued, so the linear combinations \(\alpha f_n + \beta g_n\) and \(\alpha f + \beta g\) are defined there, are measurable (linear combinations of measurable real-valued functions are measurable, Theorem 6 of Chapter 3), and are finite. Since sets of measure zero contribute nothing to the measures occurring in the definition of convergence in measure, we may and do argue on \(E \sim E_0\); equivalently, we adopt the (measure-irrelevant) convention that all functions are redefined to be \(0\) on \(E_0\).

If \(\alpha = \beta = 0\) the assertion is trivial, both sides being identically \(0\). So assume \(|\alpha| + |\beta| > 0\).

Fix \(\eta > 0\) and \(\epsilon > 0\). The heart of the matter is the following pointwise inclusion. For \(x \in E \sim E_0\),

\begin{equation*} \bigl| (\alpha f_n(x) + \beta g_n(x)) - (\alpha f(x) + \beta g(x)) \bigr| \le |\alpha| \, |f_n(x) - f(x)| + |\beta| \, |g_n(x) - g(x)| , \end{equation*}

by the triangle inequality. Consequently, if the left-hand side exceeds \(\eta\), then at least one of the two summands on the right exceeds \(\eta / 2\): were both at most \(\eta/2\), the sum would be at most \(\eta\). Therefore, writing

\begin{equation*} A_n = \Bigl\{ x \in E \sim E_0 \;\Bigm|\; |\alpha| \, |f_n(x) - f(x)| > \eta/2 \Bigr\}, \qquad B_n = \Bigl\{ x \in E \sim E_0 \;\Bigm|\; |\beta| \, |g_n(x) - g(x)| > \eta/2 \Bigr\}, \end{equation*}

we have the inclusion

\begin{equation*} \bigl\{ x \in E \sim E_0 \;\bigm|\; |(\alpha f_n + \beta g_n)(x) - (\alpha f + \beta g)(x)| > \eta \bigr\} \subseteq A_n \cup B_n . \end{equation*}

We now estimate \(m(A_n)\) and \(m(B_n)\) separately. If \(\alpha = 0\) then \(A_n = \emptyset\) and \(m(A_n) = 0\) for all \(n\). If \(\alpha \ne 0\), then

\begin{equation*} A_n = \Bigl\{ x \in E \sim E_0 \;\Bigm|\; |f_n(x) - f(x)| > \frac{\eta}{2|\alpha|} \Bigr\} , \end{equation*}

and since \(\eta / (2|\alpha|) > 0\) and \(\{f_n\} \to f\) in measure on \(E\), there is an index \(N_1\) such that \(m(A_n) < \epsilon / 2\) for all \(n \ge N_1\). In either case there is an index \(N_1\) with \(m(A_n) < \epsilon/2\) for \(n \ge N_1\). Symmetrically there is an index \(N_2\) with \(m(B_n) < \epsilon/2\) for all \(n \ge N_2\).

Set \(N = \max\{N_1, N_2\}\). For \(n \ge N\), by the monotonicity and finite subadditivity of Lebesgue measure,

\begin{equation*} m \bigl\{ x \in E \mid |(\alpha f_n + \beta g_n)(x) - (\alpha f + \beta g)(x)| > \eta \bigr\} \le m(E_0) + m(A_n) + m(B_n) < 0 + \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

Since \(\eta > 0\) and \(\epsilon > 0\) were arbitrary, \(\{\alpha f_n + \beta g_n\} \to \alpha f + \beta g\) in measure on \(E\), as claimed.

Two remarks. First, the finiteness of \(m(E)\) was nowhere used: the argument above is valid on an arbitrary measurable set \(E\), finite measure or not. Finiteness of measure is needed for Proposition 3 (pointwise a.e. convergence implies convergence in measure), not for the algebra of limits in measure. Second, the same splitting argument shows that the set of sequences converging in measure on \(E\) is a real vector space and that the map sending a sequence to its limit in measure is linear; the limit is unique up to sets of measure zero, since if \(\{h_n\} \to h\) and \(\{h_n\} \to \tilde{h}\) in measure then by the case just proved \(\{0\} = \{h_n - h_n\} \to h - \tilde{h}\) in measure, forcing \(m\{|h - \tilde{h}| > \eta\} = 0\) for every \(\eta > 0\) and hence \(h = \tilde{h}\) a.e. on \(E\).

Throughout, \(m(E) < \infty\), and \(f\) together with every \(f_n\) is measurable and finite a.e. on \(E\), so that all the sets appearing below are measurable and all the measures involved are finite.

Suppose first that \(\{f_n\} \to f\) in measure on \(E\), and let \(\{f_{n_k}\}\) be an arbitrary subsequence. We claim \(\{f_{n_k}\} \to f\) in measure on \(E\). Indeed, let \(\eta > 0\) and \(\epsilon > 0\). Choose \(N\) so that \(m\{x \in E \mid |f_n(x) - f(x)| > \eta\} < \epsilon\) for all \(n \ge N\). Since \(\{n_k\}\) is strictly increasing, \(n_k \ge k\); hence for all \(k \ge N\) we have \(n_k \ge N\) and therefore

\begin{equation*} m\{x \in E \mid |f_{n_k}(x) - f(x)| > \eta\} < \epsilon . \end{equation*}

So the subsequence converges to \(f\) in measure. By the Riesz Theorem (Theorem 4), a sequence converging in measure has a subsequence converging pointwise a.e. to the same limit; applied to \(\{f_{n_k}\}\), it produces a further subsequence \(\{f_{n_{k_j}}\}\) converging to \(f\) pointwise a.e. on \(E\). This proves the forward implication. (Note that finiteness of \(m(E)\) was not needed here.)

Conversely, suppose every subsequence of \(\{f_n\}\) has in turn a further subsequence converging to \(f\) pointwise a.e. on \(E\). We argue by contradiction: assume \(\{f_n\}\) does not converge to \(f\) in measure on \(E\).

Negating the definition of convergence in measure, there exist \(\eta_0 > 0\) and \(\epsilon_0 > 0\) with the property that for every index \(N\) there is some \(n \ge N\) for which

\begin{equation*} m\{x \in E \mid |f_n(x) - f(x)| > \eta_0\} \ge \epsilon_0 . \end{equation*}

Use this repeatedly to construct indices: take \(n_1\) to be an index (with \(n_1 \ge 1\)) for which the displayed inequality holds; having chosen \(n_1 < n_2 < \cdots < n_{k}\), apply the property with \(N = n_{k} + 1\) to obtain \(n_{k+1} > n_{k}\) for which the inequality holds. This yields a strictly increasing sequence of natural numbers \(\{n_k\}\) with

\begin{equation*} m\{x \in E \mid |f_{n_k}(x) - f(x)| > \eta_0\} \ge \epsilon_0 \quad \text{for every } k. \tag{\(\dagger\)} \end{equation*}

By hypothesis, the subsequence \(\{f_{n_k}\}\) has in turn a further subsequence \(\{f_{n_{k_j}}\}\) that converges to \(f\) pointwise a.e. on \(E\). Now \(E\) has finite measure, the functions \(f_{n_{k_j}}\) are measurable, and \(f\) is finite a.e. on \(E\); so Proposition 3 applies to the sequence \(\{f_{n_{k_j}}\}_{j=1}^{\infty}\) and yields

\begin{equation*} \{f_{n_{k_j}}\} \to f \quad \text{in measure on } E . \end{equation*}

In particular, taking \(\eta = \eta_0\) and \(\epsilon = \epsilon_0\) in the definition, there is an index \(J\) such that

\begin{equation*} m\{x \in E \mid |f_{n_{k_j}}(x) - f(x)| > \eta_0\} < \epsilon_0 \quad \text{for all } j \ge J . \end{equation*}

But \(n_{k_J}\) is one of the indices \(n_k\), so \((\dagger)\) gives \(m\{x \in E \mid |f_{n_{k_J}}(x) - f(x)| > \eta_0\} \ge \epsilon_0\). This contradiction shows that our assumption was false, and hence \(\{f_n\} \to f\) in measure on \(E\).

Two comments. The finiteness of \(m(E)\) is genuinely needed for the converse direction, and only there: it is the hypothesis of Proposition 3, which is the sole step converting a.e. pointwise convergence back into convergence in measure. On a set of infinite measure the criterion fails, as the sequence \(f_n = \chi_{[n, n+1)}\) on \(E = \mathbb{R}\) shows: it converges to \(f \equiv 0\) pointwise everywhere on \(\mathbb{R}\), so certainly every subsequence has a further subsequence converging pointwise a.e. to \(f\), yet as computed in Problem 9 it does not converge to \(f\) in measure.

Finally, observe that this criterion is precisely the mechanism used in Problem 8: it is the exact analogue, for convergence in measure, of the elementary fact that a numerical sequence converges to \(a\) provided every subsequence has a further subsequence converging to \(a\). It expresses the fact that convergence in measure on a set of finite measure is the convergence associated with a topology, indeed with the metric \(\rho\) of Problem 14, in which a.e. pointwise convergence plays the role of a “subsequential” notion.

Let \(\{a_j\}_{j=1}^{\infty}\) be a sequence of real numbers satisfying

\begin{equation*} |a_{j+1} - a_j| \le \frac{1}{2^j} \quad \text{for all } j \ge 1 . \end{equation*}

The key estimate is obtained by telescoping. Let \(n > m \ge 1\). Writing

\begin{equation*} a_n - a_m = \sum_{j=m}^{n-1} (a_{j+1} - a_j) \end{equation*}

and applying the triangle inequality together with the hypothesis,

\begin{equation*} |a_n - a_m| \le \sum_{j=m}^{n-1} |a_{j+1} - a_j| \le \sum_{j=m}^{n-1} \frac{1}{2^j} . \end{equation*}

The last sum is a finite geometric sum, which we evaluate exactly:

\begin{equation*} \sum_{j=m}^{n-1} \frac{1}{2^j} = \frac{1}{2^m} \cdot \frac{1 - (1/2)^{n-m}}{1 - 1/2} = \frac{1}{2^{m-1}} \left( 1 - \frac{1}{2^{n-m}} \right) < \frac{1}{2^{m-1}} . \end{equation*}

Hence

\begin{equation*} |a_n - a_m| < \frac{1}{2^{m-1}} \quad \text{whenever } n > m \ge 1. \tag{\(\ast\)} \end{equation*}

Now we verify the Cauchy condition. Let \(\epsilon > 0\). Since \(2^{k} \ge k+1 > k\) for every natural number \(k\) (an immediate induction), we have \(1/2^{k} \to 0\), so we may choose a natural number \(N\) with

\begin{equation*} \frac{1}{2^{N-1}} < \epsilon . \end{equation*}

Let \(m, n \ge N\) be arbitrary. If \(m = n\) then \(|a_n - a_m| = 0 < \epsilon\). If \(m \ne n\), relabel so that \(n > m\); then \(m \ge N\), and by \((\ast)\),

\begin{equation*} |a_n - a_m| < \frac{1}{2^{m-1}} \le \frac{1}{2^{N-1}} < \epsilon , \end{equation*}

using that \(k \mapsto 1/2^{k-1}\) is decreasing. In every case \(|a_n - a_m| < \epsilon\) for all \(m, n \ge N\). Therefore \(\{a_j\}\) is a Cauchy sequence of real numbers.

By the completeness of \(\mathbb{R}\), that is, by the Cauchy Convergence Criterion (Theorem 17 of Chapter 1), every Cauchy sequence of real numbers converges to a real number. Hence \(\{a_j\}\) converges to some \(a \in \mathbb{R}\).

It is worth recording the quantitative form of the conclusion, which is what makes this lemma useful in the proof of Problem 13. Fixing \(m\) and letting \(n \to \infty\) in \((\ast)\), the continuity of the absolute value gives

\begin{equation*} |a - a_m| \le \frac{1}{2^{m-1}} \quad \text{for every } m \ge 1 , \end{equation*}

so the limit is approached at a definite geometric rate. Equivalently, the hypothesis says that the telescoping series \(a_1 + \sum_{j \ge 1} (a_{j+1} - a_j)\) is absolutely convergent, since \(\sum_{j=1}^{\infty} |a_{j+1} - a_j| \le \sum_{j=1}^{\infty} 2^{-j} = 1 < \infty\), and its partial sums are exactly the \(a_n\).

Assume \(\{f_n\}\) is Cauchy in measure on \(E\); as always, each \(f_n\) is measurable and finite a.e. on \(E\). Let \(Z\) be the union of the countably many sets \(\{x \in E \mid |f_n(x)| = \infty\}\); then \(m(Z) = 0\) and every \(f_n\) is real-valued on \(E \sim Z\).

Step 1: extraction of a rapidly Cauchy subsequence. For each natural number \(j\) apply the Cauchy-in-measure hypothesis with \(\eta = 1/2^j\) and \(\epsilon = 1/2^j\) to obtain an index \(N_j\) such that

\begin{equation*} m\{x \in E \mid |f_n(x) - f_m(x)| \ge 1/2^j\} < 1/2^j \quad \text{for all } m, n \ge N_j . \end{equation*}

Define \(n_1 = N_1\) and, inductively, \(n_{j+1} = \max\{N_{j+1},\, n_j + 1\}\). Then \(\{n_j\}\) is a strictly increasing sequence of natural numbers and \(n_{j+1} > n_j \ge N_j\) for every \(j\). Consequently, setting

\begin{equation*} E_j = \{x \in E \mid |f_{n_{j+1}}(x) - f_{n_j}(x)| > 1/2^j\} , \end{equation*}

which is a measurable set contained in \(\{x \in E \mid |f_{n_{j+1}}(x) - f_{n_j}(x)| \ge 1/2^j\}\), we have

\begin{equation*} m(E_j) < \frac{1}{2^j} \quad \text{for every } j, \qquad \text{so} \qquad \sum_{j=1}^{\infty} m(E_j) < \sum_{j=1}^{\infty} \frac{1}{2^j} = 1 < \infty . \end{equation*}

Step 2: the subsequence converges pointwise a.e. Each \(E_j\) is measurable, and so is \(\limsup_j E_j = \bigcap_{k \ge 1} \bigcup_{j \ge k} E_j\), the set of points lying in infinitely many \(E_j\). By the Borel-Cantelli Lemma (Chapter 2), since \(\sum_j m(E_j) < \infty\), we have \(m(\limsup_j E_j) = 0\). Put

\begin{equation*} E^{\prime} = (E \sim Z) \sim \limsup_{j \to \infty} E_j , \end{equation*}

a measurable subset of \(E\) with \(m(E \sim E^{\prime}) \le m(Z) + m(\limsup_j E_j) = 0\). Fix \(x \in E^{\prime}\). There is an index \(J = J(x)\) with \(x \notin E_j\) for all \(j \ge J\), that is,

\begin{equation*} |f_{n_{j+1}}(x) - f_{n_j}(x)| \le \frac{1}{2^j} \quad \text{for all } j \ge J . \end{equation*}

Consider the real sequence \(b_i = f_{n_{J + i - 1}}(x)\), \(i \ge 1\). For each \(i\),

\begin{equation*} |b_{i+1} - b_i| = |f_{n_{J+i}}(x) - f_{n_{J+i-1}}(x)| \le \frac{1}{2^{J + i - 1}} \le \frac{1}{2^{i}} , \end{equation*}

since \(J \ge 1\). By the preceding problem (Problem 12), \(\{b_i\}\) is Cauchy and hence converges to a real number. Prepending the finitely many terms \(f_{n_1}(x), \ldots, f_{n_{J-1}}(x)\) does not affect convergence, so \(\lim_{j \to \infty} f_{n_j}(x)\) exists and is real for every \(x \in E^{\prime}\).

Define \(f\) on \(E\) by \(f(x) = \lim_{j \to \infty} f_{n_j}(x)\) for \(x \in E^{\prime}\) and \(f(x) = 0\) for \(x \in E \sim E^{\prime}\). Then \(f\) is real-valued on all of \(E\). It is also measurable, and no appeal to completeness of the measure is needed: set \(g_j = f_{n_j}\) on \(E^{\prime}\) and \(g_j = 0\) on \(E \sim E^{\prime}\). Since \(E^{\prime}\) is measurable, the restriction of \(g_j\) to \(E^{\prime}\) is measurable (it agrees there with the measurable function \(f_{n_j}\)) and its restriction to \(E \sim E^{\prime}\) is constant, hence measurable; so \(g_j\) is measurable on \(E\) by part (ii) of Proposition 5 of Chapter 3. By construction \(\{g_j\} \to f\) pointwise on all of \(E\): on \(E^{\prime}\) by the previous paragraph, and on \(E \sim E^{\prime}\) because every \(g_j\) and \(f\) vanish there. Therefore \(f\) is measurable on \(E\) by Proposition 9 of Chapter 3.

Step 3: the subsequence converges to \(f\) in measure. For each index \(k\) put

\begin{equation*} F_k = \bigcup_{j \ge k} E_j , \qquad \text{so} \qquad m(F_k) \le \sum_{j \ge k} m(E_j) < \sum_{j \ge k} \frac{1}{2^j} = \frac{1}{2^{k-1}} \end{equation*}

by countable subadditivity. Let \(x \in E^{\prime} \sim F_k\). Then \(|f_{n_{j+1}}(x) - f_{n_j}(x)| \le 1/2^j\) for every \(j \ge k\), so for \(i > k\) the telescoping estimate of Problem 12 gives

\begin{equation*} |f_{n_i}(x) - f_{n_k}(x)| \le \sum_{j=k}^{i-1} \frac{1}{2^j} < \frac{1}{2^{k-1}} , \end{equation*}

and letting \(i \to \infty\) yields

\begin{equation*} |f(x) - f_{n_k}(x)| \le \frac{1}{2^{k-1}} \quad \text{for all } x \in E^{\prime} \sim F_k . \end{equation*}

Hence

\begin{equation*} \Bigl\{ x \in E \;\Bigm|\; |f_{n_k}(x) - f(x)| > \tfrac{1}{2^{k-1}} \Bigr\} \subseteq F_k \cup (E \sim E^{\prime}) , \end{equation*}

a set of measure less than \(1/2^{k-1}\). Now let \(\eta > 0\) and \(\epsilon > 0\). Choose \(K\) so large that \(1/2^{K-1} < \min\{\eta, \epsilon\}\). For \(k \ge K\) we have \(1/2^{k-1} \le 1/2^{K-1} < \eta\), so

\begin{equation*} m\{x \in E \mid |f_{n_k}(x) - f(x)| > \eta\} \le m\Bigl\{ x \in E \Bigm| |f_{n_k}(x) - f(x)| > \tfrac{1}{2^{k-1}} \Bigr\} < \frac{1}{2^{k-1}} \le \frac{1}{2^{K-1}} < \epsilon . \end{equation*}

Thus \(\{f_{n_k}\} \to f\) in measure on \(E\).

Step 4: the full sequence converges to \(f\) in measure. Let \(\eta > 0\) and \(\epsilon > 0\). Since \(\{f_n\}\) is Cauchy in measure, choose \(N\) with

\begin{equation*} m\{x \in E \mid |f_n(x) - f_p(x)| \ge \eta/2\} < \epsilon/2 \quad \text{for all } n, p \ge N . \end{equation*}

By Step 3 choose an index \(k\) so large that \(n_k \ge N\) and

\begin{equation*} m\{x \in E \mid |f_{n_k}(x) - f(x)| > \eta/2\} < \epsilon/2 . \end{equation*}

Fix such a \(k\). For any \(n \ge N\) and any \(x \in E \sim Z\) (where all the values in question are real), the triangle inequality shows that if \(|f_n(x) - f(x)| > \eta\) then at least one of \(|f_n(x) - f_{n_k}(x)| \ge \eta/2\) and \(|f_{n_k}(x) - f(x)| > \eta/2\) must hold: if both failed, the sum of the two moduli would be less than \(\eta\), contradicting \(|f_n(x) - f(x)| \le |f_n(x) - f_{n_k}(x)| + |f_{n_k}(x) - f(x)|\). That is,

\begin{equation*} \{x \in E \mid |f_n(x) - f(x)| > \eta\} \subseteq \{x \in E \mid |f_n(x) - f_{n_k}(x)| \ge \eta/2\} \cup \{x \in E \mid |f_{n_k}(x) - f(x)| > \eta/2\} \cup Z , \end{equation*}

the last set being of measure zero and therefore contributing nothing to the estimate below. Since \(n \ge N\) and \(n_k \ge N\), monotonicity and finite subadditivity give

\begin{equation*} m\{x \in E \mid |f_n(x) - f(x)| > \eta\} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \quad \text{for all } n \ge N . \end{equation*}

As \(\eta > 0\) and \(\epsilon > 0\) were arbitrary, \(\{f_n\} \to f\) in measure on \(E\). This exhibits the required measurable limit function and shows that the measurable functions on \(E\) are complete with respect to convergence in measure.

Note also that Step 2 delivers, at no extra cost, the Riesz-type conclusion that the subsequence \(\{f_{n_j}\}\) converges to \(f\) pointwise almost everywhere on \(E\); and no assumption whatever was made about \(m(E)\).

Write \(\varphi(t) = t/(1+t)\) for \(t \in [0, \infty)\). Two elementary properties of \(\varphi\) carry the whole proof:

(i) \(\varphi\) is increasing on \([0, \infty)\), since \(\varphi^{\prime}(t) = 1/(1+t)^2 > 0\); equivalently, for \(0 \le s \le t\) one has \(s(1+t) \le t(1+s)\), whence \(\varphi(s) \le \varphi(t)\).

(ii) \(0 \le \varphi(t) < 1\) for all \(t \ge 0\), since \(t < 1 + t\).

First, \(\rho\) is well defined and finite. By our standing convention \(f\) and each \(f_n\) is measurable and finite a.e. on \(E\). Let \(Z\) be the union of the countably many sets \(\{x \in E \mid |f(x)| = \infty\}\) and \(\{x \in E \mid |f_n(x)| = \infty\}\); then \(Z\) is measurable and \(m(Z) = 0\), and on \(E \sim Z\) every one of these functions is real-valued, so \(|f_n - f|\) is a nonnegative real-valued measurable function there. Hence \(\varphi \circ |f_n - f|\) is measurable on \(E \sim Z\), being the composition of a continuous function with a measurable real-valued function (Proposition 7 of Chapter 3); extend it by the value \(1\) on \(Z\), which alters no integral since \(m(Z) = 0\) and leaves the extension measurable by Proposition 5 of Chapter 3. By (ii) the integrand is bounded above by \(1\) on \(E\), and since \(m(E) < \infty\) the monotonicity of integration gives

\begin{equation*} 0 \le \rho(f_n, f) = \int_E \varphi(|f_n - f|) \le \int_E 1 = m(E) < \infty . \end{equation*}

So \(\{\rho(f_n, f)\}\) is a sequence of nonnegative real numbers.

Suppose first that \(\{f_n\} \to f\) in measure on \(E\); we show \(\rho(f_n, f) \to 0\). If \(m(E) = 0\) then \(\rho(f_n, f) = 0\) for all \(n\) and there is nothing to prove, so assume \(m(E) > 0\). Let \(\epsilon > 0\). Choose

\begin{equation*} \eta = \frac{\epsilon}{2 \, m(E)} > 0 , \end{equation*}

and set \(A_n = \{x \in E \mid |f_n(x) - f(x)| > \eta\}\), a measurable set. Split the integral over \(A_n\) and \(E \sim A_n\) (Theorem 11 of Chapter 4, additivity over domains of integration for nonnegative measurable functions). On \(A_n\) we use only the crude bound (ii), \(\varphi \le 1\); on \(E \sim A_n\) we have \(|f_n - f| \le \eta\) off the null set \(Z\), so by (i), \(\varphi(|f_n - f|) \le \varphi(\eta) = \eta/(1+\eta) \le \eta\) a.e. there. Therefore

\begin{equation*} \rho(f_n, f) = \int_{A_n} \varphi(|f_n - f|) + \int_{E \sim A_n} \varphi(|f_n - f|) \le m(A_n) + \eta \cdot m(E \sim A_n) \le m(A_n) + \eta \cdot m(E) = m(A_n) + \frac{\epsilon}{2} . \end{equation*}

Since \(\{f_n\} \to f\) in measure and \(\eta > 0\), there is an index \(N\) such that \(m(A_n) < \epsilon/2\) for all \(n \ge N\). Hence \(\rho(f_n, f) < \epsilon\) for all \(n \ge N\), and \(\lim_{n \to \infty} \rho(f_n, f) = 0\).

Conversely, suppose \(\lim_{n \to \infty} \rho(f_n, f) = 0\); we show \(\{f_n\} \to f\) in measure on \(E\). Let \(\eta > 0\) and \(\epsilon > 0\) be given, and again write \(A_n = \{x \in E \mid |f_n(x) - f(x)| > \eta\}\). For a.e. \(x \in A_n\) we have \(|f_n(x) - f(x)| > \eta\), so by the monotonicity (i) of \(\varphi\),

\begin{equation*} \varphi(|f_n(x) - f(x)|) \ge \varphi(\eta) = \frac{\eta}{1 + \eta} . \end{equation*}

Since the integrand is nonnegative on all of \(E\), monotonicity of integration gives the Chebychev-type estimate

\begin{equation*} \rho(f_n, f) = \int_E \varphi(|f_n - f|) \ge \int_{A_n} \varphi(|f_n - f|) \ge \frac{\eta}{1 + \eta} \cdot m(A_n) , \end{equation*}

that is,

\begin{equation*} m(A_n) \le \frac{1 + \eta}{\eta} \, \rho(f_n, f) . \end{equation*}

Because \(\rho(f_n, f) \to 0\) and \((1+\eta)/\eta\) is a fixed constant, there is an index \(N\) such that \(\frac{1+\eta}{\eta} \rho(f_n, f) < \epsilon\), and hence \(m(A_n) < \epsilon\), for all \(n \ge N\). As \(\eta > 0\) and \(\epsilon > 0\) were arbitrary, \(\{f_n\} \to f\) in measure on \(E\).

This completes the proof of the equivalence. Observe that finiteness of \(m(E)\) was used only in the first implication, where the uniform smallness of \(\varphi(|f_n - f|)\) off \(A_n\) had to be integrated over \(E\); the second implication is valid on any measurable set. The first implication genuinely fails when \(m(E) = \infty\): on \(E = \mathbb{R}\) take \(f_n = (1/n) \chi_{[0, n]}\) and \(f \equiv 0\). For fixed \(\eta > 0\) the set \(\{x \in \mathbb{R} \mid |f_n(x)| > \eta\}\) is empty as soon as \(n > 1/\eta\), so \(\{f_n\} \to 0\) in measure on \(\mathbb{R}\); yet the integrand \(\varphi(|f_n|)\) equals the constant \((1/n)/(1 + 1/n)\) on \([0,n]\) and vanishes off it, so

\begin{equation*} \rho(f_n, 0) = n \cdot \frac{1/n}{1 + 1/n} = \frac{1}{1 + 1/n} \longrightarrow 1 \ne 0 . \end{equation*}

Finally, \(\rho\) is in fact a pseudometric on the measurable functions on \(E\), and a metric once functions equal a.e. are identified. Symmetry and nonnegativity are clear, and \(\rho(g,h) = 0\) forces \(\varphi(|g - h|) = 0\) a.e. (Proposition 9 of Chapter 4), hence \(g = h\) a.e. For the triangle inequality one uses that \(\varphi\) is increasing and subadditive: for \(s, t \ge 0\),

\begin{equation*} \varphi(s + t) = \frac{s+t}{1+s+t} = \frac{s}{1+s+t} + \frac{t}{1+s+t} \le \frac{s}{1+s} + \frac{t}{1+t} = \varphi(s) + \varphi(t) , \end{equation*}

so from \(|g - k| \le |g - h| + |h - k|\) we get \(\varphi(|g-k|) \le \varphi(|g-h|) + \varphi(|h-k|)\) pointwise, and integrating gives \(\rho(g,k) \le \rho(g,h) + \rho(h,k)\). Thus the exercise identifies convergence in measure on a set of finite measure as convergence in a metric, which is why the subsequence criterion of Problem 11 holds for it and why Problem 13 is the assertion that this metric space is complete.

Since \(f\) and \(g\) are bounded, so is \(fg\): if \(|f| \le M_1\) and \(|g| \le M_2\) on \([a,b]\), then \(|fg| \le M_1 M_2\). Put \(M = \max\{M_1, M_2\}\), so that \(|f| \le M\) and \(|g| \le M\) on \([a,b]\).

A first proof, via Theorem 8. Let \(D_f\), \(D_g\) and \(D_{fg}\) denote the sets of points of \([a,b]\) at which \(f\), \(g\) and \(fg\), respectively, fail to be continuous. If \(x_0 \in [a,b]\) is a point at which both \(f\) and \(g\) are continuous, then \(fg\) is continuous at \(x_0\): indeed, for \(x \in [a,b]\),

\begin{equation*} |f(x)g(x) - f(x_0)g(x_0)| \le |f(x)|\,|g(x) - g(x_0)| + |g(x_0)|\,|f(x) - f(x_0)| \le M\bigl(|g(x) - g(x_0)| + |f(x) - f(x_0)|\bigr), \end{equation*}

and the right-hand side is less than \(\epsilon\) once \(|x - x_0| < \delta\), where \(\delta > 0\) is chosen so small that \(|f(x) - f(x_0)| < \epsilon/(2M+1)\) and \(|g(x) - g(x_0)| < \epsilon/(2M+1)\) for such \(x\). Contrapositively,

\begin{equation*} D_{fg} \subseteq D_f \cup D_g . \end{equation*}

Now \(f\) and \(g\) are bounded and Riemann integrable over \([a,b]\), so by Theorem 8 (Lebesgue) of this chapter, \(m(D_f) = 0\) and \(m(D_g) = 0\). By the countable (here, finite) subadditivity of outer measure, \(m^*(D_f \cup D_g) = 0\), and hence \(m^*(D_{fg}) = 0\); a set of outer measure zero is measurable with measure zero. Thus \(fg\) is a bounded function on \([a,b]\) whose set of discontinuities has measure zero, and therefore, by the converse half of Theorem 8, \(fg\) is Riemann integrable over \([a,b]\).

A second proof, directly from the Darboux sums. For a nonempty set \(S \subseteq [a,b]\) and a bounded function \(h\) on \([a,b]\), write

\begin{equation*} \operatorname{osc}(h, S) = \sup_{x \in S} h(x) - \inf_{x \in S} h(x) = \sup_{x, y \in S} |h(x) - h(y)| . \end{equation*}

For \(x, y \in S\),

\begin{equation*} |f(x)g(x) - f(y)g(y)| \le |f(x)|\,|g(x) - g(y)| + |g(y)|\,|f(x) - f(y)| \le M\operatorname{osc}(g, S) + M \operatorname{osc}(f, S), \end{equation*}

and taking the supremum over \(x, y \in S\) gives

\begin{equation*} \operatorname{osc}(fg, S) \le M\bigl[\operatorname{osc}(f,S) + \operatorname{osc}(g,S)\bigr]. \end{equation*}

Let \(P = \{x_0 = a < x_1 < \cdots < x_n = b\}\) be a partition of \([a,b]\) and \(I_i = [x_{i-1}, x_i]\). Since \(U(h, P) - L(h, P) = \sum_{i=1}^n \operatorname{osc}(h, I_i)\,(x_i - x_{i-1})\), the displayed inequality gives

\begin{equation*} U(fg, P) - L(fg, P) \le M\bigl[U(f,P) - L(f,P)\bigr] + M\bigl[U(g,P) - L(g,P)\bigr]. \end{equation*}

Let \(\epsilon > 0\) and put \(\eta = \epsilon/(2M+1)\). Because \(f\) is Riemann integrable, its upper and lower Riemann integrals have the common value \(\int_a^b f\); since the upper Riemann integral is the infimum of the upper Darboux sums and the lower Riemann integral is the supremum of the lower Darboux sums, there are partitions \(Q\) and \(Q^{\prime}\) of \([a,b]\) with

\begin{equation*} U(f, Q) < \int_a^b f + \frac{\eta}{2}, \qquad L(f, Q^{\prime}) > \int_a^b f - \frac{\eta}{2}. \end{equation*}

Let \(P_1 = Q \cup Q^{\prime}\), a refinement of both. Under refinement upper Darboux sums decrease and lower Darboux sums increase (Problem 2 of Chapter 4), so \(U(f, P_1) \le U(f, Q)\) and \(L(f, P_1) \ge L(f, Q^{\prime})\), and hence \(U(f, P_1) - L(f, P_1) < \eta\). Similarly there is a partition \(P_2\) with \(U(g, P_2) - L(g, P_2) < \eta\). Let \(P\) be the common refinement \(P_1 \cup P_2\). Invoking monotonicity under refinement once more,

\begin{equation*} U(f,P) - L(f,P) \le U(f,P_1) - L(f,P_1) < \eta, \qquad U(g,P) - L(g,P) \le U(g,P_2) - L(g,P_2) < \eta, \end{equation*}

and therefore \(U(fg, P) - L(fg, P) < 2M\eta = 2M\epsilon/(2M+1) < \epsilon\). Since

\begin{equation*} 0 \le \overline{\int_a^b} fg - \underline{\int_a^b} fg \le U(fg, P) - L(fg, P) < \epsilon \end{equation*}

for every \(\epsilon > 0\), the upper and lower Riemann integrals of \(fg\) coincide, that is, \(fg\) is Riemann integrable over \([a,b]\).

Throughout, \(D\) denotes the set of points of \([a,b]\) at which \(f\) fails to be continuous, and we assume \(m(D) = 0\).

The bounded case. Suppose \(f\) is bounded on \([a,b]\). By Theorem 8 (Lebesgue) of this chapter, a bounded function on \([a,b]\) is Riemann integrable over \([a,b]\) if and only if its set of discontinuities has measure zero; since \(m(D) = 0\), the function \(f\) is Riemann integrable over \([a,b]\). By Theorem 3 of Chapter 4, a bounded function that is Riemann integrable over the closed, bounded interval \([a,b]\) is then Lebesgue integrable over \([a,b]\) and the two integrals are equal. Finally, by Theorem 7 of this chapter, a bounded function on a set of finite measure is Lebesgue integrable over that set if and only if it is measurable; since \([a,b]\) has finite measure and \(f\) is bounded and Lebesgue integrable over \([a,b]\), the function \(f\) is measurable.

It is worth recording a direct argument for the bounded case, since it uses no integration theory at all and will not need boundedness either. Let \(E = [a,b] \sim D\) be the set of points at which \(f\) is continuous. Since \(m(D) = 0\), the set \(D\) is measurable, and hence so is \(E\), being the relative complement of a measurable set in a measurable set. The restriction \(f|_E\) is continuous as a function on the metric space \(E\): given \(x_0 \in E\) and \(\epsilon > 0\), continuity of \(f\) at \(x_0\) provides \(\delta > 0\) with \(|f(x) - f(x_0)| < \epsilon\) for all \(x \in [a,b]\) with \(|x - x_0| < \delta\), and in particular for all such \(x\) lying in \(E\). Therefore, for each \(c \in \mathbb{R}\), the set

\begin{equation*} \{x \in E : f(x) > c\} \end{equation*}

is relatively open in \(E\), so it has the form \(E \cap U\) with \(U \subseteq \mathbb{R}\) open, and hence is measurable, being the intersection of two measurable sets. On the other hand

\begin{equation*} \{x \in D : f(x) > c\} \subseteq D, \end{equation*}

and \(m(D) = 0\); since Lebesgue measure is complete, every subset of a set of measure zero is measurable (of measure zero). Consequently

\begin{equation*} \{x \in [a,b] : f(x) > c\} = \{x \in E : f(x) > c\} \cup \{x \in D : f(x) > c\} \end{equation*}

is measurable for every \(c \in \mathbb{R}\), that is, \(f\) is measurable. Note that nowhere was boundedness used.

The unbounded case, by truncation. Suppose now \(f\) is an arbitrary real-valued function on \([a,b]\) whose set \(D\) of discontinuities has measure zero. For a natural number \(n\) define the truncation

\begin{equation*} f_n(x) = \max\{-n, \min\{n, f(x)\}\} = \begin{cases} n & \text{if } f(x) > n, \\ f(x) & \text{if } -n \le f(x) \le n, \\ -n & \text{if } f(x) < -n. \end{cases} \end{equation*}

Then \(f_n = \varphi_n \circ f\), where \(\varphi_n(t) = \max\{-n, \min\{n, t\}\}\) is continuous on \(\mathbb{R}\) (it is \(1\)-Lipschitz, since \(|\varphi_n(s) - \varphi_n(t)| \le |s - t|\)). If \(f\) is continuous at \(x_0\), then so is \(\varphi_n \circ f\), since \(|f_n(x) - f_n(x_0)| = |\varphi_n(f(x)) - \varphi_n(f(x_0))| \le |f(x) - f(x_0)|\); hence \(f_n\) is continuous at every point of \(E = [a,b] \sim D\), so the set of discontinuities of \(f_n\) is contained in \(D\) and therefore has measure zero. Moreover \(|f_n| \le n\), so \(f_n\) is bounded. By the bounded case already proved, each \(f_n\) is measurable on \([a,b]\).

Finally, \(\{f_n\} \to f\) pointwise on \([a,b]\): given \(x \in [a,b]\), the number \(f(x)\) is real, so for all \(n \ge |f(x)|\) we have \(f_n(x) = f(x)\). A pointwise limit of a sequence of measurable functions is measurable (Proposition 9 of Chapter 3). Therefore \(f\) is measurable.

Alternatively, and more briefly, the direct argument given above already proves the unbounded case verbatim: it shows that for every \(c \in \mathbb{R}\) the set \(\{x \in [a,b] : f(x) > c\}\) splits as a relatively open subset of the continuity set \(E\) together with a subset of the null set \(D\), both of which are measurable, and boundedness of \(f\) played no role.

An example. Define \(f\) on \([0,1]\) by

\begin{equation*} f(x) = \begin{cases} \dfrac{1}{x}\,\sin\dfrac{1}{x} & \text{if } 0 < x \le 1, \\ 0 & \text{if } x = 0. \end{cases} \end{equation*}

Then \(f\) is continuous on \((0,1]\), being built from continuous functions by composition, product and quotient with nonvanishing denominator. It is measurable on \([0,1]\). Indeed, fix \(c \in \mathbb{R}\). Since \(f\) is continuous on \((0,1]\), the set \(\{x \in (0,1] : f(x) > c\}\) is relatively open in \((0,1]\), so it equals \((0,1] \cap U\) for some open \(U \subseteq \mathbb{R}\) and is therefore measurable; and

\begin{equation*} \{x \in [0,1] : f(x) > c\} = \bigl(\{x \in (0,1] : f(x) > c\}\bigr) \cup S, \quad \text{where } S = \{0\} \text{ if } f(0) > c \text{ and } S = \emptyset \text{ otherwise}, \end{equation*}

a union of two measurable sets. Hence \(f\) is measurable on \([0,1]\). The same remark shows that any function on \([0,1]\) that is continuous on \((0,1]\) is measurable, and we use it again below.

Fix \(n \ge 2\). On the closed, bounded interval \([1/n, 1]\) the function \(f\) is continuous, hence bounded and Riemann integrable (Problem 6 of Chapter 4), so by Theorem 3 of Chapter 4 its Lebesgue integral over \([1/n,1]\) equals its Riemann integral there. The change of variable \(u = 1/x\), \(x = 1/u\), \(dx = -du/u^2\) is a \(C^1\) diffeomorphism of \([1/n, 1]\) onto \([1, n]\), and it gives

\begin{equation*} \int_{[1/n,\,1]} f = \int_{1/n}^{1} \frac{1}{x}\sin\frac{1}{x}\,dx = \int_{n}^{1} u \sin u \cdot \Bigl(-\frac{1}{u^2}\Bigr) du = \int_{1}^{n} \frac{\sin u}{u}\,du . \end{equation*}

This sequence of real numbers converges, and we verify it by the Cauchy criterion, using nothing beyond the calculus of continuous functions on closed, bounded intervals. Let \(2 \le n < m\). Integration by parts on \([n, m]\), where \(u \mapsto \sin u / u\) and \(u \mapsto \cos u / u\) are continuously differentiable, gives

\begin{equation*} \int_{n}^{m} \frac{\sin u}{u}\,du = \Bigl[\frac{-\cos u}{u}\Bigr]_{n}^{m} - \int_{n}^{m} \frac{\cos u}{u^{2}}\,du = \frac{\cos n}{n} - \frac{\cos m}{m} - \int_{n}^{m}\frac{\cos u}{u^{2}}\,du . \end{equation*}

Since \(|\cos u| \le 1\),

\begin{equation*} \Bigl| \int_{n}^{m} \frac{\sin u}{u}\,du \Bigr| \le \frac{1}{n} + \frac{1}{m} + \int_{n}^{m} \frac{du}{u^{2}} = \frac{1}{n} + \frac{1}{m} + \frac{1}{n} - \frac{1}{m} = \frac{2}{n} . \end{equation*}

Writing \(a_n = \int_{[1/n,1]} f = \int_1^n (\sin u)/u\,du\), this says \(|a_m - a_n| \le 2/n\) for all \(m > n \ge 2\), so \(\{a_n\}\) is a Cauchy sequence of real numbers and therefore converges. Thus \(\bigl\{\int_{[1/n,1]} f\bigr\}\) converges.

Nevertheless \(f\) is not Lebesgue integrable over \([0,1]\), because \(\int_{[0,1]} |f| = \infty\). The same change of variable applied to \(|f|\) gives, for each \(n \ge 2\),

\begin{equation*} \int_{[1/n,\,1]} |f| = \int_{1}^{n} \frac{|\sin u|}{u}\,du . \end{equation*}

For each natural number \(k\),

\begin{equation*} \int_{k\pi}^{(k+1)\pi} \frac{|\sin u|}{u}\,du \ge \frac{1}{(k+1)\pi}\int_{k\pi}^{(k+1)\pi} |\sin u|\,du = \frac{2}{(k+1)\pi} . \end{equation*}

Given \(n \ge 4\), let \(K\) be the largest natural number with \((K+1)\pi \le n\); then \(K \ge 1\) and \(K \to \infty\) as \(n \to \infty\). The intervals \([k\pi, (k+1)\pi]\) for \(1 \le k \le K\) are pairwise nonoverlapping and all contained in \([1, n]\), since \(\pi > 1\). As the integrand is nonnegative,

\begin{equation*} \int_{1}^{n} \frac{|\sin u|}{u}\,du \ge \sum_{k=1}^{K} \int_{k\pi}^{(k+1)\pi} \frac{|\sin u|}{u}\,du \ge \frac{2}{\pi} \sum_{k=1}^{K} \frac{1}{k+1}, \end{equation*}

and the harmonic series diverges, so the right-hand side tends to \(\infty\) with \(K\). Hence \(\int_{1}^{n} |\sin u| u^{-1}\,du \to \infty\) as \(n \to \infty\). Now \(\{|f| \cdot \chi_{[1/n,1]}\}\) is an increasing sequence of nonnegative measurable functions on \([0,1]\) converging pointwise on \([0,1]\) to \(|f| \cdot \chi_{(0,1]}\), which differs from \(|f|\) only at the single point \(x = 0\) and so equals \(|f|\) almost everywhere on \([0,1]\). Two nonnegative measurable functions that agree almost everywhere have the same integral, so by the Monotone Convergence Theorem,

\begin{equation*} \int_{[0,1]} |f| = \int_{[0,1]} |f| \cdot \chi_{(0,1]} = \lim_{n \to \infty} \int_{[0,1]} |f| \cdot \chi_{[1/n,\,1]} = \lim_{n \to \infty} \int_{[1/n,\,1]} |f| = \infty . \end{equation*}

A measurable function \(f\) is Lebesgue integrable over \([0,1]\) precisely when \(\int_{[0,1]} |f| < \infty\); hence \(f\) is not Lebesgue integrable over \([0,1]\), although \(\{\int_{[1/n,1]} f\}\) converges. This answers the first part.

The nonnegative case. No, this cannot happen if \(f \ge 0\). Suppose \(f\) is a nonnegative real-valued function on \([0,1]\) that is continuous on \((0,1]\) and for which the sequence \(\bigl\{\int_{[1/n,1]} f\bigr\}\) converges, say to the finite number \(L\). By the remark recorded at the start of this solution, \(f\) is measurable on \([0,1]\), and each \(\int_{[1/n,1]} f\) is a well-defined element of \([0, \infty]\); it is in fact finite, since \(f\) is continuous, hence bounded, on the closed, bounded interval \([1/n,1]\).

Since \(f \ge 0\), the sequence of functions \(f_n = f \cdot \chi_{[1/n,\,1]}\) is a nondecreasing sequence of nonnegative measurable functions on \([0,1]\), and it converges pointwise on \([0,1]\) to \(f \cdot \chi_{(0,1]}\), a function that agrees with \(f\) except at \(x = 0\), hence agrees with \(f\) almost everywhere on \([0,1]\). By the Monotone Convergence Theorem,

\begin{equation*} \int_{[0,1]} f = \int_{[0,1]} f \cdot \chi_{(0,1]} = \lim_{n \to \infty} \int_{[1/n,\,1]} f = L < \infty . \end{equation*}

Thus \(f\) is a nonnegative measurable function on \([0,1]\) with finite integral, and since \(|f| = f\), this says exactly that \(f\) is Lebesgue integrable over \([0,1]\). So for nonnegative \(f\) the convergence of \(\bigl\{\int_{[1/n,1]} f\bigr\}\) forces Lebesgue integrability over \([0,1]\); the phenomenon exhibited above depends essentially on the cancellation between the positive and negative parts of \(f\).

This is Proposition 2 with the open interval \((a,b)\) replaced by the closed interval \([a,b]\). The only new feature is the left endpoint: at \(x_0 = a\) continuity of a function on \([a,b]\) means continuity from the right, so a jump placed at \(a\) must be a jump to the right. We therefore use two kinds of jump functions.

If \(C\) is empty, take \(f(x) = x\), which is increasing and continuous at every point of \([a,b]\). So assume \(C \neq \emptyset\) and let \(\{q_n\}_{n \in \Lambda}\) be an enumeration of \(C\), where \(\Lambda = \{1, 2, \dots, N\}\) if \(C\) is finite and \(\Lambda = \{1,2,3,\dots\}\) if \(C\) is countably infinite. For each \(n \in \Lambda\) define a function \(\varphi_n\) on \([a,b]\) by

\begin{equation*} \varphi_n(x) = \begin{cases} 1 & \text{if } x \geq q_n \\ 0 & \text{if } x < q_n \end{cases} \quad \text{in case } q_n > a, \end{equation*}

\begin{equation*} \varphi_n(x) = \begin{cases} 1 & \text{if } x > a \\ 0 & \text{if } x = a \end{cases} \quad \text{in case } q_n = a. \end{equation*}

In either case \(\varphi_n\) takes only the values \(0\) and \(1\), is increasing on \([a,b]\), and is continuous at every point of \([a,b]\) except at \(q_n\). Define

\begin{equation*} f(x) = \sum_{n \in \Lambda} \frac{1}{2^n} \, \varphi_n(x) \quad \text{for all } x \in [a,b]. \end{equation*}

Since \(0 \le \varphi_n \le 1\) and \(\sum_n 2^{-n} \le 1\), the series converges for every \(x\), so \(f\) is a properly defined real-valued function on \([a,b]\). Moreover \(f\) is increasing: if \(a \le u < v \le b\), then \(\varphi_n(u) \le \varphi_n(v)\) for every \(n\), and hence, all the terms being nonnegative,

\begin{equation*} f(v) - f(u) = \sum_{n \in \Lambda} \frac{1}{2^n} \bigl[ \varphi_n(v) - \varphi_n(u) \bigr] \geq 0. \tag{1} \end{equation*}

\(f\) fails to be continuous at each point of \(C\). Fix \(k \in \Lambda\). Suppose first that \(q_k > a\). If \(x \in [a,b]\) and \(x < q_k\), then \(\varphi_k(q_k) - \varphi_k(x) = 1\), so by (1), applied with \(u = x\) and \(v = q_k\) and by discarding all terms but the \(k\)-th,

\begin{equation*} f(q_k) - f(x) \geq \frac{1}{2^k} \quad \text{for all } x \in [a,b] \text{ with } x < q_k . \end{equation*}

Since \(q_k > a\), there are points \(x\) of \([a,b]\) with \(x < q_k\) arbitrarily close to \(q_k\), and therefore \(f\) is not continuous at \(q_k\). Now suppose \(q_k = a\). If \(x \in (a,b]\), then \(\varphi_k(x) - \varphi_k(a) = 1\) and the same reasoning gives \(f(x) - f(a) \geq 1/2^k\) for all \(x \in (a,b]\); since \(a < b\) there are such \(x\) arbitrarily close to \(a\), and again \(f\) is not continuous at \(q_k = a\).

\(f\) is continuous at each point of \([a,b] \sim C\). Let \(x_0 \in [a,b] \sim C\) and let \(\epsilon > 0\). Choose a natural number \(M\) with \(1/2^M < \epsilon\) and put \(\Lambda_M = \{ n \in \Lambda : n \leq M \}\), a finite set. In both of the above cases the single point at which \(\varphi_n\) fails to be continuous is \(q_n\), and \(x_0 \neq q_n\) for every \(n \in \Lambda\) because \(x_0 \notin C\). Define

\begin{equation*} \delta = \min \bigl\{ |x_0 - q_n| : n \in \Lambda_M \bigr\} > 0 \end{equation*}

(with \(\delta = b-a\) if \(\Lambda_M\) is empty). Let \(x \in [a,b]\) with \(|x - x_0| < \delta\), and fix \(n \in \Lambda_M\). The interval \((x_0 - \delta, x_0 + \delta)\) does not contain \(q_n\). Each \(\varphi_n\) is constant on \((-\infty, q_n) \cap [a,b]\) and constant on \([q_n, \infty) \cap [a,b]\) (respectively on \(\{a\}\) and on \((a,b]\) in the second case), and an interval that omits \(q_n\) lies entirely in one of these two pieces; hence \(\varphi_n(x) = \varphi_n(x_0)\). Consequently all the terms with \(n \leq M\) cancel and

\begin{equation*} |f(x) - f(x_0)| = \Bigl| \sum_{n \in \Lambda, \, n > M} \frac{1}{2^n} \bigl[ \varphi_n(x) - \varphi_n(x_0) \bigr] \Bigr| \leq \sum_{n > M} \frac{1}{2^n} = \frac{1}{2^M} < \epsilon . \end{equation*}

Therefore \(f\) is continuous at \(x_0\).

Thus \(f\) is an increasing function on \([a,b]\) whose set of points of continuity is exactly \([a,b] \sim C\).

Let \(C = \mathbb{Q} \cap [0,1]\), the set of rational numbers in \([0,1]\). This is a countable subset of the nondegenerate closed, bounded interval \([0,1]\), so by the preceding problem there is an increasing function \(g\) on \([0,1]\) whose set of points of continuity is exactly \([0,1] \sim C\), that is, exactly the set of irrational numbers in \([0,1]\). Define

\begin{equation*} f(x) = g(x) + x \quad \text{for all } x \in [0,1]. \end{equation*}

\(f\) is strictly increasing. If \(0 \leq u < v \leq 1\), then \(g(u) \leq g(v)\) because \(g\) is increasing, and \(u < v\), so

\begin{equation*} f(v) - f(u) = \bigl[ g(v) - g(u) \bigr] + (v - u) \geq v - u > 0 . \end{equation*}

\(f\) has the same points of continuity as \(g\). The function \(x \mapsto x\) is continuous on \([0,1]\). Hence if \(g\) is continuous at \(x_0\), so is the sum \(f = g + \mathrm{id}\); and conversely, if \(f\) is continuous at \(x_0\), then so is \(g = f - \mathrm{id}\), being the difference of two functions continuous at \(x_0\). Therefore \(f\) is continuous at \(x_0\) if and only if \(g\) is, and so the set of points of continuity of \(f\) is exactly the set of irrational numbers in \([0,1]\).

Thus \(f\) is a strictly increasing function on \([0,1]\) that is continuous at every irrational point of \([0,1]\) and at no rational point of \([0,1]\). (Note that the endpoints \(0\) and \(1\) are rational and hence are points of discontinuity; this is why in the preceding problem the jump associated with the left endpoint had to be a jump to the right, so that the one-sided continuity of \(f\) at \(0\) also fails.)

Here continuity is continuity of \(f\) as a function on \(E\): \(f\) is continuous at \(x_0 \in E\) provided for each \(\epsilon > 0\) there is a \(\delta > 0\) such that \(|f(x) - f(x_0)| < \epsilon\) for all \(x \in E\) with \(|x - x_0| < \delta\).

Replacing \(f\) by \(-f\) changes neither monotonicity nor the set of points of continuity, so we may assume \(f\) is increasing on \(E\). Let \(D\) denote the set of points of \(E\) at which \(f\) fails to be continuous. We must show \(D\) is countable.

For \(x_0 \in E\) define

\begin{equation*} L(x_0) = \begin{cases} \sup \{ f(x) : x \in E, \ x < x_0 \} & \text{if } x_0 \text{ is a limit point of } E \cap (-\infty, x_0) \\ f(x_0) & \text{otherwise,} \end{cases} \end{equation*}

\begin{equation*} R(x_0) = \begin{cases} \inf \{ f(x) : x \in E, \ x > x_0 \} & \text{if } x_0 \text{ is a limit point of } E \cap (x_0, \infty) \\ f(x_0) & \text{otherwise.} \end{cases} \end{equation*}

In the first alternative the set over which the supremum is taken is nonempty and bounded above by \(f(x_0)\), since \(f\) is increasing; so \(L(x_0)\) is a real number and \(L(x_0) \leq f(x_0)\). Similarly \(R(x_0)\) is a real number and \(f(x_0) \leq R(x_0)\). Thus in all cases

\begin{equation*} L(x_0) \leq f(x_0) \leq R(x_0) . \end{equation*}

Claim: if \(L(x_0) = f(x_0) = R(x_0)\), then \(f\) is continuous at \(x_0\). Let \(\epsilon > 0\). We produce \(\delta_1 > 0\) controlling the left side. If \(x_0\) is not a limit point of \(E \cap (-\infty, x_0)\), choose \(\delta_1 > 0\) with \(E \cap (x_0 - \delta_1, x_0) = \emptyset\); then there is nothing to check on the left. Otherwise \(\sup \{ f(x) : x \in E, \ x < x_0 \} = f(x_0)\), so there is a \(z \in E\) with \(z < x_0\) and \(f(z) > f(x_0) - \epsilon\); put \(\delta_1 = x_0 - z\). If \(x \in E\) and \(x_0 - \delta_1 < x < x_0\), then \(z < x < x_0\) and, \(f\) being increasing, \(f(x_0) - \epsilon < f(z) \leq f(x) \leq f(x_0)\). Symmetrically there is \(\delta_2 > 0\) such that \(f(x_0) \leq f(x) < f(x_0) + \epsilon\) whenever \(x \in E\) and \(x_0 < x < x_0 + \delta_2\). With \(\delta = \min \{ \delta_1, \delta_2 \}\) we have \(|f(x) - f(x_0)| < \epsilon\) for all \(x \in E\) with \(|x - x_0| < \delta\). This proves the claim.

Consequently, for each \(x_0 \in D\) we have \(L(x_0) < f(x_0)\) or \(f(x_0) < R(x_0)\); in either case \(L(x_0) < R(x_0)\), so

\begin{equation*} J(x_0) = \bigl( L(x_0), \, R(x_0) \bigr) \end{equation*}

is a nonempty open interval of real numbers.

Claim: the intervals \(\{ J(x_0) \}_{x_0 \in D}\) are disjoint. Let \(x_0, x_1 \in D\) with \(x_0 < x_1\); it suffices to show \(R(x_0) \leq L(x_1)\). Suppose first that there is a point \(z \in E\) with \(x_0 < z < x_1\). If \(x_0\) is a limit point of \(E \cap (x_0, \infty)\), then \(R(x_0)\) is an infimum over a set containing \(f(z)\), so \(R(x_0) \leq f(z)\); if not, \(R(x_0) = f(x_0) \leq f(z)\) since \(f\) is increasing. Either way \(R(x_0) \leq f(z)\). Symmetrically \(f(z) \leq L(x_1)\). Hence \(R(x_0) \leq L(x_1)\). Suppose next that no point of \(E\) lies strictly between \(x_0\) and \(x_1\). Then \(x_0\) is not a limit point of \(E \cap (x_0, \infty)\) and \(x_1\) is not a limit point of \(E \cap (-\infty, x_1)\), so \(R(x_0) = f(x_0) \leq f(x_1) = L(x_1)\). In both cases \(R(x_0) \leq L(x_1)\), so the open intervals \(J(x_0)\) and \(J(x_1)\) are disjoint. This proves the claim.

Finally, each nonempty open interval \(J(x_0)\) contains a rational number; choose one, say \(r(x_0)\). Since the intervals are disjoint, the map \(x_0 \mapsto r(x_0)\) from \(D\) to \(\mathbb{Q}\) is one-to-one. Therefore \(D\) is equipotent to a subset of the countable set \(\mathbb{Q}\) and hence is countable.

Thus a monotone function on a subset \(E\) of \(\mathbb{R}\) is continuous at every point of \(E\) except possibly at the points of a countable subset of \(E\). Taking \(E = (a,b)\) recovers Theorem 1.

In general, no. The obstruction is the isolated points of \(E\), and in fact there is such a function if and only if \(C\) contains no point that is isolated in \(E\).

A counterexample. Let \(E = \{0\} \cup [1,2]\) and \(C = \{0\}\), a countable subset of \(E\). The point \(0\) is isolated in \(E\): taking \(\delta = 1\), the only point \(x\) of \(E\) with \(|x - 0| < \delta\) is \(x = 0\). Hence for any function \(g\) whatsoever on \(E\) and any \(\epsilon > 0\), the choice \(\delta = 1\) gives \(|g(x) - g(0)| = 0 < \epsilon\) for all \(x \in E\) with \(|x - 0| < \delta\); that is, every function on \(E\) is continuous at \(0\). So no function on \(E\), monotone or not, can have \(0\) among its points of discontinuity, and there is no monotone function on \(E\) that is continuous only at points of \(E \sim C = [1,2]\). The same argument works with \(E = \mathbb{Z}\) and \(C\) any nonempty subset of \(\mathbb{Z}\), since every point of \(\mathbb{Z}\) is isolated in \(\mathbb{Z}\).

The general criterion. Call \(x_0 \in E\) isolated in \(E\) provided there is a \(\delta > 0\) with \(E \cap (x_0 - \delta, x_0 + \delta) = \{x_0\}\). As just observed, if some point of \(C\) is isolated in \(E\), no function on \(E\) is discontinuous there, and the answer is no. We claim that if no point of \(C\) is isolated in \(E\), then such an increasing function exists.

Assume no point of \(C\) is isolated in \(E\). If \(C = \emptyset\), take \(f(x) = x\), which is increasing and continuous at every point of \(E\). Otherwise let \(\{q_n\}_{n \in \Lambda}\) be an enumeration of \(C\), with \(\Lambda\) an initial segment of the natural numbers or all of them. Fix \(n \in \Lambda\). Since \(q_n\) is not isolated in \(E\), every interval \((q_n - \delta, q_n + \delta)\) meets \(E\) in a point other than \(q_n\), so \(q_n\) is a limit point of \(E \cap (-\infty, q_n)\) or a limit point of \(E \cap (q_n, \infty)\). Define, for \(x \in E\),

\begin{equation*} \varphi_n(x) = \begin{cases} 1 & \text{if } x \geq q_n \\ 0 & \text{if } x < q_n \end{cases} \end{equation*}

in the first case, and

\begin{equation*} \varphi_n(x) = \begin{cases} 1 & \text{if } x > q_n \\ 0 & \text{if } x \leq q_n \end{cases} \end{equation*}

in the second case. In either case \(\varphi_n\) is increasing on \(E\), takes only the values \(0\) and \(1\), and is constant on every interval that omits \(q_n\). Define

\begin{equation*} f(x) = \sum_{n \in \Lambda} \frac{1}{2^n} \, \varphi_n(x) \quad \text{for all } x \in E, \end{equation*}

a properly defined real-valued function since the series is dominated by the convergent geometric series \(\sum_n 2^{-n}\). Since each \(\varphi_n\) is increasing and all terms are nonnegative, \(f\) is increasing on \(E\), and for \(u, v \in E\) with \(u < v\),

\begin{equation*} f(v) - f(u) \geq \frac{1}{2^k} \bigl[ \varphi_k(v) - \varphi_k(u) \bigr] \quad \text{for each } k \in \Lambda . \tag{2} \end{equation*}

\(f\) is discontinuous at each \(q_k \in C\). In the first case above, \(\varphi_k(q_k) - \varphi_k(x) = 1\) for every \(x \in E\) with \(x < q_k\), so by (2), \(f(q_k) - f(x) \geq 1/2^k\) for all such \(x\); and since \(q_k\) is a limit point of \(E \cap (-\infty, q_k)\) there are such \(x\) arbitrarily close to \(q_k\). In the second case, \(\varphi_k(x) - \varphi_k(q_k) = 1\) for every \(x \in E\) with \(x > q_k\), so \(f(x) - f(q_k) \geq 1/2^k\) for all such \(x\), and there are such \(x\) arbitrarily close to \(q_k\). In both cases the definition of continuity fails at \(q_k\) for \(\epsilon = 1/2^k\).

\(f\) is continuous at each \(x_0 \in E \sim C\). Given \(\epsilon > 0\), choose a natural number \(M\) with \(1/2^M < \epsilon\), and let \(\delta = \min \{ |x_0 - q_n| : n \in \Lambda, \ n \leq M \} > 0\), this minimum being over a finite set of strictly positive numbers because \(x_0 \notin C\) (take \(\delta = 1\) if the set is empty). If \(x \in E\) and \(|x - x_0| < \delta\), then for each \(n \leq M\) the interval \((x_0 - \delta, x_0 + \delta)\) omits \(q_n\) and hence \(\varphi_n\) is constant on it, so \(\varphi_n(x) = \varphi_n(x_0)\). Therefore

\begin{equation*} |f(x) - f(x_0)| \leq \sum_{n > M} \frac{1}{2^n} = \frac{1}{2^M} < \epsilon . \end{equation*}

So the answer to the question as posed is negative in general: such a monotone function exists precisely when \(C\) contains no point that is isolated in \(E\). In particular, when \(E\) is an interval, or more generally when \(E\) has no isolated points, the answer is always yes, which is consistent with Proposition 2 and with Problem 1.

A closed, bounded interval \([c,d]\) is degenerate provided \(c = d\), that is, provided it is a single point and has length zero. Suppose we drop the nondegeneracy requirement from the definition of a Vitali covering, so that a collection \(\mathcal{F}\) of closed, bounded intervals is said to cover \(E\) in the sense of Vitali provided for each \(x \in E\) and each \(\epsilon > 0\) there is an \(I \in \mathcal{F}\) that contains \(x\) and has \(\ell(I) < \epsilon\). We exhibit a set \(E\) of finite outer measure and such a collection \(\mathcal{F}\) for which the conclusion (2) of the Vitali Covering Lemma fails.

Let \(E = [0,1]\), so that \(m^*(E) = 1 < \infty\), and let

\begin{equation*} \mathcal{F} = \bigl\{ \, [x,x] : x \in [0,1] \, \bigr\}, \end{equation*}

the collection of all degenerate closed intervals with endpoint in \([0,1]\).

\(\mathcal{F}\) covers \(E\) in the sense of Vitali in the relaxed sense just described: given \(x \in E\) and \(\epsilon > 0\), the interval \(I = [x,x]\) belongs to \(\mathcal{F}\), contains \(x\), and has \(\ell(I) = 0 < \epsilon\).

Now let \(\{I_k\}_{k=1}^{n}\) be any finite subcollection of \(\mathcal{F}\), disjoint or not. Then \(A = \bigcup_{k=1}^{n} I_k\) is a set of at most \(n\) points, hence is measurable with \(m(A) = 0\). Since \(E = [0,1]\) is measurable of finite measure and \(E \sim A\) is measurable, the additivity of measure over the disjoint union \(E = (E \sim A) \cup (E \cap A)\) gives

\begin{equation*} m^* \Bigl[ E \sim \bigcup_{k=1}^{n} I_k \Bigr] = m(E) - m(E \cap A) = 1 - 0 = 1 . \end{equation*}

Hence for any \(\epsilon \leq 1\) there is no finite subcollection \(\{I_k\}_{k=1}^{n}\) of \(\mathcal{F}\) for which

\begin{equation*} m^* \Bigl[ E \sim \bigcup_{k=1}^{n} I_k \Bigr] < \epsilon . \end{equation*}

Indeed, the same computation shows that even a countable disjoint subcollection of \(\mathcal{F}\) leaves a remainder of outer measure \(1\), since a countable union of singletons has measure zero. So the Vitali Covering Lemma fails completely for such coverings, and the assumption in the definition preceding the lemma that the intervals of \(\mathcal{F}\) be nondegenerate cannot be dropped.

The reason the proof of the lemma breaks down is visible in inequality (6): the selection process there chooses at each stage an interval whose length is more than half the supremum \(s_n\) of the available lengths, and it concludes from \(\ell(I_k) \to 0\) that every point left uncovered lies in one of the enlarged intervals \(5 * I_k\). When all the intervals are degenerate, every \(5 * I_k\) is again a single point and covers nothing new; the lengths carry no information.

By a general interval we mean any interval of real numbers: open, closed, or half-open, bounded or unbounded. Such an interval \(I\) is nondegenerate provided it contains more than one point. We prove the following.

Let \(E\) be a set of finite outer measure and \(\mathcal{F}\) a collection of nondegenerate general intervals such that for each \(x \in E\) and each \(\epsilon > 0\) there is an \(I \in \mathcal{F}\) containing \(x\) with \(\ell(I) < \epsilon\). Then for each \(\epsilon > 0\) there is a finite disjoint subcollection \(\{I_k\}_{k=1}^{n}\) of \(\mathcal{F}\) for which

\begin{equation*} m^* \Bigl[ E \sim \bigcup_{k=1}^{n} I_k \Bigr] < \epsilon . \end{equation*}

Proof. Let \(\mathcal{F}_0 = \{ I \in \mathcal{F} : \ell(I) < \infty \}\) and let

\begin{equation*} \overline{\mathcal{F}} = \bigl\{ \, \overline{I} \, : \, I \in \mathcal{F}_0 \, \bigr\}, \end{equation*}

where \(\overline{I}\) denotes the closure of \(I\). If \(I\) is a nondegenerate interval of finite length with endpoints \(c < d\), then \(\overline{I} = [c,d]\) is a closed, bounded, nondegenerate interval and \(\ell(\overline{I}) = d - c = \ell(I)\). Thus \(\overline{\mathcal{F}}\) is a collection of closed, bounded, nondegenerate intervals.

\(\overline{\mathcal{F}}\) covers \(E\) in the sense of Vitali. Indeed, let \(x \in E\) and \(\epsilon > 0\). By hypothesis there is an \(I \in \mathcal{F}\) with \(x \in I\) and \(\ell(I) < \epsilon\); in particular \(\ell(I) < \infty\), so \(I \in \mathcal{F}_0\). Then \(x \in I \subseteq \overline{I} \in \overline{\mathcal{F}}\) and \(\ell(\overline{I}) = \ell(I) < \epsilon\).

Since \(m^*(E) < \infty\), the Vitali Covering Lemma applies to \(E\) and \(\overline{\mathcal{F}}\): given \(\epsilon > 0\), there is a finite disjoint subcollection of \(\overline{\mathcal{F}}\), say \(\{ \overline{I_k} \}_{k=1}^{n}\) with each \(I_k \in \mathcal{F}_0 \subseteq \mathcal{F}\), for which

\begin{equation*} m^* \Bigl[ E \sim \bigcup_{k=1}^{n} \overline{I_k} \Bigr] < \frac{\epsilon}{2} . \end{equation*}

The subcollection \(\{I_k\}_{k=1}^{n}\) of \(\mathcal{F}\) is disjoint: if \(j \neq k\), then \(I_j \cap I_k \subseteq \overline{I_j} \cap \overline{I_k} = \emptyset\). Moreover, since \(\overline{I_k} \sim I_k\) consists of at most the two endpoints of \(I_k\), the set

\begin{equation*} \Delta = \bigcup_{k=1}^{n} \bigl( \overline{I_k} \sim I_k \bigr) \end{equation*}

is finite and therefore \(m^*(\Delta) = 0\). Now

\begin{equation*} E \sim \bigcup_{k=1}^{n} I_k \subseteq \Bigl[ E \sim \bigcup_{k=1}^{n} \overline{I_k} \Bigr] \cup \Delta , \end{equation*}

since a point of \(E\) that lies in no \(I_k\) either lies in no \(\overline{I_k}\) or lies in some \(\overline{I_k} \sim I_k\). By the monotonicity and countable (here finite) subadditivity of outer measure,

\begin{equation*} m^* \Bigl[ E \sim \bigcup_{k=1}^{n} I_k \Bigr] \leq m^* \Bigl[ E \sim \bigcup_{k=1}^{n} \overline{I_k} \Bigr] + m^*(\Delta) < \frac{\epsilon}{2} + 0 < \epsilon . \end{equation*}

This is the assertion, and the proof is complete.

Two remarks. First, the intervals of \(\mathcal{F}\) of infinite length were discarded at the outset; this costs nothing, because the Vitali covering condition only ever calls on intervals of length less than a prescribed \(\epsilon\). Second, nondegeneracy is essential and is inherited by the closures, so the previous problem’s counterexample is not in conflict with what has just been proved.

No. There is a continuous function on \(\mathbb{R}\) that fails to be monotone on every open interval. This is in the same spirit as the pathology mentioned in the Remark of this section, that a continuous function on an open interval may fail to be differentiable at every point.

The example. Let \(\varphi(t) = \operatorname{dist}(t, \mathbb{Z})\) be the distance from \(t\) to the nearest integer, and define

\begin{equation*} T(x) = \sum_{m=0}^{\infty} \frac{1}{2^{m}} \, \varphi(2^{m} x) \quad \text{for all } x \in \mathbb{R}. \end{equation*}

\(T\) is continuous on \(\mathbb{R}\). Each function \(x \mapsto 2^{-m} \varphi(2^m x)\) is continuous, since \(\varphi\) is continuous (it is \(1\)-periodic and satisfies \(|\varphi(s) - \varphi(t)| \leq |s-t|\)), and \(0 \leq 2^{-m} \varphi(2^m x) \leq 2^{-m-1}\) for all \(x\). Since \(\sum_m 2^{-m-1} < \infty\), the series converges uniformly on \(\mathbb{R}\) by the Weierstrass M-test, and a uniform limit of continuous functions is continuous. So \(T\) is continuous on all of \(\mathbb{R}\).

Self-similarity on dyadic intervals. For integers \(n \geq 0\) and \(k\), let \(I_{n,k} = [k/2^n, (k+1)/2^n]\). Fix such an interval and let \(x \in I_{n,k}\).

For \(m < n\), the point \(2^m x\) ranges over \([k \cdot 2^{m-n}, (k+1) \cdot 2^{m-n}]\), an interval whose endpoints are consecutive multiples of \(2^{m-n}\) and whose length \(2^{m-n}\) is at most \(1/2\); since \(2^{m-n}\) divides \(1/2\), this interval is contained in a single interval of the form \([j/2, (j+1)/2]\), on which \(\varphi\) is affine with slope \(+1\) or \(-1\). Hence \(x \mapsto 2^{-m} \varphi(2^m x)\) is affine on \(I_{n,k}\) with slope \(+1\) or \(-1\).

For \(m \geq n\), write \(u = 2^n x - k \in [0,1]\). Then \(2^m x = 2^{m-n} u + 2^{m-n} k\), and \(2^{m-n} k\) is an integer, so by the \(1\)-periodicity of \(\varphi\) we get \(\varphi(2^m x) = \varphi(2^{m-n} u)\). Therefore

\begin{equation*} \sum_{m \geq n} \frac{1}{2^m} \varphi(2^m x) = \frac{1}{2^{n}} \sum_{j=0}^{\infty} \frac{1}{2^{j}} \varphi(2^{j} u) = \frac{1}{2^n} \, T(2^n x - k). \end{equation*}

Combining the two computations, there are a real number \(c_{n,k}\) and an integer \(s_{n,k}\) with

\begin{equation*} T(x) = c_{n,k} + s_{n,k} \cdot \Bigl( x - \frac{k}{2^n} \Bigr) + \frac{1}{2^n} \, T(2^n x - k) \quad \text{for all } x \in I_{n,k}, \tag{3} \end{equation*}

where \(s_{n,k}\) is the sum of the \(n\) slopes \(\pm 1\) contributed by the terms with \(m < n\).

The slopes can be steered to zero. On the left half \(I_{n+1,2k}\) of \(I_{n,k}\) the point \(2^n x\) ranges over \([k, k + 1/2]\), where \(\varphi\) has slope \(+1\); on the right half \(I_{n+1,2k+1}\) it ranges over \([k + 1/2, k+1]\), where \(\varphi\) has slope \(-1\). Since \(s_{n+1, \cdot}\) is \(s_{n,k}\) plus the slope contributed by the term with \(m = n\),

\begin{equation*} s_{n+1, 2k} = s_{n,k} + 1, \qquad s_{n+1, 2k+1} = s_{n,k} - 1 . \end{equation*}

Consequently, starting from any dyadic interval \(I_{n,k}\), we may pass \(|s_{n,k}|\) times to a half-interval, always taking the right half if \(s_{n,k} > 0\) and the left half if \(s_{n,k} < 0\), and arrive at a dyadic interval \(J = I_{N,K} \subseteq I_{n,k}\) with

\begin{equation*} s_{N,K} = 0 . \end{equation*}

\(T\) is monotone on no open interval. Let \((\alpha, \beta)\) be a nonempty open interval. Since a function that is monotone on an interval is monotone on every subinterval, and since \((\alpha, \beta)\) contains a nonempty bounded open interval, we may assume \(\alpha\) and \(\beta\) are finite. Choose \(n\) so large that \(2^{-n} < (\beta - \alpha)/3\) and let \(k\) be the least integer with \(k/2^n > \alpha\). Then \(k/2^n \leq \alpha + 2^{-n}\), since otherwise \(k - 1\) would already satisfy \((k-1)/2^n > \alpha\); hence

\begin{equation*} \alpha < \frac{k}{2^n} \quad \text{and} \quad \frac{k+1}{2^n} \leq \alpha + \frac{2}{2^n} < \alpha + (\beta - \alpha) = \beta , \end{equation*}

so \(I_{n,k} = [k/2^n, (k+1)/2^n] \subseteq (\alpha, \beta)\). Fix this \(I_{n,k}\) and let \(J = I_{N,K} \subseteq I_{n,k} \subseteq (\alpha,\beta)\) be a dyadic subinterval with \(s_{N,K} = 0\), as just constructed. By (3), for \(x \in J\),

\begin{equation*} T(x) = c + \frac{1}{2^N} \, T(2^N x - K), \quad \text{where } c = c_{N,K}. \end{equation*}

The map \(x \mapsto 2^N x - K\) carries the left endpoint \(p = K/2^N\) of \(J\), its midpoint \(q = (K + 1/2)/2^N\), and its right endpoint \(r = (K+1)/2^N\) onto \(0\), \(1/2\), and \(1\) respectively. Now

\begin{equation*} T(0) = 0, \qquad T(1/2) = \varphi(1/2) + \sum_{m \geq 1} \frac{1}{2^m} \varphi(2^{m-1}) = \frac{1}{2}, \qquad T(1) = \sum_{m \geq 0} \frac{1}{2^m} \varphi(2^m) = 0, \end{equation*}

since \(\varphi\) vanishes at every integer and \(2^{m-1}\) is an integer for \(m \geq 1\). Therefore

\begin{equation*} T(p) = c, \qquad T(q) = c + \frac{1}{2^{N+1}}, \qquad T( r) = c, \end{equation*}

and \(p < q < r\) are three points of \((\alpha, \beta)\) with \(T(p) < T(q)\) and \(T(q) > T( r)\). The first inequality shows \(T\) is not decreasing on \((\alpha,\beta)\) and the second shows \(T\) is not increasing on \((\alpha,\beta)\). Hence \(T\) is not monotone on \((\alpha, \beta)\).

Since \((\alpha,\beta)\) was an arbitrary nonempty open interval, the continuous function \(T\) on \(\mathbb{R}\) is monotone on no open interval. So the answer to the question is no.

Write \(L = \ell(I)\) and let \(c\) be the midpoint of \(I\), so that

\begin{equation*} I = [c - L/2,\ c + L/2], \qquad 5 * I = [c - 5L/2,\ c + 5L/2]. \end{equation*}

Consider first the case \(\gamma \geq 1/2\). Since \(I \cap J \neq \emptyset\), choose a point \(y \in I \cap J\). Let \(x\) be an arbitrary point of \(J\). Because \(x\) and \(y\) both lie in the interval \(J\),

\begin{equation*} |x - y| \leq \ell(J). \end{equation*}

The hypothesis \(\ell(I) > \gamma \cdot \ell(J)\) gives \(\ell(J) < \ell(I)/\gamma = L/\gamma\), and since \(\gamma \geq 1/2\) we have \(1/\gamma \leq 2\), so

\begin{equation*} |x - y| \leq \ell(J) < \frac{L}{\gamma} \leq 2L. \end{equation*}

On the other hand \(y \in I\), so \(|y - c| \leq L/2\). By the triangle inequality,

\begin{equation*} |x - c| \leq |x - y| + |y - c| < 2L + \frac{L}{2} = \frac{5L}{2}. \end{equation*}

Hence \(x \in [c - 5L/2,\ c + 5L/2] = 5 * I\). As \(x \in J\) was arbitrary, \(J \subseteq 5 * I\).

Now consider the case \(0 < \gamma < 1/2\). The conclusion fails for every such \(\gamma\). Indeed, fix \(\gamma\) with \(0 < \gamma < 1/2\), so that \(1/\gamma > 2\). Choose a number \(t\) with

\begin{equation*} 2 < t < \frac{1}{\gamma}, \end{equation*}

which is possible precisely because \(1/\gamma > 2\). Put

\begin{equation*} I = [0, 1], \qquad J = [-t, 0]. \end{equation*}

Then \(\ell(I) = 1\), \(\ell(J) = t\), and \(\gamma \cdot \ell(J) = \gamma t < 1 = \ell(I)\), so the hypothesis \(\ell(I) > \gamma \cdot \ell(J)\) holds. Moreover \(I \cap J = \{0\} \neq \emptyset\). The midpoint of \(I\) is \(1/2\) and \(\ell(I) = 1\), so

\begin{equation*} 5 * I = \left[\tfrac{1}{2} - \tfrac{5}{2},\ \tfrac{1}{2} + \tfrac{5}{2}\right] = [-2, 3]. \end{equation*}

Since \(t > 2\), the endpoint \(-t\) of \(J\) satisfies \(-t < -2\), so \(-t \notin 5 * I\) and therefore \(J \not\subseteq 5 * I\).

Thus the implication holds for \(\gamma \geq 1/2\) and fails for every \(\gamma \in (0, 1/2)\). The computation in the first part shows exactly why: the bound obtained is \(|x - c| < L/\gamma + L/2\), and this is at most \(5L/2\) precisely when \(1/\gamma \leq 2\); the second part shows that this restriction on \(\gamma\) cannot be relaxed.

This is exactly the geometric fact that justifies the dilation factor \(5\) in the proof of the Vitali Covering Lemma. There, in the notation of that proof, the selected intervals satisfy relation (6): \(\ell(I_{n+1}) > \ell(I)/2\) for every \(I\) in \(\mathcal{F}\) that is disjoint from \(I_1, \ldots, I_n\). That is the hypothesis of the present problem with \(\gamma = 1/2\), and it is what lets one conclude that such an \(I\) meeting \(I_N\) is contained in \(5 * I_N\).

Recall that a set has measure zero exactly when its outer measure is zero: by Proposition 4 of Chapter 2 any set of outer measure zero is measurable, and its measure is then its outer measure, namely zero.

Suppose first that such a collection \(\{I_k\}_{k=1}^{\infty}\) exists: every point of \(E\) lies in infinitely many \(I_k\) and \(\sum_{k=1}^{\infty}\ell(I_k) < \infty\). Fix a natural number \(n\) and set

\begin{equation*} E_n = \bigcup_{k = n}^{\infty} I_k . \end{equation*}

If \(x \in E\), then \(x\) belongs to \(I_k\) for infinitely many indices \(k\), and in particular for some index \(k \geq n\). Hence \(E \subseteq E_n\) for every \(n\). By the monotonicity and countable subadditivity of outer measure, together with Proposition 1 of Chapter 2, which says that the outer measure of an interval is its length,

\begin{equation*} m^*(E) \leq m^*(E_n) \leq \sum_{k=n}^{\infty} \ell(I_k). \end{equation*}

Since the series \(\sum_{k=1}^{\infty} \ell(I_k)\) converges, its tails tend to \(0\): given \(\epsilon > 0\) there is \(n\) with \(\sum_{k=n}^{\infty}\ell(I_k) < \epsilon\). Therefore \(m^*(E) < \epsilon\) for every \(\epsilon > 0\), so \(m^*(E) = 0\) and \(E\) has measure zero.

Conversely, suppose \(E\) has measure zero, so \(m^*(E) = 0\). Fix a natural number \(n\). By the definition of outer measure as an infimum, there is a countable collection \(\{J_{n,k}\}_{k}\) of open, bounded intervals with

\begin{equation*} E \subseteq \bigcup_k J_{n,k} \quad \text{and} \quad \sum_k \ell(J_{n,k}) < \frac{1}{2^{n}} . \end{equation*}

Adjoining to this collection the single extra interval \((0, 2^{-n})\) if necessary, we may assume the \(n\)-th collection is nonempty, at the cost of replacing the bound \(2^{-n}\) by \(2^{-n+1}\); this changes nothing below. Let

\begin{equation*} \mathcal{C} = \{ J_{n,k} : n \in \mathbb{N},\ k \text{ an index of the } n\text{-th collection} \}, \end{equation*}

regarded as a family indexed by the set of admissible pairs \((n,k)\). This index set is a countable union of countable sets, hence countable, and it is infinite because each of the infinitely many blocks \(n\) contributes at least one index. Enumerate the family as \(\{I_j\}_{j=1}^{\infty}\) by choosing a bijection of \(\mathbb{N}\) onto the index set. Since all terms are nonnegative, the sum of the resulting series is independent of the order of summation, and

\begin{equation*} \sum_{j=1}^{\infty} \ell(I_j) = \sum_{n=1}^{\infty} \sum_k \ell(J_{n,k}) \leq \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = 2 < \infty . \end{equation*}

Finally, let \(x \in E\). For each \(n\), the \(n\)-th collection covers \(E\), so there is an index \(k\) with \(x \in J_{n,k}\). Distinct values of \(n\) give indices \((n,k)\) in disjoint blocks of the index set, hence distinct members of the enumerated family. Therefore \(x\) belongs to \(I_j\) for infinitely many \(j\). This is the required collection.

For each natural number \(k\) define \(\varphi_k\) on \((a,b)\) by

\begin{equation*} \varphi_k(x) = \ell\big((c_k, d_k) \cap (-\infty, x)\big) = \begin{cases} 0 & \text{if } x \leq c_k, \\ x - c_k & \text{if } c_k \leq x \leq d_k, \\ d_k - c_k & \text{if } x \geq d_k, \end{cases} \end{equation*}

so that \(f = \sum_{k=1}^{\infty} \varphi_k\) pointwise on \((a,b)\). The three formulas agree at the overlaps \(x = c_k\) and \(x = d_k\), so \(\varphi_k\) is well defined, and inspection of the three pieces shows that \(\varphi_k\) is increasing and satisfies

\begin{equation*} 0 \leq \varphi_k(x) \leq d_k - c_k \quad \text{for all } x \in (a,b). \tag{\(\ast\)} \end{equation*}

First, \(f\) is well defined and finite. Indeed, by \((\ast)\) the terms of the series defining \(f(x)\) are nonnegative and dominated by the terms of the convergent series \(\sum_{k=1}^{\infty}(d_k - c_k)\), so for each \(x\) the series converges and

\begin{equation*} 0 \leq f(x) \leq \sum_{k=1}^{\infty} (d_k - c_k) < \infty . \end{equation*}

Next, \(f\) is increasing. Let \(a < x < y < b\). Each \(\varphi_k\) is increasing, so \(\varphi_k(y) - \varphi_k(x) \geq 0\) for every \(k\). Both series \(\sum_k \varphi_k(x)\) and \(\sum_k \varphi_k(y)\) converge, so we may subtract them termwise:

\begin{equation*} f(y) - f(x) = \sum_{k=1}^{\infty} \big[\varphi_k(y) - \varphi_k(x)\big] \geq 0 . \tag{\(\ast\ast\)} \end{equation*}

Hence \(f(x) \leq f(y)\), and \(f\) is increasing on \((a,b)\).

Finally, \(f\) fails to be differentiable at each point of \(E\). We show more: for each \(x_0 \in E\) the upper derivative \(\overline{D}f(x_0)\) equals \(\infty\). Since the book’s definition requires, for differentiability at \(x_0\), that \(\overline{D}f(x_0)\) and \(\underline{D}f(x_0)\) be equal and finite, this rules out differentiability at \(x_0\).

Fix \(x_0 \in E\) and fix a natural number \(n\). Since \(x_0\) belongs to infinitely many of the intervals of the collection, we may choose \(n\) distinct indices \(k_1, k_2, \ldots, k_n\) for which

\begin{equation*} x_0 \in (c_{k_i}, d_{k_i}) \quad \text{for } 1 \leq i \leq n . \end{equation*}

Each of these intervals is open and contains \(x_0\), so \(d_{k_i} - x_0 > 0\) for every \(i\). Since \(x_0 < b\), the number

\begin{equation*} \delta = \min\Big\{ b - x_0,\ \min_{1 \leq i \leq n} (d_{k_i} - x_0) \Big\} \end{equation*}

is positive. Let \(0 < t < \delta\) and put \(y = x_0 + t\); then \(y \in (a,b)\). For each \(i\) we have

\begin{equation*} c_{k_i} < x_0 < y = x_0 + t < d_{k_i}, \end{equation*}

so both \(x_0\) and \(y\) lie in \([c_{k_i}, d_{k_i}]\), and the middle formula for \(\varphi_{k_i}\) applies to each:

\begin{equation*} \varphi_{k_i}(y) - \varphi_{k_i}(x_0) = (y - c_{k_i}) - (x_0 - c_{k_i}) = y - x_0 = t . \end{equation*}

All the terms in the series \((\ast\ast)\) are nonnegative, so discarding all but the \(n\) chosen indices gives

\begin{equation*} f(x_0 + t) - f(x_0) = \sum_{k=1}^{\infty}\big[\varphi_k(y) - \varphi_k(x_0)\big] \geq \sum_{i=1}^{n} \big[\varphi_{k_i}(y) - \varphi_{k_i}(x_0)\big] = n\,t . \end{equation*}

Dividing by \(t > 0\),

\begin{equation*} \frac{f(x_0 + t) - f(x_0)}{t} \geq n \quad \text{for all } 0 < t < \delta . \end{equation*}

Consequently, for every \(h\) with \(0 < h < \delta\),

\begin{equation*} \sup_{0 < |t| \leq h} \frac{f(x_0 + t) - f(x_0)}{t} \geq n , \end{equation*}

and letting \(h \to 0\) in the definition of the upper derivative yields \(\overline{D}f(x_0) \geq n\). As \(n\) was an arbitrary natural number, \(\overline{D}f(x_0) = \infty\). Therefore \(f\) is not differentiable at \(x_0\).

Since \(x_0 \in E\) was arbitrary, \(f\) is an increasing function on \((a,b)\) that fails to be differentiable at every point of \(E\). This is exactly the assertion, made after the proof of Lebesgue’s Theorem, that the theorem is best possible: every set of measure zero in \((a,b)\) is contained in the set of nondifferentiability of some increasing function on \((a,b)\). Of course Lebesgue’s Theorem guarantees that this set of nondifferentiability itself has measure zero.

Throughout write \(I = [\alpha, \beta]\) and \(I_{\gamma} = [\alpha + \gamma, \beta + \gamma] = I + \gamma\), and for a function \(g\) on \(I_{\gamma}\) define its translate

\begin{equation*} (T g)(t) = g(t + \gamma) \quad \text{for } t \in I . \end{equation*}

The whole argument rests on the translation invariance of Lebesgue measure established in Chapter 2: outer measure is translation invariant, and by Proposition 10 of Chapter 2 the translate of a measurable set is measurable. Applying that proposition to \(A\) and to \(A + \gamma\) with the translations \(\gamma\) and \(-\gamma\), we get that \(A\) is measurable if and only if \(A + \gamma\) is measurable, and then \(m(A + \gamma) = m^*(A + \gamma) = m^*(A) = m(A)\).

Measurability of the translate. If \(g\) is measurable on \(I_{\gamma}\) then \(Tg\) is measurable on \(I\). Indeed, for \(c \in \mathbb{R}\),

\begin{equation*} \{ t \in I : (Tg)(t) > c \} = \{ s \in I_{\gamma} : g(s) > c \} - \gamma , \end{equation*}

which is a translate of a measurable set and hence measurable. The map \(T\) is a bijection from functions on \(I_{\gamma}\) onto functions on \(I\), with inverse given by translation by \(-\gamma\); by the same reasoning the inverse also preserves measurability. Note also that \(T\) is linear, order preserving, carries simple functions to simple functions and bounded functions to bounded functions, and satisfies \(T(g^{+}) = (Tg)^{+}\) and \(T(g^{-}) = (Tg)^{-}\).

Step 1: simple functions. Let \(A \subseteq I_{\gamma}\) be measurable. Then for \(t \in I\) we have \(\chi_A(t + \gamma) = 1\) exactly when \(t \in A - \gamma\), so \(T \chi_A = \chi_{A - \gamma}\) and therefore

\begin{equation*} \int_{\alpha}^{\beta} \chi_A(t + \gamma)\, dt = m(A - \gamma) = m(A) = \int_{\alpha + \gamma}^{\beta + \gamma} \chi_A . \end{equation*}

Now let \(\psi\) be simple and measurable on \(I_{\gamma}\), with canonical representation \(\psi = \sum_{i=1}^{n} a_i \chi_{A_i}\), the \(A_i\) measurable and disjoint. Then \(T\psi = \sum_{i=1}^{n} a_i \chi_{A_i - \gamma}\), and by the linearity of integration for simple functions,

\begin{equation*} \int_{\alpha}^{\beta} T\psi = \sum_{i=1}^{n} a_i\, m(A_i - \gamma) = \sum_{i=1}^{n} a_i\, m(A_i) = \int_{\alpha + \gamma}^{\beta + \gamma} \psi . \end{equation*}

Step 2: bounded measurable functions. Let \(g\) be bounded and measurable on \(I_{\gamma}\), a set of finite measure. Then \(Tg\) is bounded and measurable on \(I\), also of finite measure, so by Theorem 4 of Chapter 4 both functions are integrable, and each integral equals the lower Lebesgue integral, that is, the supremum of the integrals of the simple functions lying below the function:

\begin{equation*} \int_{\alpha + \gamma}^{\beta + \gamma} g = \sup \left\{ \int_{\alpha + \gamma}^{\beta + \gamma} \psi \ :\ \psi \text{ simple and measurable on } I_{\gamma},\ \psi \leq g \right\}, \end{equation*}

and similarly for \(Tg\) over \(I\). Since \(T\) is an order preserving bijection carrying the simple measurable functions on \(I_{\gamma}\) onto those on \(I\), it carries \(\{ \psi \text{ simple} : \psi \leq g \}\) onto \(\{ \varphi \text{ simple} : \varphi \leq Tg \}\), and by Step 1 it preserves the value of the integral. Two sets of real numbers that are in integral preserving bijection have the same supremum, so

\begin{equation*} \int_{\alpha}^{\beta} g(t + \gamma)\, dt = \int_{\alpha + \gamma}^{\beta + \gamma} g . \end{equation*}

Step 3: nonnegative measurable functions. Let \(g \geq 0\) be measurable on \(I_{\gamma}\). By the definition, in Chapter 4, of the integral of a nonnegative measurable function,

\begin{equation*} \int_{\alpha + \gamma}^{\beta + \gamma} g = \sup \left\{ \int_{\alpha + \gamma}^{\beta + \gamma} h \ :\ h \text{ bounded, measurable, of finite support},\ 0 \leq h \leq g \right\}, \end{equation*}

and likewise for \(Tg \geq 0\) over \(I\). If \(h\) is bounded, measurable, of finite support and \(0 \leq h \leq g\) on \(I_{\gamma}\), then \(Th\) is bounded, measurable, of finite support (its support is the translate by \(-\gamma\) of the support of \(h\), and translation preserves measure) and \(0 \leq Th \leq Tg\) on \(I\); and \(T\) maps this class onto the corresponding class for \(Tg\), since its inverse has the same properties. By Step 2 it preserves integrals. Hence the two suprema coincide,

\begin{equation*} \int_{\alpha}^{\beta} g(t + \gamma)\, dt = \int_{\alpha + \gamma}^{\beta + \gamma} g , \end{equation*}

both sides being finite or infinite together. In particular \(g\) is integrable over \(I_{\gamma}\) if and only if \(Tg\) is integrable over \(I\).

Step 4: general integrable functions. Let \(g\) be integrable over \(I_{\gamma}\). By Proposition 14 of Chapter 4, \(g^{+}\) and \(g^{-}\) are nonnegative measurable functions each with finite integral over \(I_{\gamma}\), and \((Tg)^{+} = T(g^{+})\), \((Tg)^{-} = T(g^{-})\). By Step 3,

\begin{equation*} \int_{\alpha}^{\beta} (Tg)^{\pm} = \int_{\alpha + \gamma}^{\beta + \gamma} g^{\pm} < \infty , \end{equation*}

so \(Tg\) is integrable over \(I\), and by the definition of the integral of an integrable function,

\begin{equation*} \int_{\alpha}^{\beta} g(t + \gamma)\, dt = \int_{\alpha}^{\beta} (Tg)^{+} - \int_{\alpha}^{\beta} (Tg)^{-} = \int_{\alpha + \gamma}^{\beta + \gamma} g^{+} - \int_{\alpha + \gamma}^{\beta + \gamma} g^{-} = \int_{\alpha + \gamma}^{\beta + \gamma} g . \end{equation*}

This is the change of variables formula. Nothing in the argument used the sign of \(\gamma\): the same proof, verbatim, gives the formula for any real \(\gamma\).

Proof of (14). Let \(f\) be integrable over \([a, b]\), extended to take the value \(f(b)\) on \((b, b + 1]\); the extension is integrable over \([a, b + 1]\), since it is integrable over \([a,b]\) and constant, hence integrable, over \([b, b + 1]\). Fix \(h\) with \(0 < h \leq 1\) and let \(a \leq u \leq v \leq b\). The function \(t \mapsto f(t + h)\) is integrable over \([u, v]\) by Step 4 applied with \(\gamma = h\) on \([u + h, v + h] \subseteq [a, b + 1]\), so \(\mathrm{Diff}_h f = \big(f(\cdot + h) - f\big)/h\) is integrable over \([u, v]\), and by Theorem 17 of Chapter 4 (linearity of integration),

\begin{equation*} \int_u^v \mathrm{Diff}_h f = \frac{1}{h}\left[ \int_u^v f(t + h)\, dt - \int_u^v f \right] = \frac{1}{h}\left[ \int_{u + h}^{v + h} f - \int_u^v f \right], \end{equation*}

the last equality by the change of variables formula. Now use Corollary 18 of Chapter 4 (Additivity Over Domains of Integration) twice, splitting \([u, v + h]\) at \(v\) and at \(u + h\); both splittings are legitimate since \(u \leq v \leq v + h\) and \(u \leq u + h \leq v + h\), and \(f\) is integrable over \([u, v+h] \subseteq [a, b+1]\):

\begin{equation*} \int_u^v f + \int_v^{v + h} f \; = \; \int_u^{v + h} f \; = \; \int_u^{u + h} f + \int_{u + h}^{v + h} f . \end{equation*}

Rearranging,

\begin{equation*} \int_{u + h}^{v + h} f - \int_u^v f = \int_v^{v + h} f - \int_u^{u + h} f . \end{equation*}

Therefore

\begin{equation*} \int_u^v \mathrm{Diff}_h f = \frac{1}{h} \int_v^{v + h} f - \frac{1}{h} \int_u^{u + h} f = \mathrm{Av}_h f(v) - \mathrm{Av}_h f(u), \end{equation*}

which is (14).

Let \(f = \chi_{\mathbb{Q}}\) on \(\mathbb{R}\), so \(f(x) = 1\) if \(x\) is rational and \(f(x) = 0\) if \(x\) is irrational. We claim that

\begin{equation*} \overline{D}f(x) = \infty \quad \text{and} \quad \underline{D}f(x) = -\infty \quad \text{for every } x \in \mathbb{R} . \end{equation*}

Recall the definitions

\begin{equation*} \overline{D}f(x) = \lim_{h \to 0}\left[\sup_{0 < |t| \leq h} \frac{f(x + t) - f(x)}{t}\right], \qquad \underline{D}f(x) = \lim_{h \to 0}\left[\inf_{0 < |t| \leq h} \frac{f(x + t) - f(x)}{t}\right]. \end{equation*}

As \(h\) decreases the supremum is taken over a smaller set, so the expression in brackets is monotone in \(h\) and the limits exist in the extended reals. Consequently, to prove \(\overline{D}f(x) = \infty\) it suffices to show that for every \(h > 0\) and every \(M > 0\) there is a \(t\) with \(0 < |t| \leq h\) and difference quotient exceeding \(M\); and symmetrically for the lower derivative.

Fix \(x\) and \(h > 0\) and \(M > 0\), and put \(\delta = \min\{h, 1/M\} > 0\).

Case 1: \(x\) is rational. Then \(f(x) = 1\) and, for \(t \neq 0\),

\begin{equation*} \frac{f(x + t) - f(x)}{t} = \frac{f(x+t) - 1}{t} = \begin{cases} 0 & \text{if } x + t \in \mathbb{Q}, \\[2pt] -\dfrac{1}{t} & \text{if } x + t \notin \mathbb{Q}. \end{cases} \end{equation*}

Since \(x\) is rational, \(x + t\) is irrational exactly when \(t\) is irrational. The irrationals are dense in \(\mathbb{R}\), so we may choose an irrational \(t\) with \(-\delta < t < 0\). For that \(t\) we have \(0 < |t| \leq h\) and

\begin{equation*} \frac{f(x + t) - f(x)}{t} = -\frac{1}{t} = \frac{1}{|t|} > \frac{1}{\delta} \geq M . \end{equation*}

Hence \(\sup_{0 < |t| \leq h}\) exceeds \(M\) for every \(h\) and every \(M\), so \(\overline{D}f(x) = \infty\). Likewise choose an irrational \(t\) with \(0 < t < \delta\); then

\begin{equation*} \frac{f(x + t) - f(x)}{t} = -\frac{1}{t} < -\frac{1}{\delta} \leq -M , \end{equation*}

so the infimum is less than \(-M\) for every \(h\) and every \(M\), and \(\underline{D}f(x) = -\infty\).

Case 2: \(x\) is irrational. Then \(f(x) = 0\) and, for \(t \neq 0\),

\begin{equation*} \frac{f(x + t) - f(x)}{t} = \frac{f(x+t)}{t} = \begin{cases} \dfrac{1}{t} & \text{if } x + t \in \mathbb{Q}, \\[2pt] 0 & \text{if } x + t \notin \mathbb{Q}. \end{cases} \end{equation*}

By the density of the rationals there is a rational number \(q\) with \(x < q < x + \delta\). Put \(t = q - x\), so \(0 < t < \delta \leq h\) and \(x + t = q\) is rational; then

\begin{equation*} \frac{f(x + t) - f(x)}{t} = \frac{1}{t} > \frac{1}{\delta} \geq M . \end{equation*}

Hence \(\overline{D}f(x) = \infty\). Similarly there is a rational \(q\) with \(x - \delta < q < x\); putting \(t = q - x\) gives \(-\delta < t < 0\), \(0 < |t| \le h\), and

\begin{equation*} \frac{f(x + t) - f(x)}{t} = \frac{1}{t} < -\frac{1}{\delta} \leq -M , \end{equation*}

so \(\underline{D}f(x) = -\infty\).

Thus at every real number \(x\),

\begin{equation*} \overline{D}f(x) = +\infty, \qquad \underline{D}f(x) = -\infty . \end{equation*}

In particular \(\chi_{\mathbb{Q}}\) fails to be differentiable at every point of \(\mathbb{R}\), and indeed the difference quotients fail to have even an infinite limit anywhere. There is no contradiction with Lebesgue’s Theorem: \(\chi_{\mathbb{Q}}\) is not monotone on any open interval, since every open interval contains both rationals and irrationals.

We construct the collection inductively, applying the Vitali Covering Lemma at each stage to what is left over, and we use the following observation to keep the newly chosen intervals disjoint from the ones already chosen.

Claim. Let \(F\) be a closed set and let \(A \subseteq E \sim F\). Then the subcollection

\begin{equation*} \mathcal{F}_F = \{ I \in \mathcal{F} : I \cap F = \emptyset \} \end{equation*}

is a Vitali covering of \(A\).

To see this, let \(x \in A\) and \(\epsilon > 0\). Since \(x \notin F\) and \(F\) is closed, its complement is open, so there is \(\delta > 0\) with \((x - \delta, x + \delta) \cap F = \emptyset\). Since \(x \in E\) and \(\mathcal{F}\) covers \(E\) in the sense of Vitali, there is an interval \(I \in \mathcal{F}\) with \(x \in I\) and \(\ell(I) < \min\{\epsilon, \delta\}\). Every point \(y \in I\) satisfies \(|y - x| \leq \ell(I) < \delta\), so \(I \subseteq (x - \delta, x + \delta)\) and hence \(I \cap F = \emptyset\). Thus \(I \in \mathcal{F}_F\), \(x \in I\) and \(\ell(I) < \epsilon\). This proves the claim.

Construction. We produce, for each natural number \(j\), a finite disjoint subcollection \(\{I_k\}_{k=1}^{n_j}\) of \(\mathcal{F}\), with \(n_1 < n_2 < \cdots\) and each stage extending the previous one, such that

\begin{equation*} m^*\left[E \sim \bigcup_{k=1}^{n_j} I_k\right] < \frac{1}{j} . \tag{\(\dagger\)} \end{equation*}

For \(j = 1\): \(E\) has finite outer measure and \(\mathcal{F}\) is a Vitali covering of \(E\), so the Vitali Covering Lemma with \(\epsilon = 1\) provides a finite disjoint subcollection \(\{I_k\}_{k=1}^{n_1}\) of \(\mathcal{F}\) with \(m^*[E \sim \bigcup_{k=1}^{n_1} I_k] < 1\).

Inductive step: suppose \(\{I_k\}_{k=1}^{n_j}\) has been chosen, disjoint and satisfying \((\dagger)\). Put

\begin{equation*} F_j = \bigcup_{k=1}^{n_j} I_k, \qquad E_j = E \sim F_j . \end{equation*}

As a finite union of closed intervals, \(F_j\) is closed. By \((\dagger)\), \(m^*(E_j) < 1/j < \infty\). If \(E_j = \emptyset\), then \(m^*[E \sim \bigcup_{k=1}^{n_j} I_k] = 0\) and the finite disjoint collection \(\{I_k\}_{k=1}^{n_j}\), which is countable, already satisfies the conclusion; so we may assume \(E_j \neq \emptyset\). By the Claim applied with \(F = F_j\) and \(A = E_j\), the collection \(\mathcal{F}_{F_j}\) is a Vitali covering of \(E_j\). Apply the Vitali Covering Lemma to the set \(E_j\) of finite outer measure, the covering \(\mathcal{F}_{F_j}\), and \(\epsilon = 1/(j+1)\): there is a finite disjoint subcollection of \(\mathcal{F}_{F_j}\), which we label \(I_{n_j + 1}, \ldots, I_{n_{j+1}}\), for which

\begin{equation*} m^*\left[E_j \sim \bigcup_{k = n_j + 1}^{n_{j+1}} I_k\right] < \frac{1}{j+1} . \end{equation*}

(If the lemma returns the empty subcollection we may adjoin one further interval of \(\mathcal{F}_{F_j}\) containing a point of \(E_j\), which only shrinks the left side; so we may take \(n_{j+1} > n_j\).)

The intervals \(I_{n_j + 1}, \ldots, I_{n_{j+1}}\) are pairwise disjoint, and each belongs to \(\mathcal{F}_{F_j}\), hence is disjoint from \(F_j\) and therefore from each of \(I_1, \ldots, I_{n_j}\). So \(\{I_k\}_{k=1}^{n_{j+1}}\) is a disjoint collection. Moreover, since \(E_j = E \sim F_j\),

\begin{equation*} E \sim \bigcup_{k=1}^{n_{j+1}} I_k = \left(E \sim F_j\right) \sim \bigcup_{k = n_j + 1}^{n_{j+1}} I_k = E_j \sim \bigcup_{k = n_j + 1}^{n_{j+1}} I_k , \end{equation*}

so \(m^*[E \sim \bigcup_{k=1}^{n_{j+1}} I_k] < 1/(j+1)\), which is \((\dagger)\) for \(j + 1\). The induction is complete.

Conclusion. If the process terminates at some stage because \(E_j = \emptyset\), we are done as noted. Otherwise the stages nest, and taking the union over all \(j\) we obtain a countable collection \(\{I_k\}_{k=1}^{\infty} \subseteq \mathcal{F}\). It is disjoint: any two of its members lie in some common finite stage \(\{I_k\}_{k=1}^{n_j}\), which is disjoint. Finally, for each natural number \(j\),

\begin{equation*} E \sim \bigcup_{k=1}^{\infty} I_k \subseteq E \sim \bigcup_{k=1}^{n_j} I_k , \end{equation*}

so by the monotonicity of outer measure and \((\dagger)\),

\begin{equation*} m^*\left[E \sim \bigcup_{k=1}^{\infty} I_k\right] \leq m^*\left[E \sim \bigcup_{k=1}^{n_j} I_k\right] < \frac{1}{j} . \end{equation*}

Since this holds for every \(j\), \(m^*[E \sim \bigcup_{k=1}^{\infty} I_k] = 0\), as required.

Let \(\mathcal{C}\) be a collection of closed, bounded, nondegenerate intervals and let

\begin{equation*} U = \bigcup_{I \in \mathcal{C}} I . \end{equation*}

Since \(U = \bigcup_{n=1}^{\infty} \big(U \cap (-n, n)\big)\) and the measurable sets form a \(\sigma\)-algebra, it suffices to prove that

\begin{equation*} A_n = U \cap (-n, n) \end{equation*}

is measurable for each natural number \(n\). Fix \(n\); we may assume \(A_n \neq \emptyset\), since the empty set is measurable. Note \(m^*(A_n) \leq 2n < \infty\).

A Vitali covering. The key point is that although the intervals of \(\mathcal{C}\) may all be long, every point of \(U\) lies in arbitrarily short closed subintervals of members of \(\mathcal{C}\). Define

\begin{equation*} \mathcal{F}_n = \{ J : J \text{ is a closed, bounded, nondegenerate interval},\ J \subseteq [-n,n],\ \text{and } J \subseteq I \text{ for some } I \in \mathcal{C} \} . \end{equation*}

We check that \(\mathcal{F}_n\) covers \(A_n\) in the sense of Vitali. Let \(x \in A_n\) and \(\epsilon > 0\). Since \(x \in U\), there is an interval \(I = [c, d] \in \mathcal{C}\) with \(c \leq x \leq d\) and \(c < d\). Since \(x \in (-n, n)\), there is \(\delta > 0\) with \([x - \delta, x + \delta] \subseteq (-n, n)\).

If \(x < d\), put \(\eta = \min\{\epsilon/2,\ \delta,\ d - x\} > 0\) and \(J = [x, x + \eta]\). Then \(c \leq x\) and \(x + \eta \leq d\), so \(J \subseteq I\); also \(J \subseteq [x - \delta, x + \delta] \subseteq [-n, n]\); and \(J\) is nondegenerate with \(x \in J\) and \(\ell(J) = \eta < \epsilon\).

If \(x = d\), then \(c < x\); put \(\eta = \min\{\epsilon/2,\ \delta,\ x - c\} > 0\) and \(J = [x - \eta, x]\). The same three verifications apply.

In either case \(J \in \mathcal{F}_n\), \(x \in J\) and \(\ell(J) < \epsilon\). So \(\mathcal{F}_n\) is a Vitali covering of \(A_n\).

Applying the lemma. The set \(A_n\) has finite outer measure and \(\mathcal{F}_n\) covers it in the sense of Vitali, so for each natural number \(j\) the Vitali Covering Lemma provides a finite disjoint subcollection \(\{J^{\,j}_1, \ldots, J^{\,j}_{m_j}\}\) of \(\mathcal{F}_n\) for which

\begin{equation*} m^*\left[A_n \sim \bigcup_{k=1}^{m_j} J^{\,j}_k \right] < \frac{1}{j} . \end{equation*}

Let

\begin{equation*} S = \bigcup_{j=1}^{\infty} \bigcup_{k=1}^{m_j} J^{\,j}_k . \end{equation*}

This is a countable union of closed intervals, hence a Borel set and in particular measurable. Moreover every \(J^{\,j}_k\) is contained in some member of \(\mathcal{C}\) and in \([-n,n]\), so

\begin{equation*} S \subseteq U \cap [-n, n] . \end{equation*}

Now decompose \(A_n\). First,

\begin{equation*} A_n \cap S = S \cap (-n, n) . \end{equation*}

Indeed, \(A_n \cap S \subseteq S \cap (-n,n)\) because \(A_n \subseteq (-n,n)\); conversely, if \(y \in S \cap (-n,n)\) then \(y \in U\) and \(y \in (-n,n)\), so \(y \in A_n\) and hence \(y \in A_n \cap S\). The set \(S \cap (-n,n)\) is measurable, being the intersection of two measurable sets.

Second, for every \(j\),

\begin{equation*} A_n \sim S \subseteq A_n \sim \bigcup_{k=1}^{m_j} J^{\,j}_k , \end{equation*}

so by the monotonicity of outer measure \(m^*(A_n \sim S) < 1/j\) for every \(j\), whence \(m^*(A_n \sim S) = 0\). By Proposition 4 of Chapter 2, a set of outer measure zero is measurable, so \(A_n \sim S\) is measurable.

Therefore

\begin{equation*} A_n = (A_n \cap S) \cup (A_n \sim S) = \big[S \cap (-n,n)\big] \cup (A_n \sim S) \end{equation*}

is the union of two measurable sets and hence measurable. Since this holds for every \(n\),

\begin{equation*} U = \bigcup_{n=1}^{\infty} A_n \end{equation*}

is measurable, being a countable union of measurable sets.

Observe where nondegeneracy was used: it is what guarantees that each \(x \in U\) has, on at least one side, room inside a member of \(\mathcal{C}\) for arbitrarily short closed nondegenerate subintervals, and hence that \(\mathcal{F}_n\) is a genuine Vitali covering. For degenerate intervals the statement is false: if \(E\) is any nonmeasurable set, then \(E = \bigcup_{x \in E} [x, x]\) is a union of degenerate closed intervals.

Since \(f(0) = 0\), for \(t \neq 0\) the difference quotient at \(0\) is

\begin{equation*} \frac{f(0+t) - f(0)}{t} = \frac{t\,\sin(1/t)}{t} = \sin(1/t). \end{equation*}

So, with the definitions of Section 6.2,

\begin{equation*} \overline{D}f(0) = \lim_{h \to 0}\left[\sup_{0 < |t| \le h} \sin(1/t)\right], \qquad \underline{D}f(0) = \lim_{h \to 0}\left[\inf_{0 < |t| \le h} \sin(1/t)\right]. \end{equation*}

Fix \(h > 0\). Choose a natural number \(k\) so large that

\begin{equation*} t_k = \frac{1}{\pi/2 + 2k\pi} \le h . \end{equation*}

Then \(0 < t_k \le h\) and \(\sin(1/t_k) = \sin(\pi/2 + 2k\pi) = 1\). Since \(|\sin| \le 1\) always,

\begin{equation*} \sup_{0 < |t| \le h} \sin(1/t) = 1 \quad \text{for every } h > 0 , \end{equation*}

and therefore \(\overline{D}f(0) = 1\).

Similarly, for \(h > 0\) choose \(k\) so large that

\begin{equation*} s_k = \frac{1}{-\pi/2 + 2k\pi} \le h . \end{equation*}

Then \(0 < s_k \le h\) and \(\sin(1/s_k) = \sin(2k\pi - \pi/2) = -1\), so

\begin{equation*} \inf_{0 < |t| \le h} \sin(1/t) = -1 \quad \text{for every } h > 0 , \end{equation*}

and therefore \(\underline{D}f(0) = -1\).

Thus

\begin{equation*} \underline{D}f(0) = -1 < 1 = \overline{D}f(0), \end{equation*}

and in particular \(f\) fails to be differentiable at \(x = 0\), although both derivates are finite there. (Observe also that \(f\) is continuous at \(0\), since \(|f(x)| \le |x|\) for all \(x\); continuity at a point is of course no guarantee of differentiability there.)

Write \(g = g^{+} - g^{-}\), where \(g^{+} = \max\{g, 0\}\) and \(g^{-} = \max\{-g, 0\}\) are nonnegative measurable functions. Since \(|g^{\pm}| \le |g|\) and \(g\) is integrable over \([a,b]\), both \(g^{+}\) and \(g^{-}\) are integrable over \([a, b]\), and hence over every measurable subset of \([a,b]\). Define, for \(x \in [a,b]\),

\begin{equation*} f_1(x) = \int_a^x g^{+}, \qquad f_2(x) = \int_a^x g^{-} . \end{equation*}

By the linearity of integration, \(f = f_1 - f_2\) on \([a,b]\).

Each \(f_i\) is increasing on \([a, b]\). Indeed, let \(a \le x < y \le b\). By the additivity of integration over domains (Chapter 4),

\begin{equation*} f_1(y) - f_1(x) = \int_a^y g^{+} - \int_a^x g^{+} = \int_x^y g^{+} \ge 0 , \end{equation*}

because \(g^{+} \ge 0\) and the integral of a nonnegative integrable function is nonnegative. Hence \(f_1(x) \le f_1(y)\), and the same computation with \(g^{-}\) in place of \(g^{+}\) shows \(f_2\) is increasing.

Now apply Lebesgue’s Theorem to each of the monotone functions \(f_1\) and \(f_2\) on the open interval \((a,b)\): there are sets \(E_1, E_2 \subseteq (a,b)\) of measure zero such that \(f_1\) is differentiable at every point of \((a,b) \sim E_1\) and \(f_2\) is differentiable at every point of \((a,b) \sim E_2\). The set \(E = E_1 \cup E_2\) has measure zero, by the countable subadditivity of measure.

Let \(x \in (a,b) \sim E\). Then both limits

\begin{equation*} f_1^{\prime}(x) = \lim_{t \to 0} \frac{f_1(x+t) - f_1(x)}{t}, \qquad f_2^{\prime}(x) = \lim_{t \to 0} \frac{f_2(x+t) - f_2(x)}{t} \end{equation*}

exist and are finite, and therefore, since

\begin{equation*} \frac{f(x+t) - f(x)}{t} = \frac{f_1(x+t) - f_1(x)}{t} - \frac{f_2(x+t) - f_2(x)}{t} \end{equation*}

for all \(t \neq 0\) with \(x + t \in [a,b]\), the limit as \(t \to 0\) of the left-hand side exists and equals \(f_1^{\prime}(x) - f_2^{\prime}(x)\). Thus \(f\) is differentiable at \(x\), with

\begin{equation*} f^{\prime}(x) = f_1^{\prime}(x) - f_2^{\prime}(x) . \end{equation*}

Since \(E\) has measure zero, \(f\) is differentiable at almost every point of \((a, b)\).

Recall that (15) is the conclusion of Corollary 4: for an increasing function on a closed, bounded interval \([a,b]\), the derivative \(f^{\prime}\) is integrable over \([a, b]\) and

\begin{equation*} \int_a^b f^{\prime} \le f(b) - f(a). \end{equation*}

For \(f\) increasing and bounded on the open interval \((a,b)\) the endpoint values are to be read as the one-sided limits, and the assertion to be verified is: \(f^{\prime}\), which by Lebesgue’s Theorem is defined almost everywhere on \((a,b)\), is integrable over \((a,b)\) and

\begin{equation*} \int_a^b f^{\prime} \le \lim_{x \to b^{-}} f(x) - \lim_{x \to a^{+}} f(x). \end{equation*}

Since \(f\) is bounded on \((a,b)\), the numbers

\begin{equation*} A = \inf_{x \in (a,b)} f(x), \qquad B = \sup_{x \in (a,b)} f(x) \end{equation*}

are finite, and since \(f\) is increasing, monotone convergence of the values gives

\begin{equation*} A = \lim_{x \to a^{+}} f(x), \qquad B = \lim_{x \to b^{-}} f(x). \end{equation*}

Indeed, for \(\epsilon > 0\) there is \(x_0 \in (a,b)\) with \(f(x_0) < A + \epsilon\), and then \(A \le f(x) \le f(x_0) < A + \epsilon\) for all \(x \in (a, x_0)\); the argument at \(b\) is symmetric.

Define \(\tilde{f}\) on the closed interval \([a,b]\) by

\begin{equation*} \tilde{f}(a) = A, \qquad \tilde{f}(x) = f(x) \ \text{ for } x \in (a,b), \qquad \tilde{f}(b) = B . \end{equation*}

Then \(\tilde{f}\) is increasing on \([a,b]\): if \(a < x < y < b\) then \(\tilde{f}(x) = f(x) \le f(y) = \tilde{f}(y)\) since \(f\) is increasing; and \(\tilde{f}(a) = A \le f(x) \le B = \tilde{f}(b)\) for every \(x \in (a,b)\) by the definition of infimum and supremum; finally \(A \le B\).

Apply Corollary 4 to the increasing function \(\tilde{f}\) on \([a,b]\): \(\tilde{f}\,’\) is integrable over \([a,b]\) and

\begin{equation*} \int_a^b \tilde{f}\,’ \le \tilde{f}(b) - \tilde{f}(a) = B - A . \end{equation*}

Finally, \(\tilde{f}\) and \(f\) agree on \((a,b)\), and differentiability at a point \(x \in (a,b)\) together with the value of the derivative there depends only on the values of the function in a neighbourhood of \(x\) contained in \((a,b)\). Hence for every \(x \in (a,b)\), \(\tilde{f}\) is differentiable at \(x\) if and only if \(f\) is, and then \(\tilde{f}\,’(x) = f^{\prime}(x)\). Both derivatives exist for almost all \(x \in (a,b)\), by Lebesgue’s Theorem. Since \(\{a, b\}\) has measure zero, the two integrals agree, and we conclude that \(f^{\prime}\) is integrable over \((a,b)\) with

\begin{equation*} \int_a^b f^{\prime} = \int_a^b \tilde{f}\,’ \le B - A = \sup_{x \in (a,b)} f(x) - \inf_{x \in (a,b)} f(x) = \lim_{x \to b^{-}} f(x) - \lim_{x \to a^{+}} f(x). \end{equation*}

This is (15) for \(f\) on \((a,b)\); note that, since \(A\) and \(B\) are respectively the largest and the smallest values that an increasing extension of \(f\) to \([a,b]\) may take at \(a\) and at \(b\), the bound obtained is the sharp form (18) of the inequality.

That \(c\) is a local minimizer for \(f\) means that there is a \(\delta > 0\) such that

\begin{equation*} (c - \delta, c + \delta) \subseteq (a,b) \quad \text{and} \quad f(x) \ge f( c) \ \text{ for all } x \in (c-\delta, c+\delta). \end{equation*}

For \(0 < h \le \delta/2\) put

\begin{equation*} S(h) = \sup_{0 < |t| \le h} \frac{f(c+t) - f( c)}{t}, \qquad I(h) = \inf_{0 < |t| \le h} \frac{f(c+t) - f( c)}{t}, \end{equation*}

so that \(\overline{D}f( c) = \lim_{h \to 0} S(h)\) and \(\underline{D}f( c) = \lim_{h \to 0} I(h)\). As \(h\) decreases the supremum is taken over a smaller set of values of \(t\), so \(S\) is nondecreasing in \(h\) and \(I\) is nonincreasing in \(h\); consequently both limits exist in the extended real numbers, and

\begin{equation*} \overline{D}f( c) = \inf_{0 < h \le \delta/2} S(h), \qquad \underline{D}f( c) = \sup_{0 < h \le \delta/2} I(h) . \end{equation*}

Fix \(h\) with \(0 < h \le \delta/2\).

Take \(t = h\). Then \(0 < |t| \le h\) and \(c + t \in (c - \delta, c+\delta)\), so \(f(c+t) - f( c) \ge 0\); since \(t > 0\),

\begin{equation*} \frac{f(c+h) - f( c)}{h} \ge 0, \quad \text{hence } S(h) \ge 0 . \end{equation*}

As this holds for every such \(h\), taking the infimum over \(h\) gives \(\overline{D}f( c) \ge 0\).

Now take \(t = -h\). Then \(0 < |t| \le h\) and \(c + t \in (c-\delta, c+\delta)\), so again \(f(c+t) - f( c) \ge 0\); since now \(t = -h < 0\),

\begin{equation*} \frac{f(c-h) - f( c)}{-h} \le 0, \quad \text{hence } I(h) \le 0 . \end{equation*}

As this holds for every such \(h\), taking the supremum over \(h\) gives \(\underline{D}f( c) \le 0\).

Therefore

\begin{equation*} \underline{D}f( c) \le 0 \le \overline{D}f( c), \end{equation*}

as asserted. (In particular, if \(f\) happens to be differentiable at the local minimizer \(c\), then \(\underline{D}f( c) = \overline{D}f( c) = f^{\prime}( c)\), and the two inequalities force \(f^{\prime}( c) = 0\): this is the familiar interior extremum theorem. The same argument applied to \(-f\) shows that at a local maximizer one again has \(\underline{D}f( c) \le 0 \le \overline{D}f( c)\), since \(\overline{D}(-f) = -\underline{D}f\) and \(\underline{D}(-f) = -\overline{D}f\).)

Throughout, “increasing” means \(f(u) \le f(v)\) whenever \(u < v\), as in Section 6.1.

It is convenient to split the upper derivative into its one-sided parts. For \(c \in (a,b)\) set

\begin{equation*} D^{+}f( c) = \limsup_{t \to 0^{+}} \frac{f(c+t) - f( c)}{t}, \qquad D^{-}f( c) = \limsup_{s \to 0^{+}} \frac{f( c) - f(c-s)}{s}. \end{equation*}

For \(h > 0\),

\begin{equation*} \sup_{0 < |t| \le h} \frac{f(c+t)-f( c)}{t} = \max\left\{ \sup_{0 < t \le h}\frac{f(c+t)-f( c)}{t},\ \sup_{0 < s \le h}\frac{f( c)-f(c-s)}{s}\right\}, \end{equation*}

because the substitution \(t = -s\) turns a quotient with \(t < 0\) into \(\left(f( c)-f(c-s)\right)/s\). Each of the two suprema is nondecreasing in \(h\), so letting \(h \to 0\) gives

\begin{equation*} \overline{D}f( c) = \max\{D^{+}f( c),\ D^{-}f( c)\}. \end{equation*}

The proof below establishes the theorem for the hypothesis \(D^{+}f \ge 0\) on \((a,b)\) (and, by reflection, for \(D^{-}f \ge 0\)); a closing remark shows that the two-sided derivate \(\overline{D}f = \max\{D^{+}f, D^{-}f\}\), which is what Section 6.2 denotes by \(\overline{D}f\), does not suffice, so that the hypothesis has to be read one-sidedly.

Step 1, the reduction supplied by the hint. Suppose it is known that

(L) if \(g\) is continuous on \([a,b]\) and \(D^{+}g \ge \epsilon > 0\) on \((a,b)\), then \(g\) is increasing on \([a,b]\).

Let \(f\) be continuous on \([a,b]\) with \(D^{+}f \ge 0\) on \((a,b)\), and let \(\epsilon > 0\). Put \(g(x) = f(x) + \epsilon x\); \(g\) is continuous on \([a,b]\). For \(c \in (a,b)\) and \(t \neq 0\),

\begin{equation*} \frac{g(c+t) - g( c)}{t} = \frac{f(c+t) - f( c)}{t} + \epsilon , \end{equation*}

so the supremum over \(0 < t \le h\) of the left side is \(\epsilon\) more than that of the first term on the right; letting \(h \to 0\),

\begin{equation*} D^{+}g( c) = D^{+}f( c) + \epsilon \ge \epsilon > 0 . \end{equation*}

By (L), \(g\) is increasing on \([a,b]\): if \(a \le u < v \le b\) then \(f(u) + \epsilon u \le f(v) + \epsilon v\), that is,

\begin{equation*} f(v) - f(u) \ge \epsilon(u - v) \ge -\epsilon (b-a). \end{equation*}

Since \(\epsilon > 0\) is arbitrary, \(f(v) - f(u) \ge 0\). Thus \(f\) is increasing on \([a,b]\), and it remains only to prove (L).

Step 2, the crossing-point argument, which proves (L). Assume \(g\) is continuous on \([a,b]\) and \(D^{+}g \ge \epsilon > 0\) on \((a,b)\), and suppose, to get a contradiction, that \(g\) is not increasing: there are \(u, v\) with \(a \le u < v \le b\) and \(g(u) > g(v)\). Fix \(\lambda\) with \(g(v) < \lambda < g(u)\) and set

\begin{equation*} S = \{x \in [u,v] : g(x) \ge \lambda\}, \qquad c = \sup S . \end{equation*}

\(S\) is nonempty (\(u \in S\)) and closed in \([u,v]\) by the continuity of \(g\), so \(c \in S\), i.e. \(g( c) \ge \lambda\); and \(c < v\) since \(g(v) < \lambda\). By the definition of \(c\),

\begin{equation*} g(x) < \lambda \quad \text{for all } x \in (c, v]. \end{equation*}

Letting \(x \to c^{+}\) and using continuity gives \(g( c) \le \lambda\), hence \(g( c) = \lambda\); in particular \(c \neq u\), since \(g(u) > \lambda\), so \(c > u\) and \(c \in (u,v) \subseteq (a,b)\). Consequently, for \(0 < t \le v - c\),

\begin{equation*} \frac{g(c+t) - g( c)}{t} = \frac{g(c+t) - \lambda}{t} < 0 , \end{equation*}

so that

\begin{equation*} D^{+}g( c) \le 0 . \end{equation*}

Since \(c \in (a,b)\), the hypothesis gives \(D^{+}g( c) \ge \epsilon > 0\), and this contradicts \(D^{+}g( c) \le 0\). Hence no such pair \(u < v\) exists and \(g\) is increasing on \([a,b]\). This proves (L), and with Step 1 it proves

if \(f\) is continuous on \([a,b]\) and \(D^{+}f \ge 0\) on \((a,b)\), then \(f\) is increasing on \([a,b]\).

The same holds with \(D^{-}\) in place of \(D^{+}\). Indeed, put \(\tilde{f}(x) = -f(-x)\) on \([-b,-a]\); for \(s > 0\),

\begin{equation*} \frac{\tilde{f}(x+s) - \tilde{f}(x)}{s} = \frac{f(-x) - f(-x-s)}{s}, \end{equation*}

so \(D^{+}\tilde{f}(x) = D^{-}f(-x)\), and \(\tilde{f}\) is increasing precisely when \(f\) is; so the statement for \(D^{-}\) follows from the one just proved.

Why the hypothesis must be read one-sidedly. With the two-sided derivate of Section 6.2 the assertion is false. Observe first where the argument above breaks down: at the crossing point \(c\) we obtained only \(D^{+}g( c) \le 0\), and \(\overline{D}g( c) = \max\{D^{+}g( c), D^{-}g( c)\} \ge \epsilon\) then forces \(D^{-}g( c) \ge \epsilon\), which merely says that \(g\) dips below \(\lambda\) arbitrarily near \(c\) on the left as well. That is not absurd, and in fact no argument can rule it out.

Take \(\alpha = 1/2\) and \(\beta = 13\), so that \(\beta\) is an odd integer and \(\alpha\beta = 13/2 > 1 + 3\pi/2\), and let

\begin{equation*} W(x) = \sum_{n=0}^{\infty} \alpha^{n} \cos\left(\beta^{n}\pi x\right). \end{equation*}

The series is dominated by the convergent geometric series \(\sum \alpha^{n}\), so it converges uniformly on \(\mathbb{R}\) and \(W\) is continuous. Because \(\beta^{n}\) is odd, \(\cos(\beta^{n}\pi) = -1\), so

\begin{equation*} W(0) = \sum_{n=0}^{\infty} \alpha^{n} = 2, \qquad W(1) = -\sum_{n=0}^{\infty}\alpha^{n} = -2 , \end{equation*}

and \(W\) is not increasing on \([0,1]\). We claim that nevertheless \(\overline{D}W(x) = +\infty\) for every \(x \in (0,1)\), so that \(W\) satisfies the hypothesis as printed.

Fix \(x\) and a natural number \(m\). Write \(\beta^{m}x = k_m + \xi_m\) with \(k_m\) an integer and \(-1/2 < \xi_m \le 1/2\), and put

\begin{equation*} y_m^{+} = \frac{k_m + 1}{\beta^{m}}, \qquad y_m^{-} = \frac{k_m - 1}{\beta^{m}} . \end{equation*}

Then \(y_m^{+} - x = (1 - \xi_m)/\beta^{m}\) and \(y_m^{-} - x = -(1+\xi_m)/\beta^{m}\), so

\begin{equation*} \frac{1}{2\beta^{m}} \le y_m^{+} - x \le \frac{3}{2\beta^{m}}, \qquad -\frac{3}{2\beta^{m}} \le y_m^{-} - x \le -\frac{1}{2\beta^{m}} , \end{equation*}

and both points tend to \(x\); discard the finitely many \(m\) for which they fall outside \((0,1)\). Split the series at \(n = m\).

For \(n < m\), the Mean Value Theorem applied to the cosine gives \(|\cos(\beta^{n}\pi y) - \cos(\beta^{n}\pi x)| \le \beta^{n}\pi|y - x|\), so for either choice of sign,

\begin{equation*} \left| \frac{1}{y_m^{\pm} - x}\sum_{n=0}^{m-1}\alpha^{n}\left[\cos(\beta^{n}\pi y_m^{\pm}) - \cos(\beta^{n}\pi x)\right] \right| \le \pi \sum_{n=0}^{m-1}(\alpha\beta)^{n} \le \frac{\pi}{\alpha\beta - 1}\,(\alpha\beta)^{m}. \end{equation*}

For \(n \ge m\), use that \(\beta\) is odd: \(\beta^{n-m}(k_m \pm 1)\) has the parity of \(k_m + 1\) and \(\beta^{n-m}k_m\) has the parity of \(k_m\), whence

\begin{equation*} \cos(\beta^{n}\pi y_m^{\pm}) = (-1)^{k_m+1}, \qquad \cos(\beta^{n}\pi x) = (-1)^{k_m}\cos\left(\beta^{n-m}\pi \xi_m\right). \end{equation*}

Therefore, for both choices of sign,

\begin{equation*} \sum_{n=m}^{\infty}\alpha^{n}\left[\cos(\beta^{n}\pi y_m^{\pm}) - \cos(\beta^{n}\pi x)\right] = -(-1)^{k_m}\,\Sigma_m, \qquad \Sigma_m = \sum_{n=m}^{\infty}\alpha^{n}\left[1 + \cos(\beta^{n-m}\pi\xi_m)\right]. \end{equation*}

Every term of \(\Sigma_m\) is nonnegative, and its \(n = m\) term is \(\alpha^{m}(1 + \cos(\pi\xi_m)) \ge \alpha^{m}\) because \(|\xi_m| \le 1/2\); hence \(\Sigma_m \ge \alpha^{m}\). Dividing by \(y_m^{+} - x > 0\) produces a quantity of sign \(-(-1)^{k_m}\) and absolute value at least \(\alpha^{m}\cdot \frac{2\beta^{m}}{3} = \frac{2}{3}(\alpha\beta)^{m}\); dividing by \(y_m^{-} - x < 0\) produces a quantity of the opposite sign \((-1)^{k_m}\) and again of absolute value at least \(\frac{2}{3}(\alpha\beta)^{m}\).

Set \(\kappa = \frac{2}{3} - \frac{\pi}{\alpha\beta - 1} = \frac{2}{3} - \frac{\pi}{5.5} > 0\). Combining the two estimates, the difference quotients

\begin{equation*} \frac{W(y_m^{+}) - W(x)}{y_m^{+} - x} \qquad \text{and} \qquad \frac{W(y_m^{-}) - W(x)}{y_m^{-} - x} \end{equation*}

have opposite signs and each has absolute value at least \(\kappa\,(\alpha\beta)^{m}\). In particular, for each \(m\) at least one of the two is \(\ge \kappa\,(\alpha\beta)^{m}\). Since \(y_m^{\pm} \to x\) and \(\alpha\beta > 1\), this gives

\begin{equation*} \overline{D}W(x) = \limsup_{t \to 0}\frac{W(x+t) - W(x)}{t} = +\infty \end{equation*}

at every \(x\). (The other member of each pair shows likewise that \(\underline{D}W(x) = -\infty\); this is Weierstrass’s classical proof that \(W\) is nowhere differentiable.)

So \(W\) is continuous on \([0,1]\), satisfies \(\overline{D}W \ge 0\) on \((0,1)\), and is not increasing. Accordingly the hypothesis of the problem is to be read as a hypothesis on one fixed one-sided derivate, and under that reading Steps 1 and 2 above are a complete proof.

Fix \(c \in (a,b)\) and choose \(h_0 > 0\) with \((c - h_0, c + h_0) \subseteq (a,b)\). For \(0 < h \le h_0\) and for a real-valued function \(\varphi\) on \((a,b)\) write

\begin{equation*} Q_{\varphi}(t) = \frac{\varphi(c+t) - \varphi( c)}{t} \quad (0 < |t| \le h_0), \qquad S_{\varphi}(h) = \sup_{0 < |t| \le h} Q_{\varphi}(t), \qquad I_{\varphi}(h) = \inf_{0 < |t| \le h} Q_{\varphi}(t), \end{equation*}

so that, by definition, \(\overline{D}\varphi( c) = \lim_{h \to 0} S_{\varphi}(h)\) and \(\underline{D}\varphi( c) = \lim_{h \to 0} I_{\varphi}(h)\). Each \(Q_{\varphi}(t)\) is a real number and the index set is nonempty, so \(S_{\varphi}(h) \in (-\infty, +\infty]\) and \(I_{\varphi}(h) \in [-\infty, +\infty)\); in particular the sums \(S_f(h) + S_g(h)\) and \(I_f(h) + I_g(h)\) are unambiguously defined in the extended reals. Moreover \(S_{\varphi}\) is nondecreasing and \(I_{\varphi}\) is nonincreasing in \(h\), so the two limits above exist in \([-\infty, +\infty]\); as usual, the outer inequalities of the problem are asserted whenever the sums on their outer sides are defined, i.e. are not of the form \(+\infty + (-\infty)\).

The whole proof rests on the identity

\begin{equation*} Q_{f+g}(t) = \frac{\left(f(c+t) + g(c+t)\right) - \left(f( c) + g( c)\right)}{t} = Q_f(t) + Q_g(t), \qquad 0 < |t| \le h_0 . \end{equation*}

The upper estimate. Fix \(h \in (0, h_0]\). For every \(t\) with \(0 < |t| \le h\),

\begin{equation*} Q_{f+g}(t) = Q_f(t) + Q_g(t) \le S_f(h) + S_g(h), \end{equation*}

since each term is dominated by the corresponding supremum (the inequality is trivially true if the right-hand side is \(+\infty\)). Taking the supremum over such \(t\),

\begin{equation*} S_{f+g}(h) \le S_f(h) + S_g(h) \quad \text{for all } 0 < h \le h_0 . \end{equation*}

Now let \(h \to 0\). All three quantities are monotone in \(h\), so all three limits exist in \([-\infty, +\infty]\), and if \(\overline{D}f( c) + \overline{D}g( c)\) is defined then \(\lim_{h \to 0}\left[S_f(h) + S_g(h)\right] = \overline{D}f( c) + \overline{D}g( c)\). Hence

\begin{equation*} \overline{D}(f+g)( c) \le \overline{D}f( c) + \overline{D}g( c). \end{equation*}

The lower estimate. Fix \(h \in (0,h_0]\). For every \(t\) with \(0 < |t| \le h\),

\begin{equation*} Q_{f+g}(t) = Q_f(t) + Q_g(t) \ge I_f(h) + I_g(h), \end{equation*}

and taking the infimum over such \(t\),

\begin{equation*} I_{f+g}(h) \ge I_f(h) + I_g(h) \quad \text{for all } 0 < h \le h_0 . \end{equation*}

Letting \(h \to 0\) exactly as before,

\begin{equation*} \underline{D}(f+g)( c) \ge \underline{D}f( c) + \underline{D}g( c). \end{equation*}

The middle estimate. For each \(h \in (0,h_0]\) the infimum of a nonempty set of real numbers does not exceed its supremum, so \(I_{f+g}(h) \le S_{f+g}(h)\); letting \(h \to 0\) gives

\begin{equation*} \underline{D}(f+g)( c) \le \overline{D}(f+g)( c) . \end{equation*}

(This is the inequality \(\underline{D}\varphi \le \overline{D}\varphi\) recorded in Section 6.2 for every \(\varphi\).)

Since \(c \in (a,b)\) was arbitrary, the chain

\begin{equation*} \underline{D}f + \underline{D}g \le \underline{D}(f+g) \le \overline{D}(f+g) \le \overline{D}f + \overline{D}g \end{equation*}

holds at every point of \((a,b)\).

Both outer inequalities may be strict. First observe that for any real-valued \(\varphi\) and any \(h > 0\), \(Q_{-\varphi}(t) = -Q_{\varphi}(t)\), so \(S_{-\varphi}(h) = -I_{\varphi}(h)\) and \(I_{-\varphi}(h) = -S_{\varphi}(h)\); letting \(h \to 0\),

\begin{equation*} \overline{D}(-\varphi) = -\underline{D}\varphi, \qquad \underline{D}(-\varphi) = -\overline{D}\varphi . \end{equation*}

Now take \((a,b) = (-1,1)\), let \(f\) be the function of Problem 15, \(f(x) = x \sin(1/x)\) for \(x \neq 0\) and \(f(0) = 0\), let \(g = -f\), and evaluate at \(c = 0\). By Problem 15, \(\overline{D}f(0) = 1\) and \(\underline{D}f(0) = -1\), so by the displayed identities \(\overline{D}g(0) = 1\) and \(\underline{D}g(0) = -1\). Since \(f + g \equiv 0\) we have \(\underline{D}(f+g)(0) = \overline{D}(f+g)(0) = 0\). Thus the right-hand end of the chain is \(2\) and the left-hand end is \(-2\), while both middle terms are \(0\).

Throughout put

\begin{equation*} \sigma(t) = g(\gamma + t) - g(\gamma) = g(\gamma+t) - c , \qquad Q(\tau) = \frac{f(c + \tau) - f( c)}{\tau} \quad (\tau \neq 0,\ c + \tau \in [a,b]), \end{equation*}

and, writing \(r_0 = \min\{c - a,\ b - c\} > 0\) (positive because \(c\) is an interior point of \([a,b]\)), put \(S( r) = \sup_{0 < |\tau| \le r} Q(\tau)\) for \(0 < r \le r_0\); every \(\tau\) occurring here has \(c + \tau \in [a,b]\), so \(S( r)\) is well defined in \((-\infty, +\infty]\), \(S\) is nondecreasing in \(r\), and \(\overline{D}f( c) = \lim_{r \to 0^{+}} S( r) = \inf_{0 < r \le r_0} S( r)\). Since \(c\) is interior to \([a,b]\) and \(g\) is continuous at \(\gamma\), there is \(\delta_1 > 0\) with \((\gamma - \delta_1, \gamma + \delta_1) \subseteq (\alpha, \beta)\) and \(g\left((\gamma-\delta_1, \gamma+\delta_1)\right) \subseteq (a,b)\); so \(f \circ g\) is defined on \((\gamma - \delta_1, \gamma+\delta_1)\) and its upper derivative at \(\gamma\) makes sense. Note that \(\sigma\) is continuous, \(\sigma(0) = 0\), and \(\sigma(t)/t \to g^{\prime}(\gamma)\) as \(t \to 0\). Whenever \(\sigma(t) \neq 0\),

\begin{equation*} \frac{(f \circ g)(\gamma+t) - (f\circ g)(\gamma)}{t} = \frac{f(c + \sigma(t)) - f( c)}{\sigma(t)} \cdot \frac{\sigma(t)}{t} = Q(\sigma(t)) \cdot \frac{\sigma(t)}{t}. \qquad (\dagger) \end{equation*}

We use twice the following elementary fact about upper limits.

Lemma. Let \(u\) and \(v\) be real-valued on a punctured neighbourhood of \(0\), with \(v(t) \to m > 0\) as \(t \to 0\), and let \(L = \limsup_{t \to 0} u(t) \in [-\infty, +\infty]\). Then \(\limsup_{t\to0} u(t)v(t) = mL\).

Proof of the Lemma. Fix \(\eta \in (0, m)\); for all small \(t \neq 0\) we have \(m - \eta < v(t) < m + \eta\), and in particular \(v(t) > 0\).

We first show \(\limsup_{t \to 0} u(t)v(t) \le mL\). Suppose \(L\) is finite. For all small \(t \neq 0\) we also have \(u(t) < L + \eta\), and multiplying by \(v(t) > 0\) gives

\begin{equation*} u(t)v(t) < (L+\eta)\,v(t) \le \max\left\{ (L+\eta)(m-\eta),\ (L+\eta)(m+\eta) \right\}, \end{equation*}

the last step because \((L+\eta)v(t)\) lies between the two displayed products according to the sign of \(L + \eta\). Hence \(\limsup_{t\to0} uv \le \max\{(L+\eta)(m-\eta), (L+\eta)(m+\eta)\}\), and both products tend to \(mL\) as \(\eta \to 0\), so \(\limsup_{t\to0} uv \le mL\). If \(L = -\infty\), then for each \(M > 0\) we have \(u(t) < -M\) for all small \(t \neq 0\), whence \(u(t)v(t) < -M(m-\eta)\); as \(M\) is arbitrary and \(m - \eta > 0\), \(\limsup_{t\to0} uv = -\infty = mL\). If \(L = +\infty\) the inequality \(\limsup uv \le mL\) is vacuous.

For the reverse inequality choose \(t_n \to 0\) with \(t_n \neq 0\) and \(u(t_n) \to L\). Since \(v(t_n) \to m > 0\), we get \(u(t_n)v(t_n) \to mL\), this also when \(L = \pm\infty\). Hence \(\limsup_{t \to 0} uv \ge mL\), and the Lemma is proved.

Part (i). Let \(m = g^{\prime}(\gamma) > 0\). Since \(\sigma(t)/t \to m\), there is \(\delta_2 \in (0, \delta_1]\) such that

\begin{equation*} \frac{m}{2} \le \frac{\sigma(t)}{t} \le 2m \quad \text{for } 0 < |t| \le \delta_2 , \end{equation*}

and, shrinking \(\delta_2\) further, we may also require \(2m\delta_2 \le r_0\), so that \(S(2mh)\) is defined for every \(h \in (0, \delta_2]\). In particular \(\sigma(t) \neq 0\) for \(0 < |t| \le \delta_2\), \(\sigma(t)\) has the same sign as \(t\), and \(|\sigma(t)| \le 2m|t|\). Thus \((\dagger)\) is available for all such \(t\), and by the Lemma (with \(u(t) = Q(\sigma(t))\) and \(v(t) = \sigma(t)/t \to m > 0\)),

\begin{equation*} \overline{D}(f\circ g)(\gamma) = \limsup_{t \to 0} Q(\sigma(t)) \cdot \frac{\sigma(t)}{t} = m \cdot \limsup_{t \to 0} Q(\sigma(t)) . \end{equation*}

It remains to prove that

\begin{equation*} \limsup_{t \to 0} Q(\sigma(t)) = \overline{D}f( c), \end{equation*}

and for this we compare the suprema of the two families of difference quotients.

First, let \(0 < h \le \delta_2\). For \(0 < |t| \le h\) we have \(0 < |\sigma(t)| \le 2mh\), so \(Q(\sigma(t))\) belongs to the family \(\{Q(\tau) : 0 < |\tau| \le 2mh\}\), and therefore

\begin{equation*} \sup_{0 < |t| \le h} Q(\sigma(t)) \le S(2mh). \end{equation*}

Letting \(h \to 0^{+}\) gives \(\limsup_{t \to 0} Q(\sigma(t)) \le \overline{D}f( c)\).

Second, again let \(0 < h \le \delta_2\) and put

\begin{equation*} \rho(h) = \min\{\sigma(h),\ -\sigma(-h)\} . \end{equation*}

Both entries are positive, because \(\sigma(h) \ge (m/2)h > 0\) and \(\sigma(-h) \le (m/2)(-h) < 0\); also \(\rho(h) \le 2mh\), so \(\rho(h) \to 0^{+}\) as \(h \to 0^{+}\). The function \(\sigma\) is continuous on \([-h, h]\) with \(\sigma(-h) < 0 < \sigma(h)\), so by the Intermediate Value Theorem it attains every value in \([\sigma(-h), \sigma(h)] \supseteq [-\rho(h), \rho(h)]\). Hence every \(\tau\) with \(0 < |\tau| \le \rho(h)\) equals \(\sigma(t)\) for some \(t \in [-h,h]\), and such \(t\) is nonzero because \(\sigma(0) = 0 \neq \tau\). Consequently

\begin{equation*} \sup_{0 < |t| \le h} Q(\sigma(t)) \ge \sup_{0 < |\tau| \le \rho(h)} Q(\tau) = S(\rho(h)). \end{equation*}

Since \(\rho(h) \to 0^{+}\) as \(h \to 0^{+}\) and \(S( r)\) has the limit \(\overline{D}f( c)\) as \(r \to 0^{+}\), letting \(h \to 0^{+}\) gives \(\limsup_{t\to 0} Q(\sigma(t)) \ge \overline{D}f( c)\).

Combining the two estimates, \(\limsup_{t \to 0} Q(\sigma(t)) = \overline{D}f( c)\), and therefore

\begin{equation*} \overline{D}(f \circ g)(\gamma) = \overline{D}f( c) \cdot g^{\prime}(\gamma), \end{equation*}

this being an equality in the extended reals (both sides are simultaneously \(+\infty\) or \(-\infty\), since \(g^{\prime}(\gamma) > 0\)).

Part (ii). Now let \(g^{\prime}(\gamma) = 0\) and suppose \(\overline{D}f( c)\) and \(\underline{D}f( c)\) are finite. Since \(S( r) \downarrow \overline{D}f( c)\) as \(r \to 0^{+}\) and the limit is finite, there is \(r_1 \in (0, r_0]\) with \(S(r_1) \le \overline{D}f( c) + 1 < \infty\); likewise, since the infima \(\inf_{0<|\tau| \le r}Q(\tau)\) increase to the finite number \(\underline{D}f( c)\) as \(r\) decreases to \(0\), there is \(r_2 \in (0, r_0]\) with \(\inf_{0 < |\tau| \le r_2} Q(\tau) \ge \underline{D}f( c) - 1 > -\infty\). Put \(\delta_0 = \min\{r_1, r_2\}\) and

\begin{equation*} M = \max\left\{ |\overline{D}f( c)| + 1,\ |\underline{D}f( c)| + 1 \right\}, \end{equation*}

so that

\begin{equation*} |Q(\tau)| \le M \quad \text{for all } \tau \text{ with } 0 < |\tau| \le \delta_0 . \end{equation*}

Let \(\eta > 0\). Since \(g\) is continuous at \(\gamma\), there is \(\delta^{\prime} \in (0, \delta_1]\) with \(|\sigma(t)| \le \delta_0\) for \(|t| \le \delta^{\prime}\); and since \(\sigma(t)/t \to g^{\prime}(\gamma) = 0\), we may shrink \(\delta^{\prime}\) so that in addition

\begin{equation*} \frac{|\sigma(t)|}{|t|} < \frac{\eta}{M} \quad \text{for } 0 < |t| \le \delta^{\prime} . \end{equation*}

Fix \(t\) with \(0 < |t| \le \delta^{\prime}\). If \(\sigma(t) = 0\) then \((f\circ g)(\gamma + t) = f( c) = (f \circ g)(\gamma)\) and the difference quotient of \(f \circ g\) at \(\gamma\) vanishes. If \(\sigma(t) \neq 0\) then, by \((\dagger)\),

\begin{equation*} \left| \frac{(f\circ g)(\gamma+t) - (f\circ g)(\gamma)}{t} \right| = |Q(\sigma(t))| \cdot \frac{|\sigma(t)|}{|t|} \le M \cdot \frac{\eta}{M} = \eta . \end{equation*}

In both cases the difference quotient of \(f \circ g\) at \(\gamma\) has absolute value at most \(\eta\) for \(0 < |t| \le \delta^{\prime}\). As \(\eta > 0\) was arbitrary,

\begin{equation*} \lim_{t \to 0} \frac{(f\circ g)(\gamma+t) - (f\circ g)(\gamma)}{t} = 0 , \end{equation*}

so \(f \circ g\) is in fact differentiable at \(\gamma\) with \((f\circ g)^{\prime}(\gamma) = 0\); in particular

\begin{equation*} \overline{D}(f \circ g)(\gamma) = 0 = \underline{D}(f\circ g)(\gamma). \end{equation*}

The finiteness hypothesis in (ii) cannot be dropped. Take \([a,b] = [\alpha,\beta] = [-1,1]\), \(\gamma = 0\), \(g(t) = t^2\), and

\begin{equation*} f(s) = \begin{cases} \sqrt{s} & \text{if } 0 \le s \le 1,\\ 0 & \text{if } -1 \le s < 0.\end{cases} \end{equation*}

Then \(g\) is continuous on \([-1,1]\) and differentiable at \(0\) with \(g^{\prime}(0) = 0\), and \(c = g(0) = 0 \in (-1,1)\). The difference quotients of \(f\) at \(0\) are \(Q(\tau) = 1/\sqrt{\tau}\) for \(\tau > 0\) and \(Q(\tau) = 0\) for \(\tau < 0\), so \(\overline{D}f(0) = +\infty\) (while \(\underline{D}f(0) = 0\)). Here \((f \circ g)(t) = \sqrt{t^2} = |t|\), whose difference quotients at \(0\) are \(|t|/t = \pm 1\), so

\begin{equation*} \overline{D}(f\circ g)(0) = 1 \neq 0 . \end{equation*}

Thus the conclusion of (ii) fails when the upper derivative of \(f\) at \(c\) is infinite: the unboundedness of \(Q\) near \(0\) is picked up exactly along the values \(\tau = \sigma(t)\), at a rate that compensates for \(\sigma(t)/t \to 0\).

Throughout, \(I\) denotes an interval of real numbers (bounded or unbounded, open, closed or half-open) and a real-valued function \(f\) on \(I\) is measurable provided \(\{x \in I : f(x) > c\}\) is a measurable set for every \(c \in \mathbb{R}\); this is the criterion of Proposition 1 of Chapter 3. Every interval is a Borel set, hence measurable.

Step 1: a strictly increasing function on an interval is measurable.

Let \(f\) be strictly increasing on \(I\) and fix \(c \in \mathbb{R}\). Put

\begin{equation*} E_c = \{x \in I : f(x) > c\}. \end{equation*}

If \(E_c = \emptyset\) it is measurable, so assume \(E_c \neq \emptyset\) and set \(\alpha = \inf E_c\), an element of \([-\infty, \infty)\).

By the definition of infimum, \(E_c \subseteq [\alpha, \infty) \cap I\) (interpreting \([-\infty,\infty)\) as \(\mathbb{R}\) when \(\alpha = -\infty\)).

Conversely, let \(x \in I\) with \(x > \alpha\). Since \(\alpha\) is the infimum of \(E_c\), there is a point \(y \in E_c\) with \(\alpha \leq y < x\). As \(f\) is strictly increasing and \(y < x\), we have \(f(x) > f(y) > c\), so \(x \in E_c\). Thus

\begin{equation*} (\alpha, \infty) \cap I \subseteq E_c \subseteq [\alpha, \infty) \cap I . \end{equation*}

Consequently \(E_c\) is either \((\alpha, \infty) \cap I\) or \(\big( (\alpha,\infty) \cap I \big) \cup \{\alpha\}\), and the latter can occur only when \(\alpha \in I\), in which case it equals \([\alpha, \infty) \cap I\). In either case \(E_c\) is the intersection of two intervals and hence is an interval, therefore a Borel set and so measurable. Since \(c \in \mathbb{R}\) was arbitrary, \(f\) is measurable on \(I\).

Step 2: a monotone function on an interval is measurable.

First suppose \(f\) is increasing on \(I\), that is, \(f(u) \leq f(v)\) whenever \(u, v \in I\) and \(u < v\). Define

\begin{equation*} g(x) = f(x) + x \quad \text{for all } x \in I . \end{equation*}

If \(u < v\) are points of \(I\), then \(f(u) \leq f(v)\) and \(u < v\), so \(g(u) = f(u) + u < f(v) + v = g(v)\). Hence \(g\) is strictly increasing on \(I\), and therefore measurable by Step 1. The identity function \(x \mapsto x\) is continuous on \(I\), hence measurable. By Theorem 6 of Chapter 3 (linear combinations of measurable real-valued functions are measurable),

\begin{equation*} f(x) = g(x) - x \quad \text{for all } x \in I \end{equation*}

is measurable on \(I\).

Now suppose \(f\) is decreasing on \(I\). Then \(-f\) is increasing on \(I\), hence measurable by the preceding paragraph, and therefore \(f = -(-f)\) is measurable.

Thus every monotone real-valued function defined on an interval is measurable.

Recall the definitions of Section 6.2: for \(x\) in \((a,b)\),

\begin{equation*} \overline{D}f(x) = \lim_{h \to 0^{+}} \left[ \sup_{0 < |t| \le h} \frac{f(x+t) - f(x)}{t} \right], \qquad \underline{D}f(x) = \lim_{h \to 0^{+}} \left[ \inf_{0 < |t| \le h} \frac{f(x+t) - f(x)}{t} \right], \end{equation*}

the supremum and infimum being taken over those \(t\) with \(x + t\) in \([a,b]\). By hypothesis there is a number \(M \geq 0\) with

\begin{equation*} |\overline{D}f(x)| \le M \quad \text{and} \quad |\underline{D}f(x)| \le M \quad \text{for all } x \in (a,b). \end{equation*}

In particular \(-M \le \underline{D}f(x)\) and \(\overline{D}f(x) \le M\) on \((a,b)\). We show \(|f(u) - f(v)| \le M |u - v|\) for all \(u, v\) in \([a,b]\), so that \(f\) is Lipschitz with Lipschitz constant \(M\).

Step 1: a local estimate at each interior point.

Fix \(\epsilon > 0\) and put \(K = M + \epsilon\). Let \(x \in (a,b)\). Since

\begin{equation*} \limsup_{t \to 0} \frac{f(x+t)-f(x)}{t} = \overline{D}f(x) \le M < K \quad\text{and}\quad \liminf_{t \to 0} \frac{f(x+t)-f(x)}{t} = \underline{D}f(x) \ge -M > -K, \end{equation*}

there is a \(\delta_x > 0\) such that

\begin{equation*} -K < \frac{f(x+t) - f(x)}{t} < K \quad \text{whenever } 0 < |t| < \delta_x \text{ and } x+t \in [a,b]. \end{equation*}

The two inequalities together say \(\left| \frac{f(x+t)-f(x)}{t} \right| < K\), and multiplying by \(|t|\) (which is legitimate for \(t\) of either sign) gives \(|f(x+t) - f(x)| < K|t|\). Writing \(y = x + t\), and noting that the case \(y = x\) is trivial, this says exactly

\begin{equation*} |f(y) - f(x)| \le K\,|y - x| \quad \text{for all } y \in [a,b] \text{ with } |y - x| < \delta_x. \tag{\(\ast\)} \end{equation*}

Step 2: from the local estimate to a global one on \((a,b)\).

Let \(u, w\) satisfy \(a < u < w < b\). Put

\begin{equation*} S = \{ v \in [u,w] : |f(v) - f(u)| \le K (v-u) \}. \end{equation*}

Then \(u \in S\), so \(S \neq \emptyset\), and \(S\) is closed in \([u,w]\) because \(f\) is continuous and both sides of the defining inequality are continuous in \(v\). Let \(s = \sup S\); since \(S\) is closed and bounded, \(s \in S\).

Suppose \(s < w\). Then \(s \in [u, w) \subseteq (a,b)\), so \((\ast)\) holds at \(s\) with some \(\delta_s > 0\). Choose \(v\) with \(s < v < \min\{w,\ s + \delta_s\}\). Then, using \((\ast)\) at \(s\) and \(s \in S\),

\begin{equation*} |f(v) - f(u)| \le |f(v) - f(s)| + |f(s) - f(u)| \le K(v - s) + K(s - u) = K(v-u), \end{equation*}

so \(v \in S\) and \(v > s\), contradicting \(s = \sup S\). Hence \(s = w\), that is,

\begin{equation*} |f(w) - f(u)| \le K (w - u) \quad \text{for all } a < u < w < b. \tag{\(\ast\ast\)} \end{equation*}

Step 3: including the endpoints, and letting \(\epsilon\) tend to \(0\).

Let \(u, v \in [a,b]\) with \(u < v\). For each natural number \(n\) set

\begin{equation*} u_n = \max\Big\{ u,\ a + \tfrac{1}{n} \Big\}, \qquad v_n = \min\Big\{ v,\ b - \tfrac{1}{n} \Big\}. \end{equation*}

Then \(a < u_n\) and \(v_n < b\) for every \(n\) with \(1/n < b-a\), while \(u_n \to u\) and \(v_n \to v\) (if \(u > a\) then \(u_n = u\) once \(1/n < u - a\), and if \(u = a\) then \(u_n = a + 1/n\); similarly for \(v_n\)). Since \(u < v\), we have \(u_n < v_n\) for all sufficiently large \(n\). For such \(n\), \((\ast\ast)\) applies and gives \(|f(v_n) - f(u_n)| \le K(v_n - u_n)\). Since \(f\) is continuous on \([a,b]\), letting \(n \to \infty\) gives

\begin{equation*} |f(v) - f(u)| \le K (v-u) = (M+\epsilon)(v-u). \end{equation*}

This holds for every \(\epsilon > 0\), so letting \(\epsilon \to 0^{+}\) we obtain

\begin{equation*} |f(v) - f(u)| \le M\,|v - u| \quad \text{for all } u, v \in [a,b]. \end{equation*}

Therefore \(f\) is Lipschitz on \([a,b]\), with Lipschitz constant \(M\).

The function is differentiable on all of \([0,1]\).

For \(0 < x \le 1\) the chain and product rules give

\begin{equation*} f^{\prime}(x) = 2x \sin\!\left(\frac{1}{x^{2}}\right) - \frac{2}{x}\cos\!\left(\frac{1}{x^{2}}\right), \end{equation*}

and this expression is continuous on \((0,1]\). At \(x = 0\), for \(0 < x \le 1\),

\begin{equation*} \left| \frac{f(x) - f(0)}{x - 0} \right| = \left| x \sin\!\left(\frac{1}{x^{2}}\right) \right| \le x \to 0 , \end{equation*}

so \(f^{\prime}(0)\) exists and equals \(0\). Also \(|f(x)| \le x^2\) shows \(f\) is continuous at \(0\), so \(f\) is continuous on \([0,1]\) and differentiable at every point of \([0,1]\), as asserted in the remark. Since \(f^{\prime}\) is continuous on \((0,1]\) it is measurable there, and \(\{0\}\) has measure zero, so \(f^{\prime}\) is a measurable function on \([0,1]\). Thus \(f^{\prime}\) is integrable over \([0,1]\) if and only if \(\int_{0}^{1} |f^{\prime}| < \infty\).

A sequence of disjoint intervals on which \(f\) oscillates.

For each natural number \(n\) set

\begin{equation*} a_n = \left[ \left(2n + \tfrac{1}{2}\right)\pi \right]^{-1/2}, \qquad b_n = \left[ \left(2n - \tfrac{1}{2}\right)\pi \right]^{-1/2}. \end{equation*}

Since \(2n - \tfrac12 < 2n + \tfrac12\) we have \(0 < a_n < b_n\), and \(b_1 = (3\pi/2)^{-1/2} < 1\), so \(b_n \le b_1 < 1\) for all \(n\) and each \([a_n, b_n]\) is contained in \((0,1)\). Moreover

\begin{equation*} b_{n+1} = \left[ \left(2n + \tfrac{3}{2}\right)\pi \right]^{-1/2} < \left[ \left(2n + \tfrac{1}{2}\right)\pi \right]^{-1/2} = a_n , \end{equation*}

so the intervals \([a_n, b_n]\), \(n = 1, 2, \dots\), are pairwise disjoint.

Because \(1/a_n^{2} = (2n + \tfrac12)\pi\) and \(1/b_n^{2} = (2n - \tfrac12)\pi\),

\begin{equation*} \sin\!\left(\frac{1}{a_n^{2}}\right) = \sin\frac{\pi}{2} = 1, \qquad \sin\!\left(\frac{1}{b_n^{2}}\right) = \sin\!\left(-\frac{\pi}{2}\right) = -1 , \end{equation*}

and therefore

\begin{equation*} f(a_n) = a_n^{2} = \frac{1}{\left(2n + \tfrac12\right)\pi}, \qquad f(b_n) = -\,b_n^{2} = -\,\frac{1}{\left(2n - \tfrac12\right)\pi}. \end{equation*}

Lower bound for the integral.

On the closed interval \([a_n, b_n] \subset (0,1]\) the derivative \(f^{\prime}\) is continuous, so its Riemann and Lebesgue integrals agree (by Theorem 3 of Chapter 4) and the Fundamental Theorem of Calculus applies:

\begin{equation*} \int_{a_n}^{b_n} f^{\prime} = f(b_n) - f(a_n). \end{equation*}

Hence

\begin{equation*} \int_{a_n}^{b_n} |f^{\prime}| \ \ge\ \left| \int_{a_n}^{b_n} f^{\prime} \right| = |f(b_n) - f(a_n)| = a_n^{2} + b_n^{2} = \frac{1}{\left(2n+\tfrac12\right)\pi} + \frac{1}{\left(2n-\tfrac12\right)\pi} > \frac{2}{(2n+1)\pi}. \end{equation*}

Now \(|f^{\prime}|\) is nonnegative and measurable on \([0,1]\), and the intervals \([a_n, b_n]\) are disjoint measurable subsets of \([0,1]\). By Theorem 11 of Chapter 4 (Additivity Over Domains of Integration for nonnegative measurable functions), applied \(N-1\) times, \(\int_{A_N} |f^{\prime}| = \sum_{n=1}^{N} \int_{a_n}^{b_n} |f^{\prime}|\) where \(A_N = \bigcup_{n=1}^{N} [a_n,b_n]\); and, again by Theorem 11 of Chapter 4 applied to the disjoint decomposition \([0,1] = A_N \cup ([0,1] \setminus A_N)\) together with the nonnegativity of \(\int_{[0,1] \setminus A_N} |f^{\prime}|\), we get \(\int_{0}^{1}|f^{\prime}| \ge \int_{A_N} |f^{\prime}|\). Hence, for every natural number \(N\),

\begin{equation*} \int_{0}^{1} |f^{\prime}| \ \ge\ \sum_{n=1}^{N} \int_{a_n}^{b_n} |f^{\prime}| \ >\ \frac{2}{\pi} \sum_{n=1}^{N} \frac{1}{2n+1} . \end{equation*}

The series \(\sum_{n \ge 1} 1/(2n+1)\) diverges, being a positive multiple, term by term, of a tail of the harmonic series: \(1/(2n+1) \ge 1/(3n)\) for \(n \ge 1\). Letting \(N \to \infty\),

\begin{equation*} \int_{0}^{1} |f^{\prime}| = \infty . \end{equation*}

Therefore \(f^{\prime}\) is not integrable over \([0,1]\), even though \(f\) is continuous on \([0,1]\) and differentiable at every point of \([0,1]\). This is exactly the phenomenon the remark asserts: without a monotonicity hypothesis, differentiability at every point does not force the derivative to be integrable. Corollary 4 of this section, which does give the integrability of \(f^{\prime}\), is therefore not merely an artefact of its proof: its hypothesis that \(f\) be increasing cannot simply be dropped.

No. There are continuous functions on \([0,1]\) whose restriction to every nondegenerate closed subinterval fails to be of bounded variation.

The example. As recorded in the Remark that follows the proof of Lebesgue’s Theorem in Section 6.2, Weierstrass exhibited in 1872 a function that is continuous on an open interval and differentiable at no point of that interval; the footnote there points to the simpler example of Bartel van der Waerden, examined in Chapter 8 of Fitzpatrick’s Advanced Calculus. Let \(w\) be such a function, continuous and nowhere differentiable on an open interval \((\alpha, \beta)\), and fix \(\alpha < \alpha^{\prime} < \beta^{\prime} < \beta\). Define

\begin{equation*} f(x) = w\big( \alpha^{\prime} + (\beta^{\prime} - \alpha^{\prime}) x \big) \quad \text{for } x \in [0,1]. \end{equation*}

Then \(f\) is continuous on \([0,1]\), being the composition of a continuous function with an affine map. Moreover \(f\) is differentiable at no point \(x_0\) of \((0,1)\). Indeed, write \(c = \beta^{\prime} - \alpha^{\prime} > 0\) and \(p = \alpha^{\prime} + c x_0 \in (\alpha^{\prime}, \beta^{\prime}) \subseteq (\alpha, \beta)\). For \(s \neq 0\) small enough that \(x_0 + s \in (0,1)\),

\begin{equation*} \frac{w(p + cs) - w(p)}{cs} = \frac{1}{c} \cdot \frac{f(x_0 + s) - f(x_0)}{s}. \end{equation*}

As \(s\) ranges over the punctured neighbourhoods of \(0\), \(t = cs\) ranges over the punctured neighbourhoods of \(0\) as well. So if the right-hand side had a limit as \(s \to 0\), the difference quotients of \(w\) at \(p\) would have a limit as \(t \to 0\), that is, \(w\) would be differentiable at \(p\), which is false. Hence \(f\) is not differentiable at \(x_0\).

Why no subinterval works. Suppose, to get a contradiction, that \([a,b] \subseteq [0,1]\) is nondegenerate, so \(a < b\), and that the restriction \(g = f|_{[a,b]}\) is of bounded variation on \([a,b]\). By Corollary 6 of this section, \(g\) is then differentiable almost everywhere on the open interval \((a,b)\). Since \(m\big( (a,b) \big) = b - a > 0\), the set of points of \((a,b)\) at which \(g\) is differentiable is nonempty; choose such a point \(x_0 \in (a,b)\).

Differentiability of the restriction \(g\) at the interior point \(x_0\) is precisely differentiability of \(f\) at \(x_0\): the difference quotients

\begin{equation*} \frac{g(x_0 + t) - g(x_0)}{t} = \frac{f(x_0+t)-f(x_0)}{t} \end{equation*}

coincide for all \(t \neq 0\) with \(|t| < \min\{x_0 - a,\ b - x_0\}\), and the derivative is a limit as \(t \to 0\), so only such \(t\) matter. Hence \(f\) is differentiable at \(x_0 \in (0,1)\), contradicting the fact that \(f\) is nowhere differentiable.

Therefore no nondegenerate closed subinterval \([a,b]\) of \([0,1]\) has \(f|_{[a,b]}\) of bounded variation, and the answer to the question is negative.

Remark. Continuity by itself is therefore of no help here. What the argument uses is only Corollary 6: if \(f\) restricted to some nondegenerate \([a,b]\) were of bounded variation, then \(f\) would be differentiable at almost every point of \((a,b)\), hence at some point, which a nowhere differentiable \(f\) forbids. Compare Problem 7 of Section 6.2, where the same kind of example answers the analogous question about monotonicity.

No: \(TV(f) = \infty\), so \(f\) is not of bounded variation on \([0,1]\).

Here \(f(x) = 1\) if \(x \in [0,1]\) is rational and \(f(x) = 0\) if \(x \in [0,1]\) is irrational.

Fix a natural number \(n\). We construct a partition \(P_n = \{x_0, x_1, \dots, x_{2n}\}\) of \([0,1]\) whose points alternate between rational and irrational. For \(0 \le k \le n\) put

\begin{equation*} x_{2k} = \frac{k}{n}, \end{equation*}

so that \(x_0 = 0\), \(x_{2n} = 1\), and every \(x_{2k}\) is rational. For \(0 \le k \le n-1\), the open interval \(\left( \tfrac{k}{n},\ \tfrac{k+1}{n} \right)\) is nonempty and the irrational numbers are dense in \(\mathbb{R}\), so we may choose an irrational number

\begin{equation*} x_{2k+1} \in \left( \frac{k}{n},\ \frac{k+1}{n} \right). \end{equation*}

Then

\begin{equation*} 0 = x_0 < x_1 < x_2 < \cdots < x_{2n-1} < x_{2n} = 1 \end{equation*}

is a genuine partition of \([0,1]\), with \(f(x_i) = 1\) for even \(i\) and \(f(x_i) = 0\) for odd \(i\).

Consequently every one of the \(2n\) consecutive differences has absolute value \(1\):

\begin{equation*} V(f, P_n) = \sum_{i=1}^{2n} |f(x_i) - f(x_{i-1})| = \sum_{i=1}^{2n} 1 = 2n . \end{equation*}

By the definition of total variation as the supremum of \(V(f,P)\) over all partitions \(P\) of \([0,1]\),

\begin{equation*} TV(f) \ge V(f, P_n) = 2n \quad \text{for every natural number } n, \end{equation*}

and hence \(TV(f) = \infty\). By the definition of bounded variation, \(f\) is not of bounded variation on \([0,1]\).

Remark. Two independent confirmations. First, Corollary 6 of this section: were \(f\) of bounded variation on \([0,1]\), it would be differentiable at almost every point of \((0,1)\); but the density of both the rationals and the irrationals makes \(f\) discontinuous at every point of \([0,1]\), hence differentiable nowhere, and \((0,1)\) does not have measure zero. Second, Jordan’s Theorem: an increasing function on \([0,1]\) has at most countably many points of discontinuity, since the open intervals \(\big( \lim_{x \to c^{-}} g(x),\ \lim_{x \to c^{+}} g(x) \big)\) associated with distinct discontinuity points \(c\) are nonempty and pairwise disjoint, and each contains a rational. So the difference of two increasing functions is continuous off a countable set, whereas \(f\) is continuous nowhere.

A first, simple answer. Define, for \(x \in [0, 2\pi]\),

\begin{equation*} h(x) = x + \sin x, \qquad g(x) = x . \end{equation*}

Then \(h - g = \sin = f\) on \([0,2\pi]\). The function \(g\) is clearly increasing. As for \(h\), it is differentiable on \([0,2\pi]\) with

\begin{equation*} h^{\prime}(x) = 1 + \cos x \ge 0 \quad \text{for all } x \in [0, 2\pi], \end{equation*}

since \(\cos x \ge -1\). A function that is continuous on a closed, bounded interval and has nonnegative derivative there is increasing: for \(u < v\) the Mean Value Theorem provides \(\xi \in (u,v)\) with \(h(v) - h(u) = h^{\prime}(\xi)(v-u) \ge 0\). Hence \(h\) is increasing, and \(f = h - g\) is a decomposition of the required kind.

The canonical Jordan decomposition. It is instructive to write down the decomposition supplied by Lemma 5, namely

\begin{equation*} f(x) = \big[ f(x) + TV\big(f_{[0,x]}\big) \big] - TV\big(f_{[0,x]}\big) \quad \text{for all } x \in [0,2\pi]. \end{equation*}

Write \(V(x) = TV(f_{[0,x]})\) for the total variation function. Since \(\sin\) is increasing on \([0, \pi/2]\), decreasing on \([\pi/2, 3\pi/2]\) and increasing on \([3\pi/2, 2\pi]\), and since the total variation of a monotone function on an interval is the absolute value of the difference of its endpoint values (the first Example of this section), additivity (20) of the total variation over adjacent intervals gives

\begin{equation*} V(x) = \begin{cases} \sin x & \text{if } 0 \le x \le \pi/2, \\[2pt] 2 - \sin x & \text{if } \pi/2 \le x \le 3\pi/2, \\[2pt] 4 + \sin x & \text{if } 3\pi/2 \le x \le 2\pi . \end{cases} \end{equation*}

Indeed \(V(\pi/2) = 1\), \(V(3\pi/2) = 1 + 2 = 3\) and \(V(2\pi) = 3 + 1 = 4\), and the three formulas agree at the junction points \(\pi/2\) and \(3\pi/2\). Each piece is increasing on its subinterval, so \(V\) is increasing on \([0,2\pi]\).

Therefore, taking \(g = V\) and \(h = f + V\),

\begin{equation*} g(x) = \begin{cases} \sin x & \text{if } 0 \le x \le \pi/2, \\[2pt] 2 - \sin x & \text{if } \pi/2 \le x \le 3\pi/2, \\[2pt] 4 + \sin x & \text{if } 3\pi/2 \le x \le 2\pi, \end{cases} \qquad h(x) = \begin{cases} 2\sin x & \text{if } 0 \le x \le \pi/2, \\[2pt] 2 & \text{if } \pi/2 \le x \le 3\pi/2, \\[2pt] 4 + 2\sin x & \text{if } 3\pi/2 \le x \le 2\pi . \end{cases} \end{equation*}

Both are increasing: \(g\) rises from \(0\) to \(1\) on \([0,\pi/2]\), from \(1\) to \(3\) on \([\pi/2, 3\pi/2]\) (as \(\sin\) decreases there), and from \(3\) to \(4\) on \([3\pi/2, 2\pi]\); while \(h\) rises from \(0\) to \(2\) on \([0,\pi/2]\), is constant equal to \(2\) on \([\pi/2, 3\pi/2]\), and rises from \(2\) to \(4\) on \([3\pi/2, 2\pi]\). The values match at \(\pi/2\) and \(3\pi/2\), so both functions are well defined and increasing on all of \([0,2\pi]\). Finally, on each of the three subintervals,

\begin{equation*} h - g = 2\sin x - \sin x = \sin x, \qquad 2 - (2 - \sin x) = \sin x, \qquad (4 + 2\sin x) - (4 + \sin x) = \sin x, \end{equation*}

so \(f = h - g\) on \([0,2\pi]\), as required.

Set-up and notation. Recall that \(f\) is a step function on \([a,b]\) provided there is a partition

\begin{equation*} a = x_0 < x_1 < \cdots < x_n = b \end{equation*}

of \([a,b]\) such that \(f\) is constant on each open interval \((x_{i-1}, x_i)\). Write

\begin{equation*} c_i = \text{the value of } f \text{ on } (x_{i-1}, x_i) \quad (1 \le i \le n), \qquad d_i = f(x_i) \quad (0 \le i \le n). \end{equation*}

The values \(d_i\) at the partition points are unconstrained; they need not agree with either neighbouring constant. We claim

\begin{equation*} TV(f) = \sum_{i=1}^{n} \Big( |c_i - d_{i-1}| + |d_i - c_i| \Big). \tag{\(\dagger\)} \end{equation*}

Written out, this is the sum of the absolute jumps encountered as one runs through the values

\begin{equation*} d_0,\ c_1,\ d_1,\ c_2,\ d_2,\ \dots,\ c_n,\ d_n \end{equation*}

in order. In particular \(TV(f) < \infty\), so every step function is of bounded variation.

The lower bound. For each \(i\) with \(1 \le i \le n\) pick a point \(y_i \in (x_{i-1}, x_i)\); this is possible since \(x_{i-1} < x_i\). Let

\begin{equation*} P^{*} = \{x_0 < y_1 < x_1 < y_2 < x_2 < \cdots < y_n < x_n\}, \end{equation*}

a partition of \([a,b]\). Since \(f(y_i) = c_i\) and \(f(x_i) = d_i\),

\begin{equation*} V(f, P^{*}) = \sum_{i=1}^{n} \Big( |f(y_i) - f(x_{i-1})| + |f(x_i) - f(y_i)| \Big) = \sum_{i=1}^{n} \Big( |c_i - d_{i-1}| + |d_i - c_i| \Big). \end{equation*}

Since \(TV(f)\) is the supremum of \(V(f, P)\) over all partitions \(P\) of \([a,b]\), the right-hand side of \((\dagger)\) is a lower bound for \(TV(f)\).

The upper bound. Let \(P\) be any partition of \([a,b]\) and let \(P^{\prime}\) be the refinement of \(P\) obtained by adjoining to it all the points \(x_0, x_1, \dots, x_n\). By the observation made in this section just before formula (19) – if \(c\) belongs to \((a,b)\), \(P\) is a partition of \([a,b]\) and \(P^{\prime}\) is the refinement of \(P\) obtained by adjoining \(c\) to \(P\), then \(V(f,P) \le V(f,P^{\prime})\), by the triangle inequality – adjoining a point to a partition does not decrease the variation, so, adjoining the missing \(x_i\) one at a time,

\begin{equation*} V(f, P) \le V(f, P^{\prime}). \end{equation*}

It therefore suffices to bound \(V(f, P^{\prime})\) for an arbitrary partition \(P^{\prime}\) that contains every \(x_i\).

Fix such a \(P^{\prime}\) and fix \(i\) with \(1 \le i \le n\). The points of \(P^{\prime}\) lying in \([x_{i-1}, x_i]\) form a finite set

\begin{equation*} x_{i-1} = s_0 < s_1 < \cdots < s_k = x_i , \end{equation*}

and each of \(s_1, \dots, s_{k-1}\) lies in the open interval \((x_{i-1}, x_i)\), so \(f(s_j) = c_i\) for \(1 \le j \le k-1\). There are two cases.

If \(k \ge 2\), then

\begin{equation*} \sum_{j=1}^{k} |f(s_j) - f(s_{j-1})| = |c_i - d_{i-1}| + \underbrace{0 + \cdots + 0}_{k-2 \text{ terms}} + |d_i - c_i| = |c_i - d_{i-1}| + |d_i - c_i| . \end{equation*}

If \(k = 1\), then, by the triangle inequality,

\begin{equation*} \sum_{j=1}^{k} |f(s_j) - f(s_{j-1})| = |d_i - d_{i-1}| \le |d_i - c_i| + |c_i - d_{i-1}| . \end{equation*}

In either case the contribution to \(V(f,P^{\prime})\) of the consecutive pairs of \(P^{\prime}\) lying inside \([x_{i-1},x_i]\) is at most \(|c_i - d_{i-1}| + |d_i - c_i|\).

Every pair of consecutive points of \(P^{\prime}\) lies inside exactly one of the intervals \([x_{i-1}, x_i]\), because \(P^{\prime}\) contains all the \(x_i\). Summing over \(i\),

\begin{equation*} V(f, P^{\prime}) \le \sum_{i=1}^{n} \Big( |c_i - d_{i-1}| + |d_i - c_i| \Big). \end{equation*}

Hence the same bound holds for \(V(f,P)\) for every partition \(P\), and so the right-hand side of \((\dagger)\) is an upper bound for \(TV(f)\).

Combining the two bounds gives \((\dagger)\).

Two checks. Suppose the values at the partition points are chosen so that \(d_0 = c_1\) and \(d_i = c_i\) for \(1 \le i \le n\), that is, \(f\) is left-continuous at each \(x_i\). Then \(|d_i - c_i| = 0\) for every \(i\), and \(|c_i - d_{i-1}| = |c_i - c_{i-1}|\) for \(i \ge 2\) while \(|c_1 - d_0| = 0\), so \((\dagger)\) reduces to \(\sum_{i=2}^{n} |c_i - c_{i-1}|\), the sum of the absolute jumps of \(f\), which is the expected answer. And for \(f = \chi_{\{x_1\}}\) on \([a,b]\) with \(a < x_1 < b\) we get \(c_1 = c_2 = 0\), \(d_0 = d_2 = 0\), \(d_1 = 1\), so \((\dagger)\) gives \(TV(f) = (0 + 1) + (1 + 0) = 2\), as it must be: \(f\) rises by \(1\) and falls by \(1\).

(a) No: \(f\) is not of bounded variation on \([-1,1]\).

Observe first that \(f\) is continuous on \([-1,1]\): away from \(0\) it is a composition and product of continuous functions, and \(|f(x)| \le x^2\) for all \(x\), so \(f(x) \to 0 = f(0)\) as \(x \to 0\). Thus \(f\) is a perfectly respectable continuous function; nonetheless its total variation is infinite.

For each natural number \(n\) put

\begin{equation*} x_n = \frac{1}{\sqrt{n\pi}}, \end{equation*}

so that \(1/x_n^2 = n\pi\) and hence

\begin{equation*} f(x_n) = \frac{1}{n\pi}\cos(n\pi) = \frac{(-1)^n}{n\pi}. \end{equation*}

The sequence \(\{x_n\}\) is strictly decreasing to \(0\), and \(x_1 = 1/\sqrt{\pi} < 1\) since \(\pi > 1\). Therefore, for each natural number \(N \ge 2\),

\begin{equation*} P_N = \{-1,\ 0,\ x_N,\ x_{N-1},\ \ldots,\ x_2,\ x_1,\ 1\} \end{equation*}

is a genuine partition of \([-1,1]\): its points are listed in strictly increasing order.

Since every term in the sum defining \(V(f,P_N)\) is nonnegative, discarding all the terms except those coming from consecutive pairs \(x_{n+1} < x_n\) gives

\begin{equation*} V(f, P_N) \ \ge\ \sum_{n=1}^{N-1} \bigl| f(x_n) - f(x_{n+1}) \bigr|. \end{equation*}

Because \(f(x_n)\) and \(f(x_{n+1})\) have opposite signs, \(|f(x_n) - f(x_{n+1})| = |f(x_n)| + |f(x_{n+1})|\), and therefore

\begin{equation*} V(f,P_N) \ \ge\ \sum_{n=1}^{N-1}\left[\frac{1}{n\pi} + \frac{1}{(n+1)\pi}\right] \ \ge\ \frac{1}{\pi}\sum_{n=1}^{N-1}\frac{1}{n}. \end{equation*}

The harmonic series diverges, so the right-hand side tends to \(\infty\) as \(N \to \infty\). Hence the set of variations of \(f\) with respect to partitions of \([-1,1]\) is unbounded above, that is,

\begin{equation*} TV(f) = \sup\{V(f,P) : P \text{ a partition of } [-1,1]\} = \infty, \end{equation*}

and \(f\) is not of bounded variation on \([-1,1]\).

(b) Yes: \(g\) is of bounded variation on \([-1,1]\).

We show \(g\) is Lipschitz on \([-1,1]\) and then invoke the observation made in Section 6.3 immediately after the definition of total variation: if \(h\) is Lipschitz on \([a,b]\) with Lipschitz constant \(c\), then for every partition \(P = \{x_0,\ldots,x_k\}\) of \([a,b]\),

\begin{equation*} V(h,P) = \sum_{i=1}^{k}|h(x_i)-h(x_{i-1})| \le c\sum_{i=1}^{k}(x_i - x_{i-1}) = c\,(b-a), \end{equation*}

so \(c\,(b-a)\) bounds the set of all variations of \(h\) and hence \(TV(h) \le c\,(b-a) < \infty\).

For \(x \neq 0\) the product and chain rules give

\begin{equation*} g^{\prime}(x) = 2x\cos(1/x) + x^2 \cdot \sin(1/x)\cdot \frac{1}{x^2} = 2x\cos(1/x) + \sin(1/x). \end{equation*}

At \(x = 0\) the difference quotient is

\begin{equation*} \frac{g(h) - g(0)}{h} = \frac{h^2\cos(1/h)}{h} = h\cos(1/h), \end{equation*}

whose absolute value is at most \(|h|\); hence it tends to \(0\) and \(g^{\prime}(0) = 0\). So \(g\) is differentiable at every point of \([-1,1]\), and for \(x \neq 0\) in \([-1,1]\),

\begin{equation*} |g^{\prime}(x)| \le 2|x|\,|\cos(1/x)| + |\sin(1/x)| \le 2 + 1 = 3, \end{equation*}

while \(|g^{\prime}(0)| = 0 \le 3\). Thus \(|g^{\prime}| \le 3\) on all of \([-1,1]\).

Now let \(u < v\) belong to \([-1,1]\). The function \(g\) is continuous on \([u,v]\) and differentiable on \((u,v)\), so by the Mean Value Theorem there is a point \(c \in (u,v)\) with

\begin{equation*} g(v) - g(u) = g^{\prime}( c)\,(v-u), \end{equation*}

whence \(|g(v)-g(u)| \le 3\,|v-u|\). Therefore \(g\) is Lipschitz on \([-1,1]\) with constant \(3\), and consequently

\begin{equation*} TV(g) \le 3\cdot\bigl(1 - (-1)\bigr) = 6 < \infty. \end{equation*}

So \(g\) is of bounded variation on \([-1,1]\).

The contrast between (a) and (b) is exactly the competition between the amplitude \(x^2\) and the frequency of oscillation: in (b) the oscillation \(\cos(1/x)\) is slow enough that the amplitude \(x^2\) damps it out, while in (a) the far faster oscillation \(\cos(1/x^2)\) forces the graph up and down across a total height comparable to \(\sum 1/n\).

Throughout, \(f\) and \(g\) are real-valued functions of bounded variation on the closed, bounded interval \([a,b]\), and \(P = \{x_0, \ldots, x_k\}\) denotes an arbitrary partition of \([a,b]\).

Linear combinations. Let \(\alpha, \beta\) be real numbers and set \(h = \alpha f + \beta g\). For any partition \(P\),

\begin{equation*} \begin{aligned} V(h,P) &= \sum_{i=1}^{k} \bigl| \alpha[f(x_i)-f(x_{i-1})] + \beta[g(x_i)-g(x_{i-1})] \bigr| \\ &\le |\alpha| \sum_{i=1}^{k}|f(x_i)-f(x_{i-1})| + |\beta| \sum_{i=1}^{k}|g(x_i)-g(x_{i-1})| \\ &= |\alpha|\,V(f,P) + |\beta|\,V(g,P) \\ &\le |\alpha|\,TV(f) + |\beta|\,TV(g), \end{aligned} \end{equation*}

the first inequality being the triangle inequality applied termwise, and the last following from the definition of \(TV\) as a supremum over partitions. Thus the number \(|\alpha|\,TV(f) + |\beta|\,TV(g)\), which is finite because \(f\) and \(g\) are of bounded variation, is an upper bound for the set of variations of \(h\) with respect to partitions of \([a,b]\). Taking the supremum over all partitions \(P\),

\begin{equation*} TV(\alpha f + \beta g) \le |\alpha|\,TV(f) + |\beta|\,TV(g) < \infty, \end{equation*}

so \(\alpha f + \beta g\) is of bounded variation on \([a,b]\).

Products. Yes, the product of two functions of bounded variation on \([a,b]\) is again of bounded variation on \([a,b]\).

First observe that a function of bounded variation on \([a,b]\) is bounded. Indeed, if \(x\) belongs to \((a,b)\), then \(\{a, x, b\}\) is a partition of \([a,b]\) and therefore

\begin{equation*} |f(x) - f(a)| \le |f(x)-f(a)| + |f(b)-f(x)| = V(f, \{a,x,b\}) \le TV(f); \end{equation*}

this inequality also holds trivially at \(x = a\), and at \(x=b\) it follows from the partition \(\{a,b\}\). Hence

\begin{equation*} |f(x)| \le |f(a)| + TV(f) =: M \quad \text{for all } x \in [a,b], \end{equation*}

and \(M < \infty\). Let \(N\) be the analogous bound for \(g\).

Now fix a partition \(P = \{x_0,\ldots,x_k\}\). For each index \(i\), add and subtract \(f(x_i)g(x_{i-1})\):

\begin{equation*} \begin{aligned} |f(x_i)g(x_i) - f(x_{i-1})g(x_{i-1})| &= \bigl| f(x_i)[g(x_i)-g(x_{i-1})] + g(x_{i-1})[f(x_i)-f(x_{i-1})] \bigr| \\ &\le M\,|g(x_i)-g(x_{i-1})| + N\,|f(x_i)-f(x_{i-1})|. \end{aligned} \end{equation*}

Summing over \(i\) from \(1\) to \(k\) gives

\begin{equation*} V(fg, P) \le M\,V(g,P) + N\,V(f,P) \le M\cdot TV(g) + N\cdot TV(f). \end{equation*}

The right-hand side is a finite constant independent of \(P\), so taking the supremum over all partitions of \([a,b]\) yields

\begin{equation*} TV(fg) \le M\cdot TV(g) + N\cdot TV(f) < \infty. \end{equation*}

Therefore \(fg\) is of bounded variation on \([a,b]\).

Two remarks. First, both conclusions used only that the interval \([a,b]\) is closed and bounded; boundedness of \(f\) and \(g\), which the product argument needs, was not assumed but deduced from bounded variation. Second, the estimates show that the functions of bounded variation on \([a,b]\) form an algebra of bounded functions.

Recall that \(P\) refines \(P^{\prime}\) means that \(P^{\prime} \subseteq P\) as sets of points of \([a,b]\) (both containing \(a\) and \(b\)). Since \(P\) and \(P^{\prime}\) are finite sets, \(P\) is obtained from \(P^{\prime}\) by adjoining finitely many points, say \(m = \#(P \setminus P^{\prime}) \ge 0\) of them. We argue by induction on \(m\).

If \(m = 0\) then \(P = P^{\prime}\) and there is nothing to prove.

Consider first the case \(m = 1\), that is, \(P\) is obtained from \(P^{\prime} = \{x_0, x_1, \ldots, x_k\}\) by adjoining a single point \(c\) not in \(P^{\prime}\). Since \(a = x_0 \le c \le b = x_k\) and \(c\) differs from every \(x_i\), there is exactly one index \(j\) with \(1 \le j \le k\) for which

\begin{equation*} x_{j-1} < c < x_j . \end{equation*}

Thus \(P = \{x_0, \ldots, x_{j-1}, c, x_j, \ldots, x_k\}\). The two variations \(V(f,P^{\prime})\) and \(V(f,P)\) have all their terms in common except that the single term \(|f(x_j) - f(x_{j-1})|\) of \(V(f,P^{\prime})\) is replaced in \(V(f,P)\) by the two terms \(|f( c)-f(x_{j-1})|\) and \(|f(x_j)-f( c)|\). By the triangle inequality,

\begin{equation*} |f(x_j) - f(x_{j-1})| = \bigl| [f(x_j) - f( c)] + [f( c) - f(x_{j-1})] \bigr| \le |f(x_j)-f( c)| + |f( c)-f(x_{j-1})|. \end{equation*}

Adding to both sides the common terms \(\sum_{i \neq j} |f(x_i) - f(x_{i-1})|\) gives

\begin{equation*} V(f,P^{\prime}) \le V(f,P). \end{equation*}

Now suppose the assertion holds whenever the refinement adjoins \(m\) points, and let \(P\) be obtained from \(P^{\prime}\) by adjoining \(m+1\) points \(c_1, \ldots, c_{m+1}\). Put \(Q = P^{\prime} \cup \{c_1, \ldots, c_m\}\). Then \(Q\) is a refinement of \(P^{\prime}\) adjoining \(m\) points, so by the inductive hypothesis \(V(f,P^{\prime}) \le V(f,Q)\); and \(P = Q \cup \{c_{m+1}\}\) is a refinement of \(Q\) adjoining a single point, so by the case \(m=1\) already established, \(V(f,Q) \le V(f,P)\). Combining,

\begin{equation*} V(f,P^{\prime}) \le V(f,Q) \le V(f,P). \end{equation*}

By induction, \(V(f,P^{\prime}) \le V(f,P)\) whenever \(P\) is a refinement of \(P^{\prime}\), as required.

Note that no hypothesis whatever is needed on \(f\): the argument uses only the triangle inequality in \(\mathbb{R}\). In particular the inequality holds even when \(TV(f) = \infty\).

Here \(TV(f,P_n)\) denotes the variation \(V(f,P_n)\) of \(f\) with respect to the partition \(P_n\), as defined in this section; we write \(V(f,P)\) for it below. “Increasing” is meant in the weak sense, \(V(f,P_n) \le V(f,P_{n+1})\) for all \(n\).

Since \(f\) is of bounded variation, the number

\begin{equation*} TV(f) = \sup\{V(f,P) : P \text{ a partition of } [a,b]\} \end{equation*}

is finite. By the definition of the supremum, for each natural number \(n\) there is a partition \(Q_n\) of \([a,b]\) with

\begin{equation*} TV(f) - \frac{1}{n} < V(f, Q_n) \le TV(f). \end{equation*}

Call this inequality \((*)\).

Define

\begin{equation*} P_n = Q_1 \cup Q_2 \cup \cdots \cup Q_n \qquad (n = 1, 2, \ldots), \end{equation*}

the union being taken as sets of points of \([a,b]\). Each \(Q_j\) is a finite subset of \([a,b]\) containing \(a\) and \(b\), so \(P_n\) is a finite subset of \([a,b]\) containing \(a\) and \(b\); listing its points in increasing order exhibits \(P_n\) as a partition of \([a,b]\).

Two containments are immediate from the definition of \(P_n\):

\begin{equation*} P_n \subseteq P_{n+1} \quad \text{and} \quad Q_n \subseteq P_n . \end{equation*}

The first says \(P_{n+1}\) is a refinement of \(P_n\), so by the preceding problem

\begin{equation*} V(f,P_n) \le V(f,P_{n+1}) \quad \text{for every } n, \end{equation*}

that is, the sequence \(\{V(f,P_n)\}\) is increasing. The second says \(P_n\) is a refinement of \(Q_n\), so again by the preceding problem, together with \((*)\),

\begin{equation*} V(f,P_n) \ \ge\ V(f,Q_n) \ >\ TV(f) - \frac{1}{n}. \end{equation*}

On the other hand \(V(f,P_n) \le TV(f)\) for every \(n\), since \(TV(f)\) is the supremum of the variations of \(f\) over all partitions. Hence

\begin{equation*} TV(f) - \frac{1}{n} < V(f,P_n) \le TV(f) \quad \text{for every } n, \end{equation*}

and therefore \(|V(f,P_n) - TV(f)| < 1/n\) for every \(n\). Since \(1/n \to 0\), we conclude that

\begin{equation*} \lim_{n \to \infty} V(f,P_n) = TV(f). \end{equation*}

Thus \(\{P_n\}\) is a sequence of partitions of \([a,b]\) for which \(\{V(f,P_n)\}\) is increasing and converges to \(TV(f)\).

The inequality is to be read in the extended reals: no boundedness of variation is assumed of the \(f_n\), and if \(\liminf_n TV(f_n) = \infty\) there is nothing to prove. So the content of the assertion is that if the \(f_n\) have variations that do not blow up along some subsequence, then neither does the variation of the limit.

Fix an arbitrary partition \(P = \{x_0, x_1, \ldots, x_k\}\) of \([a,b]\). The key point is that \(V(f,P)\) involves only the finitely many values \(f(x_0), \ldots, f(x_k)\), and at each of these finitely many points the numerical sequence \(\{f_n(x_i)\}\) converges to \(f(x_i)\) by hypothesis.

Consequently, for each index \(i\) with \(1 \le i \le k\),

\begin{equation*} \lim_{n\to\infty} \bigl| f_n(x_i) - f_n(x_{i-1}) \bigr| = \bigl| f(x_i) - f(x_{i-1}) \bigr|, \end{equation*}

by the algebra of limits of real sequences together with the continuity of the absolute value function. Since a sum of finitely many convergent sequences converges to the sum of the limits,

\begin{equation*} \lim_{n \to \infty} V(f_n, P) = \lim_{n\to\infty} \sum_{i=1}^{k} \bigl| f_n(x_i) - f_n(x_{i-1}) \bigr| = \sum_{i=1}^{k} \bigl| f(x_i) - f(x_{i-1}) \bigr| = V(f,P). \end{equation*}

On the other hand, by the definition of total variation as a supremum over partitions,

\begin{equation*} V(f_n, P) \le TV(f_n) \quad \text{for every } n. \end{equation*}

Since the sequence \(\{V(f_n,P)\}\) converges, its limit equals its limit inferior, and the limit inferior is monotone with respect to termwise inequality of sequences. Hence

\begin{equation*} V(f,P) = \lim_{n\to\infty} V(f_n,P) = \liminf_{n \to \infty} V(f_n, P) \le \liminf_{n\to\infty} TV(f_n). \end{equation*}

The number \(\liminf_n TV(f_n)\) therefore is an upper bound for \(V(f,P)\), and this holds for every partition \(P\) of \([a,b]\), since \(P\) was arbitrary. Taking the supremum over all such \(P\),

\begin{equation*} TV(f) = \sup_{P} V(f,P) \le \liminf_{n\to\infty} TV(f_n), \end{equation*}

which is the desired inequality.

In particular, if there is a constant \(M\) with \(TV(f_n) \le M\) for all \(n\), then \(TV(f) \le M\) and the pointwise limit \(f\) is of bounded variation. The inequality may be strict, as a moving spike shows. On \([0,1]\) let \(f_n\) be the piecewise linear function that vanishes outside the interval \((1/(n+1), 1/n)\) and rises to the value \(1\) at the midpoint of that interval. Then \(TV(f_n) = 2\) for every \(n\). Yet \(f_n \to 0\) pointwise on \([0,1]\): we have \(f_n(0) = 0\) for all \(n\), and if \(0 < x \le 1\) then \(1/n < x\) for all \(n\) large, so \(f_n(x) = 0\) for all such \(n\). Here \(TV(f) = TV(0) = 0 < 2 = \liminf_n TV(f_n)\).

Let \(P = \{x_0, x_1, \ldots, x_k\}\) be an arbitrary partition of \([a,b]\).

Subadditivity. By the triangle inequality in \(\mathbb{R}\), applied to each term,

\begin{equation*} \begin{aligned} V(f+g, P) &= \sum_{i=1}^{k} \bigl| [f(x_i)+g(x_i)] - [f(x_{i-1})+g(x_{i-1})] \bigr| \\ &= \sum_{i=1}^{k} \bigl| [f(x_i)-f(x_{i-1})] + [g(x_i)-g(x_{i-1})] \bigr| \\ &\le \sum_{i=1}^{k} |f(x_i)-f(x_{i-1})| + \sum_{i=1}^{k} |g(x_i)-g(x_{i-1})| \\ &= V(f,P) + V(g,P). \end{aligned} \end{equation*}

Since \(V(f,P) \le TV(f)\) and \(V(g,P) \le TV(g)\) by the definition of total variation, we obtain

\begin{equation*} V(f+g,P) \le TV(f) + TV(g). \end{equation*}

The right-hand side is a finite number independent of \(P\); as \(P\) was an arbitrary partition of \([a,b]\), it is an upper bound for the set of variations of \(f+g\) with respect to partitions of \([a,b]\). Taking the supremum over all partitions,

\begin{equation*} TV(f+g) \le TV(f) + TV(g). \end{equation*}

In particular \(f + g\) is of bounded variation on \([a,b]\).

Positive homogeneity. Let \(\alpha\) be a real number. For the partition \(P\) above,

\begin{equation*} V(\alpha f, P) = \sum_{i=1}^{k} \bigl| \alpha f(x_i) - \alpha f(x_{i-1}) \bigr| = \sum_{i=1}^{k} |\alpha|\,\bigl|f(x_i)-f(x_{i-1})\bigr| = |\alpha|\,V(f,P). \end{equation*}

Call this identity \((*)\); it holds for every partition \(P\) of \([a,b]\).

If \(\alpha = 0\), then \(\alpha f\) is identically zero, so \(V(\alpha f, P) = 0\) for every \(P\) and hence \(TV(\alpha f) = 0 = |\alpha|\,TV(f)\).

Suppose \(\alpha \neq 0\). Let \(S = \{V(f,P) : P \text{ a partition of } [a,b]\}\), so that \(TV(f) = \sup S\). By \((*)\), the set of variations of \(\alpha f\) with respect to partitions of \([a,b]\) is exactly

\begin{equation*} \{V(\alpha f, P) : P\} = \{|\alpha| \, s : s \in S\} = |\alpha| \cdot S . \end{equation*}

For a nonempty set \(S\) of real numbers and a positive constant \(c = |\alpha|\), one has \(\sup(c\cdot S) = c \sup S\): indeed \(s \le \sup S\) gives \(cs \le c\sup S\) for all \(s \in S\), so \(\sup(c\cdot S) \le c \sup S\); and applying this same inequality to the set \(c \cdot S\) with constant \(1/c\) gives \(\sup S = \sup\bigl( (1/c)(c\cdot S) \bigr) \le (1/c)\sup(c\cdot S)\), that is, \(c\sup S \le \sup(c\cdot S)\). Therefore

\begin{equation*} TV(\alpha f) = \sup\{V(\alpha f, P) : P\} = |\alpha| \sup\{V(f,P) : P\} = |\alpha|\,TV(f). \end{equation*}

Both computations show, incidentally, that the functions of bounded variation on \([a,b]\) form a linear space on which \(TV\) is a seminorm; it fails to be a norm only because \(TV(f) = 0\) exactly when \(f\) is constant.

Note at the outset that \(f\) is continuous on \([0,1]\). On \((0,1]\) it is built from continuous functions by composition and multiplication, and since \(|f(x)| \le x^{\alpha}\) for \(0 < x \le 1\) with \(\alpha > 0\), we have \(f(x) \to 0 = f(0)\) as \(x \to 0^{+}\).

The case \(\alpha > \beta\). On the open set \((0,1)\) the product and chain rules give, for \(0 < x < 1\),

\begin{equation*} f^{\prime}(x) = \alpha x^{\alpha-1}\sin(1/x^{\beta}) + x^{\alpha}\cos(1/x^{\beta})\cdot\bigl(-\beta x^{-\beta-1}\bigr) = \alpha x^{\alpha-1}\sin(1/x^{\beta}) - \beta x^{\alpha-\beta-1}\cos(1/x^{\beta}), \end{equation*}

and this formula persists at \(x = 1\) (one-sided). Thus \(f^{\prime}\) exists at every point of \((0,1]\) and is continuous there, hence measurable; \(f^{\prime}\) is therefore defined except possibly at the single point \(x=0\), a set of measure zero, which is all that is needed to speak of its integral over \([0,1]\).

Since \(|\sin| \le 1\) and \(|\cos| \le 1\),

\begin{equation*} |f^{\prime}(x)| \le \alpha x^{\alpha-1} + \beta x^{\alpha-\beta-1} \quad \text{for } 0 < x \le 1. \end{equation*}

Both bounding functions are nonnegative and continuous on \((0,1]\), and for \(0 < \epsilon < 1\) elementary calculus gives

\begin{equation*} \int_{\epsilon}^{1} \alpha x^{\alpha-1}\,dx = 1 - \epsilon^{\alpha} \le 1, \qquad \int_{\epsilon}^{1} \beta x^{\alpha-\beta-1}\,dx = \frac{\beta}{\alpha-\beta}\bigl(1 - \epsilon^{\alpha-\beta}\bigr) \le \frac{\beta}{\alpha-\beta}, \end{equation*}

where the second computation uses \(\alpha - \beta > 0\), so that the exponent \(\alpha-\beta-1\) exceeds \(-1\) and the antiderivative \(x^{\alpha-\beta}/(\alpha-\beta)\) is legitimate and tends to \(0\) as \(\epsilon \to 0^{+}\). On each interval \([1/n, 1]\) the function \(|f^{\prime}|\) is continuous, hence its Riemann and Lebesgue integrals agree, and monotonicity of the integral applied to the pointwise bound displayed above gives

\begin{equation*} \int_{1/n}^{1}|f^{\prime}| \ \le\ \int_{1/n}^{1}\alpha x^{\alpha-1}\,dx + \int_{1/n}^{1}\beta x^{\alpha-\beta-1}\,dx \ \le\ 1 + \frac{\beta}{\alpha-\beta} \qquad \text{for every } n . \end{equation*}

The sequence \(\{|f^{\prime}|\cdot \chi_{[1/n,1]}\}\) consists of nonnegative measurable functions and increases pointwise on \((0,1]\) to \(|f^{\prime}|\), so the Monotone Convergence Theorem (Chapter 4) yields

\begin{equation*} \int_{0}^{1} |f^{\prime}| \ = \ \lim_{n\to\infty}\int_{1/n}^{1}|f^{\prime}| \ \le\ 1 + \frac{\beta}{\alpha-\beta} \ <\ \infty . \end{equation*}

Since \(|f^{\prime}|\) has finite integral and \(f^{\prime}\) is measurable, \(f^{\prime}\) is integrable over \([0,1]\).

We now deduce bounded variation. Let \(P = \{t_0, t_1, \ldots, t_k\}\) be a partition of \([0,1]\), so \(0 = t_0 < t_1 < \cdots < t_k = 1\). For each index \(i \ge 2\) the interval \([t_{i-1}, t_i]\) is a compact subinterval of \((0,1]\) on which \(f\) is continuously differentiable, so the Fundamental Theorem of Calculus of elementary analysis gives

\begin{equation*} f(t_i) - f(t_{i-1}) = \int_{t_{i-1}}^{t_i} f^{\prime} , \qquad\text{hence}\qquad |f(t_i)-f(t_{i-1})| \le \int_{t_{i-1}}^{t_i} |f^{\prime}| ; \end{equation*}

here the Riemann and Lebesgue integrals of the continuous, bounded function \(f^{\prime}\) over \([t_{i-1},t_i]\) agree.

For the first subinterval, let \(0 < s < t_1\). The same argument on \([s,t_1]\) yields

\begin{equation*} |f(t_1) - f(s)| \le \int_{s}^{t_1}|f^{\prime}| \le \int_{0}^{t_1}|f^{\prime}| . \end{equation*}

Letting \(s \to 0^{+}\) and using the continuity of \(f\) at \(0\) established above, together with \(f(0)=0\), we get

\begin{equation*} |f(t_1) - f(t_0)| \le \int_{0}^{t_1}|f^{\prime}| . \end{equation*}

Summing the \(k\) inequalities and using the additivity of the integral over the disjoint intervals \((t_{i-1}, t_i)\) whose union is \((0,1)\) up to a finite set,

\begin{equation*} V(f,P) = \sum_{i=1}^{k}|f(t_i)-f(t_{i-1})| \le \sum_{i=1}^{k}\int_{t_{i-1}}^{t_i}|f^{\prime}| = \int_{0}^{1}|f^{\prime}| . \end{equation*}

The right-hand side is finite and independent of \(P\), so taking the supremum over all partitions of \([0,1]\),

\begin{equation*} TV(f) \le \int_{0}^{1}|f^{\prime}| \le 1 + \frac{\beta}{\alpha-\beta} < \infty, \end{equation*}

and \(f\) is of bounded variation on \([0,1]\).

The case \(\alpha \le \beta\). We exhibit partitions whose variations are unbounded. For each integer \(n \ge 0\) set

\begin{equation*} x_n = \Bigl(\frac{\pi}{2} + n\pi\Bigr)^{-1/\beta}, \end{equation*}

so that \(1/x_n^{\beta} = \pi/2 + n\pi\) and consequently

\begin{equation*} \sin(1/x_n^{\beta}) = (-1)^n, \qquad f(x_n) = (-1)^n x_n^{\alpha}. \end{equation*}

Since \(t \mapsto t^{-1/\beta}\) is strictly decreasing on \((0,\infty)\), the sequence \(\{x_n\}\) is strictly decreasing to \(0\); and \(x_0 = (\pi/2)^{-1/\beta} < 1\) because \(\pi/2 > 1\). Hence, for each integer \(N \ge 1\),

\begin{equation*} P_N = \{0,\ x_N,\ x_{N-1},\ \ldots,\ x_1,\ x_0,\ 1\} \end{equation*}

is a partition of \([0,1]\).

Discarding the nonnegative terms of \(V(f,P_N)\) other than those coming from consecutive pairs \(x_{n+1} < x_n\), and using that \(f(x_n)\) and \(f(x_{n+1})\) have opposite signs,

\begin{equation*} V(f,P_N) \ \ge\ \sum_{n=0}^{N-1}\bigl|f(x_n)-f(x_{n+1})\bigr| = \sum_{n=0}^{N-1}\bigl(x_n^{\alpha} + x_{n+1}^{\alpha}\bigr) \ \ge\ \sum_{n=0}^{N-1} x_n^{\alpha} = \sum_{n=0}^{N-1}\Bigl(\frac{\pi}{2}+n\pi\Bigr)^{-\alpha/\beta}. \end{equation*}

Write \(p = \alpha/\beta\); the hypothesis \(\alpha \le \beta\) says \(0 < p \le 1\). For \(t \ge 1\) and \(0 < p \le 1\) we have \(t^{p} \le t\) and hence \(t^{-p} \ge t^{-1}\). Since \(\pi/2 + n\pi \ge 1\) for every \(n \ge 0\), it follows that

\begin{equation*} V(f,P_N) \ \ge\ \sum_{n=0}^{N-1} \frac{1}{\pi/2 + n\pi} \ \ge\ \sum_{n=0}^{N-1}\frac{1}{\pi(n+1)} = \frac{1}{\pi}\sum_{m=1}^{N}\frac{1}{m}, \end{equation*}

using \(\pi/2 + n\pi \le \pi + n\pi = \pi(n+1)\). Because the harmonic series diverges, the right-hand side tends to \(\infty\) as \(N \to \infty\). Therefore the set of variations of \(f\) with respect to partitions of \([0,1]\) is unbounded above, so

\begin{equation*} TV(f) = \infty \end{equation*}

and \(f\) is not of bounded variation on \([0,1]\).

The dividing line is exactly \(\alpha = \beta\): the amplitude \(x^{\alpha}\) must decay strictly faster than the reciprocal of the number of oscillations accumulated near \(0\), which is governed by \(x^{-\beta}\).

Throughout we use the additivity formula (20) of Section 6.3,

\begin{equation*} TV(f_{[a,b]}) \;=\; TV(f_{[a,c]}) + TV(f_{[c,b]}) \qquad (a \le c \le b). \end{equation*}

No finiteness hypothesis is needed here, and both sides are to be read in \([0,\infty]\): as the text observes, adjoining the point \(c\) to a partition of \([a,b]\) does not decrease the associated variation, so the supremum defining \(TV(f_{[a,b]})\) may be taken over partitions \(P\) containing \(c\); each such \(P\) induces partitions \(P_1\) of \([a,c]\) and \(P_2\) of \([c,b]\) with \(V(f_{[a,b]}, P) = V(f_{[a,c]}, P_1) + V(f_{[c,b]}, P_2)\) by (19), and taking suprema over all such \(P\) yields (20). In particular \(f\) is of bounded variation on \([a,b]\) if and only if it is of bounded variation on both \([a,c]\) and \([c,b]\); contrapositively, if \(f\) fails to be of bounded variation on \([a,b]\) then it fails to be of bounded variation on at least one of the two halves \([a, (a+b)/2]\), \([(a+b)/2, b]\).

Construction of the point. Put \([a_0, b_0] = [0,1]\). Given \([a_n, b_n]\) of length \(2^{-n}\) on which \(f\) fails to be of bounded variation, bisect it; by the remark just made, \(f\) fails to be of bounded variation on at least one of the two closed halves, and we let \([a_{n+1}, b_{n+1}]\) be such a half (say the left one if both qualify). This produces a descending sequence

\begin{equation*} [0,1] = [a_0,b_0] \supseteq [a_1,b_1] \supseteq [a_2,b_2] \supseteq \cdots, \qquad b_n - a_n = 2^{-n}, \end{equation*}

of nonempty closed bounded intervals, on each of which \(f\) fails to be of bounded variation. By the Nested Set Theorem the intersection \(\bigcap_{n\ge 0}[a_n,b_n]\) is nonempty, and since the lengths tend to \(0\) it consists of a single point \(x_0 \in [0,1]\).

The conclusion. Fix \(\delta > 0\) and set \(J_\delta = [x_0 - \delta, x_0 + \delta] \cap [0,1]\), a nondegenerate closed subinterval of \([0,1]\). Choose \(n\) with \(2^{-n} < \delta\). Since \(a_n \le x_0 \le b_n\) and \(b_n - a_n < \delta\), every point of \([a_n,b_n]\) lies within \(\delta\) of \(x_0\); as \([a_n,b_n] \subseteq [0,1]\) we conclude \([a_n, b_n] \subseteq J_\delta\). If \(f\) were of bounded variation on \(J_\delta\), then two applications of (20) (splitting \(J_\delta\) at \(a_n\) and then at \(b_n\)) would show \(f\) to be of bounded variation on \([a_n,b_n]\), a contradiction. Hence

for every \(\delta > 0\), \(f\) fails to be of bounded variation on \([x_0-\delta, x_0+\delta] \cap [0,1]\). \((\ast)\)

Consequently, if \([c,d] \subseteq [0,1]\) is any nondegenerate closed interval containing \(x_0\) in its interior relative to \([0,1]\) (that is, \(c < x_0 < d\), or \(x_0 = c = 0\), or \(x_0 = d = 1\)), then \([c,d] \supseteq J_\delta\) for all small \(\delta > 0\), and by \((\ast)\) together with (20) the function \(f\) fails to be of bounded variation on \([c,d]\).

A remark on the literal statement: the conclusion cannot be upgraded to every nondegenerate closed subinterval containing \(x_0\), once \(x_0\) is allowed to be an endpoint of that subinterval. Indeed, let \(h(t) = t\cos(\pi/2t)\) for \(0 < t \le 1\) and \(h(0) = 0\); the Example of Section 6.3 shows \(h\) is continuous, and using the partitions \(P_n = \{0, 1/2n, 1/[2n-1], \dots, 1/3, 1/2\}\) of \([0,1/2]\), that

\begin{equation*} V(h_{[0,1/2]}, P_n) = \tfrac12 + \sum_{j=2}^{n} \tfrac1j \longrightarrow \infty , \end{equation*}

so \(h\) is not of bounded variation on \([0, 1/2]\). Define \(g\) on \([0,1]\) by \(g(x) = h(1/2 - x)\) for \(0 \le x < 1/2\) and \(g(x) = 0\) for \(1/2 \le x \le 1\). Since \(g(x) = h(1/2 - x)\) holds for every \(x \in [0,1/2]\) (at \(x = 1/2\) both sides are \(0\)) and \(h\) is continuous, \(g\) is continuous on \([0,1/2]\); it is identically \(0\) on \([1/2,1]\), so \(g\) is continuous on \([0,1]\). Being the reflection \(x \mapsto h(1/2 - x)\) of \(h_{[0,1/2]}\), which carries partitions to partitions and leaves the associated variations unchanged, \(g\) fails to be of bounded variation on \([0, 1/2]\), and hence, by (20), on \([0,1]\). On the other hand \(g\) vanishes identically on \([1/2, 1]\), so it is of bounded variation there; and for \([c,d] \subseteq [0, 1/2)\) the numbers \(1/2 - x\), \(x \in [c,d]\), range over \([1/2 - d, 1/2 - c] \subseteq (0, 1/2]\), on which \(h\) is continuously differentiable, so \(g\) is continuously differentiable on \([c,d]\) with bounded derivative, hence Lipschitz there by the Mean Value Theorem and therefore of bounded variation. So for each \(x_0 \in [0,1]\) there is a nondegenerate closed subinterval of \([0,1]\) containing \(x_0\) on which \(g\) is of bounded variation: take \([1/2, 1]\) itself if \(x_0 \ge 1/2\), and take \([0, (x_0 + 1/2)/2]\) if \(x_0 < 1/2\), this last interval being a nondegenerate closed subinterval of \([0,1/2)\) that contains \(x_0\). Thus \((\ast)\), failure on every relative neighborhood of \(x_0\), is the correct form of the assertion, and it is exactly what the bisection argument delivers.

(i) Define

\begin{equation*} f(x) = \begin{cases} x\cos(\pi/2x) & \text{if } 0 < x \le 1,\\ 0 & \text{if } x = 0.\end{cases} \end{equation*}

As observed in the Example of Section 6.3, \(f\) is continuous on \([0,1]\) and is not of bounded variation on \([0,1]\): for the partition \(P_n = \{0, 1/2n, 1/[2n-1], \dots, 1/3, 1/2, 1\}\) one computes \(V(f, P_n) = 1 + 1/2 + \cdots + 1/n\), and the harmonic series diverges.

Fix \(0 < \epsilon < 1\). On \([\epsilon, 1]\) the function \(f\) is continuously differentiable, with

\begin{equation*} f^{\prime}(x) = \cos(\pi/2x) + \frac{\pi}{2x}\sin(\pi/2x), \end{equation*}

so \(|f^{\prime}|\) is bounded on the compact interval \([\epsilon, 1]\), say by \(c\). By the Mean Value Theorem \(|f(u) - f(v)| \le c\,|u-v|\) for \(u, v \in [\epsilon,1]\), so \(f\) is Lipschitz there and hence, by Proposition 7, absolutely continuous on \([\epsilon,1]\).

But \(f\) is not absolutely continuous on \([0,1]\): by Theorem 8, an absolutely continuous function on a closed bounded interval is of bounded variation there, and \(f\) is not of bounded variation on \([0,1]\).

(ii) Assume in addition that \(f\) is increasing on \([0,1]\). We first record an elementary fact: if \(\{(\alpha_k, \beta_k)\}_{k=1}^{n}\) is a disjoint collection of open subintervals of \((u, v)\) and \(g\) is increasing on \([u,v]\), then

\begin{equation*} \sum_{k=1}^{n}\bigl[g(\beta_k) - g(\alpha_k)\bigr] \le g(v) - g(u). \end{equation*}

Indeed, relabel so that \(u \le \alpha_1 < \beta_1 \le \alpha_2 < \beta_2 \le \cdots \le \alpha_n < \beta_n \le v\); since \(g\) is increasing, each of the complementary differences \(g(\alpha_1) - g(u)\), \(g(\alpha_{k+1}) - g(\beta_k)\), \(g(v) - g(\beta_n)\) is nonnegative, and adding them to the displayed sum telescopes to \(g(v) - g(u)\).

Now let \(\epsilon > 0\). Since \(f\) is continuous at \(0\), choose \(c \in (0,1)\) with

\begin{equation*} f( c) - f(0) < \epsilon/2 . \end{equation*}

Since \(f\) is absolutely continuous on \([c, 1]\), choose \(\delta > 0\) that responds to the \(\epsilon/2\) challenge for the absolute continuity of \(f\) on \([c,1]\).

We claim this \(\delta\) responds to the \(\epsilon\) challenge for the absolute continuity of \(f\) on \([0,1]\). Let \(\{(a_k, b_k)\}_{k=1}^{n}\) be a disjoint collection of open subintervals of \((0,1)\) with \(\sum_{k=1}^{n}[b_k - a_k] < \delta\). For each \(k\) write

\begin{equation*} a_k^{\prime} = \max\{a_k, c\},\quad b_k^{\prime} = \max\{b_k, c\}, \qquad a_k^{\prime\prime} = \min\{a_k, c\},\quad b_k^{\prime\prime} = \min\{b_k, c\}. \end{equation*}

Since \(f\) is increasing, an inspection of the three possible positions of \((a_k,b_k)\) relative to \(c\) (entirely right of \(c\), entirely left of \(c\), straddling \(c\)) shows in each case that

\begin{equation*} |f(b_k) - f(a_k)| = f(b_k) - f(a_k) = \bigl[f(b_k^{\prime}) - f(a_k^{\prime})\bigr] + \bigl[f(b_k^{\prime\prime}) - f(a_k^{\prime\prime})\bigr]. \end{equation*}

The intervals \((a_k^{\prime}, b_k^{\prime})\), after discarding the degenerate ones, form a disjoint collection of open subintervals of \((c, 1)\) whose lengths sum to at most \(\sum_k [b_k - a_k] < \delta\); hence by the choice of \(\delta\),

\begin{equation*} \sum_{k=1}^{n}\bigl[f(b_k^{\prime}) - f(a_k^{\prime})\bigr] < \epsilon/2 . \end{equation*}

The intervals \((a_k^{\prime\prime}, b_k^{\prime\prime})\), after discarding degenerate ones, form a disjoint collection of open subintervals of \((0, c)\), so by the elementary fact above,

\begin{equation*} \sum_{k=1}^{n}\bigl[f(b_k^{\prime\prime}) - f(a_k^{\prime\prime})\bigr] \le f( c) - f(0) < \epsilon/2 . \end{equation*}

Adding, \(\sum_{k=1}^{n}|f(b_k) - f(a_k)| < \epsilon\). Therefore \(f\) is absolutely continuous on \([0,1]\).

(iii) Let \(f(x) = \sqrt{x}\) on \([0,1]\). It is continuous and increasing on \([0,1]\). For \(0 < \epsilon < 1\), \(f\) is differentiable on \([\epsilon, 1]\) with \(f^{\prime}(x) = 1/(2\sqrt{x}) \le 1/(2\sqrt{\epsilon})\), so by the Mean Value Theorem \(f\) is Lipschitz on \([\epsilon, 1]\) and therefore, by Proposition 7, absolutely continuous on \([\epsilon,1]\). Since \(f\) is continuous on \([0,1]\) and increasing, part (ii) shows that \(f\) is absolutely continuous on \([0,1]\).

However \(f\) is not Lipschitz on \([0,1]\). If \(c \ge 0\) were a Lipschitz constant, then for \(0 < x \le 1\),

\begin{equation*} \sqrt{x} = |f(x) - f(0)| \le c\,|x - 0| = c\,x, \quad\text{so}\quad c \ge \frac{1}{\sqrt{x}} . \end{equation*}

Letting \(x \to 0^{+}\) makes the right-hand side unbounded, a contradiction. Hence \(f\) is absolutely continuous but not Lipschitz on \([0,1]\); this justifies the assertion made after Proposition 7.

(A word on the transcription. The book prints the display as \(\sum_{k=1}^{\infty}|f(b_k) - f(a_k)| < \sum_{k=1}^{\infty}[b_k - a_k] < \delta\), in which the \(\epsilon\) announced in the preceding clause never appears; the words “\(\epsilon\) if” have been lost in typesetting, exactly as the parallel display in the next problem reads “\(m^{*}(f(E)) < \epsilon\) if \(m(E) < \delta\)”. The statement above is the intended one, and is what we prove.)

Suppose first that \(f\) satisfies the countable criterion. Let \(\epsilon > 0\) and let \(\delta > 0\) respond to it; replacing \(\delta\) by \(\min\{\delta, b-a\}\) we may assume \(\delta \le b - a\). Let \(\{(a_k, b_k)\}_{k=1}^{n}\) be a finite disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^{n}[b_k - a_k] < \delta\). Put \(\sigma = \sum_{k=1}^{n}[b_k - a_k]\). The set \(K = \bigcup_{k=1}^{n}[a_k, b_k]\) is closed, being a finite union of closed intervals, and \(m(K) = \sigma\), since adjoining the finitely many endpoints to the disjoint open intervals adds nothing to the measure. Hence \((a,b) \sim K\) is open and, by the excision property of measure, \(m\bigl((a,b) \sim K\bigr) \ge (b-a) - \sigma > 0\) because \(\sigma < \delta \le b-a\). In particular \((a,b) \sim K\) is nonempty, and being open it contains some open interval \((\alpha, \beta)\) with \(\alpha < \beta\). Choose a disjoint sequence of open intervals \(\{(a_k, b_k)\}_{k=n+1}^{\infty}\) contained in \((\alpha, \beta)\) with

\begin{equation*} \sum_{k=n+1}^{\infty}[b_k - a_k] < \delta - \sigma \end{equation*}

(for instance intervals of length \(2^{-j}\eta\) marching to the right inside \((\alpha,\beta)\), with \(\eta\) small). Then \(\{(a_k, b_k)\}_{k=1}^{\infty}\) is a countable disjoint collection of open intervals in \((a,b)\) of total length less than \(\delta\), so by hypothesis

\begin{equation*} \sum_{k=1}^{n} |f(b_k) - f(a_k)| \le \sum_{k=1}^{\infty} |f(b_k) - f(a_k)| < \epsilon . \end{equation*}

Thus \(\delta\) responds to the \(\epsilon\) challenge in the definition of absolute continuity, and \(f\) is absolutely continuous on \([a,b]\).

Conversely, suppose \(f\) is absolutely continuous on \([a,b]\). Let \(\epsilon > 0\) and choose \(\delta > 0\) as a response to the \(\epsilon/2\) challenge regarding the criterion for the absolute continuity of \(f\), that is, for every finite disjoint collection \(\{(a_k, b_k)\}_{k=1}^{n}\) of open intervals in \((a,b)\),

\begin{equation*} \text{if } \sum_{k=1}^{n}[b_k - a_k] < \delta, \text{ then } \sum_{k=1}^{n}|f(b_k) - f(a_k)| < \epsilon/2 . \end{equation*}

Now let \(\{(a_k, b_k)\}_{k=1}^{\infty}\) be a countable disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^{\infty}[b_k - a_k] < \delta\). For each natural number \(n\), the first \(n\) of these intervals form a finite disjoint collection of open intervals in \((a,b)\) with

\begin{equation*} \sum_{k=1}^{n}[b_k - a_k] \le \sum_{k=1}^{\infty}[b_k - a_k] < \delta , \end{equation*}

so that \(\sum_{k=1}^{n}|f(b_k) - f(a_k)| < \epsilon/2\). The partial sums of the series of nonnegative terms \(\sum |f(b_k)-f(a_k)|\) are therefore bounded above by \(\epsilon/2\), so the series converges and

\begin{equation*} \sum_{k=1}^{\infty}|f(b_k) - f(a_k)| \le \epsilon/2 < \epsilon . \end{equation*}

Hence the same \(\delta\) responds to the \(\epsilon\) challenge in the countable criterion.

Assume first that \(f\) is increasing and absolutely continuous on \([a,b]\), and let \(\epsilon > 0\). By the preceding problem choose \(\delta > 0\) so that for every countable disjoint collection \(\{(a_k, b_k)\}_{k=1}^{\infty}\) of open intervals in \((a,b)\),

\begin{equation*} \text{if } \sum_{k=1}^{\infty}[b_k - a_k] < \delta, \text{ then } \sum_{k=1}^{\infty}|f(b_k) - f(a_k)| < \epsilon/2 . \end{equation*}

Let \(E\) be a measurable subset of \([a,b]\) with \(m(E) < \delta\). Apply assertion (i) of Theorem 11 of Chapter 2 with the number \(\delta - m(E) > 0\): there is an open set \(\mathcal{O} \supseteq E\) with \(m^{*}(\mathcal{O} \sim E) < \delta - m(E)\), and hence, by the countable additivity of measure applied to the disjoint union \(\mathcal{O} = E \cup (\mathcal{O} \sim E)\),

\begin{equation*} m(\mathcal{O}) = m(E) + m(\mathcal{O} \sim E) < \delta . \end{equation*}

Put \(U = \mathcal{O} \cap (a,b)\), an open subset of \((a,b)\) with \(E \sim \{a,b\} \subseteq U\) and \(m(U) < \delta\). By Proposition 9 of Chapter 1, \(U\) is the disjoint union of a countable collection \(\{(a_k, b_k)\}_{k}\) of open intervals, necessarily contained in \((a,b)\), and by countable additivity \(\sum_k [b_k - a_k] = m(U) < \delta\). If this collection is infinite, the choice of \(\delta\) gives \(\sum_k [f(b_k) - f(a_k)] < \epsilon/2\) directly; if it is finite, the same bound follows from the definition of absolute continuity together with the fact that the \(\delta\) furnished by the preceding problem also responds to the \(\epsilon/2\) challenge for finite collections (a finite collection may be enlarged to an infinite one of total length still less than \(\delta\), exactly as in the first half of the solution to Problem 38). Since \(f\) is increasing, \(f((a_k, b_k)) \subseteq [f(a_k), f(b_k)]\), so by the monotonicity and countable subadditivity of outer measure,

\begin{equation*} m^{*}(f(U)) \le \sum_{k} m^{*}\bigl(f((a_k,b_k))\bigr) \le \sum_{k}\bigl[f(b_k) - f(a_k)\bigr] < \epsilon/2 . \end{equation*}

Since \(f(E) \subseteq f(U) \cup \{f(a), f(b)\}\) and a two-point set has outer measure zero, \(m^{*}(f(E)) \le \epsilon/2 < \epsilon\). This proves the forward implication.

Conversely, assume \(f\) is increasing on \([a,b]\), that \(f\) is continuous, and that the stated condition holds. Let \(\epsilon > 0\) and choose \(\delta > 0\) so that \(m^{*}(f(E)) < \epsilon\) whenever \(E \subseteq [a,b]\) is measurable with \(m(E) < \delta\). Let \(\{(a_k, b_k)\}_{k=1}^{n}\) be a finite disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^{n}[b_k - a_k] < \delta\), and set \(E = \bigcup_{k=1}^{n}[a_k, b_k]\), a measurable set with \(m(E) = \sum_{k=1}^{n}[b_k - a_k] < \delta\). Since \(f\) is continuous and increasing, the Intermediate Value Theorem gives \(f([a_k, b_k]) = [f(a_k), f(b_k)]\). Relabel so that \(b_k \le a_{k+1}\); then \(f(b_k) \le f(a_{k+1})\), so the intervals \([f(a_k), f(b_k)]\), \(1 \le k \le n\), are non-overlapping. Hence

\begin{equation*} \sum_{k=1}^{n}|f(b_k) - f(a_k)| = \sum_{k=1}^{n}\bigl[f(b_k) - f(a_k)\bigr] = m\Bigl(\bigcup_{k=1}^{n}[f(a_k), f(b_k)]\Bigr) = m^{*}(f(E)) < \epsilon , \end{equation*}

the second equality because finitely many non-overlapping closed bounded intervals have union whose measure is the sum of their lengths (their pairwise intersections are finite sets, hence of measure zero), and the third because \(f(E) = \bigcup_{k=1}^{n} f([a_k,b_k]) = \bigcup_{k=1}^{n}[f(a_k), f(b_k)]\). The same remark justifies \(m(E) = \sum_{k=1}^{n}[b_k - a_k]\) above. Therefore \(\delta\) responds to the \(\epsilon\) challenge and \(f\) is absolutely continuous on \([a,b]\).

The continuity assumption in the converse direction cannot be dropped, and it is harmless in the sense that absolutely continuous functions are continuous. That it cannot be dropped is seen from the increasing function \(f\) on \([0,1]\) defined by \(f(x) = 0\) for \(0 \le x < 1/2\) and \(f(x) = 1\) for \(1/2 \le x \le 1\): every subset of \([0,1]\) has image contained in the two-point set \(\{0,1\}\), so \(m^{*}(f(E)) = 0\) for all \(E\) and the displayed condition holds vacuously with any \(\delta\); yet \(f\) is not continuous and therefore is not absolutely continuous on \([0,1]\). Thus the intended reading of the problem is that \(f\) is an increasing continuous function, or equivalently that the criterion is being used to decide absolute continuity among continuous increasing functions – which is exactly how it is applied in Problems 40 and 41.

Let \(f\) be increasing and absolutely continuous on \([a,b]\) and let \(E \subseteq [a,b]\) have measure zero. Let \(\epsilon > 0\). The forward implication of the preceding problem – which uses only that \(f\) is increasing and absolutely continuous – provides \(\delta > 0\) such that \(m^{*}(f(A)) < \epsilon\) for every measurable \(A \subseteq [a,b]\) with \(m(A) < \delta\). Since \(E\) is measurable with \(m(E) = 0 < \delta\), we conclude \(m^{*}(f(E)) < \epsilon\). As \(\epsilon > 0\) was arbitrary, \(m^{*}(f(E)) = 0\). By Proposition 4 of Chapter 2, any set of outer measure zero is measurable, so \(f(E)\) is a set of measure zero. Thus \(f\) maps sets of measure zero onto sets of measure zero.

Now consider the Cantor-Lebesgue function \(\varphi\) on \([0,1]\), which is increasing and continuous, and let \(\psi(x) = x + \varphi(x)\) for \(0 \le x \le 1\). Suppose, to get a contradiction, that \(\varphi\) is absolutely continuous on \([0,1]\). The identity function \(x \mapsto x\) is Lipschitz on \([0,1]\), hence absolutely continuous by Proposition 7. Since linear combinations of absolutely continuous functions are absolutely continuous (see also Problem 42), \(\psi\) would then be absolutely continuous on \([0,1]\). Moreover \(\psi\) is increasing, being the sum of an increasing function and a strictly increasing one.

Therefore, by what was just proved, \(\psi\) would map sets of measure zero onto sets of measure zero. The Cantor set \(C\) has measure zero, by Proposition 19 of Chapter 2. But Proposition 21 of Chapter 2 (page 52) asserts that this very function \(\psi\) maps \(C\) onto a measurable set of positive measure; indeed its proof, decomposing \([0,2] = \psi(\mathcal{O}) \cup \psi( C)\) for \(\mathcal{O} = [0,1] \sim C\) and computing \(m(\psi(\mathcal{O})) = m(\mathcal{O}) = 1\), gives \(m(\psi( C)) = 1\). Since \(1 \ne 0\), we have a contradiction. Hence \(\varphi\) is not absolutely continuous on \([0,1]\).

(i) Since \(f\) is absolutely continuous on \([a,b]\) it is continuous there (the criterion for absolute continuity, applied to a collection consisting of a single interval, is the criterion for uniform continuity). Let \(F \subseteq [a,b]\) be closed. Then \(F\) is closed and bounded, hence compact by the Heine-Borel Theorem. The continuous image of a compact set is compact, so \(f(F)\) is compact and in particular closed.

Now let \(E \subseteq [a,b]\) be an \(F_{\sigma}\) set, say \(E = \bigcup_{n=1}^{\infty} F_n\) with each \(F_n\) closed. Each \(F_n\) is a closed subset of \([a,b]\), and since images preserve unions,

\begin{equation*} f(E) = f\Bigl(\bigcup_{n=1}^{\infty} F_n\Bigr) = \bigcup_{n=1}^{\infty} f(F_n), \end{equation*}

a countable union of closed sets. Hence \(f(E)\) is an \(F_{\sigma}\) set, and in particular is measurable, since every \(F_{\sigma}\) set is Borel and every Borel set is measurable.

(ii) By the preceding problem, since \(f\) is increasing and absolutely continuous, \(f\) carries every subset of \([a,b]\) of measure zero onto a set of measure zero.

Conclusion. Let \(E \subseteq [a,b]\) be measurable. By assertion (iv) of Theorem 11 of Chapter 2 there is an \(F_{\sigma}\) set \(F\) with \(F \subseteq E\) and \(m^{*}(E \sim F) = 0\). Write \(F = \bigcup_{n=1}^{\infty} F_n\) with each \(F_n\) closed; since \(F_n \subseteq F \subseteq E \subseteq [a,b]\), every \(F_n\) is a closed subset of \([a,b]\), so part (i) applies and \(f(F)\) is an \(F_{\sigma}\) set, hence measurable. The set \(E \sim F\) is a subset of \([a,b]\) with \(m^{*}(E \sim F) = 0\), hence is measurable of measure zero by Proposition 4 of Chapter 2, so by part (ii) \(f(E \sim F)\) has measure zero and is therefore measurable. Finally, from \(E = F \cup (E \sim F)\) we get

\begin{equation*} f(E) = f(F) \cup f(E \sim F), \end{equation*}

the union of two measurable sets. Hence \(f(E)\) is measurable. Therefore \(f\) maps measurable subsets of \([a,b]\) to measurable sets.

Let \(f\) and \(g\) be absolutely continuous on the closed, bounded interval \([a,b]\).

The sum. Let \(\epsilon > 0\). Choose \(\delta_1 > 0\) as a response to the \(\epsilon/2\) challenge regarding the criterion for the absolute continuity of \(f\), and \(\delta_2 > 0\) as a response to the \(\epsilon/2\) challenge for \(g\). Put \(\delta = \min\{\delta_1, \delta_2\} > 0\). Let \(\{(a_k, b_k)\}_{k=1}^{n}\) be a finite disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^{n}[b_k - a_k] < \delta\). By the triangle inequality,

\begin{equation*} \sum_{k=1}^{n}\bigl|(f+g)(b_k) - (f+g)(a_k)\bigr| \le \sum_{k=1}^{n}|f(b_k) - f(a_k)| + \sum_{k=1}^{n}|g(b_k) - g(a_k)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

Hence \(f + g\) is absolutely continuous on \([a,b]\). (The same argument, with \(\epsilon/\bigl(2(|\alpha| + |\beta| + 1)\bigr)\) in place of \(\epsilon/2\) and the estimate \(|\alpha| \sum_k |f(b_k) - f(a_k)| + |\beta| \sum_k |g(b_k) - g(a_k)|\) in place of the one displayed, shows that any linear combination \(\alpha f + \beta g\) is absolutely continuous, as asserted in the text preceding Proposition 7.)

The product. Absolutely continuous functions are continuous, and a continuous real-valued function on the compact interval \([a,b]\) is bounded by the Extreme Value Theorem. So choose \(M \ge 1\) with

\begin{equation*} |f(x)| \le M \quad\text{and}\quad |g(x)| \le M \quad \text{for all } x \in [a,b]. \end{equation*}

Let \(\epsilon > 0\). Choose \(\delta_1 > 0\) as a response to the \(\epsilon/(2M)\) challenge for the absolute continuity of \(f\), and \(\delta_2 > 0\) as a response to the \(\epsilon/(2M)\) challenge for \(g\); put \(\delta = \min\{\delta_1, \delta_2\} > 0\). Let \(\{(a_k, b_k)\}_{k=1}^{n}\) be a finite disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^{n}[b_k - a_k] < \delta\). For each \(k\), adding and subtracting \(f(a_k)g(b_k)\),

\begin{equation*} \begin{aligned} |f(b_k)g(b_k) - f(a_k)g(a_k)| &= \bigl|\,[f(b_k) - f(a_k)]\,g(b_k) + f(a_k)\,[g(b_k) - g(a_k)]\,\bigr| \\ &\le M\,|f(b_k) - f(a_k)| + M\,|g(b_k) - g(a_k)| . \end{aligned} \end{equation*}

Summing over \(k\) and using the choice of \(\delta\),

\begin{equation*} \sum_{k=1}^{n}\bigl|(fg)(b_k) - (fg)(a_k)\bigr| \le M \sum_{k=1}^{n}|f(b_k) - f(a_k)| + M \sum_{k=1}^{n}|g(b_k) - g(a_k)| < M\cdot\frac{\epsilon}{2M} + M\cdot\frac{\epsilon}{2M} = \epsilon . \end{equation*}