Solutions to Royden & Fitzpatrick’s Real Analysis

The witness is \(a^{-1}b^{-1}\): commutativity and associativity of multiplication give

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

Two remarks make this conclusive. First, \(ab \neq 0\), so \((ab)^{-1}\) is defined: were \(ab = 0\), then \(b = (a^{-1}a)b = a^{-1}(ab) = a^{-1}\cdot 0 = 0\), against \(b \neq 0\), where \(c \cdot 0 = 0\) follows from \(c\cdot 0 = c(0+0) = c\cdot 0 + c \cdot 0\) on adding an additive inverse of \(c \cdot 0\). Second, inverses are unique: if \(x \neq 0\) and \(xy = 1 = xz\), then \(y = (zx)y = z(xy) = z\). Hence the display forces \((ab)^{-1} = a^{-1}b^{-1}\).

All three follow from P1 and P2 once we record \((-a)b = -(ab)\), which holds since \(ab + (-a)b = (a + (-a))b = 0 \cdot b = 0\) and additive inverses are unique; applied twice it gives \((-a)(-b) = ab\).

(i) By P2 either \(a \in \mathcal{P}\) or \(-a \in \mathcal{P}\), and

\begin{equation*} a^2 = a \cdot a = (-a)(-a) , \end{equation*}

so P1 puts \(a^2 \in \mathcal{P}\) in either case. Since \(1 \neq 0\) by Nontriviality and \(1 = 1^2\), we get \(1 > 0\).

(ii) Here \(a \neq 0\) by P2, so \(a^{-1}\) exists, and \(a^{-1} \neq 0\) (else \(1 = a a^{-1} = a \cdot 0 = 0\)). Were \(-a^{-1} \in \mathcal{P}\), P1 would give

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

contradicting P2 since \(1 \in \mathcal{P}\) by (i). So P2 leaves \(a^{-1} \in \mathcal{P}\).

(iii) Distributivity and \((-b)c = -(bc)\) give \((a-b)c = ac - bc\), and likewise \((a-b)(-c) = bc - ac\); by hypothesis \(a - b \in \mathcal{P}\). If \(c > 0\), then P1 gives \(ac - bc \in \mathcal{P}\), i.e. \(ac > bc\); if \(c < 0\), then \(-c \in \mathcal{P}\) and P1 gives \(bc - ac \in \mathcal{P}\), i.e. \(ac < bc\).

Both directions come from the single sandwich \(\inf E \leq x \leq \sup E\), valid for every \(x \in E\) because an infimum is a lower bound and a supremum an upper bound. (Suprema and infima are taken in the extended reals, so they exist for every nonempty \(E\).)

(i) Suppose \(\inf E = \sup E =: c\). Picking any \(x_0 \in E\) gives \(-\infty < \inf E \leq x_0 \leq \sup E < \infty\), so \(c\) is real; and for each \(x \in E\),

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

whence \(x = c\) by antisymmetry (P2 forbids both \(x - c\) and \(c - x\) positive). Thus \(E = \{c\}\).

(ii) Suppose \(E = \{x_0\}\). Then \(x_0\) is an upper bound lying in \(E\), hence the least one, and likewise the greatest lower bound; so \(\inf E = x_0 = \sup E\).

(i) If \(a \neq 0\), then \(a^{-1}\) exists and \(b = a^{-1}(ab) = a^{-1} \cdot 0 = 0\), using \(a^{-1}\cdot 0 = 0\) from Problem 1.

(ii) Distributivity together with \((-b)x = -(bx)\) (Problem 2) gives

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

If \(a^2 = b^2\), then \((a-b)(a+b) = 0\), so part (i) forces \(a - b = 0\) or \(a + b = 0\), i.e. \(a = b\) or \(a = -b\).

(iii) \(E \neq \varnothing\) since \(0^2 = 0 < c\), and \(1 + c\) is an upper bound: if \(x \in E\) had \(x > 1 + c > 1 > 0\), two applications of Problem 2(iii) would give

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

against \(x^2 < c\); so \(x \leq 1 + c\) for every \(x \in E\) by P2. The Completeness Axiom yields \(x_0 = \sup E\). Two preliminaries: \(t = c(1+c)^{-1}\) satisfies \(0 < t < 1\) and \(t \leq c\) (Problem 2(ii),(iii)), hence \(t^2 < t \leq c\), so \(t \in E\) and \(x_0 \geq t > 0\); and \(0 < u < v\) implies \(u^2 < uv < v^2\) by Problem 2(iii). We exclude the two nonzero alternatives of P2 for \(x_0^2 - c\).

(i) If \(x_0^2 < c\), put \(\epsilon = c - x_0^2 > 0\) and \(h = \min\{\tfrac12, \epsilon(2(2x_0+1))^{-1}\} > 0\), so that \(h < 1\) (hence \(h^2 < h\)) and \(h(2x_0+1) < \epsilon\). Then

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

so \(x_0 + h \in E\) exceeds the upper bound \(x_0\), which is impossible.

(ii) If \(x_0^2 > c\), put \(\epsilon = x_0^2 - c > 0\) and \(k = \min\{\tfrac12 x_0, \epsilon(4x_0)^{-1}\} > 0\), so that \(0 < k < x_0\) and \(2x_0 k < \epsilon\). Then

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

Every \(x \in E\) satisfies \(x < x_0 - k\): this is clear when \(x \leq 0\) since \(x_0 - k > 0\), while \(x \geq x_0 - k > 0\) would give \(x^2 \geq (x_0-k)^2 > c\) by monotonicity. So \(x_0 - k\) is an upper bound below \(\sup E\), which is impossible.

Hence \(x_0^2 = c\) with \(x_0 > 0\). If also \(x > 0\) and \(x^2 = c = x_0^2\), the conclusion of part (ii) gives \(x = x_0\) or \(x = -x_0\), and \(-x_0 < 0\) rules out the latter; so \(x = x_0 = \sqrt{c}\).

Everything follows from the completed square

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

an identity of the Field Axioms. Since \(1 > 0\) (Problem 2(i)) forces \(2, 4 > 0\) by P1, and \(a \neq 0\), Problem 4(i) gives \(2a \neq 0\) and \(4a \neq 0\); cancelling the factor \(4a\) (again Problem 4(i)) yields, 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) Write \(d = b^2 - 4ac > 0\), so Problem 4(iii) supplies \(\sqrt d > 0\) with \((\sqrt d)^2 = d\). By \((\ast)\) and Problem 4(ii), \(x\) solves the equation iff \(2ax + b = \pm \sqrt d\), i.e., multiplying by \((2a)^{-1}\), iff

\begin{equation*} x = \frac{-b + \sqrt{b^2-4ac}}{2a} \qquad \text{or} \qquad x = \frac{-b - \sqrt{b^2-4ac}}{2a} . \end{equation*}

Every step is an equivalence, so these are exactly the solutions, and they are distinct: equality would give \(2\sqrt d = 0\), hence \(\sqrt d = 0\) by Problem 4(i), against \(\sqrt d > 0\).

(ii) If \(b^2 - 4ac < 0\), a solution \(x\) would give \((2ax+b)^2 = b^2 - 4ac < 0\) by \((\ast)\), whereas every square is \(0\) or positive by Problem 2(i), contradicting P2.

The greatest lower bound of \(E\) is \(-s\), where \(s = \sup(-E)\) and \(-E = \{-x \mid x \in E\}\). Everything rests on the reflection rule

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

which holds because \((-u) - (-v) = v - u\). If \(m\) is a lower bound for \(E\), reflection makes \(-m\) an upper bound for \(-E\); as \(-E\) is nonempty, the Completeness Axiom furnishes \(s = \sup(-E)\). Now \(-s\) is a lower bound for \(E\), since \(x \in E\) gives \(-x \leq s\) and hence \(-s \leq x\); and it is the greatest one, since any lower bound \(\ell\) of \(E\) makes \(-\ell\) an upper bound of \(-E\), whence \(s \leq -\ell\) and so \(\ell \leq -s\). Therefore \(E\) has an infimum and \(\inf E = -\sup\{-x \mid x \in E\}\).

Directly from \(|x| = x\) for \(x \geq 0\) and \(|x| = -x\) for \(x < 0\) we read off three facts used below: \(|x|\) is one of \(x\), \(-x\); \(|-x| = |x|\); and \(-|x| \leq x \leq |x|\). (Check!) We also use that inequalities add, since \((v+v’) - (u+u’)\) is a sum of two nonnegative terms and P1 applies, and the sign rules \((-x)y = -(xy)\), \((-x)(-y) = xy\) of Problem 2.

(i) P2 gives four cases.

(a) \(a = 0\) or \(b = 0\): both sides vanish.

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

(c) \(a > 0 > b\) (and symmetrically \(b > 0 > a\)): then \(a(-b) = -(ab) > 0\) by P1, so \(ab < 0\) and \(|ab| = -(ab) = a(-b) = |a|\,|b|\).

(d) \(a, b < 0\): then \(ab = (-a)(-b) > 0\) by P1, so \(|ab| = (-a)(-b) = |a|\,|b|\).

(ii) Adding \(a \leq |a|\) and \(b \leq |b|\) gives \(a + b \leq |a| + |b|\); adding \(-|a| \leq a\) and \(-|b| \leq b\) gives \(-(|a|+|b|) \leq a+b\), which reflects (Problem 6) to \(-(a+b) \leq |a| + |b|\). Since \(|a+b|\) is one of \(a+b\), \(-(a+b)\),

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

(iii) Both \(x - a\) and \(-(x-a)\) are at most \(|x-a| = |-(x-a)|\), and \(|x-a|\) is one of the two; hence

\begin{equation*} |x-a| < \epsilon \iff x - a < \epsilon \text{ and } a - x < \epsilon \iff a - \epsilon < x < a + \epsilon . \end{equation*}

Induct on \(n\) on the statement \(S(n)\): \((n, n+1) \cap \mathbb{N} = \varnothing\). Two properties of \(\mathbb{N}\), the intersection of all inductive subsets of \(\mathbb{R}\), are used: every natural number is at least \(1\), because \(\{x \mid x \geq 1\}\) is inductive; and a natural number \(n \neq 1\) has \(n - 1 \in \mathbb{N}\), which is the first assertion of Problem 9 (proved there from the definition of \(\mathbb{N}\) alone, so there is no circularity).

Base \(S(1)\): the set \(E = \{1\} \cup \{x \in \mathbb{R} \mid x \geq 2\}\) is inductive (Check!), so \(\mathbb{N} \subseteq E\), and \(E\) meets \((1,2)\) nowhere.

Step: suppose \(S(k)\) and let \(n \in \mathbb{N}\) satisfy \(k + 1 < n < k + 2\). Since \(k \geq 1\) we get \(n > 2 > 1\), so \(n \neq 1\) and \(n - 1 \in \mathbb{N}\); subtracting \(1\) throughout,

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

contradicting \(S(k)\). Hence \(S(k+1)\) holds, and induction finishes the proof.

The witness for the first assertion is the set

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

which is inductive: \(1 \in F\), and for \(x \in F\) either \(x = 1\), so \((x+1) - 1 = 1 \in \mathbb{N}\), or \(x - 1 \in \mathbb{N}\), so \(x = (x-1) + 1 \in \mathbb{N}\) by inductiveness of \(\mathbb{N}\) and again \((x+1) - 1 \in \mathbb{N}\). Hence \(\mathbb{N} \subseteq F\), and a natural number \(n > 1\) is in \(F\) with \(n \neq 1\), so \(n - 1 \in \mathbb{N}\).

For the second, induct on \(m\) on the statement \(S(m)\): every natural \(n > m\) has \(n - m \in \mathbb{N}\). The base \(S(1)\) is the first assertion. Assuming \(S(m)\), let \(n \in \mathbb{N}\) with \(n > m + 1 > m\). Then \(n - m \in \mathbb{N}\) by \(S(m)\), and \(n - m > 1\), so the first assertion gives

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

which is \(S(m+1)\).

The integer is \(n\) itself, which lies in \([n, n+1)\). Uniqueness follows from the difference lemma: for integers \(n > m\), the number \(n - m\) is a natural number, hence \(n - m \geq 1\). Indeed, an integer \(m \in [n, n+1)\) with \(m \neq n\) satisfies \(m > n\), so \(m \geq n + 1\), contradicting \(m < n + 1\).

Proof of the difference lemma. Write \(\mathbb{Z} = \mathbb{N} \cup \{0\} \cup (-\mathbb{N})\). Since every natural number is at least \(1\) (Problem 8), every member of \(-\mathbb{N}\) is at most \(-1\); consequently a positive integer is a natural number. Also \(\mathbb{N}\) is closed under addition, since for fixed \(m \in \mathbb{N}\) the set \(\{x \in \mathbb{R} \mid x + m \in \mathbb{N}\}\) is inductive (Check!). Now split on \(m\).

(a) \(m \in \mathbb{N}\): then \(n > m \geq 1 > 0\) forces \(n \in \mathbb{N}\), and \(n - m \in \mathbb{N}\) by Problem 9.

(b) \(m = 0\): then \(n > 0\) gives \(n \in \mathbb{N}\) and \(n - m = n\).

(c) \(m = -q\) with \(q \in \mathbb{N}\): for \(n \in \mathbb{N}\), \(n - m = n + q \in \mathbb{N}\) by additive closure; for \(n = 0\), \(n - m = q\); and for \(n = -p\) with \(p \in \mathbb{N}\), \(-p > -q\) gives \(q > p\), so \(n - m = q - p \in \mathbb{N}\) by Problem 9.

(The lemma is used again in Problems 11 and 12.)

The largest member is any \(m \in E\) with \(m > c - 1\), where \(c = \sup E\) is supplied by the Completeness Axiom (\(E\) is nonempty and bounded above); such an \(m\) exists because \(c - 1 < c\) is not an upper bound for \(E\). Were some \(n \in E\) to satisfy \(n > m\), the difference lemma (Problem 10) would make \(n - m\) a natural number, hence \(n - m \geq 1\), and

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

contradicting that \(c\) bounds \(E\) above. So \(n \leq m\) for every \(n \in E\).

Given \(a < b\), the witness is \(x = r + \sqrt 2\), where Theorem 2 (density of \(\mathbb{Q}\)) supplies a rational \(r\) with

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

adding \(\sqrt 2\) throughout gives \(a < x < b\). Here \(\sqrt 2\) is the positive square root furnished by Problem 4(iii), and it is irrational by the argument of Section 1.2. So \(x\) is irrational too: were it rational, \(\sqrt 2 = x - r\) would be rational, since \(\mathbb{Q}\) is closed under subtraction.

That last closure is the one point needing proof. For rationals \(a_1/a_2\) and \(b_1/b_2\),

\begin{equation*} \frac{a_1}{a_2} - \frac{b_1}{b_2} = \frac{a_1 b_2 - b_1 a_2}{a_2 b_2} , \end{equation*}

with \(a_2 b_2 \neq 0\) by Problem 4(i), so it suffices that \(\mathbb{Z}\) be closed under multiplication and subtraction. Products: a zero factor is trivial, and otherwise \((\pm p)(\pm q) = \pm(pq)\) with \(pq \in \mathbb{N}\), because for fixed \(m \in \mathbb{N}\) the set \(\{x \in \mathbb{R} \mid xm \in \mathbb{N}\}\) is inductive (using additive closure of \(\mathbb{N}\), Problem 10). Differences: \(n - m\) lies in \(\mathbb{N}\), \(\{0\}\), or \(-\mathbb{N}\) according as \(n > m\), \(n = m\), or \(n < m\), by the difference lemma of Problem 10.

The two witnesses are

\begin{equation*} E = \{r \in \mathbb{Q} \mid r < x\} \qquad \text{and} \qquad E’ = \{t \notin \mathbb{Q} \mid t < x\} , \end{equation*}

each of which has supremum \(x\). Both are nonempty, since density (Theorem 2 for \(\mathbb{Q}\), Problem 12 for the irrationals) puts a member of each strictly between \(x - 1\) and \(x\); and both are bounded above by \(x\) by construction. So the Completeness Axiom gives \(c = \sup E \leq x\). If \(c < x\), density supplies a rational \(r\) with \(c < r < x\), so \(r \in E\) exceeds its upper bound \(c\), which is impossible. Hence \(\sup E = x\), and the same argument with Problem 12 in place of Theorem 2 gives \(\sup E’ = x\).

Induct on \(n\). The base \(n = 1\) is the equality \((1+r)^1 = 1 + 1 \cdot r\). Assuming \((1+r)^k \geq 1 + kr\) and multiplying by \(1 + r > 0\), which preserves the inequality by Problem 2(iii),

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

the last step because \(k \geq 1 > 0\) and \(r^2 > 0\) by Problem 2(i). This is the statement for \(k+1\).

Each part is an induction; in (i) and (ii) the base case at \(n = 1\) reads \(1 = 1\) (Check!), so only the successor steps are recorded.

(i) Assuming the formula at \(n\),

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

which is the formula at \(n+1\).

(ii) The companion identity \(1 + \ldots + n = n(n+1)/2\) is itself an induction, its step being \(n(n+1)/2 + (n+1) = (n+1)(n+2)/2\). Assuming the cube identity at \(n\) and using it,

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

(iii) The printed numerator \(1 - r^{n-1}\) is a misprint, since it already fails at \(n = 1\), giving \(0\) instead of \(1 + r\); the intended identity, which we prove for \(r \neq 1\), is

\begin{equation*} 1 + r + \ldots + r^n = \frac{1 - r^{n+1}}{1-r} . \end{equation*}

At \(n = 1\) the right side is \((1-r)(1+r)/(1-r) = 1+r\), and the step is

\begin{equation*} \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+2}}{1-r} . \end{equation*}

The enumeration \(0, -1, 1, -2, 2, \ldots\) of \(\mathbb{Z}\) is the map \(f \colon \mathbb{N} \to \mathbb{Z}\),

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

which carries odd arguments onto \(\{0,1,2,\ldots\}\) and even ones onto \(\{-1,-2,\ldots\}\). It rests on the dichotomy that every natural number is odd or even but not both: existence by induction (\(1 = 2 \cdot 1 - 1\), and \(2k - 1\) is followed by \(2k\), which is followed by \(2(k+1) - 1\)); exclusivity because \(2k - 1 = 2k’\) would make the integer \(k - k’\) equal \(1/2\), while \(0\) is the only integer in \([0,1)\) by Problem 10.

\(f\) is one-to-one, since \(f \geq 0\) on odd and \(f \leq -1\) on even arguments, ruling out mixed collisions, while \(f(m) = f(n)\) within a parity class gives \(m = n\) on multiplying by \(2\). And \(f\) is onto: \(p \geq 0\) equals \(f(2p+1)\), and \(p < 0\) equals \(f(-2p)\), with \(-2p = 2(-p)\) an even natural number. Hence \(\mathbb{Z}\) is equipotent to \(\mathbb{N}\), so countable.

Here “to \(\mathbb{N}\)” means into \(\mathbb{N}\), not onto.

(i) Let \(A\) be countable. If \(A\) is countably infinite, a one-to-one correspondence \(A \to \mathbb{N}\) is already such a map. If \(A\) is finite, then \(A = \varnothing\), whose empty map serves, or \(A\) is in one-to-one correspondence \(g\) with some \(\{1, \ldots, n\}\), and \(\iota \circ g\) is one-to-one into \(\mathbb{N}\) for \(\iota\) the inclusion (Problem 20).

(ii) Let \(f \colon A \to \mathbb{N}\) be one-to-one. Regarded as a map onto \(B = f(A) \subseteq \mathbb{N}\) it is a one-to-one correspondence, so \(A\) is equipotent to \(B\); and \(B\), a set of natural numbers, is countable by Theorem 3. A set equipotent to a countable set is countable (Section 1.3, following the definition), so \(A\) is countable.

Countability of \(\mathbb{N}^n\) is an induction on \(n\), the case \(n = 1\) being \(\mathbb{N}\) itself. Given \(\mathbb{N}^n\) countable, Problem 17 supplies a one-to-one \(h \colon \mathbb{N}^n \to \mathbb{N}\); pairing it with the one-to-one map \(g(m,n) = (m+n)^2 + n\) of the text’s proof of Corollary 4 gives

\begin{equation*} G(k_1, \ldots, k_{n+1}) = g\big(h(k_1, \ldots, k_n),\, k_{n+1}\big) , \end{equation*}

which is one-to-one on \(\mathbb{N}^{n+1}\): equal values force the two \(g\)-arguments to agree as ordered pairs, hence \(k_{n+1} = k’_{n+1}\) and, since \(h\) is one-to-one, \((k_1, \ldots, k_n) = (k’_1, \ldots, k’_n)\). So \(\mathbb{N}^{n+1}\) is countable, again by Problem 17.

\(\mathbb{N}^n\) is not finite, since \(\varphi(k) = (k, 1, \ldots, 1)\) maps \(\mathbb{N}\) one-to-one into it and no finite set \(A\) admits such a map: a one-to-one correspondence \(\psi\) of \(A\) with \(\{1, \ldots, p\}\) would make \(\psi \circ \varphi\) (one-to-one by Problem 20) restrict to a one-to-one map of \(\{1, \ldots, p+1\}\) onto a nonempty subset \(C\) of \(\{1, \ldots, p\}\); Theorem 3 makes \(C\) equipotent to some \(\{1, \ldots, q\}\) with \(q \leq p\), so with \(m = (p+1) - q \in \mathbb{N}\) (Problem 9) the sets \(\{1, \ldots, q+m\}\) and \(\{1, \ldots, q\}\) would be equipotent, against the pigeonhole principle.

Hence \(\mathbb{N}^n\) is countable and infinite, i.e. countably infinite.

Run the text’s proof of Corollary 6 with the initial segment \(\{1, \ldots, M\}\) in place of \(\mathbb{N}\) as index range. If \(\Lambda\) or \(E = \bigcup_{\lambda \in \Lambda} E_\lambda\) is empty, then \(E\) is finite and so countable; otherwise finiteness of \(\Lambda\) lets us relabel \(\Lambda = \{\lambda_1, \ldots, \lambda_M\}\). For each \(n \leq M\) with \(E_{\lambda_n} \neq \varnothing\), countability leaves exactly two cases, and we choose \(f_n\) mapping \(\{1, \ldots, N(n)\}\) onto \(E_{\lambda_n}\) when it is finite and \(\mathbb{N}\) onto \(E_{\lambda_n}\) when it is countably infinite. Set

\begin{equation*} E’ = \big\{ (n,k) \in \mathbb{N} \times \mathbb{N} \;\big|\; n \leq M,\ E_{\lambda_n} \neq \varnothing,\ k \leq N(n) \text{ if } E_{\lambda_n} \text{ is finite} \big\} \end{equation*}

and \(f(n,k) = f_n(k)\). Then \(f\) maps \(E’\) onto \(E\), since each \(x \in E\) lies in some \(E_{\lambda_n}\), which \(f_n\) covers. Now \(E’ \subseteq \mathbb{N} \times \mathbb{N}\) is countable by Corollary 4(i) with \(n = 2\) and Theorem 3, and it is nonempty because \(E\) is; Theorem 5 therefore makes \(E\) countable.

Method (2): induct on \(M\), which reduces to the union of two countable sets \(A\) and \(B\). Take one-to-one maps \(g \colon A \to \mathbb{N}\) and \(h \colon B \to \mathbb{N}\) (Problem 17) and set \(\Phi = 2g\) on \(A\) and \(\Phi = 2h - 1\) on \(B \sim A\). Even and odd values keep the two pieces apart, so \(\Phi\) is one-to-one and Problem 17 applies.

The printed target of the composition is a misprint: \(g \circ f\) maps \(A\) to \(C\), and that is what we prove.

The composition is one-to-one, since \(g(f(x)) = g(f(x’))\) forces \(f(x) = f(x’)\) as \(g\) is one-to-one, hence \(x = x’\) as \(f\) is; and it is onto, since \(z \in C\) is \(g(y)\) for some \(y \in B\) and \(y = f(x)\) for some \(x \in A\), whence \((g \circ f)(x) = z\).

For the inverse, each \(y \in B\) has exactly one preimage under \(f\) (at least one because \(f\) is onto, at most one because \(f\) is one-to-one), so \(f^{-1} \colon B \to A\) is a mapping, characterized by

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

It is one-to-one: if \(f^{-1}(y) = f^{-1}(y’) = x\), then \(y = f(x) = y’\). It is onto: every \(x \in A\) is \(f^{-1}(f(x))\).

Induct on \(n\) on the statement that \(\{1, \ldots, n+m\}\) is equipotent to \(\{1, \ldots, n\}\) for no natural number \(m\); by symmetry of equipotence (Problem 20) the direction of the correspondence is immaterial.

Base \(n = 1\): a one-to-one map on \(\{1, \ldots, 1+m\}\), which contains the distinct members \(1\) and \(2\) since \(1 + m \geq 2\), cannot take values in the singleton \(\{1\}\).

Step: assume the statement at \(n\) and let \(f\) be a one-to-one map of \(\{1, \ldots, q\}\) onto \(\{1, \ldots, n+1\}\), where \(q = (n+1) + m = (n+m) + 1\). Pick \(p\) with \(f(p) = n+1\) and let \(\tau\) transpose \(p\) and \(q\), fixing all else (the identity if \(p = q\)); as \(\tau \circ \tau\) is the identity, \(\tau\) is one-to-one and onto, so

\begin{equation*} h = f \circ \tau \quad \text{satisfies} \quad h(q) = f(p) = n+1 \end{equation*}

and is again a one-to-one correspondence (Problem 20). Restrict \(h\) to \(\{1, \ldots, n+m\} = \{1, \ldots, q\} \sim \{q\}\): its values avoid \(n+1\) because \(h\) is one-to-one, and every \(i \leq n\) has an \(h\)-preimage, necessarily distinct from \(q\). So the restriction is a one-to-one correspondence of \(\{1, \ldots, n+m\}\) with \(\{1, \ldots, n\}\), contradicting the statement at \(n\).

The diagonal set \(D = \{n \in \mathbb{N} \mid n \notin f(n)\}\) defeats every candidate enumeration \(f\). Were \(2^{\mathbb{N}}\) countable, it would be a nonempty countable set (it contains \(\varnothing\)) and hence the image of a mapping \(f\) of \(\mathbb{N}\) onto it, since for nonempty countable \(E\), a one-to-one correspondence of \(\mathbb{N}\) or of \(\{1, \ldots, n\}\) onto \(E\) extends to such a surjection by repeating one value (Check!); we call this the Observation and use it again in Problem 23. Since \(D \in 2^{\mathbb{N}}\), we could write \(D = f(m)\), and then

\begin{equation*} m \in D \iff m \notin f(m) = D , \end{equation*}

which is impossible. Hence \(2^{\mathbb{N}}\) is uncountable.

For the product, induct on \(n\); the case \(n = 2\) carries the argument. Let \(A\) and \(B\) be countable, both nonempty (else \(A \times B = \varnothing\)). The Observation of Problem 22 supplies surjections \(f\) of \(\mathbb{N}\) onto \(A\) and \(g\) of \(\mathbb{N}\) onto \(B\), so

\begin{equation*} h(m,k) = \big(f(m), g(k)\big) \end{equation*}

maps \(\mathbb{N} \times \mathbb{N}\) onto \(A \times B\); as \(\mathbb{N} \times \mathbb{N}\) is a nonempty countable set by Corollary 4(i), Theorem 5 makes \(A \times B\) countable. For the step, \(((x_1,\ldots,x_n), x_{n+1}) \mapsto (x_1, \ldots, x_{n+1})\) is a one-to-one correspondence of \((A_1 \times \cdots \times A_n) \times A_{n+1}\) with \(A_1 \times \cdots \times A_{n+1}\), and a set equipotent to a countable set is countable.

For \(\mathbb{N}^{\mathbb{N}}\), embed \(2^{\mathbb{N}}\) in it by \(E \mapsto \chi_E\), where

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

This is one-to-one, since \(E \neq F\) differ at some \(n\), where \(\chi_E\) and \(\chi_F\) take the values \(1\) and \(2\). Were \(\mathbb{N}^{\mathbb{N}}\) countable, Theorem 3 would make its subset \(\{\chi_E \mid E \in 2^{\mathbb{N}}\}\) countable, hence \(2^{\mathbb{N}}\) countable, contradicting Problem 22.

The copy of \(\mathbb{N}\) inside \(I\) is \(X = \{x_k \mid k \in \mathbb{N}\}\), where

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

for two points \(a < b\) of \(I\) furnished by nondegeneracy. Indeed \([a,b] \subseteq I\) because \(I\) is an interval, and \(0 < 1/k \leq 1\) gives \(a < x_k \leq b\); the map \(k \mapsto x_k\) is one-to-one since \(b - a \neq 0\).

Everything else is the finite case of Theorem 3: a nonempty subset \(A\) of a set equipotent to \(\{1, \ldots, n\}\) is equipotent to \(\{1, \ldots, N\}\) for some \(N \leq n\). First, \(\mathbb{N}\) is not finite: were \(\psi\) a one-to-one correspondence of \(\mathbb{N}\) with \(\{1, \ldots, m\}\), the image \(\psi(\{1, \ldots, m+1\})\) would be equipotent both to \(\{1, \ldots, m+1\}\) and to some \(\{1, \ldots, j\}\) with \(j \leq m\), so \(\{1, \ldots, j\}\) and \(\{1, \ldots, j+p\}\) with \(p = m+1-j \in \mathbb{N}\) would be equipotent, against the pigeonhole principle (Problem 21). Second, were \(I\) finite, say equipotent to \(\{1, \ldots, n\}\) via \(\varphi\), the same fact applied to the nonempty subset \(\varphi(X)\) would make \(\mathbb{N}\), equipotent to \(X\) and so to \(\varphi(X)\), equipotent to some \(\{1, \ldots, N\}\), that is, finite.

It is enough to show every nondegenerate interval is equipotent to \((0,1)\), since equipotence is an equivalence relation (Problem 20).

The tool is an absorbing lemma: if \(C \subseteq X\) is countable and \(X \sim C\) contains a countably infinite set \(D\), then \(X\) and \(X \sim C\) are equipotent. For \(E = C \cup D\) is countable by Corollary 6 and is infinite, containing \(D\) (Problem 24), hence countably infinite, while \(E \sim C = D\) because \(D\) and \(C\) are disjoint. Enumerating \(E = \{e_k\}\) and \(D = \{d_k\}\), the map

\begin{equation*} h(e_k) = d_k \quad (k \in \mathbb{N}), \qquad h(x) = x \quad (x \in X \sim E) \end{equation*}

is one-to-one (the \(d_k\) are distinct, \(h\) is the identity off \(E\), and the images \(D \subseteq E\) and \(X \sim E\) are disjoint), with image \(D \cup (X \sim E) = X \sim C\).

Reduction to an open interval. With \(\alpha = \inf I\) and \(\beta = \sup I\) in the extended reals, any \(z\) with \(\alpha < z < \beta\) lies between two members of \(I\), hence in \(I\); so for \(J = (\alpha, \beta)\),

\begin{equation*} J \subseteq I \subseteq J \cup \big(\{\alpha, \beta\} \cap \mathbb{R}\big) , \end{equation*}

and \(C = I \sim J\) has at most two points. Nondegeneracy forces \(\alpha < \beta\), so \(J\) is a nondegenerate interval, not finite by Problem 24 and hence (as in that solution) containing a countably infinite \(D\). The lemma makes \(I\) equipotent to \(J\).

Since \(\alpha < \beta\), the interval \(J\) has exactly one of four forms.

(a) \((a,b)\) with \(a < b\) real: \(x \mapsto (x-a)/(b-a)\), inverse \(s \mapsto a + s(b-a)\).

(b) \((a, \infty)\): \(x \mapsto x - a\) onto \((0,\infty)\), followed by \(t \mapsto t/(1+t)\), which lands in \((0,1)\) as \(0 < t < 1+t\) and has the two-sided inverse \(s \mapsto s/(1-s)\), since \(s = t/(1+t)\) gives \(1 - s = 1/(1+t)\). (Check!)

(c) \((-\infty, b)\): \(x \mapsto b - x\) onto \((0, \infty)\), then (b).

(d) \(\mathbb{R}\): \(x \mapsto x/(1+|x|)\) onto \((-1,1)\), its values being less than \(1\) in absolute value, with two-sided inverse \(y \mapsto y/(1-|y|)\) (both maps are odd, so the check for \(x \geq 0\) suffices), followed by \(y \mapsto (y+1)/2\) onto \((0,1)\).

Yes. Let \(\Sigma\) be the set of sequences \(a=(a_n)\) of \(0\)’s and \(1\)’s having infinitely many zero terms, and let \(B\) be the set of those members of \(\Sigma\) having infinitely many terms equal to \(1\) as well.

(i) \(\sigma(a)=\sum_{n\ge1}a_n2^{-n}\) is a one-to-one correspondence of \(\Sigma\) onto \([0,1)\). Problem 44 with \(p=2\) gives each \(x\in(0,1)\) a binary expansion, unique except when \(x=q/2^n\), in which case the two expansions are one eventually \(0\) and one eventually \(1\); so membership in \(\Sigma\) selects exactly one expansion of each \(x\in(0,1)\), and \(0=\sigma(0,0,\ldots)\).

(ii) \(B\) is equipotent to \(\mathbb{R}\). A member of \(\Sigma\sim B\) has only finitely many terms equal to \(1\), and for each \(N\) exactly \(2^{N}\) such sequences vanish past index \(N\), so \(\Sigma\sim B\) is countable (Corollary 6) and \(C=\sigma(\Sigma\sim B)\) is a countable subset of \([0,1)\) with \(\sigma(B)=[0,1)\sim C\). The sequences \(b^{(m)}\), where \(b^{(m)}_n=1\) exactly when \(n\) is even and \(n\ge2m\), lie in \(B\) and are distinct, so \([0,1)\sim C\) has a countably infinite subset; the Lemma of Problem 25, applied with \(X=[0,1)\), makes \(\sigma(B)\) equipotent to \([0,1)\), which is equipotent to \(\mathbb{R}\) by Problem 25.

(iii) \(B\) is equipotent to \(\mathbb{N}^{\mathbb{N}}\). Let \(\Psi(m)\) be the concatenation of the blocks consisting of \(m_1-1\) zeros, then \(m_2\) ones, then \(m_3\) zeros, and so on alternating; every block past the first is nonempty, so \(\Psi(m)\) has infinitely many ones and infinitely many zeros, that is \(\Psi(m)\in B\). Conversely each \(a\in B\) takes both values infinitely often, so its maximal constant runs are infinitely many and each finite; reading off their lengths \(\ell_1,\ell_2,\ldots\) and setting \(m_1=\ell_1+1,\ m_j=\ell_j\ (j\ge2)\) when \(a_1=0\), and \(m_1=1,\ m_{j+1}=\ell_j\) when \(a_1=1\), inverts \(\Psi\) on both sides. (Check!)

(iv) \(\mathbb{N}^{\mathbb{N}}\times\mathbb{N}^{\mathbb{N}}\) is equipotent to \(\mathbb{N}^{\mathbb{N}}\): interleaving, \(\Theta(u,v)_{2k-1}=u_k\) and \(\Theta(u,v)_{2k}=v_k\), has the de-interleaving map as a two-sided inverse.

Equipotence is an equivalence relation (Problem 20), and \(X\sim X’\), \(Y\sim Y’\) give \(X\times Y\sim X’\times Y’\) through \((x,y)\mapsto(f(x),g(y))\). Hence, by (ii), (iii), (iv), (iii), (ii) in turn,

\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*}

Neither. \(\mathbb{Q}\) is not open: for \(x\in\mathbb{Q}\) and \(r>0\) the interval \((x-r,x+r)\) contains an irrational number by Problem 12, so it is not contained in \(\mathbb{Q}\), and the defining property of openness fails at every point of the nonempty set \(\mathbb{Q}\).

\(\mathbb{Q}\) is not closed: given \(x\in\mathbb{R}\), every open interval containing \(x\) contains an interval \((x-r,x+r)\) with \(r>0\), which contains a rational number by the density of \(\mathbb{Q}\) (Theorem 2). Hence

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

the inequality because \(\sqrt{2}\) is irrational, so \(\mathbb{Q}\) fails to contain all its points of closure.

Only \(\emptyset\) and \(\mathbb{R}\). Both are open by Proposition 8 and closed by Proposition 12.

Conversely, let \(E\) be both open and closed with \(\emptyset\neq E\neq\mathbb{R}\), and fix \(x\in E\). The proof of Proposition 9 produces, from the openness of \(E\), the extended real numbers

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

for which \(x\in I_x=(a_x,b_x)\subseteq E\) and \(a_x\notin E\), \(b_x\notin E\) (assertion (2) of that proof). Since \(I_x\subseteq E\neq\mathbb{R}\), at least one of \(a_x,b_x\) is real.

(i) \(b_x\) real. Given \(r>0\), the number \(t=\max\{x,\,b_x-r/2\}\) satisfies \(a_x<t<b_x\) and \(b_x-r<t\), so \((b_x-r,b_x+r)\) meets \(I_x\subseteq E\); hence \(b_x\in\overline{E}=E\), contradicting \(b_x\notin E\).

(ii) \(b_x=\infty\), so \(a_x\) is real. Symmetrically \(t=\min\{x,\,a_x+r/2\}\) puts a point of \(E\) in \((a_x-r,a_x+r)\), giving \(a_x\in\overline{E}=E\), again a contradiction.

Take

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

which are disjoint since no real number is both \(<1\) and \(>1\).

Each of \([0,1]\) and \([1,2]\) is closed, its complement being a union of open intervals and so open (Propositions 8 and 11), whence \(\overline{A}\subseteq[0,1]\) and \(\overline{B}\subseteq[1,2]\) by the minimality in Proposition 10. Moreover \(1\) is a point of closure of \(A\): given \(r>0\), the Archimedean Property provides \(n\) with \(1/n<\min\{r,1\}\), and \(1-1/n\in A\cap(1-r,1+r)\). Symmetrically \(1\) is a point of closure of \(B\), using \(1+1/n\). Therefore

\begin{equation*} 1\in\overline{A}\cap\overline{B}\neq\emptyset . \end{equation*}

Here \(x\in E^{\prime}\) means: every open interval containing \(x\) contains a point of \(E\) other than \(x\).

(i) Let \(x\) be a point of closure of \(E^{\prime}\) and let \(I\) be an open interval containing \(x\); choose \(y\in I\cap E^{\prime}\).

(a) \(y=x\). Then \(x\in E^{\prime}\) already.

(b) \(y\neq x\). The set \(I\sim\{x\}\) is open (an intersection of the open sets \(I\) and \(\mathbb{R}\sim\{x\}\), Proposition 8), so \(J=(y-r,y+r)\subseteq I\sim\{x\}\) for some \(r>0\); since \(y\in E^{\prime}\), the interval \(J\) contains a point \(z\in E\sim\{y\}\), and \(z\neq x\) because \(J\) misses \(x\). So \(z\in(E\sim\{x\})\cap I\).

In either case \(I\) meets \(E\sim\{x\}\), so \(x\in E^{\prime}\) and \(E^{\prime}\) contains all its points of closure, that is, \(E^{\prime}\) is closed.

(ii) \((\supseteq)\) Always \(E\subseteq\overline{E}\); and if \(x\in E^{\prime}\), every open interval containing \(x\) contains a point of \(E\sim\{x\}\subseteq E\), so \(x\in\overline{E}\).

\((\subseteq)\) Let \(x\in\overline{E}\). If \(x\in E\) we are done; otherwise \(E\sim\{x\}=E\), so every open interval containing \(x\) meets \(E\sim\{x\}\) and \(x\in E^{\prime}\). Hence \(\overline{E}=E\cup E^{\prime}\).

Map \(E\) one-to-one into \(\mathbb{Q}\times\mathbb{Q}\). For each \(x\in E\) fix \(r_x>0\) with \((x-r_x,x+r_x)\cap E=\{x\}\), and use the density of \(\mathbb{Q}\) (Theorem 2, once on each side of \(x\)) to choose rationals

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

and set \(\varphi(x)=(p_x,q_x)\).

\(\varphi\) is one-to-one: if \(\varphi(y)=\varphi(x)\), 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 \(y\in(x-r_x,x+r_x)\cap E=\{x\}\), that is \(y=x\).

Finally \(\mathbb{Q}\times\mathbb{Q}=\bigcup_{q\in\mathbb{Q}}(\{q\}\times\mathbb{Q})\) is a countable union of countable sets, hence countable by Corollary 6 (each factor being countable by Corollary 4). So \(E\) is equipotent to the subset \(\varphi(E)\) of a countable set, and is countable by Theorem 3.

Both equivalences are immediate from the definitions, once one notes \(\operatorname{int}E\subseteq E\) for every \(E\) (a point of \((x-r,x+r)\subseteq E\) is \(x\) itself).

(i) If \(E\) is open, each \(x\in E\) has an \(r>0\) with \((x-r,x+r)\subseteq E\), which says \(x\in\operatorname{int}E\); so \(E\subseteq\operatorname{int}E\) and \(E=\operatorname{int}E\). Conversely if \(E=\operatorname{int}E\), each \(x\in E\) lies in \(\operatorname{int}E\) and so has such an \(r\), which is openness.

(ii) If \(E\) is dense and \(x\in\operatorname{int}(\mathbb{R}\sim E)\), pick \(r>0\) with \((x-r,x+r)\cap E=\emptyset\); density puts a member of \(E\) between \(x-r\) and \(x+r\), a contradiction, so \(\operatorname{int}(\mathbb{R}\sim E)=\emptyset\).

Conversely, if \(\operatorname{int}(\mathbb{R}\sim E)=\emptyset\) and \(a<b\) had \((a,b)\cap E=\emptyset\), then with

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

we get \((x-r,x+r)=(a,b)\subseteq\mathbb{R}\sim E\), so \(x\in\operatorname{int}(\mathbb{R}\sim E)\), a contradiction. Hence every \((a,b)\) meets \(E\), that is, \(E\) is dense.

Take \(F_n=[n,\infty)\).

Each \(F_n\) is nonempty (it contains \(n\)) and closed, its complement \((-\infty,n)\) being open (Proposition 11); the collection is descending since \(x\ge n+1\) forces \(x\ge n\); and \(F_1=[1,\infty)\) is unbounded above, since by the Archimedean Property no real \(b\) exceeds every natural number. Yet the intersection is empty: given \(x\in\mathbb{R}\), either \(x\le0\) and \(x\notin F_1\), or \(x>0\) and the Archimedean Property supplies \(n>x\), so \(x\notin F_n\). Thus

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

the only hypothesis of the Nested Set Theorem that fails being the boundedness of \(F_1\).

Over the Field and Positivity Axioms, (C), (HB) and (NST) are equivalent, where (C) is the Completeness Axiom, (HB) says every open cover of a closed bounded set has a finite subcover, and (NST) is the conclusion of the Nested Set Theorem. The forward implications are the two theorems of the text: the Heine-Borel proof uses (C) exactly once, to form \(c=\sup E\), and the Nested Set proof uses (C) only through Heine-Borel, its other ingredients being Proposition 11 and De Morgan’s Identities. Two converses remain.

(HB) implies (C). Let \(E\) be nonempty and bounded above with no least upper bound. Then \(E\) has no largest member, so no member of \(E\) is an upper bound of \(E\). Fix \(a\in E\) and an upper bound \(b\); then \(a<b\), and \([a,b]\) is bounded and closed (Propositions 8, 11). Cover it: for \(x\in[a,b]\),

(i) \(x\) is not an upper bound of \(E\): choose \(e\in E\) with \(e>x\) and put \(\mathcal{O}_x=(-\infty,e)\), no point of which is an upper bound of \(E\);

(ii) \(x\) is an upper bound of \(E\): since \(x\) is not the least one, choose an upper bound \(y<x\) and put \(\mathcal{O}_x=(y,\infty)\), every point of which is an upper bound of \(E\).

By (HB) finitely many \(\mathcal{O}_{x_1},\ldots,\mathcal{O}_{x_m}\) cover \([a,b]\). If all are of type (i), their union is \((-\infty,e^{*})\) with \(e^{*}=\max\{e_1,\ldots,e_m\}\in E\); it contains \(b\), so \(b<e^{*}\le b\), absurd. So some index is of type (ii); let \(y^{*}\) be the least of the corresponding numbers \(y\), an upper bound of \(E\), and put \(c=\min\{y^{*},b\}\). Then \(c\) is an upper bound of \(E\) and \(a\le c\le b\), so \(c\in\mathcal{O}_{x_i}\) for some \(i\): type (i) gives \(c<e_i\in E\), and type (ii) gives \(c>y_j\ge y^{*}\ge c\). Both are absurd, so \(E\) has a least upper bound.

(NST) implies (C). First, \(\mathbb{N}\) is unbounded above. Otherwise let \(M\) bound it and put \(F_n=\overline{T_n}\), \(T_n=\{m\in\mathbb{N} \mid m\ge n\}\); these are nonempty, closed and descending by Proposition 10, and \(F_1\subseteq[1,M]\) by the minimality in Proposition 10, so \(F_1\) is bounded. Take \(x\in\bigcap_nF_n\) by (NST); since \(x\) is a point of closure of each \(T_n\), each \(n\) admits \(m\ge n\) with \(|x-m|<1/2\). Taking \(n=1\), then \(n=m_1+1\), gives \(m_2>m_1\) with

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

which is absurd. Hence the Archimedean Property holds.

Now let \(E\) be nonempty and bounded above. If \(E\) has a largest member that member is the least upper bound, so assume not; then no member of \(E\) is an upper bound. Choose \(a_1\in E\) and an upper bound \(b_1>a_1\), and bisect: with \(m_n=(a_n+b_n)/2\), put \((a_{n+1},b_{n+1})=(a_n,m_n)\) if \(m_n\) is an upper bound of \(E\) and \((m_n,b_n)\) otherwise. An induction gives, for every \(n\), that \(a_n\) is not an upper bound, \(b_n\) is, \(a_n\le a_{n+1}<b_{n+1}\le b_n\), and \(b_n-a_n=(b_1-a_1)/2^{\,n-1}\). The sets \(F_n=[a_n,b_n]\) are nonempty, closed, descending, with \(F_1\) bounded, so (NST) supplies \(c\) with \(a_n\le c\le b_n\) for all \(n\). Since \(2^{\,n-1}\ge n\) (induction), the Archimedean Property gives, for each \(\epsilon>0\), an \(n\) with \(b_n-a_n\le(b_1-a_1)/n<\epsilon\).

\(c\) is an upper bound: were \(e\in E\) with \(e>c\), taking \(\epsilon=e-c\) would give \(b_n\le c+(b_n-a_n)<e\), contradicting that \(b_n\) is an upper bound. \(c\) is the least: were \(u<c\) an upper bound, taking \(\epsilon=c-u\) would give \(a_n\ge c-(b_n-a_n)>u\), and since \(a_n\) is not an upper bound some \(e\in E\) exceeds \(a_n>u\), a contradiction. So \(c=\sup E\).

Let \(\mathcal{B}\) be the collection of Borel sets, by definition the smallest \(\sigma\)-algebra containing the open sets, and let \(\mathcal{C}\) be the smallest \(\sigma\)-algebra containing the closed sets; both exist by Proposition 13, since \(2^{\mathbb{R}}\) is one such \(\sigma\)-algebra. The whole point is Proposition 11: \(E\) is closed if and only if \(\mathbb{R}\sim E\) is open.

\(\mathcal{C}\subseteq\mathcal{B}\): if \(F\) is closed then \(\mathbb{R}\sim F\) is open, so it lies in \(\mathcal{B}\), and \(\sigma\)-algebras are closed under complements, so

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

thus \(\mathcal{B}\) is a \(\sigma\)-algebra containing the closed sets and minimality of \(\mathcal{C}\) applies.

\(\mathcal{B}\subseteq\mathcal{C}\): if \(\mathcal{O}\) is open then \(\mathbb{R}\sim\mathcal{O}\) is closed, so it lies in \(\mathcal{C}\), and hence

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

now minimality of \(\mathcal{B}\) applies. Therefore \(\mathcal{B}=\mathcal{C}\).

Let \(\mathcal{I}=\{\,[a,b) \mid a<b\,\}\), let \(\mathcal{A}\) be the smallest \(\sigma\)-algebra containing \(\mathcal{I}\) (it exists by Proposition 13), and let \(\mathcal{B}\) be the Borel sets. We show each contains the other’s generators.

\(\mathcal{A}\subseteq\mathcal{B}\): for \(a<b\),

\begin{equation*} [a,b)=\bigcap_{n=1}^{\infty}\Big(a-\tfrac1n,\,b\Big), \end{equation*}

since \(x>a-1/n\) for all \(n\) forces \(x\ge a\) by the Archimedean Property. Each \((a-1/n,b)\) is open, hence Borel, and \(\sigma\)-algebras are closed under countable intersections, so \(\mathcal{I}\subseteq\mathcal{B}\); minimality of \(\mathcal{A}\) gives the inclusion.

\(\mathcal{B}\subseteq\mathcal{A}\): every open interval lies in \(\mathcal{A}\). Indeed, with \(N\) chosen so that \(1/N<b-a\),

\begin{equation*} (a,b)=\bigcup_{n=N}^{\infty}\Big[a+\tfrac1n,\,b\Big), \end{equation*}

each member lying in \(\mathcal{I}\) because \(a+1/n\le a+1/N<b\), and the reverse inclusion holding since \(a<x<b\) admits \(n\ge N\) with \(1/n<x-a\). Likewise

\begin{equation*} (a,\infty)=\bigcup_{n=1}^{\infty}(a,a+n),\quad (-\infty,b)=\bigcup_{n=1}^{\infty}(b-n,b),\quad \mathbb{R}=\bigcup_{n=1}^{\infty}(-n,n). \end{equation*}

A nonempty open set is a countable disjoint union of such intervals (Proposition 9) and \(\emptyset\in\mathcal{A}\), so every open set lies in \(\mathcal{A}\); minimality of \(\mathcal{B}\) finishes. Hence \(\mathcal{A}=\mathcal{B}\).

Every open interval is a countable union of closed bounded intervals: choosing \(N\) with \(2/N<b-a\),

\begin{equation*} (a,b)=\bigcup_{n\ge N}\Big[a+\tfrac1n,\,b-\tfrac1n\Big], \end{equation*}

the inclusion \(\supseteq\) being clear and \(\subseteq\) holding because the Archimedean Property supplies \(n\ge N\) with \(1/n<\min\{x-a,\,b-x\}\) for each \(x\in(a,b)\); similarly

\begin{equation*} (a,\infty)=\bigcup_{n}\Big[a+\tfrac1n,\,a+n\Big],\quad (-\infty,b)=\bigcup_{n}\Big[b-n,\,b-\tfrac1n\Big], \end{equation*}

and \(\mathbb{R}=\bigcup_n[-n,n]\). Each such interval is closed, its complement being a union of open intervals (Propositions 8 and 11).

Now let \(\mathcal{O}\) be open. If \(\mathcal{O}=\emptyset\) it is closed (Proposition 12) and so is an \(F_\sigma\) set. Otherwise Proposition 9 writes \(\mathcal{O}\) as the union of a countable collection of open intervals, each an \(F_\sigma\) set by the above; a countable union of countable unions is a countable union (Corollary 6 applied to the index sets), so \(\mathcal{O}\) is a countable union of closed sets, that is, an \(F_\sigma\) set.

Put \(s_n=\sup\{a_k \mid k\ge n\}\) and \(i_n=\inf\{a_k \mid k\ge n\}\), so \(\{s_n\}\) is decreasing, \(\{i_n\}\) is increasing, and by the Lemma of Problem 39 (which does not use the present result)

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

\(L\) is a cluster point, in three cases.

(i) \(L=\infty\). Then \(\{a_n\}\) is unbounded above (a bound \(M\) would give \(s_n\le M\)), and deleting finitely many terms leaves it so; choose recursively \(n_k>n_{k-1}\) with \(a_{n_k}>k\), whence \(\{a_{n_k}\}\to\infty\).

(ii) \(L=-\infty\). Then \(\{s_n\}\to-\infty\); pick \(N_k\) with \(s_{N_k}<-k\) and recursively \(n_k>n_{k-1}\) with \(n_k\ge N_k\), so that \(a_{n_k}\le s_{n_k}\le s_{N_k}<-k\) and \(\{a_{n_k}\}\to-\infty\).

(iii) \(L\in\mathbb{R}\). For each \(\epsilon>0\) and each index \(N\) there is \(k\ge N\) with \(|a_k-L|<\epsilon\): choose \(n\ge N\) with \(s_n<L+\epsilon\), note \(s_n\ge\inf_m s_m=L\), and use the definition of supremum to get \(k\ge n\) with \(a_k>s_n-\epsilon\ge L-\epsilon\), while \(a_k\le s_n<L+\epsilon\). Recursing with \(\epsilon=1/k\) and \(N=n_{k-1}+1\) builds a subsequence converging to \(L\), by the Archimedean Property.

Every cluster point \(c\) satisfies \(c\le L\). Assume \(L<\infty\) and let \(t>L\) be real. Since \(\inf_n s_n<t\), there is \(N\) with \(a_n\le s_N<t\) for all \(n\ge N\). Let \(\{a_{n_k}\}\to c\); a strictly increasing sequence of indices has \(n_k\ge k\) (induction), so \(a_{n_k}<t\) for \(k\ge N\). Hence \(c\neq\infty\), and \(c\le t\) either trivially (\(c=-\infty\)) or by Proposition 14 applied to \(\{a_{n_k}\}_{k\ge N}\). As \(t>L\) was an arbitrary real, \(c\le L\) when \(L\in\mathbb{R}\), and \(c=-\infty=L\) when \(L=-\infty\).

So the cluster points form a nonempty set with largest element \(L=\limsup\{a_n\}\).

For \(\liminf\): \(\{a_{n_k}\}\to c\) if and only if \(\{-a_{n_k}\}\to-c\) (Theorem 18 with \(\alpha=-1,\beta=0\) for finite \(c\), the definition for \(c=\pm\infty\)), so negation carries the cluster points of \(\{a_n\}\) onto those of \(\{-a_n\}\) and reverses order. By the above and Proposition 19(iii), the largest cluster point of \(\{-a_n\}\) is \(\limsup\{-a_n\}=-\liminf\{a_n\}\); hence \(\liminf\{a_n\}\) is the smallest cluster point of \(\{a_n\}\).

Put \(s_n=\sup\{a_k \mid k\ge n\}\) and \(i_n=\inf\{a_k \mid k\ge n\}\); the sets shrink as \(n\) grows, so \(\{s_n\}\) is decreasing, \(\{i_n\}\) is increasing, and \(i_n\le a_n\le s_n\).

Lemma. A decreasing sequence of extended real numbers converges in the extended sense to its infimum, an increasing one to its supremum. Indeed, in the decreasing case: if every \(s_n=\infty\) both sides are \(\infty\); if some \(s_m=-\infty\) then \(s_n=-\infty\) for \(n\ge m\) and both sides are \(-\infty\); otherwise some tail \(\{s_n\}_{n\ge N}\) is real with \(\inf_n s_n=\inf_{n\ge N}s_n\), and that tail converges to its infimum by Theorem 15 if bounded below, while if not, each real \(c\) admits \(M\ge N\) with \(s_n\le s_M<c\) for \(n\ge M\), so \(\{s_n\}\to-\infty=\inf_n s_n\). Altering finitely many initial terms changes neither convergence nor limit. The increasing case follows by negation. Hence \(\limsup\{a_n\}=\inf_n s_n\) and \(\liminf\{a_n\}=\sup_n i_n\), used freely below.

(i) Let \(\limsup\{a_n\}=\ell\in\mathbb{R}\) and \(\epsilon>0\). From \(\inf_n s_n=\ell<\ell+\epsilon\) there is \(N\) with \(a_n\le s_N<\ell+\epsilon\) for \(n\ge N\), so only the indices \(n<N\) can have \(a_n>\ell+\epsilon\). And every \(s_N\ge\ell>\ell-\epsilon\), so by the definition of supremum some \(k\ge N\) has \(a_k>\ell-\epsilon\); as \(N\) is arbitrary, infinitely many indices do.

Conversely, assume the two conditions. With \(\epsilon=1\) only finitely many terms exceed \(\ell+1\), so \(\{a_n\}\) is bounded above and each \(s_n\) is real. Given \(\epsilon>0\), let \(N\) exceed the finitely many indices \(n\) with \(a_n>\ell+\epsilon\); then \(s_n\le s_N\le\ell+\epsilon\) for \(n\ge N\), while for every \(n\) some \(k\ge n\) has \(a_k>\ell-\epsilon\), so \(s_n>\ell-\epsilon\). Thus \(|s_n-\ell|\le\epsilon\) for \(n\ge N\); running this with \(\epsilon/2\) yields strict inequality, so \(\{s_n\}\to\ell\).

(ii) If \(\{a_n\}\) is unbounded above, so is each set \(\{a_k \mid k\ge n\}\) (it differs from the whole by finitely many terms), so every \(s_n=\infty\) and \(\limsup\{a_n\}=\infty\). If instead \(a_n\le M\) for all \(n\), then \(s_n\le M\) and \(\limsup\{a_n\}\le M<\infty\).

(iii) For a set \(S\), \(\sup S=-\inf(-S)\) in the extended sense; with \(S=\{a_k \mid k\ge n\}\) this reads \(s_n=-i_n^{-}\), where \(i_n^{-}=\inf\{-a_k \mid k\ge n\}\). Since negation interchanges suprema and infima,

\begin{equation*} \limsup\{a_n\}=\inf_n\big(-i_n^{-}\big)=-\sup_n i_n^{-}=-\liminf\{-a_n\} . \end{equation*}

(iv) Suppose \(\liminf\{a_n\}=\limsup\{a_n\}=a\). If \(a\in\mathbb{R}\), choose \(N\) with \(|s_n-a|<\epsilon\) and \(|i_n-a|<\epsilon\) for \(n\ge N\); then \(a-\epsilon<i_n\le a_n\le s_n<a+\epsilon\). If \(a=\infty\), then \(\{i_n\}\to\infty\) and \(a_n\ge i_n\); if \(a=-\infty\), then \(\{s_n\}\to-\infty\) and \(a_n\le s_n\). In each case \(\{a_n\}\to a\).

Conversely let \(\{a_n\}\to a\). If \(a\in\mathbb{R}\) and \(\epsilon>0\), choose \(N\) with \(a-\epsilon<a_k<a+\epsilon\) for \(k\ge N\); then \(a-\epsilon\le i_n\le s_n\le a+\epsilon\) for \(n\ge N\), so both monotone limits lie in \([a-\epsilon,a+\epsilon]\), and \(\epsilon\) is arbitrary. If \(a=\infty\), each real \(c\) has \(i_n\ge c\) for large \(n\), so \(\{i_n\}\to\infty\) and \(s_n\ge i_n\); if \(a=-\infty\), each real \(c\) has \(s_n\le c\) for large \(n\), so \(\{s_n\}\to-\infty\) and \(i_n\le s_n\).

(v) The printed inequality is a misprint: \(a_n=b_n=(-1)^n\) satisfies \(a_n\le b_n\) yet \(\limsup\{a_n\}=1>-1=\liminf\{b_n\}\). The intended assertion is monotonicity of both limits. For it, fix \(n\): every \(k\ge n\) has \(a_k\le b_k\le s_n^{b}\), so \(s_n^{a}\le s_n^{b}\), and \(i_n^{a}\le a_k\le b_k\), so \(i_n^{a}\le i_n^{b}\). Taking \(\inf_n\) and \(\sup_n\),

\begin{equation*} \limsup\{a_n\}\le\limsup\{b_n\},\qquad \liminf\{a_n\}\le\liminf\{b_n\} . \end{equation*}

Suppose \(\{a_n\}\to a\). Every subsequence then converges to \(a\): a strictly increasing sequence of indices has \(n_k\ge k\) (induction), so the threshold \(N\) witnessing \(|a_n-a|<\epsilon\), or \(a_n>c\), or \(a_n<c\) for \(n\ge N\) serves for \(\{a_{n_k}\}\) as well. In particular \(a\) is a cluster point, \(\{a_n\}\) being a subsequence of itself. If \(c\) is a cluster point, say \(\{a_{n_k}\}\to c\), then also \(\{a_{n_k}\}\to a\), and a sequence has at most one extended limit: two real limits coincide by Proposition 14, a real limit forces boundedness (Proposition 14) whereas \(\pm\infty\) limits force unboundedness above or below, and convergence to \(\infty\) and to \(-\infty\) would make the terms eventually both positive and negative. So \(c=a\), the unique cluster point.

Conversely, let \(c\) be the only cluster point. By Problem 38, \(\liminf\{a_n\}\) and \(\limsup\{a_n\}\) are cluster points, so

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

and \(\{a_n\}\to c\) by Proposition 19(iv).

With \(i_n=\inf\{a_k \mid k\ge n\}\) increasing and \(s_n=\sup\{a_k \mid k\ge n\}\) decreasing, the Lemma of Problem 39 gives \(\liminf a_n=\sup_n i_n\) and \(\limsup a_n=\inf_n s_n\). Fix indices \(m,n\) and put \(p=\max\{m,n\}\); then

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

the outer inequalities by monotonicity and the inner two because \(a_p\in\{a_k \mid k\ge p\}\). So each \(s_n\) bounds \(\{i_m\}\) above, giving \(\sup_m i_m\le s_n\), and then \(\sup_m i_m\) bounds \(\{s_n\}\) below:

\begin{equation*} \liminf a_n \;=\; \sup_m i_m \;\le\; \inf_n s_n \;=\; \limsup a_n . \end{equation*}

Method (2): by Problem 38 the set of cluster points of \(\{a_n\}\) is nonempty with smallest element \(\liminf a_n\) and largest element \(\limsup a_n\).

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

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

so that \(A=\inf_n s_n^a\), \(B=\inf_n s_n^b\) and \(\limsup[a_nb_n]=\inf_n s_n^{ab}\) by the Lemma of Problem 39. Since \(s_n^a\ge a_n>0\) and \(s_n^b\ge b_n\ge0\), both \(A\) and \(B\) lie in \([0,\infty]\), so the proviso says exactly that neither \(A=0\) with \(B=\infty\) nor \(A=\infty\) with \(B=0\) occurs; likewise \(a_nb_n\ge0\) keeps the left-hand side out of \(-\infty\).

(i) \(A=\infty\) or \(B=\infty\). The proviso makes the other factor nonzero, hence positive, so \(A\cdot B=\infty\) and the inequality is trivial.

(ii) \(A,B\) finite. Given \(\epsilon>0\), choose \(N_1\) with \(s_{N_1}^a<A+\epsilon\) and \(N_2\) with \(s_{N_2}^b<B+\epsilon\), and put \(N=\max\{N_1,N_2\}\). For \(k\ge N\) we have \(0<a_k<A+\epsilon\) and \(0\le b_k<B+\epsilon\), and \(B+\epsilon>0\), so

\begin{equation*} a_kb_k \;<\; a_k(B+\epsilon) \;<\; (A+\epsilon)(B+\epsilon), \end{equation*}

whence

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

Were \(\limsup[a_nb_n]>AB\), a small enough \(\epsilon>0\) would contradict this; hence \(\limsup[a_nb_n]\le AB\).

Call an index \(n\) a peak index of \(\{a_n\}\) provided \(a_m\le a_n\) for every \(m\ge n\), and let \(P\) be the set of peak indices.

(i) \(P\) infinite. List its members increasingly as \(n_1<n_2<\cdots\), which is possible because every nonempty set of natural numbers has a least member (Theorem 1). Since \(n_k\in P\) and \(n_{k+1}>n_k\), we get \(a_{n_{k+1}}\le a_{n_k}\): a decreasing subsequence.

(ii) \(P\) finite. Choose \(N\) exceeding every member of \(P\) (take \(N=1\) if \(P=\emptyset\)) and put \(n_1=N\). Given \(n_k\ge N\), the index \(n_k\) is not a peak index, so some \(m>n_k\) has \(a_m>a_{n_k}\); set \(n_{k+1}=m\). This is an increasing subsequence.

For the Bolzano-Weierstrass Theorem: a bounded sequence \(\{a_n\}\) has a monotone subsequence \(\{a_{n_k}\}\) by the above, and this subsequence inherits the bound, so it converges by the Monotone Convergence Criterion (Theorem 15).

For \(y\ge0\) let \(\lfloor y\rfloor\) be the largest integer not exceeding \(y\); it exists because the set of natural numbers exceeding \(y\) is nonempty (Archimedean Property) and so has a least member \(k_0\) (Theorem 1), with \(\lfloor y\rfloor=k_0-1\), and it satisfies \(\lfloor y\rfloor\le y<\lfloor y\rfloor+1\).

Convergence. For any integers \(0\le a_n\le p-1\) the partial sums \(s_N=\sum_{n\le N}a_n/p^n\) increase and

\begin{equation*} s_N\le\sum_{n=1}^{N}\frac{p-1}{p^n}=1-\frac{1}{p^N}<1 , \end{equation*}

so the series is summable by Proposition 20(iii) (Problem 45); its sum \(x\) satisfies \(0\le x\le1\) by Proposition 14 applied to \(\{s_N\}\) and to \(\{-s_N\}\).

Existence. Given \(0<x<1\), put \(x_0=x\) and recursively

\begin{equation*} a_n=\lfloor p\,x_{n-1}\rfloor,\qquad x_n=p\,x_{n-1}-a_n\qquad(n\ge1). \end{equation*}

An induction gives \(0\le x_n<1\), hence \(0\le p\,x_{n-1}<p\) and so \(a_n\) is an integer with \(0\le a_n\le p-1\). From \(x_{n-1}=(a_n+x_n)/p\), induction gives \(x=s_N+x_N/p^N\), so

\begin{equation*} 0\le x-s_N<\frac{1}{p^N}\le\frac1N , \end{equation*}

using \(p^N\ge2^N\ge N\) (induction), and the Archimedean Property yields \(s_N\to x\).

Two expansions differ in one fixed way. Let \(\{a_n\}\neq\{b_n\}\) expand the same \(x\), let \(N\) be the least index with \(a_N\neq b_N\) (Theorem 1), and say \(a_N>b_N\), so \(a_N-b_N\ge1\). Subtracting the series (Theorem 18 on partial sums) and using \(a_n=b_n\) for \(n<N\),

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

So equality holds throughout: \(a_N-b_N=1\), and the series of nonnegative terms \(\bigl((p-1)-(b_n-a_n)\bigr)/p^n\) has sum \(0\), so every term vanishes. Since \(0\le a_n,b_n\le p-1\), this says \(b_n=p-1\) and \(a_n=0\) for all \(n>N\): one expansion is eventually \(0\), the other eventually \(p-1\).

Uniqueness except for \(q/p^n\). If \(x\in(0,1)\) has two expansions then, by the last paragraph, \(x=\sum_{n=1}^{N}a_n/p^n=q/p^N\) with \(q=\sum_{n\le N}a_np^{N-n}\) an integer.

Conversely let \(x=q/p^N\in(0,1)\). Every integer \(0\le q\le p^N-1\) is \(\sum_{n=1}^{N}c_np^{N-n}\) with integers \(0\le c_n\le p-1\): induct on \(N\), taking \(c_1=\lfloor q/p^N\rfloor\le p-1\) and applying the inductive hypothesis to \(q-c_1p^N\in[0,p^N-1]\). Dividing by \(p^N\) gives \(x=\sum_{n=1}^{M}c_n/p^n\), where \(M\le N\) is the largest index with \(c_M\neq0\) (some \(c_n\neq0\) since \(x>0\)). Then \(a_n=c_n\ (n\le M)\), \(a_n=0\ (n>M)\) is one expansion, and, because \(\sum_{n>M}(p-1)/p^n=1/p^M\),

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

is a second. There are never three: pairwise distinct expansions split each pair into one eventually \(0\) and one eventually \(p-1\), and \(p-1\neq0\) since \(p\ge2\), so two of the three would both be eventually \(p-1\), contradicting the paragraph above for that pair.

Write \(s_n=\sum_{k=1}^{n}a_k\) and \(s_0=0\), so that \(s_i-s_j=\sum_{k=j+1}^{i}a_k\) for \(j<i\); the series is summable exactly when \(\{s_n\}\) converges.

(i) If the series is summable then \(\{s_n\}\) is Cauchy (Theorem 17): given \(\epsilon>0\) pick \(N^{\prime}\) with \(|s_i-s_j|<\epsilon\) for \(i,j\ge N^{\prime}\) and put \(N=N^{\prime}+1\); for \(n\ge N\) and any natural \(m\), both \(n-1\ge N^{\prime}\) 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, assume the condition and choose \(N\) so that these sums are \(<\epsilon/2\) for \(n\ge N\). Individual terms are then controlled: 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*}

both sums being of the permitted form, so \(|a_k|<\epsilon\) for \(k\ge N+2\). Now let \(i>j\ge N+1\). If \(i\ge j+2\), then \(|s_i-s_j|=\bigl|\sum_{k=n}^{n+m}a_k\bigr|\) with \(n=j+1\ge N\) and \(m=i-j-1\ge1\), so \(|s_i-s_j|<\epsilon/2\); if \(i=j+1\), then \(|s_i-s_j|=|a_{j+1}|<\epsilon\) since \(j+1\ge N+2\). Hence \(\{s_n\}\) is Cauchy and converges by Theorem 17.

(ii) Let \(\epsilon>0\). Applying (i) to the summable series \(\sum|a_k|\) gives an \(N\) with \(\sum_{k=n}^{n+m}|a_k|<\epsilon\) for \(n\ge N\) and every natural \(m\), so 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*}

By the converse half of (i), \(\sum_{k=1}^{\infty}a_k\) is summable.

(iii) With \(a_k\ge0\) the partial sums satisfy \(s_{n+1}-s_n=a_{n+1}\ge0\), so \(\{s_n\}\) is increasing, and a monotone sequence converges if and only if it is bounded (Theorem 15). That is exactly the assertion.

Over the Field and Positivity Axioms, (C), (BW) and (MC) are equivalent, where (C) is the Completeness Axiom, (BW) says every bounded sequence has a convergent subsequence, and (MC) says every bounded monotone sequence converges. The forward implications are the two theorems of the text: (C) implies (BW) is Theorem 16 and (C) implies (MC) is Theorem 15. Proposition 14 and Theorem 18 are available throughout, their proofs using only the ordered-field axioms and the definition of convergence.

Lemma A. Each of (BW) and (MC) makes \(\mathbb{N}\) unbounded above. Were \(\mathbb{N}\) bounded above, the sequence \(a_n=n\) would be increasing and bounded. Under (BW) it has a convergent subsequence \(\{n_k\}\), which is therefore Cauchy (the first half of the proof of Theorem 17 uses only the definition of convergence), so \(n_{k+1}-n_k<1\) for large \(k\); but \(n_{k+1}-n_k\) is a natural number (induction from Problem 9), hence at least \(1\). Under (MC) it converges to some \(a\), and choosing \(N\) with \(|a_n-a|<1/2\) for \(n\ge N\) gives \(1=|a_{N+1}-a_N|<1\). Both are absurd. Consequently \(\{1/2^n\}\to0\), since \(2^n\ge n\) (induction).

Lemma B (bisection). Let \(E\) be nonempty and bounded above by \(b_0\); fix \(e\in E\) and put \(a_0=e-1\), which is not an upper bound and satisfies \(a_0<b_0\). Recursively, with \(c=(a_n+b_n)/2\),

\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{otherwise.} \end{cases} \end{equation*}

An induction gives, for every \(n\): \(a_n\) is not an upper bound for \(E\), \(b_n\) is, \(\{a_n\}\) increases, \(\{b_n\}\) decreases, \(a_0\le a_n<b_n\le b_0\), and \(b_n-a_n=(b_0-a_0)/2^{n}\), so both sequences are bounded and \(\{b_n-a_n\}\to0\) by Lemma A.

(BW) implies (C). Form \(\{a_n\},\{b_n\}\) as in Lemma B and take \(\{a_{n_k}\}\to L\) by (BW). Since \(n_k\ge k\), also \(\{b_{n_k}-a_{n_k}\}\to0\), so \(b_{n_k}=a_{n_k}+(b_{n_k}-a_{n_k})\to L\) by Theorem 18. \(L\) is an upper bound: each \(x\in E\) has \(x\le b_{n_k}\) for all \(k\), so \(x\le L\) by the monotonicity of convergence (Theorem 18, formula (7)) applied against the constant sequence \(x\). \(L\) is least: an upper bound \(c<L\) would admit \(k\) with \(a_{n_k}>c\), and since \(a_{n_k}\) is not an upper bound some \(y\in E\) exceeds \(a_{n_k}>c\). Hence \(L=\sup E\).

(MC) implies (C). Form \(\{a_n\},\{b_n\}\) as in Lemma B; \(\{a_n\}\) is increasing and bounded, so \(\{a_n\}\to L\) by (MC), and \(b_n=a_n+(b_n-a_n)\to L\) by Lemma A and Theorem 18. The two arguments just given, with \(a_n,b_n\) in place of \(a_{n_k},b_{n_k}\), show \(L=\sup E\).

Extend \(f\) affinely across the gaps. If \(E=\emptyset\) take \(g\equiv0\) and if \(E=\mathbb{R}\) take \(g=f\); otherwise \(\mathcal{O}=\mathbb{R}\sim E\) is open and nonempty (Proposition 11) and so is the disjoint union of a countable collection of open intervals, its components (Proposition 9), none of which is all of \(\mathbb{R}\).

A finite endpoint of a component lies in \(E\): if the right endpoint \(b\) of \(I=(a,b)\) lay in \(\mathcal{O}\), it would lie in another component \(J=(c,d)\), and any \(z\) with \(\max\{a,c\}<z<b\) (read \(c<z<b\) when \(a=-\infty\)) would lie in both \(I\) and \(J\), contradicting disjointness; left endpoints are symmetric.

Define \(g=f\) on \(E\); on a bounded component \((a,b)\), so \(a,b\in E\),

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

on a component \((a,\infty)\), \(g\equiv f(a)\); on a component \((-\infty,b)\), \(g\equiv f(b)\). The two coefficients above are nonnegative and sum to \(1\), so for real \(c\),

\begin{equation*} |f(a)-c|<\epsilon \ \text{ and } \ |f(b)-c|<\epsilon \ \Longrightarrow\ |g(x)-c|\le\epsilon \quad\text{on } (a,b). \tag{\(\ast\)} \end{equation*}

At a point of \(\mathcal{O}\), \(g\) agrees near that point with an affine function on its component, hence is continuous there.

Let \(x_0\in E\) and \(\epsilon>0\), and take \(\delta>0\) with \(|f(y)-f(x_0)|<\epsilon\) for \(y\in E\), \(|y-x_0|<\delta\). We produce \(\delta_+>0\) with \(|g(x)-g(x_0)|\le\epsilon\) for \(x_0\le x<x_0+\delta_+\); the left-hand side is mirror-image.

(i) Some \((x_0,x_0+\eta)\) misses \(E\). Then it lies in \(\mathcal{O}\); let \(I\) be the component containing \(x_0+\eta/2\) and \(a\) its left endpoint. If \(a<x_0\) (including \(a=-\infty\)) then \(x_0\in I\subseteq\mathcal{O}\), impossible; if \(a>x_0\) then \(a\in E\) by the endpoint fact while \(x_0<a<x_0+\eta\), also impossible. So \(a=x_0\), and on \(I\) the function \(g\) agrees with an affine \(\ell\) having \(\ell(x_0)=f(x_0)=g(x_0)\) (the constant \(f(x_0)\) in the unbounded case). Continuity of \(\ell\) gives \(\delta_+>0\), shrunk so that \((x_0,x_0+\delta_+)\subseteq I\), with \(|g(x)-g(x_0)|<\epsilon\) there.

(ii) Every \((x_0,x_0+\eta)\) meets \(E\). Choose \(x_1\in E\) with \(x_0<x_1<x_0+\delta\) and put \(\delta_+=x_1-x_0\). Let \(x_0\le x<x_1\). If \(x\in E\), then \(|g(x)-g(x_0)|=|f(x)-f(x_0)|<\epsilon\). If not, the component \(I\) containing \(x\) omits \(x_0\) and \(x_1\), so \(I=(a,b)\) with \(x_0\le a<x<b\le x_1\); then \(a,b\in E\) lie within \(\delta\) of \(x_0\), so \((\ast)\) with \(c=f(x_0)=g(x_0)\) gives \(|g(x)-g(x_0)|\le\epsilon\).

Thus \(|g(x)-g(x_0)|\le\epsilon\) for \(|x-x_0|<\min\{\delta_+,\delta_-\}\); running the argument with \(\epsilon/2\) makes the inequality strict, so \(g\) is continuous at \(x_0\), hence on \(\mathbb{R}\), and \(g=f\) on \(E\).

Answer: exactly at the irrational numbers and at \(x=0\).

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

\begin{equation*} f(x)=q\,x\,\sin\frac1q=x\cdot\frac{\sin(1/q)}{1/q}, \tag{1} \end{equation*}

and \(0=0/1\) gives \(f(0)=0\). We use \(|\sin t|\le|t|\), \(\sin t<t\) for \(t>0\), and \(\sin t/t\to1\) as \(t\to0\), together with Proposition 21 (continuity at \(x_*\) is equivalent to \(f(x_n)\to f(x_*)\) for every sequence \(\{x_n\}\to x_*\)).

At \(0\). For every \(x\) we have \(|f(x)|\le|x|\): this is an equality at irrational \(x\), and at \(x=p/q\) in lowest terms \(|f(x)|=|p||\sin(1/q)|\le|p|/q=|x|\). So \(\{x_n\}\to0\) forces \(f(x_n)\to0=f(0)\).

At an irrational \(x_*\). Let \(\{x_n\}\to x_*\) and \(\epsilon>0\). The sequence is bounded, say \(|x_n|\le M\) with \(M>0\), and \(\sin t/t\to1\) supplies \(\eta>0\) with \(|\sin t/t-1|<\epsilon/M\) for \(0<t<\eta\). The denominators of the rational terms tend to infinity: otherwise infinitely many rational terms \(x_n=p_n/q_n\) have \(q_n\le Q\), and then \(|p_n|=q_n|x_n|\le QM\), so those terms take only finitely many values and one value repeats infinitely often, giving a constant subsequence, whose value must be \(x_*\) – making \(x_*\) rational. So there is \(N\) beyond which every rational term has \(1/q_n<\eta\), and then by (1)

\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*}

while \(|f(x_n)-x_n|=0\) at irrational terms. Hence \(f(x_n)-x_n\to0\), so \(f(x_n)\to x_*=f(x_*)\) by Theorem 18.

At a nonzero rational \(x_*=p/q\) in lowest terms. Here \(0<1/q\le1\), so \(\sin(1/q)<1/q\) and the factor in (1) is strictly less than \(1\); since \(x_*\neq0\), this gives \(f(x_*)\neq x_*\). The irrationals are dense (Problem 12), so choose irrational \(x_n\) with \(|x_n-x_*|<1/n\); then \(\{x_n\}\to x_*\) by the Archimedean Property while \(f(x_n)=x_n\to x_*\neq f(x_*)\).

Throughout we use the sequential criterion, Proposition 21.

(i) Fix \(x_*\in E\) and a sequence \(\{x_n\}\) in \(E\) with \(\{x_n\}\to x_*\), so that \(f(x_n)\to f(x_*)\) and \(g(x_n)\to g(x_*)\). By the linearity of convergence (Theorem 18 with \(\alpha=\beta=1\)),

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

For the product, first let \(a_n\to a\) and \(b_n\to b\); the sequence \(\{a_n\}\) is bounded (Proposition 14), so choose \(M>0\) with \(|a_n|\le M\) for all \(n\) and \(M\ge|b|\), and then

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

which is less than \(\epsilon\) once \(|a_n-a|\) and \(|b_n-b|\) are less than \(\epsilon/(2M)\). With \(a_n=f(x_n)\) and \(b_n=g(x_n)\) this gives \((fg)(x_n)\to(fg)(x_*)\). So \(f+g\) and \(fg\) are continuous on \(E\).

(ii) Let \(h\) be continuous on \(D\) with \(h(D)\subseteq E\), let \(x_*\in D\) and \(\{x_n\}\to x_*\) in \(D\). Then \(h(x_n)\to h(x_*)\), and \(\{h(x_n)\}\) is a sequence in \(E\) converging to the point \(h(x_*)\) of \(E\), so continuity of \(f\) there gives

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

(iii) For real \(u,v,u^{\prime},v^{\prime}\),

\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*}

assuming without loss of generality \(\max\{u,v\}\ge\max\{u^{\prime},v^{\prime}\}\) and, say, \(\max\{u,v\}=u\), the left side is at most \(u-u^{\prime}\le|u-u^{\prime}|\), since \(\max\{u^{\prime},v^{\prime}\}\ge u^{\prime}\). Now given \(x_*\in E\) and \(\epsilon>0\), continuity of \(f\) and of \(g\) at \(x_*\) supplies \(\delta_1,\delta_2>0\) forcing \(|f(x)-f(x_*)|<\epsilon\) and \(|g(x)-g(x_*)|<\epsilon\) respectively; for \(x\in E\) with \(|x-x_*|<\min\{\delta_1,\delta_2\}\), (2) gives \(|\max\{f,g\}(x)-\max\{f,g\}(x_*)|<\epsilon\).

(iv) The triangle inequality (Problem 7) gives \(\bigl||s|-|t|\bigr|\le|s-t|\), so for \(x\in E\) within the \(\delta\) provided by the continuity of \(f\) at \(x_*\),

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

Let \(f\) be Lipschitz on \(E\) with constant \(c\ge0\), and given \(\epsilon>0\) take \(\delta=\epsilon/(c+1)>0\), which depends on neither point: for \(x,x^{\prime}\in E\) with \(|x-x^{\prime}|<\delta\),

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

So \(f\) is uniformly continuous on \(E\).

The converse fails for \(f(x)=\sqrt{x}\) on \(E=[0,1]\). This \(f\) is continuous: at \(x>0\),

\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 there, while at \(x=0\) the choice \(\delta=\epsilon^2\) does. Since \([0,1]\) is closed and bounded, Theorem 23 makes \(f\) uniformly continuous on it.

But \(f\) is not Lipschitz: a constant \(c\ge0\) would give, at \(x=0\) and \(x^{\prime}=t\in(0,1]\),

\begin{equation*} \sqrt{t}\le c\,t, \qquad\text{that is}\qquad \frac{1}{\sqrt{t}}\le c , \end{equation*}

and \(t=1/(c+1)^{2}\in(0,1]\) then yields \(c+1\le c\).

Take \(\varphi\) to be the piecewise linear interpolant of \(f\) at a uniform partition of mesh \(< \delta\), where \(\delta\) responds to the \(\epsilon\) challenge for the uniform continuity of \(f\) on the closed bounded interval \([a,b]\) (Theorem 23). Explicitly, fix \(n > (b-a)/\delta\), put \(x_i = a + i(b-a)/n\), and for \(x \in [x_i, x_{i+1}]\) 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*}

The two prescriptions at a shared endpoint \(x_i\) both return \(f(x_i)\), so \(\varphi\) is well defined, continuous, and linear on each \([x_i, x_{i+1}]\): it is piecewise linear. (Check!)

Given \(x \in [x_i, x_{i+1}]\), write \(\lambda = (x_{i+1} - x)/(x_{i+1} - x_i) \in [0,1]\), so that \(\varphi(x) = \lambda f(x_i) + (1-\lambda) f(x_{i+1})\). Since \(|x - x_i| \le (b-a)/n < \delta\) and \(|x - x_{i+1}| < \delta\),

\begin{equation*} \begin{aligned} |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})| \\ &< \lambda\,\epsilon + (1-\lambda)\,\epsilon = \epsilon , \end{aligned} \end{equation*}

the strict inequality persisting when \(\lambda \in \{0,1\}\) because then one term is absent and the other is already \(< \epsilon\). Hence \(|f - \varphi| < \epsilon\) on \([a,b]\).

Necessity is the Extreme Value Theorem, applied to the nonempty closed bounded set \(E\); for sufficiency the two witnesses are \(f(x) = |x|\) and \(f(x) = 1/|x - x_0|\).

(i) \(E\) is bounded. The function \(f(x) = |x|\) is Lipschitz with constant \(1\) on \(E\) by the reverse triangle inequality, hence continuous, so by hypothesis it attains a maximum at some \(x_{*} \in E\). Then \(E \subseteq [-|x_{*}|, |x_{*}|]\).

(ii) \(E\) is closed. Suppose \(x_0\) is a point of closure of \(E\) with \(x_0 \notin E\), and set \(f(x) = 1/|x - x_0|\) for \(x \in E\); this is continuous on \(E\), being the composition of the continuous nowhere-zero \(x \mapsto |x - x_0|\) with \(t \mapsto 1/t\) on \((0,\infty)\). By hypothesis \(f\) attains a maximum \(M > 0\), so

\begin{equation*} |x - x_0| \ge \frac{1}{M} \quad \text{for all } x \in E , \end{equation*}

whereas \(x_0\) being a point of closure supplies an \(x \in E\) with \(|x - x_0| < 1/M\). This contradiction forces \(x_0 \in E\), so \(E\) contains all its points of closure.

Necessity is the Heine-Borel Theorem; for sufficiency the two covers are \(\{(-n,n)\}\) and \(\{|x - x_0| > 1/n\}\). Both are ascending in \(n\), so a finite subcover collapses to a single index \(N\). (The empty set is trivially closed and bounded, so assume \(E \ne \emptyset\).)

(i) \(E\) is bounded. The open intervals \(\mathcal{O}_n = (-n,n)\) cover \(\mathbb{R}\) by the Archimedean Property, hence cover \(E\); a finite subcover gives \(E \subseteq \mathcal{O}_N = (-N, N)\).

(ii) \(E\) is closed. Suppose \(x_0\) is a point of closure of \(E\) with \(x_0 \notin E\), and put

\begin{equation*} \begin{aligned} \mathcal{O}_n &= \Bigl(-\infty,\ x_0 - \tfrac{1}{n}\Bigr) \cup \Bigl(x_0 + \tfrac{1}{n},\ \infty\Bigr) , \end{aligned} \end{equation*}

an open set. Each \(x \in E\) has \(|x - x_0| > 0\), so lies in some \(\mathcal{O}_n\) by the Archimedean Property; a finite subcover gives \(E \subseteq \mathcal{O}_N\), that is, \(|x - x_0| > 1/N\) for all \(x \in E\). Then the open interval \((x_0 - 1/N, x_0 + 1/N)\) misses \(E\), contradicting \(x_0\) being a point of closure. Hence \(x_0 \in E\).

Sufficiency is the identity function \(f(x) = x\), continuous on \(E\) with image \(E\); necessity is the Intermediate Value Theorem, via the following characterization, used again in Problem 55.

Lemma. A nonempty \(I \subseteq \mathbb{R}\) is an interval if and only if it has the betweenness property: \(x, y \in I\) and \(x < z < y\) imply \(z \in I\).

Every interval is cut out by inequalities on the point itself, so has the property. Conversely put \(a = \inf I\), \(b = \sup I\) in the extended reals, so \(I \subseteq [a,b]\). If \(a < z < b\) then \(z\) is neither a lower nor an upper bound for \(I\), so there are \(x, y \in I\) with \(x < z < y\), whence \(z \in I\). Thus \((a,b) \subseteq I \subseteq [a,b]\) and \(I\) is an interval, degenerating to \(\{a\}\) when \(a = b\). \(\square\)

Now let \(E\) be an interval, \(f\) continuous on \(E\), and \(u = f(x)\), \(v = f(y)\) in \(f(E)\) with \(u < w < v\); note \(x \ne y\).

(i) \(x < y\). Betweenness gives \([x,y] \subseteq E\), and \(f(x) = u < w < v = f(y)\), so the Intermediate Value Theorem on \([x,y]\) yields \(x_0\) with \(f(x_0) = w\).

(ii) \(y < x\). Here \([y,x] \subseteq E\) and \(g = -f\) satisfies \(g(y) = -v < -w < -u = g(x)\), so the Intermediate Value Theorem applied to \(g\) yields \(x_0\) with \(f(x_0) = w\).

Either way \(w \in f(E)\), so \(f(E)\) has the betweenness property and the Lemma makes it an interval.

Forward is Problem 54 (continuous image of an interval is an interval, no monotonicity needed); the converse is the observation that a jump at \(x_0\) leaves the gap \((f(x_0^-), f(x_0^+))\) free of values of \(f\) save \(f(x_0)\) itself, violating the betweenness property of the interval \(f(I)\).

Replacing \(f\) by \(-f\) if necessary, assume \(f\) increasing: \(-f\) is increasing exactly when \(f\) is decreasing, continuity is unaffected, and \(-S\) has the betweenness property exactly when \(S\) does, so both halves transfer.

Let \(f(I)\) be an interval and \(x_0 \in I\). Since \(I\) is open there are points of \(I\) on both sides of \(x_0\), so monotonicity gives one-sided limits

\begin{equation*} \begin{aligned} f(x_0^-) = \sup_{x \in I,\, x < x_0} f(x) &\le f(x_0) \\ &\le \inf_{x \in I,\, x > x_0} f(x) = f(x_0^+) , \end{aligned} \end{equation*}

the sup and inf being finite because \(f(x_0)\) bounds each set on the relevant side; \(f\) is continuous at \(x_0\) precisely when the two ends agree. Suppose instead \(f(x_0^-) < f(x_0^+)\). By the displayed formulas,

\begin{equation*} \begin{aligned} f(x) \le f(x_0^-) \text{ for } x < x_0 , \\ f(x) \ge f(x_0^+) \text{ for } x > x_0 , \end{aligned} \end{equation*}

so \(f(x_0)\) is the only value of \(f\) in \(\bigl(f(x_0^-), f(x_0^+)\bigr)\); choose \(c\) in that open interval with \(c \ne f(x_0)\), so \(c \notin f(I)\). Picking \(x_1 < x_0 < x_2\) in \(I\) gives

\begin{equation*} f(x_1) \le f(x_0^-) < c < f(x_0^+) \le f(x_2) , \end{equation*}

with \(f(x_1), f(x_2) \in f(I)\) and \(c \notin f(I)\), contradicting betweenness. Hence \(f\) is continuous at every \(x_0 \in I\).

The continuity set is \(C = \bigcap_{n=1}^{\infty} \mathcal{O}_n\), where \(\mathcal{O}_n = \{x : \omega(x) < 1/n\}\) is open and \(\omega\) is the oscillation

\begin{equation*} \begin{aligned} \omega(x) &= \inf_{\delta > 0} \operatorname{osc}(f, x, \delta) \in [0,\infty] , \\ \operatorname{osc}(f, x, \delta) &= \sup \{ |f(y) - f(z)| : y, z \in (x-\delta,\ x+\delta) \} . \end{aligned} \end{equation*}

(The sup may be \(+\infty\), which is harmless; it is nondecreasing in \(\delta\).) Three points.

(i) \(f\) is continuous at \(x\) iff \(\omega(x) = 0\). If \(f\) is continuous at \(x\), pick \(\delta\) with \(|f(y) - f(x)| < \epsilon/3\) on \((x-\delta, x+\delta)\); the triangle inequality gives \(\operatorname{osc}(f,x,\delta) \le 2\epsilon/3 < \epsilon\), so \(\omega(x) < \epsilon\) for every \(\epsilon > 0\). Conversely if \(\omega(x) = 0\) and \(\epsilon > 0\), some \(\delta\) has \(\operatorname{osc}(f,x,\delta) < \epsilon\), and then \(|y - x| < \delta\) forces \(|f(y) - f(x)| \le \operatorname{osc}(f,x,\delta) < \epsilon\).

(ii) \(\mathcal{O}_n\) is open. Given \(x \in \mathcal{O}_n\), choose \(\delta\) with \(\operatorname{osc}(f,x,\delta) < 1/n\). If \(|x^{\prime} - x| < \delta/2\) then \((x^{\prime} - \delta/2, x^{\prime} + \delta/2) \subseteq (x - \delta, x + \delta)\), so the supremum runs over fewer pairs and

\begin{equation*} \begin{aligned} \omega(x^{\prime}) &\le \operatorname{osc}(f, x^{\prime}, \delta/2) \\ &\le \operatorname{osc}(f, x, \delta) < \tfrac{1}{n} , \end{aligned} \end{equation*}

putting \((x - \delta/2, x + \delta/2) \subseteq \mathcal{O}_n\).

(iii) By (i), \(x \in C\) iff \(\omega(x) = 0\) iff \(\omega(x) < 1/n\) for every \(n\) (Archimedean Property), that is, iff \(x \in \bigcap_n \mathcal{O}_n\). Hence \(C\) is a \(G_\delta\) set.

The convergence set is \(X = \bigcap_{k=1}^{\infty} F_k\) with \(F_k = \bigcup_{N=1}^{\infty} A_{N,k}\), where

\begin{equation*} A_{N,k} = \bigcap_{n \ge N} \bigcap_{m \ge N} E_{n,m,k} , \qquad E_{n,m,k} = \Big\{ x : |f_n(x) - f_m(x)| \le \tfrac{1}{k} \Big\} . \end{equation*}

Indeed, by Theorem 17 (the Cauchy Convergence Criterion) \(x \in X\) iff \(\{f_n(x)\}\) is Cauchy, that is, iff for every \(k\) there is an \(N\) with \(|f_n(x) - f_m(x)| \le 1/k\) for all \(n, m \ge N\); the closed inequality is equivalent to the strict one, by choosing \(k\) with \(1/k < \epsilon\). Reading the three quantifiers as \(\bigcap_k \bigcup_N \bigcap_{n,m \ge N}\) gives the display.

Each \(E_{n,m,k}\) is closed: \(g = |f_n - f_m|\) is continuous by Problem 49 (sums and absolute values of continuous functions are continuous), so its complement \(g^{-1}((1/k, \infty))\) is open by Proposition 22 applied on \(E = \mathbb{R}\), and Proposition 11 applies. Hence \(A_{N,k}\) is closed by Proposition 12, \(F_k\) is a countable union of closed sets, that is, an \(F_\sigma\) set, and \(X\) is the intersection of the countable collection \(\{F_k\}_{k=1}^{\infty}\) of \(F_\sigma\) sets.

Open sets: for \(O\) open, Proposition 22 applied with \(E = \mathbb{R}\) gives \(f^{-1}(O) = \mathbb{R} \cap U = U\) for some open \(U\), so \(f^{-1}(O)\) is open.

Closed sets: for \(C\) closed, \(\mathbb{R} \sim C\) is open (Proposition 11), so

\begin{equation*} \mathbb{R} \sim f^{-1}( C) = f^{-1}(\mathbb{R} \sim C) \end{equation*}

is open by the preceding paragraph, and \(f^{-1}( C)\) is closed, again by Proposition 11.

Borel sets: the collection \(\mathcal{A} = \{A \subseteq \mathbb{R} : f^{-1}(A) \text{ is Borel}\}\) is a \(\sigma\)-algebra containing the open sets, hence contains \(\mathcal{B}\). Indeed \(f^{-1}(\emptyset) = \emptyset\) is Borel; the identities

\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*}

transport complements and countable unions of members of \(\mathcal{A}\) back into \(\mathcal{A}\), because \(\mathcal{B}\) is closed under both; and every open \(O\) lies in \(\mathcal{A}\) since \(f^{-1}(O)\) is open, hence Borel. As \(\mathcal{B}\) is by definition the smallest \(\sigma\)-algebra containing the open sets (Proposition 13), \(\mathcal{B} \subseteq \mathcal{A}\), which says \(f^{-1}(A)\) is Borel for every Borel \(A\).

The \(\epsilon/3\) argument. Fix \(x \in E\) and \(\epsilon > 0\), and use uniform convergence to choose \(N\) with \(|f_n(y) - f(y)| < \epsilon/3\) for all \(y \in E\) and all \(n \ge N\); the single index \(N\) serving every point of \(E\) is exactly what uniformity supplies. Since \(f_N\) is continuous at \(x\), choose \(\delta > 0\) so that \(y \in E\) and \(|y - x| < \delta\) imply \(|f_N(y) - f_N(x)| < \epsilon/3\). For such \(y\),

\begin{equation*} \begin{aligned} |f(y) - f(x)| &\le |f(y) - f_N(y)| + |f_N(y) - f_N(x)| + |f_N(x) - f(x)| \\ &< \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} + \tfrac{\epsilon}{3} = \epsilon , \end{aligned} \end{equation*}

the outer terms being controlled by the uniform estimate at the points \(y, x \in E\) with index \(n = N\). Hence \(f\) is continuous at each \(x \in E\).

Proposition 21. If \(f\) is continuous at \(x_*\) and \(\{x_n\} \to x_*\) in \(E\), pick \(\delta\) for the \(\epsilon\) challenge at \(x_*\) and then \(N\) with \(|x_n - x_*| < \delta\) for \(n \ge N\); since each \(x_n \in E\), \(|f(x_n) - f(x_*)| < \epsilon\) for \(n \ge N\). Conversely, if \(f\) is not continuous at \(x_*\) there is \(\epsilon_0 > 0\) such that for each \(n\) some \(x_n \in E\) has

\begin{equation*} |x_n - x_*| < \frac{1}{n} \qquad \text{and} \qquad |f(x_n) - f(x_*)| \ge \epsilon_0 ; \end{equation*}

then \(\{x_n\} \to x_*\) by the Archimedean Property while \(\{f(x_n)\} \not\to f(x_*)\), so the sequential condition fails. Contrapositively, it implies continuity.

Lemma (used twice below). A closed \(E\) contains the limit \(x_*\) of any sequence in \(E\): every open interval about \(x_*\) contains some \(x_n \in E\), so \(x_*\) is a point of closure of \(E\).

Extreme Value Theorem. Let \(E\) be nonempty, closed and bounded and \(f\) continuous on \(E\).

(i) \(f(E)\) is bounded above. Otherwise choose \(x_n \in E\) with \(f(x_n) > n\); the Bolzano-Weierstrass Theorem (Theorem 16) applies to this bounded sequence to give \(\{x_{n_k}\} \to x_*\), with \(x_* \in E\) by the Lemma, so \(\{f(x_{n_k})\} \to f(x_*)\) by Proposition 21 and is therefore bounded (Proposition 14). Yet \(n_k \ge k\) gives \(f(x_{n_k}) > k\), a contradiction.

(ii) \(m = \sup f(E)\) is attained. The supremum exists by (i) and the Completeness Axiom. Choose \(x_n \in E\) with \(m - 1/n < f(x_n) \le m\), pass to a subsequence \(\{x_{n_k}\} \to x_* \in E\) as in (i), so \(\{f(x_{n_k})\} \to f(x_*)\) by Proposition 21. But \(n_k \ge k\) gives

\begin{equation*} |f(x_{n_k}) - m| = m - f(x_{n_k}) < \frac{1}{n_k} \le \frac{1}{k} , \end{equation*}

so \(\{f(x_{n_k})\} \to m\), and uniqueness of limits (Proposition 14) forces \(f(x_*) = m\).

Applying (i) and (ii) to \(-f\), which is continuous on \(E\), produces \(x_{**} \in E\) with \(f(x_{**}) \le f(x)\) for all \(x \in E\); \(f\) attains its minimum there.

Apply countable additivity to the disjoint collection \(E_1 = A\), \(E_2 = B \setminus A\), \(E_k = \emptyset\) for \(k \ge 3\), whose union is \(B\) because \(A \subseteq B\). Each \(E_k\) lies in \(\mathcal{A}\), since a \(\sigma\)-algebra contains \(\emptyset = \mathbb{R} \setminus \mathbb{R}\) and \(B \setminus A = B \cap (\mathbb{R} \setminus A)\). Hence

\begin{equation*} \begin{aligned} m(B) = \sum_{k=1}^{\infty} m(E_k) &= m(A) + m(B \setminus A) + \sum_{k=3}^{\infty} m(\emptyset) \\ &\ge m(A) , \end{aligned} \end{equation*}

the last step because \(m\) takes values in \([0, \infty]\), so a series of its values dominates its first term.

Apply countable additivity to the disjoint collection \(E_1 = A\), \(E_k = \emptyset\) for \(k \ge 2\), whose union is \(A\) (and \(\emptyset = \mathbb{R} \setminus \mathbb{R}\) lies in \(\mathcal{A}\)):

\begin{equation*} m(A) = \sum_{k=1}^{\infty} m(E_k) = m(A) + \sum_{k=2}^{\infty} m(\emptyset) . \end{equation*}

Since \(m(A) < \infty\) it may be subtracted from both sides, leaving \(\sum_{k=2}^{\infty} m(\emptyset) = 0\); a series of identical nonnegative terms vanishes only if the term does, so \(m(\emptyset) = 0\).

Disjointify: put \(F_1 = E_1\) and \(F_k = E_k \setminus \bigcup_{j=1}^{k-1} E_j\) for \(k \ge 2\), each in \(\mathcal{A}\) since a \(\sigma\)-algebra is closed under finite unions, complements and intersections.

The \(F_k\) are disjoint: for \(j < k\), every point of \(F_k\) misses \(E_j \supseteq F_j\). They have the same union as the \(E_k\): \(F_k \subseteq E_k\) gives one inclusion, and if \(x \in \bigcup_k E_k\) then the smallest index \(k_0\) with \(x \in E_{k_0}\) (Well-Ordering Principle) puts \(x \in F_{k_0}\). Countable additivity and monotonicity (Problem 1, applicable since \(F_k \subseteq E_k\) with both in \(\mathcal{A}\)) now give

\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) \le \sum_{k=1}^{\infty} m(E_k). \end{equation*}

Translation invariance is immediate: \(x \mapsto x + y\) is a bijection of \(E\) onto \(E + y\), so the two sets are simultaneously empty, finite with equal cardinality, or infinite, and each clause of the definition of \(c\) returns the same value. For countable additivity let \(\{E_k\}\) be disjoint with union \(E\); write \(\#F\) for the cardinality of a finite \(F\), and note that \(F \subseteq E\) finite forces \(c(E) \ge \#F\).

(i) \(c(E) \ge \sum_{k} c(E_k)\). If some \(E_{k_0}\) is infinite then so is \(E \supseteq E_{k_0}\) and both sides are \(\infty\). Otherwise each \(E_k\) is finite, and for each \(n\) the disjoint finite union \(F = \bigcup_{k=1}^n E_k\) satisfies

\begin{equation*} c(E) \ge \#F = \sum_{k=1}^{n} \#E_k = \sum_{k=1}^{n} c(E_k) ; \end{equation*}

take the supremum over \(n\).

(ii) \(c(E) \le \sum_k c(E_k)\). Assume the right side is \(M < \infty\), so every \(E_k\) is finite. If \(E\) had more than \(M\) elements, pick distinct \(x_1, \dots, x_{M+1} \in E\), say \(x_i \in E_{k_i}\), and set \(n = \max_i k_i\); then by disjointness

\begin{equation*} M + 1 \le \#\left(\bigcup_{k=1}^{n} E_k\right) = \sum_{k=1}^{n} c(E_k) \le M , \end{equation*}

a contradiction. So \(E\) is finite with \(c(E) = \#E \le M\).

A countable set has outer measure zero, whereas \(m^{*}([0,1]) = \ell([0,1]) = 1\) by Proposition 1; so \([0,1]\) is uncountable.

For the first fact (the Example preceding Proposition 1): if \(C = \{c_k\}_{k=1}^{\infty}\) and \(\epsilon > 0\), the open bounded intervals \(I_k = (c_k - \epsilon/2^{k+1},\ c_k + \epsilon/2^{k+1})\) cover \(C\), so the infimum defining outer measure gives

\begin{equation*} \begin{aligned} 0 \le m^{*}( C) \le \sum_{k=1}^{\infty} \ell(I_k) = \sum_{k=1}^{\infty} \frac{\epsilon}{2^{k}} = \epsilon \end{aligned} \end{equation*}

for every \(\epsilon > 0\), whence \(m^{*}( C) = 0\). Were \([0,1]\) countable we would get \(1 = 0\).

Write \(B = [0,1] \setminus A\), the rationals in \([0,1]\), a countable set and so of outer measure zero (the Example of Section 2.2). Since \([0,1] = A \cup B\), finite subadditivity (Proposition 3) and Proposition 1 give

\begin{equation*} 1 = m^{*}([0,1]) \le m^{*}(A) + m^{*}(B) = m^{*}(A) , \end{equation*}

while monotonicity applied to \(A \subseteq [0,1]\) gives \(m^{*}(A) \le 1\). Hence \(m^{*}(A) = 1\).

Take \(G = \bigcap_{n=1}^{\infty} \mathcal{O}_n\), where \(\mathcal{O}_n\) is the union of a countable cover of \(E\) by open bounded intervals of total length within \(1/n\) of \(m^{*}(E)\).

Such covers exist because \(m^{*}(E) < \infty\): boundedness gives \(E \subseteq (-M,M)\) for some \(M\), so \(m^{*}(E) \le 2M\) by monotonicity and Proposition 1, and hence \(m^{*}(E) + 1/n\) fails to be a lower bound for the sums in the infimum defining \(m^{*}(E)\). Fixing such a cover \(\{I_{n,k}\}_k\) of \(E\) and putting \(\mathcal{O}_n = \bigcup_k I_{n,k}\), countable subadditivity (Proposition 3) with Proposition 1 gives

\begin{equation*} m^{*}(\mathcal{O}_n) \le \sum_{k=1}^{\infty} \ell(I_{n,k}) < m^{*}(E) + \frac{1}{n} . \end{equation*}

Each \(\mathcal{O}_n\) is open and contains \(E\), so \(G\) is a \(G_\delta\) set containing \(E\). Monotonicity on both sides of \(E \subseteq G \subseteq \mathcal{O}_n\) gives

\begin{equation*} m^{*}(E) \le m^{*}(G) \le m^{*}(\mathcal{O}_n) < m^{*}(E) + \frac{1}{n} \end{equation*}

for every \(n\), and since \(m^{*}(E)\) is finite the difference \(m^{*}(G) - m^{*}(E)\) is a nonnegative number below every \(1/n\), hence \(0\).

Pass to closures, which is legitimate exactly because the collection is finite: a finite union of closed sets is closed, so \(\bigcup_{k=1}^n \overline{I_k}\) is a closed set containing \(\bigcup_{k=1}^n I_k\) and hence its closure, while the reverse inclusion is monotonicity of closure. (If some \(I_k\) is unbounded then \(m^*(I_k) = \infty\) by Proposition 1 and there is nothing to prove; so take \(I_k = (a_k, b_k)\) bounded.)

Since \(B\) is dense in the closed set \([0,1]\) and \(B \subseteq \bigcup_{k=1}^n I_k\),

\begin{equation*} \begin{aligned} [0,1] = \overline{B} \subseteq \overline{\bigcup_{k=1}^{n} I_k} = \bigcup_{k=1}^{n} \overline{I_k} . \end{aligned} \end{equation*}

Monotonicity of \(m^*\), then finite subadditivity (Proposition 3), then Proposition 1 twice, using \(\ell(\overline{I_k}) = b_k - a_k = \ell(I_k)\), give

\begin{equation*} \begin{aligned} 1 = m^*([0,1]) &\le \sum_{k=1}^{n} m^*\big(\overline{I_k}\big) \\ &= \sum_{k=1}^{n} \ell(I_k) = \sum_{k=1}^{n} m^*(I_k) . \end{aligned} \end{equation*}

Monotonicity of \(m^*\) on \(B \subseteq A \cup B\) and finite subadditivity (Proposition 3) sandwich the two sides:

\begin{equation*} m^*(B) \le m^*(A \cup B) \le m^*(A) + m^*(B) = 0 + m^*(B) = m^*(B) . \end{equation*}

Refine any cover of \(A \cup B\) into one by intervals of length \(< \alpha\), which then splits: no such interval meets both \(A\) and \(B\), since two of its points differ by less than \(\alpha\). The inequality \(\le\) is finite subadditivity (Proposition 3); boundedness gives \(m^*(A \cup B) < \infty\).

Subdivision. A nonempty bounded open \(I = (c,d)\) of length \(L\) and any \(\eta > 0\) admit finitely many open intervals of length \(< \alpha\) covering \(I\) with total length \(< \ell(I) + \eta\): take \(N\) with \(h = L/N < \alpha/2\), the points \(t_i = c + ih\), and

\begin{equation*} \begin{aligned} J_i = (t_{i-1} - \delta,\ t_i + \delta) , \qquad \delta = \min\{\alpha/8,\ \eta/(4N)\} , \end{aligned} \end{equation*}

so \(I \subseteq [c,d] \subseteq \bigcup_i J_i\), each \(\ell(J_i) = h + 2\delta < \alpha/2 + \alpha/4 < \alpha\), and \(\sum_i \ell(J_i) = L + 2N\delta \le \ell(I) + \eta/2\).

Now fix \(\epsilon > 0\) and take a cover \(\{I_k\}\) of \(A \cup B\) by nonempty open bounded intervals with \(\sum_k \ell(I_k) < m^*(A \cup B) + \epsilon\). Subdividing \(I_k\) with \(\eta = \epsilon/2^k\) yields a countable collection \(\mathcal{J}\) of open intervals, each of length \(< \alpha\), covering \(A \cup B\), with

\begin{equation*} \begin{aligned} \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{aligned} \end{equation*}

Put \(\mathcal{J}_A = \{J \in \mathcal{J} : J \cap A \ne \emptyset\}\) and \(\mathcal{J}_B = \mathcal{J} \setminus \mathcal{J}_A\). By the separation, no \(J \in \mathcal{J}_A\) meets \(B\), so \(\mathcal{J}_B\) contains every member meeting \(B\) and hence covers \(B\), while \(\mathcal{J}_A\) covers \(A\). The definition of outer measure and disjointness of the two subcollections give

\begin{equation*} \begin{aligned} 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{aligned} \end{equation*}

and \(\epsilon > 0\) was arbitrary.

Every interval is built from the rays \((a,\infty)\) by one complement, one countable intersection and one finite intersection, all of which \(\mathcal{A}\) admits (countable intersections via De Morgan’s Identities, finite ones by padding with \(\emptyset\) or \(\mathbb{R}\)). Fix \(a < b\).

(i) \((-\infty, a] = (a,\infty)^{C} \in \mathcal{A}\).

(ii) \([a, \infty) = \bigcap_{n=1}^{\infty} (a - 1/n,\ \infty) \in \mathcal{A}\), since \(x > a - 1/n\) for all \(n\) iff \(x \ge a\); hence also \((-\infty, a) = [a,\infty)^{C} \in \mathcal{A}\).

(iii) The four bounded types are finite intersections of members of \(\mathcal{A}\):

\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*}

(iv) The degenerate intervals: \(\emptyset, \mathbb{R} \in \mathcal{A}\) and \(\{a\} = [a,\infty) \cap (-\infty,a]\).

Every interval is of one of these forms, so \(\mathcal{A}\) contains all intervals.

Apply Problem 11 to the Borel \(\sigma\)-algebra \(\mathcal{B}\). It qualifies: \(\mathcal{B}\) is by definition (Section 2.3) the intersection of all \(\sigma\)-algebras of subsets of \(\mathbb{R}\) containing the open sets, hence is itself such a \(\sigma\)-algebra (an intersection of \(\sigma\)-algebras is one, and the family is nonempty since the power set of \(\mathbb{R}\) belongs to it), and each ray \((a,\infty)\) is open, so lies in \(\mathcal{B}\). Problem 11 then puts every interval in \(\mathcal{B}\).

Translation is a bijection of \(\mathbb{R}\), so \(x \in E + y\) iff \(x - y \in E\); consequently it commutes with arbitrary unions, intersections and complements, and carries open sets to open sets (translate the interval \((x - y - r, x - y + r) \subseteq \mathcal{O}\) by \(y\)) and hence closed sets to closed sets.

(i) \(F = \bigcup_{k} C_k\) with \(C_k\) closed gives \(F + y = \bigcup_k (C_k + y)\), a countable union of closed sets.

(ii) \(G = \bigcap_k \mathcal{O}_k\) with \(\mathcal{O}_k\) open gives \(G + y = \bigcap_k (\mathcal{O}_k + y)\), a countable intersection of open sets.

(iii) Outer measure is translation invariant (Proposition 2), so \(m^*(E + y) = m^*(E) = 0\); a set of outer measure zero is measurable (Proposition 4), so \(m(E + y) = 0\).

Take \(E_n = E \cap [-n,n]\) for a suitable \(n\). These are bounded subsets of \(E\) with \(E = \bigcup_{n=1}^{\infty} E_n\), by the Archimedean Property. Were every \(m^*(E_n) = 0\), countable subadditivity (Proposition 3) would give

\begin{equation*} m^*(E) \le \sum_{n=1}^{\infty} m^*(E_n) = 0 , \end{equation*}

contradicting \(m^*(E) > 0\). So \(m^*(E_n) > 0\) for some \(n\).

Slice \(E\) by the half-open intervals of length \(\epsilon\) and sweep the tail into one piece: \(E = B_1 \cup \cdots \cup B_N \cup C\) below.

For \(k \in \mathbb{Z}\) put \(A_k = E \cap (k\epsilon,\ (k+1)\epsilon]\), measurable as an intersection of measurable sets (every interval is measurable, Proposition 8). These intervals partition \(\mathbb{R}\), so the \(A_k\) are disjoint with union \(E\), and monotonicity with Proposition 1 gives \(m(A_k) \le \epsilon\). Re-index by a bijection \(\mathbb{N} \to \mathbb{Z}\) as \(\{B_k\}_{k=1}^{\infty}\). The series \(\sum_k m(B_k)\) converges, since for each \(n\) additivity over the finite disjoint collection \(\{B_k\}_{k=1}^n\) (Proposition 6 with \(A = \mathbb{R}\)) and monotonicity give

\begin{equation*} \sum_{k=1}^{n} m(B_k) = m\Big( \bigcup_{k=1}^{n} B_k \Big) \le m(E) < \infty . \end{equation*}

So choose \(N\) with \(\sum_{k > N} m(B_k) \le \epsilon\) and set \(C = \bigcup_{k > N} B_k\), measurable by Proposition 7, with

\begin{equation*} m( C) \le \sum_{k=N+1}^{\infty} m(B_k) \le \epsilon \end{equation*}

by countable subadditivity (Proposition 3). Then \(B_1, \dots, B_N, C\) are disjoint measurable sets of measure at most \(\epsilon\) whose union is \(E\).

Both equivalences are the already proved ones, (i) and (ii), read at \(E^{C}\). Three symmetries carry them across: \(E\) is measurable exactly when \(E^{C}\) is (the Carathéodory condition is symmetric in \(E\) and \(E^{C}\)); \(\mathcal{O}\) is open exactly when \(\mathcal{O}^{C}\) is closed; and \(G\) is \(G_\delta\) exactly when \(G^{C}\) is \(F_\sigma\) (De Morgan’s Identities). The single computation needed is that for \(F = \mathcal{O}^{C}\),

\begin{equation*} \mathcal{O} \sim E^{C} = \mathcal{O} \cap E = E \cap F^{C} = E \sim F , \end{equation*}

so an outer difference for \(E^{C}\) is literally an inner difference for \(E\).

(iii) Since \(F \mapsto F^{C}\) is a bijection from the closed sets contained in \(E\) onto the open sets containing \(E^{C}\), and the display converts \(m^*(\mathcal{O} \sim E^{C})\) into \(m^*(E \sim F)\) term by term, assertion (i) for \(E^{C}\) says word for word that (iii) holds for \(E\). Hence \(E\) measurable \(\iff\) \(E^{C}\) measurable \(\iff\) \(E^{C}\) satisfies (i) \(\iff\) \(E\) satisfies (iii).

(iv) Identically, with \(G \supseteq E^{C}\) a \(G_\delta\) set and \(F = G^{C} \subseteq E\) the corresponding \(F_\sigma\) set: \(G \sim E^{C} = E \sim F\), so (ii) for \(E^{C}\) says that (iv) holds for \(E\).

Method (2) for the sufficiency half of (iv): given an \(F_\sigma\) set \(F \subseteq E\) with \(m^*(E \sim F) = 0\), the set \(E \sim F\) is measurable by Proposition 4 and \(F\) is measurable by Theorem 9 (an \(F_\sigma\) set is Borel), so \(E = F \cup [E \sim F]\) is measurable by Proposition 5.

Both directions rest on the set identity \(\mathcal{O} \sim F = [\mathcal{O} \sim E] \cup [E \sim F]\), valid whenever \(F \subseteq E \subseteq \mathcal{O}\).

Necessity. Let \(E\) be measurable and \(\epsilon > 0\). Assertion (i) of Theorem 11 supplies an open \(\mathcal{O} \supseteq E\) with \(m^*(\mathcal{O} \sim E) < \epsilon/2\), and assertion (iii) (Problem 2.16) a closed \(F \subseteq E\) with \(m^*(E \sim F) < \epsilon/2\). A point of \(\mathcal{O}\) off \(F\) either misses \(E\) or lies in \(E \sim F\), which gives the identity, so by the subadditivity of outer measure (Proposition 3),

\begin{equation*} m^*(\mathcal{O} \sim F) \le m^*(\mathcal{O} \sim E) + m^*(E \sim F) < \epsilon . \end{equation*}

Sufficiency. Given such \(F \subseteq E \subseteq \mathcal{O}\) with \(m^*(\mathcal{O} \sim F) < \epsilon\), the inclusion \(F \subseteq E\) gives \(\mathcal{O} \sim E \subseteq \mathcal{O} \sim F\), so monotonicity yields

\begin{equation*} m^*(\mathcal{O} \sim E) \le m^*(\mathcal{O} \sim F) < \epsilon . \end{equation*}

As \(\epsilon > 0\) was arbitrary, \(E\) satisfies assertion (i) of Theorem 11 and hence is measurable.

The \(G_\delta\) half holds for every set of finite outer measure, but the \(F_\sigma\) half requires \(E\) measurable (for a choice set \(V \subseteq [0,1]\), Lemma 16 forces every closed subset of \(V\) to have measure zero, while \(m^*(V) > 0\)); we prove the statement as intended, for measurable \(E\) with \(m^*(E) < \infty\).

Outer half. For each natural number \(k\), finiteness of \(m^*(E)\) supplies a countable cover of \(E\) by open, bounded intervals \(\{I_j^{k}\}_{j}\) with \(\sum_j \ell(I_j^{k}) < m^*(E) + 1/k\); set \(\mathcal{O}_k = \bigcup_j I_j^{k}\) and \(G = \bigcap_k \mathcal{O}_k\), a \(G_\delta\) set containing \(E\). By monotonicity together with Propositions 1 and 3, \[ m^*(E) \le m^*(G) \le m^*(\mathcal{O}_k) \le \sum_j \ell(I_j^{k}) < m^*(E) + 1/k \] for every \(k\), and letting \(k \to \infty\) gives \(m^*(G) = m^*(E)\).

Inner half. Assertion (iv) of Theorem 11 (Problem 2.16) gives an \(F_\sigma\) set \(F \subseteq E\) with \(m^*(E \sim F) = 0\), whence by \(E = F \cup [E \sim F]\), subadditivity (Proposition 3) and monotonicity, \[ m^*(E) \le m^*(F) + m^*(E \sim F) = m^*(F) \le m^*(E) . \]

Take \(\mathcal{O}\) to be the union of a cover of \(E\) by open, bounded intervals whose lengths sum to less than \(m^*(E) + \epsilon_0\), where \(\epsilon_0\) is the gap supplied by nonmeasurability.

Since \(E\) is not measurable, assertion (i) of Theorem 11 fails for \(E\): there is an \(\epsilon_0 > 0\) such that every open \(\mathcal{U} \supseteq E\) satisfies \(m^*(\mathcal{U} \sim E) \ge \epsilon_0\). Finiteness of \(m^*(E)\) supplies a countable collection \(\{I_k\}\) of open, bounded intervals covering \(E\) with \(\sum_k \ell(I_k) < m^*(E) + \epsilon_0\); put \(\mathcal{O} = \bigcup_k I_k\), an open set containing \(E\). By Propositions 1 and 3, \[ m^*(\mathcal{O}) \le \sum_{k} \ell(I_k) < m^*(E) + \epsilon_0 < \infty , \] so \(\mathcal{O}\) has finite outer measure and, the subtraction being legitimate since \(m^*(E) < \infty\), \(m^*(\mathcal{O}) - m^*(E) < \epsilon_0\). Applying the gap to this \(\mathcal{O}\), \[ m^*(\mathcal{O} \sim E) \ge \epsilon_0 > m^*(\mathcal{O}) - m^*(E) . \]

Necessity is the Carathéodory condition at the test set \(A = (a,b)\), together with \(m^*((a,b)) = b - a\) (Proposition 1): \[ b - a = m^*\big( (a,b) \cap E \big) + m^*\big( (a,b) \sim E \big) . \]

Sufficiency. Assume \(\ell(I) = m^*(I \cap E) + m^*(I \sim E)\) for every nonempty open, bounded interval \(I\), and let \(\epsilon > 0\); we verify assertion (i) of Theorem 11. Finiteness of \(m^*(E)\) supplies a cover of \(E\) by open, bounded intervals of total length less than \(m^*(E) + \epsilon\), whose union \(\mathcal{O}\) is an open set containing \(E\) with \(m^*(\mathcal{O}) < m^*(E) + \epsilon < \infty\) (Propositions 1 and 3). Write \(\mathcal{O} = \bigcup_k I_k\) as a countable disjoint union of nonempty open intervals (page 17); each \(I_k\) is bounded, since monotonicity gives \(\ell(I_k) = m^*(I_k) \le m^*(\mathcal{O}) < \infty\), so the hypothesis applies to it. Two bounds:

(a) \(\sum_k \ell(I_k) \le m^*(\mathcal{O})\): the \(I_k\) are disjoint and measurable (Proposition 8), so Proposition 6 with \(A = \mathbb{R}\) and monotonicity give \(\sum_{k=1}^{n} \ell(I_k) = m^*(\bigcup_{k=1}^{n} I_k) \le m^*(\mathcal{O})\) for every \(n\).

(b) \(m^*(E) \le \sum_k m^*(I_k \cap E)\), from \(E = \bigcup_k [I_k \cap E]\) and countable subadditivity (Proposition 3).

Both series have finite sum, being dominated by \(m^*(\mathcal{O})\), so they may be subtracted term by term. Hence, applying subadditivity to \(\mathcal{O} \sim E = \bigcup_k [I_k \sim E]\) and then the hypothesis on each \(I_k\),

\begin{equation*} \begin{aligned} m^*(\mathcal{O} \sim E) &\le \sum_{k} m^*(I_k \sim E) \\ &= \sum_{k} \ell(I_k) - \sum_{k} m^*(I_k \cap E) \\ &\le m^*(\mathcal{O}) - m^*(E) < \epsilon . \end{aligned} \end{equation*}

So \(E\) satisfies assertion (i) of Theorem 11 and is therefore measurable.

In each case the witness for \(E_1 \cup E_2\) is the union of the witnesses for \(E_1\) and \(E_2\); only monotonicity and countable subadditivity of outer measure (Proposition 3) are used.

Property (ii) as the definition. Take \(G_\delta\) sets \(G_i \supseteq E_i\) with \(m^*(G_i \sim E_i) = 0\) for \(i = 1, 2\), and put \(G = G_1 \cup G_2 \supseteq E_1 \cup E_2\). Writing \(G_1 = \bigcap_n U_n\) and \(G_2 = \bigcap_n V_n\) with all \(U_n, V_n\) open, \[ G_1 \cup G_2 = \bigcap_{n,m} \big[ U_n \cup V_m \big] , \] a countable intersection of open sets and hence \(G_\delta\): the inclusion \(\subseteq\) is immediate, and if \(x\) lies in the right-hand side but not in \(G_1\), say \(x \notin U_{n_0}\), then \(x \in V_m\) for every \(m\), that is, \(x \in G_2\). Since a point of \(G\) outside \(E_1 \cup E_2\) lies in some \(G_i\) and so in \(G_i \sim E_i\), \[ G \sim [E_1 \cup E_2] \subseteq [G_1 \sim E_1] \cup [G_2 \sim E_2] , \] whence \(m^*(G \sim [E_1 \cup E_2]) \le 0 + 0 = 0\) and \(E_1 \cup E_2\) is measurable.

Property (iv) as the definition. Take \(F_\sigma\) sets \(F_i \subseteq E_i\) with \(m^*(E_i \sim F_i) = 0\), and put \(F = F_1 \cup F_2 \subseteq E_1 \cup E_2\); with \(F_1 = \bigcup_n C_n\) and \(F_2 = \bigcup_n D_n\) all \(C_n, D_n\) closed, \(F = \bigcup_n [C_n \cup D_n]\) is \(F_\sigma\), the union of two closed sets being closed. A point of \(E_1 \cup E_2\) outside \(F\) lies in some \(E_i\) and in neither \(F_1\) nor \(F_2\), so \[ [E_1 \cup E_2] \sim F \subseteq [E_1 \sim F_1] \cup [E_2 \sim F_2] , \] whence \(m^*([E_1 \cup E_2] \sim F) = 0\) and \(E_1 \cup E_2\) is measurable.

They are the same set function: \(m^{**}(A) = m^*(A)\) for every set \(A\) of real numbers, measurable or not.

Monotonicity gives \(m^*(A) \le m^*(\mathcal{O})\) for every open \(\mathcal{O} \supseteq A\), and taking the infimum over such \(\mathcal{O}\) yields \(m^*(A) \le m^{**}(A)\). For the reverse inequality we may assume \(m^*(A) < \infty\), since otherwise the inequality just proved forces \(m^{**}(A) = \infty = m^*(A)\). Let \(\epsilon > 0\). By the definition of \(m^*(A)\) as an infimum there is a countable collection \(\{I_k\}\) of open, bounded intervals covering \(A\) with \(\sum_k \ell(I_k) < m^*(A) + \epsilon\), and \(\mathcal{O} = \bigcup_k I_k\) is then an open set containing \(A\), one of the competitors in the infimum defining \(m^{**}(A)\). So by Propositions 1 and 3, \[ m^{**}(A) \le m^*(\mathcal{O}) \le \sum_{k} \ell(I_k) < m^*(A) + \epsilon , \] and as \(\epsilon > 0\) was arbitrary, \(m^{**}(A) \le m^*(A)\).

\(m^{***}\) is the inner measure determined by closed sets: \(m^{***} \le m^*\) always, with equality on every measurable set, and among sets of finite outer measure equality holds only for the measurable ones.

(1) \(m^{***}(A) \le m^*(A)\): each closed \(F \subseteq A\) has \(m^*(F) \le m^*(A)\) by monotonicity, and we take the supremum.

(2) Equality for measurable \(A\). Let \(\epsilon > 0\). Assertion (iii) of Theorem 11 gives a closed \(F \subseteq A\) with \(m^*(A \sim F) < \epsilon\), so from \(A = F \cup [A \sim F]\) and subadditivity, \[ m^*(A) \le m^*(F) + m^*(A \sim F) < m^{***}(A) + \epsilon . \] Letting \(\epsilon \downarrow 0\) and combining with (1) gives equality; the case \(m^*(A) = \infty\) is covered too, the display then forcing \(m^*(F) = \infty\).

(3) Conversely, let \(m^*(A) < \infty\) with \(m^{***}(A) = m^*(A)\). For each \(n\) the supremum supplies a closed \(F_n \subseteq A\) with \(m(F_n) > m^*(A) - 1/n\); closed sets are measurable and \(\mathcal{M}\) is a \(\sigma\)-algebra (Theorem 9), so \(F = \bigcup_n F_n \subseteq A\) is measurable and monotonicity applied to \(F_n \subseteq F \subseteq A\) gives \(m(F) = m^*(A)\). By Problem 2.22 there are open \(\mathcal{O}_n \supseteq A\) with \(m^*(\mathcal{O}_n) < m^*(A) + 1/n\), so \(G = \bigcap_n \mathcal{O}_n\) is a measurable set containing \(A\) with \(m(G) = m^*(A)\). Now \(F \subseteq A \subseteq G\) with \(m(F) = m(G) = m^*(A) < \infty\), so excision gives \(m(G \sim F) = 0\); hence \(m^*(A \sim F) = 0\) by monotonicity, \(A \sim F\) is measurable (Proposition 4), and \(A = F \cup [A \sim F]\) is measurable.

(4) The inequality in (1) can be strict. Let \(V \subseteq [0,1]\) be a choice set for the rational equivalence relation (Section 2.6) and \(\Lambda = [-1,1] \cap \mathbb{Q}\), a bounded, countably infinite set. Any closed \(F \subseteq V\) is bounded and measurable, and \(\{\lambda + F\}_{\lambda \in \Lambda}\) is disjoint by property (16), so \(m(F) = 0\) by Lemma 16. Hence \(m^{***}(V) = 0 < m^*(V)\), the strict inequality because \(m^*(V) = 0\) would make \(V\) measurable.

Both sides equal \(m(E_1) + m(E_1 \cap E_2) + m(E_2 \sim E_1)\); since no subtraction occurs, this is an identity in \([0, \infty]\) and no finiteness is needed.

All four sets are measurable, \(\mathcal{M}\) being a \(\sigma\)-algebra (Theorem 9). Splitting off \(E_2 \sim E_1\) gives the two disjoint decompositions

\begin{equation*} \begin{aligned} E_1 \cup E_2 &= E_1 \cup [E_2 \sim E_1], \\ E_2 &= [E_1 \cap E_2] \cup [E_2 \sim E_1] \end{aligned} \end{equation*}

(Check!), so finite additivity of Lebesgue measure over disjoint measurable sets (Proposition 6 with \(A = \mathbb{R}\)) yields

\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) + m(E_2). \end{aligned} \end{equation*}

Take \(B_k = [k, \infty)\).

Each \(B_k\) is an interval, hence measurable (Proposition 8), the collection is descending, and \(m(B_k) = \ell([k,\infty)) = \infty\) by Proposition 1; so \(m(B_1) = \infty\), the hypothesis in question fails, and \(\lim_{k \to \infty} m(B_k) = \infty\). On the other hand \(\bigcap_{k=1}^{\infty} [k, \infty) = \emptyset\) by the Archimedean property of \(\mathbb{R}\), so \[ m\Big( \bigcap_{k} B_k \Big) = 0 \ne \infty = \lim_{k \to \infty} m(B_k) , \] and the conclusion of Theorem 15 (ii) fails. Descent being the only other hypothesis, \(m(B_1) < \infty\) cannot be omitted.

The inequality \(\le\) is countable subadditivity of outer measure (Proposition 3) applied to \(A \cap \bigcup_{k} E_k = \bigcup_{k} [A \cap E_k]\).

For \(\ge\), fix \(n\). Then \(\{E_k\}_{k=1}^{n}\) is a finite disjoint collection of measurable sets, so Proposition 6 at the test set \(A\), followed by monotonicity, gives

\begin{equation*} \begin{aligned} \sum_{k=1}^{n} m^*(A \cap E_k) &= m^*\Big( A \cap \bigcup_{k=1}^{n} E_k \Big) \\ &\le m^*\Big( A \cap \bigcup_{k=1}^{\infty} E_k \Big) . \end{aligned} \end{equation*}

The right-hand side is independent of \(n\), and the sum of a series of nonnegative terms is the supremum in \([0, \infty]\) of its partial sums, so the full series is bounded by it as well, with no finiteness assumption anywhere.

Part (i). Since \(\mathcal{M}^{\prime}\) is a \(\sigma\)-algebra it contains \(\emptyset = \mathbb{R}^{C}\) and, by De Morgan’s Identities, is closed under finite unions, finite intersections and relative complements \(B \sim A = B \cap A^{C}\), so every set named below lies in \(\mathcal{M}^{\prime}\).

Finite additivity: pad a finite disjoint collection \(\{E_k\}_{k=1}^{n}\) with \(E_k = \emptyset\) for \(k > n\) and apply countable additivity, the added terms contributing \(m^{\prime}(\emptyset) = 0\).

Monotonicity and excision: for \(A \subseteq B\) the sets \(A\) and \(B \sim A\) are disjoint with union \(B\), so finite additivity gives \[ m^{\prime}(B) = m^{\prime}(A) + m^{\prime}(B \sim A) \ge m^{\prime}(A) , \] and when \(m^{\prime}(A) < \infty\) the identity rearranges to \(m^{\prime}(B \sim A) = m^{\prime}(B) - m^{\prime}(A)\).

Countable monotonicity: let \(E \subseteq \bigcup_{k} E_k\). Disjointify as in the discussion preceding Proposition 7, \(E_1^{\prime} = E_1\) and \(E_k^{\prime} = E_k \sim \bigcup_{i < k} E_i\), so that \(\{E_k^{\prime}\}\) is disjoint with \(E_k^{\prime} \subseteq E_k\) and the same union; then \(F_k = E \cap E_k^{\prime}\) are disjoint with \(\bigcup_k F_k = E\), and countable additivity followed by monotonicity gives \[ m^{\prime}(E) = \sum_{k} m^{\prime}(F_k) \le \sum_{k} m^{\prime}(E_k) . \]

Part (ii). Both continuity properties of Theorem 15 hold, by the proofs given there, which use only the countable additivity, monotonicity and excision just established.

(a) Ascending \(\{A_k\}\). If \(m^{\prime}(A_{k_0}) = \infty\) for some \(k_0\), monotonicity makes both sides \(\infty\). Otherwise set \(A_0 = \emptyset\) and \(C_k = A_k \sim A_{k-1}\); the \(C_k\) are disjoint with \(\bigcup_k C_k = \bigcup_k A_k\) (for \(x\) in the union take the least \(k\) with \(x \in A_k\); for \(j < k\), \(C_j \subseteq A_{k-1}\) while \(C_k\) misses \(A_{k-1}\)), so countable additivity, excision (legitimate as each \(m^{\prime}(A_{k-1}) < \infty\)) and telescoping give

\begin{equation*} \begin{aligned} m^{\prime}\Big( \bigcup_{k=1}^{\infty} A_k \Big) &= \sum_{k=1}^{\infty} \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*}

(b) Descending \(\{B_k\}\) with \(m^{\prime}(B_1) < \infty\). Then \(m^{\prime}(B_k)\) and \(m^{\prime}(\bigcap_k B_k)\) are all finite, so every excision below is legitimate. The sets \(D_k = B_1 \sim B_k\) ascend with \(\bigcup_k D_k = B_1 \sim \bigcap_k B_k\) (De Morgan’s Identities), so (a) and excision give

\begin{equation*} \begin{aligned} m^{\prime}(B_1) - m^{\prime}\Big( \bigcap_{k=1}^{\infty} B_k \Big) &= \lim_{k \to \infty} \big[ m^{\prime}(B_1) - m^{\prime}(B_k) \big] , \end{aligned} \end{equation*}

and the finite term \(m^{\prime}(B_1)\) cancels.

Continuity for ascending collections turns the finitely additive partial sums into the series; this is the converse of Problem 2.27.

Let \(m^{\prime}\) be a finitely additive, continuous \([0,\infty]\)-valued set function on a \(\sigma\)-algebra \(\mathcal{M}^{\prime}\) with \(m^{\prime}(\emptyset) = 0\), let \(\{E_k\}_{k=1}^{\infty}\) be a countable disjoint collection in \(\mathcal{M}^{\prime}\), and put \(E = \bigcup_k E_k \in \mathcal{M}^{\prime}\) and \(A_n = \bigcup_{k=1}^{n} E_k\). The collection \(\{A_n\}\) is ascending with \(\bigcup_n A_n = E\), and finite additivity gives \(m^{\prime}(A_n) = \sum_{k=1}^{n} m^{\prime}(E_k)\) for every \(n\). Hence, by continuity for ascending collections,

\begin{equation*} \begin{aligned} m^{\prime}(E) &= \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{aligned} \end{equation*}

the last equality being the definition of the sum of a series of nonnegative terms, whose partial sums are nondecreasing and so converge in \([0, \infty]\). Only the ascending form of continuity was used.

(i) Rational equivalence, \(x \sim y\) provided \(x - y \in \mathbb{Q}\), is an equivalence relation on any set \(E\) of real numbers because \(\mathbb{Q}\) is a subgroup of the additive group \(\mathbb{R}\): \(x - x = 0 \in \mathbb{Q}\) gives reflexivity; \(y - x = -(x-y)\) gives symmetry; and \(x - z = (x-y) + (y-z)\) gives transitivity.

(ii) \(\mathcal{C}_{\mathbb{Q}} = \{0\}\). Any two rationals differ by a rational, so \(\mathbb{Q}\) is a single rational equivalence class and a one-point subset is a choice set. Its two characteristic properties hold: no two distinct points to compare, and each \(x \in \mathbb{Q}\) is \(0 + x\) with \(x\) rational. (No appeal to the Axiom of Choice is needed.)

(iii) No in both cases, since \(x \approx y\) provided \(x - y\) irrational is never reflexive: \(x - x = 0\) is rational. On \(\mathbb{R}\) transitivity fails as well, \(0 \approx \sqrt{2}\) and \(\sqrt{2} \approx 2\) while \(0 - 2\) is rational; on \(\mathbb{Q}\) the relation is empty, the difference of two rationals being rational.

Were the choice set \(\mathcal{C}_E\) countable, \(E\) would be covered by countably many countable sets and so have outer measure zero.

Indeed, by the second characteristic property of a choice set every \(x \in E\) is \(c + q\) with \(c \in \mathcal{C}_E\) and \(q\) rational, so \[ E \subseteq \bigcup_{q \in \mathbb{Q}} (q + \mathcal{C}_E) , \] a countable union since \(\mathbb{Q}\) is countable. Each translate \(q + \mathcal{C}_E\) is countable and hence of outer measure zero (the Example in Section 2.2), so monotonicity and countable subadditivity (Proposition 3) give \(m^*(E) = 0\), contradicting \(m^*(E) > 0\). Thus \(\mathcal{C}_E\) is uncountable, and in particular infinite.

A set \(E\) with \(m^*(E) > 0\) has \(m^*(E \cap [-n,n]) > 0\) for some \(n\), and a nonmeasurable subset of that bounded piece is a nonmeasurable subset of \(E\); so the bounded case of Theorem 17 gives the general case.

Indeed, \(E = \bigcup_{n=1}^{\infty} E_n\) with \(E_n = E \cap [-n,n]\) bounded, so if every \(m^*(E_n)\) were zero, countable subadditivity of outer measure (Proposition 3) would force \(m^*(E) = 0\). Fix \(n\) with \(m^*(E_n) > 0\); by the bounded case \(E_n\) contains a nonmeasurable set \(W\), and \(W \subseteq E_n \subseteq E\).

No for finite \(\Lambda\), yes for uncountably infinite bounded \(\Lambda\), no for unbounded \(\Lambda\).

(i) Finite \(\Lambda\): take \(E = [0,1)\) and \(\Lambda = \{0,1\}\), bounded and measurable with disjoint translates \([0,1)\) and \([1,2)\), yet \(m(E) = 1\). With \(\Lambda\) finite the right-hand side of (15) is \((\#\Lambda) \cdot m(E) < \infty\) whatever \(m(E)\) may be, so no contradiction arises.

(ii) Uncountably infinite (bounded) \(\Lambda\): the lemma persists. Being infinite, \(\Lambda\) has a countably infinite subset \(\Lambda_0\) (choose distinct points recursively), which is bounded because \(\Lambda\) is, and \(\{\lambda + E\}_{\lambda \in \Lambda_0}\) is disjoint as a subcollection of a disjoint collection; Lemma 16 applied to \(E\) and \(\Lambda_0\) gives \(m(E) = 0\).

(iii) Unbounded \(\Lambda\): take \(E = [0,1)\) and \(\Lambda = \mathbb{Z}\), whose translates \([n, n+1)\) are disjoint, yet \(m(E) = 1\). Boundedness of \(\Lambda\) and \(E\) is what makes \(\bigcup_{\lambda \in \Lambda} (\lambda + E)\) bounded and so of finite measure in (15); here that union is \(\mathbb{R}\) and (15) reads \(\infty = \sum_{n \in \mathbb{Z}} 1\).

Take \(G = \bigcap_{n=1}^{\infty} \mathcal{O}_n\), where \(\mathcal{O}_n\) is an open set containing \(E\) with \(m(\mathcal{O}_n) < \alpha + 1/n\) and \(\alpha = m^*(E) < \infty\).

Such \(\mathcal{O}_n\) exist by Problem 2.22: the union of a cover of \(E\) by open, bounded intervals of total length less than \(\alpha + 1/n\) will do. Then \(G\) is a \(G_\delta\) set containing \(E\), measurable by Theorem 9, and monotonicity gives \[ \alpha = m^*(E) \le m^*(G) \le m(\mathcal{O}_n) < \alpha + \tfrac{1}{n} \] for every \(n\), so \(m^*(G) = m^*(E)\).

Were \(m^*(G \sim E) = 0\), the set \(G \sim E\) would be measurable (Proposition 4) and hence so would \[ E = G \sim [G \sim E] , \] an identity valid because \(E \subseteq G\), the measurable sets forming a \(\sigma\)-algebra. This contradicts the nonmeasurability of \(E\), so \(m^*(G \sim E) > 0\).

Take \(f = \psi^{-1}\), where \(\psi(x) = [\varphi(x) + x]/2\) and \(\varphi\) is the Cantor-Lebesgue function: \(f\) carries \(A = \psi(\mathbf{C})\), of measure \(1/2\), onto the Cantor set \(\mathbf{C}\), of measure zero.

\(\psi\) is continuous, and strictly increasing because \(\varphi\) is increasing (Proposition 20) while the identity is strictly increasing; since \(\psi(0) = 0\) and \(\psi(1) = 1\), the Intermediate Value Theorem makes \(\psi\) a one-to-one map of \([0,1]\) onto \([0,1]\).

Let \(\mathcal{O} = [0,1] \sim \mathbf{C} = \bigcup_k I_k\), the union of the disjoint open intervals deleted in the Cantor construction; \(m(\mathbf{C}) = 0\) by Proposition 19, so \(\sum_k \ell(I_k) = m(\mathcal{O}) = 1\). On \(I_k\) the function \(\varphi\) is constant, say \(\varphi \equiv c_k\), so \(\psi(x) = (c_k + x)/2\) there and \(\psi(I_k)\) is an open interval of length \(\ell(I_k)/2\); these images are disjoint, \(\psi\) being one-to-one, so \(\psi(\mathcal{O})\) is open and countable additivity gives \[ m(\psi(\mathcal{O})) = \tfrac{1}{2} \sum_k \ell(I_k) = \tfrac{1}{2} . \] The set \(\psi(\mathbf{C})\) is compact, hence measurable, and \(\psi\) being a bijection of \([0,1]\) carries the disjoint decomposition \([0,1] = \mathbf{C} \cup \mathcal{O}\) to the disjoint decomposition \([0,1] = \psi(\mathbf{C}) \cup \psi(\mathcal{O})\), so \[ m(A) = m(\psi(\mathbf{C})) = 1 - \tfrac{1}{2} = \tfrac{1}{2} > 0 . \] Finally \(f = \psi^{-1}\) is strictly increasing (Check!) and continuous by Problem 2.45, and \(f(A) = \mathbf{C}\) has measure zero.

Necessity. Since \(I\) is open there is an \(r > 0\) with \((x_0 - r, x_0 + r) \subseteq I\); take \(a_n = x_0 - r/(n+1)\) and \(b_n = x_0 + r/(n+1)\), which lie in \(I\) with \(a_n < x_0 < b_n\) and both tend to \(x_0\). Continuity at \(x_0\), in its sequential form, gives \[ \lim_{n \to \infty} [f(b_n) - f(a_n)] = f(x_0) - f(x_0) = 0 . \]

Sufficiency. Let \(\varepsilon > 0\), choose \(N\) with \(f(b_N) - f(a_N) < \varepsilon\), and put \(\delta = \min\{x_0 - a_N,\, b_N - x_0\} > 0\). If \(|x - x_0| < \delta\) then \(a_N < x < b_N\), so \(x \in I\) (as \(a_N, b_N \in I\)), and since \(f\) is increasing both \(f(x)\) and \(f(x_0)\) lie in \([f(a_N), f(b_N)]\), whence \[ |f(x) - f(x_0)| \le f(b_N) - f(a_N) < \varepsilon . \] As \(\varepsilon > 0\) was arbitrary, \(f\) is continuous at \(x_0\).

Monotonicity traps \(f(x)\) between values of \(\varphi\) at points of \(\mathcal{O} = [0,1] \sim \mathbf{C}\) arbitrarily near \(x\); at the two endpoints the printed statement needs the normalization that \(f\), like \(\varphi\), take its values in \([0,1]\), since \(g(0) = -1\) with \(g = \varphi\) on \((0,1]\) is increasing and agrees with \(\varphi\) on \(\mathcal{O}\) (note \(0 \in \mathbf{C}\)).

\(\mathcal{O}\) is dense in \([0,1]\): a nondegenerate interval \(J \subseteq \mathbf{C}\) would give \(0 < \ell(J) = m(J) \le m(\mathbf{C}) = 0\) by Proposition 1, monotonicity and Proposition 19.

Let \(x \in \mathbf{C} \cap (0,1)\). For all large \(n\), density supplies \(u_n \in \mathcal{O} \cap (x - 1/n, x)\) and \(v_n \in \mathcal{O} \cap (x, x + 1/n)\). Since \(f\) is increasing and equals \(\varphi\) on \(\mathcal{O}\), and \(\varphi\) is increasing (Proposition 20), both \(f(x)\) and \(\varphi(x)\) lie in \([\varphi(u_n), \varphi(v_n)]\), so \[ |f(x) - \varphi(x)| \le \varphi(v_n) - \varphi(u_n) \longrightarrow 0 \] by the continuity of \(\varphi\) (Proposition 20). Hence \(f = \varphi\) on \((0,1)\). At the endpoints \(\varphi(0) = 0\) and \(\varphi(1) = 1\), \(\varphi\) being increasing onto \([0,1]\), while \(f(0) \le f(v_n) = \varphi(v_n) \to 0\) and symmetrically \(f(1) \ge 1\); with the normalization these inequalities are equalities, so \(f = \varphi\) on \([0,1]\).

No. Take \(f = \psi^{-1}\) on \(E = [0,2]\), where \(\psi(x) = \varphi(x) + x\) is the strictly increasing continuous map of \([0,1]\) onto \([0,2]\) of Proposition 21, and let \(A \subseteq \mathbf{C}\) be the measurable set with \(\psi(A)\) nonmeasurable furnished by assertion (ii) of that proposition.

The inverse \(f\) is defined on all of \([0,2]\), \(\psi\) being one-to-one and onto, and is continuous by Problem 2.45. Since \(\psi\) is a bijection of \([0,1]\) onto \([0,2]\), \[ f^{-1}(A) = \{ y \in [0,2] : \psi^{-1}(y) \in A \} = \psi(A) , \] which is nonmeasurable, whereas \(A\) is measurable: \(A \subseteq \mathbf{C}\) and \(m(\mathbf{C}) = 0\) (Proposition 19) give \(m^*(A) = 0\) by monotonicity, so Proposition 4 applies.

Everything follows from the estimate \(m^{*}(f(J)) \le c\,\ell\) for a set \(J \subseteq [a,b]\) contained in an interval of length \(\ell\): all \(u, v \in J\) have \(|f(u) - f(v)| \le c\,|u-v| \le c\,\ell\), so \(f(J)\) lies in an interval of length \(c\,\ell\) and monotonicity with Proposition 1 applies. The Lipschitz condition also makes \(f\) continuous.

Measure zero. Let \(m^{*}(E) = 0\) and \(\epsilon > 0\). Take open intervals \(\{I_k\}\) covering \(E\) with \(\sum_k \ell(I_k) < \epsilon\) and put \(J_k = I_k \cap [a,b]\), so that \(f(E) \subseteq \bigcup_k f(J_k)\) and each \(J_k\) lies in an interval of length \(\ell(I_k)\). By countable subadditivity (Proposition 3) and the estimate, \[ m^{*}(f(E)) \le \sum_{k} m^{*}(f(J_k)) \le c \sum_{k} \ell(I_k) \le c\,\epsilon , \] so \(m^{*}(f(E)) = 0\) and \(f(E)\) is measurable of measure zero (Proposition 4).

\(F_\sigma\) sets. Let \(F = \bigcup_k F_k \subseteq [a,b]\) with each \(F_k\) closed; replacing \(F_k\) by \(F_k \cap [a,b]\), which changes neither closedness nor the union, we may take each \(F_k\) closed and bounded. Then each \(f(F_k)\) is compact, hence closed, \(f\) being continuous, and \(f(F) = \bigcup_k f(F_k)\) is an \(F_\sigma\) set.

Measurable sets. For measurable \(E \subseteq [a,b]\), assertion (iv) of Theorem 11 gives an \(F_\sigma\) set \(F \subseteq E\) with \(m^{*}(E \sim F) = 0\), and then \[ f(E) = f(F) \cup f(E \sim F) \] is measurable, since \(f(F)\) is \(F_\sigma\), hence Borel, hence measurable (Theorem 9), while \(f(E \sim F)\) has measure zero by the first part and \(\mathcal{M}\) is a \(\sigma\)-algebra.

Write \(F = \bigcap_{n=0}^{\infty} F_n\), where \(F_0 = [0,1]\) and \(F_n\) is the disjoint union of the \(2^n\) closed intervals left after removing from each of the \(2^{n-1}\) intervals of \(F_{n-1}\) its open middle interval of length \(\alpha 3^{-n}\); the \(2^n\) intervals share the common length \[ \ell_n = \frac{\ell_{n-1} - \alpha 3^{-n}}{2} = (1-\alpha)2^{-n} + \alpha 3^{-n} , \] the closed formula following by induction from \(\ell_0 = 1\) (Check!). In particular \(\ell_n > 0\), so the construction never runs out of room, and \(\ell_n \to 0\).

\(F\) is closed, being an intersection of the sets \(F_n\), each a finite union of closed intervals.

Measure. Finite additivity of Lebesgue measure and Proposition 1 give \[ m(F_n) = 2^n \ell_n = (1-\alpha) + \alpha (2/3)^{n} , \] and \(\{F_n\}\) is a descending collection of measurable sets with \(m(F_0) = 1 < \infty\), so continuity of measure (Theorem 15) gives \[ m(F) = \lim_{n \to \infty} m(F_n) = 1 - \alpha . \]

Density of \([0,1] \sim F\). It suffices that \(F\) contain no nondegenerate interval \(J\). Choose \(n\) with \(\ell_n < \ell(J)\); then \(J \subseteq F \subseteq F_n\), and since the components of \(F_n\) are separated by nonempty deleted open gaps, an interval inside \(F_n\) meeting two components would contain a point of a gap between them. So \(J\) lies in a single component and \(\ell(J) \le \ell_n < \ell(J)\), a contradiction.

Take \(\mathcal{O} = (0,1) \sim F\), where \(F \subseteq [0,1]\) is the generalized Cantor set of Problem 2.39 for a fixed \(\alpha \in (0,1)\); then \(m(\partial\mathcal{O}) \ge 1 - \alpha > 0\).

\(\mathcal{O} = (0,1) \cap (\mathbb{R} \sim F)\) is open, \(F\) being closed. Every \(x \in F \cap (0,1)\) lies in \(\partial\mathcal{O} = \overline{\mathcal{O}} \sim \mathcal{O}\), the interior of the open set \(\mathcal{O}\) being \(\mathcal{O}\) itself: it fails to lie in \(\mathcal{O}\) because \(x \in F\), while for \(\epsilon > 0\) small enough that \((x - \epsilon, x + \epsilon) \subseteq (0,1)\), this nondegenerate subinterval meets the dense set \([0,1] \sim F\) (Problem 2.39) in a point \(y\), and \(y \in (0,1)\) puts \(y \in \mathcal{O}\). Hence \(F \cap (0,1) \subseteq \partial\mathcal{O}\).

Since \(0\) and \(1\) are never deleted in the construction, \(F \cap (0,1) = F \sim \{0,1\}\), and a two-point set has measure zero (Proposition 4), so excision gives \(m(F \cap (0,1)) = m(F) = 1 - \alpha\). The boundary \(\partial\mathcal{O}\) is closed, hence measurable (Theorem 9), so monotonicity gives \[ m(\partial\mathcal{O}) \ge m\big( F \cap (0,1) \big) = 1 - \alpha > 0 . \]

The Cantor set \(\mathbf{C} = \bigcap_{k \ge 1} C_k\) is closed and nonempty by Proposition 19, so only the density clause is at issue.

Every endpoint persists: if \(e\) is an endpoint of a component \(J\) of \(C_k\), then \(C_{k+1}\) retains the two closed outer thirds of \(J\), of which \(e\) is again an endpoint, so by induction \(e \in C_n\) for all \(n \ge k\), while \(e \in C_n\) for \(n < k\) because \(\{C_n\}\) descends. Hence every endpoint of every constituent interval lies in \(\mathbf{C}\).

Let \(x \in \mathbf{C}\) and let \(\mathcal{U}\) be a neighborhood of \(x\); pick \(\epsilon > 0\) with \((x - \epsilon, x + \epsilon) \subseteq \mathcal{U}\) and \(k\) with \(3^{-k} < \epsilon\). The component \(J\) of \(C_k\) containing \(x\) has \(\ell(J) = 3^{-k}\), so

\begin{equation*} |y - x| \le \ell(J) = 3^{-k} < \epsilon \qquad (y \in J), \end{equation*}

whence \(J \subseteq \mathcal{U}\). For each \(j \ge 1\) exactly \(2^{j}\) components of \(C_{k+j}\) lie inside \(J\) (each component inside \(J\) splits into two; induct on \(j\)); they are pairwise disjoint and nondegenerate, so their \(2^{j+1}\) endpoints are distinct points of \(J \cap \mathbf{C}\). Thus \(J \cap \mathbf{C} \subseteq \mathcal{U} \cap \mathbf{C}\) is infinite, and \(\mathbf{C}\) is perfect.

Suppose \(X\) is perfect and countable, and enumerate \(X = \{x_n\}_{n=1}^{\infty}\); the list is infinite because the neighborhood \((x-1,x+1)\) of any \(x \in X\) already contains infinitely many points of \(X\). Write \(B(y,r) = (y-r, y+r)\) and \(\bar B(y,r) = [y-r,y+r]\). We construct \(y_n \in X\) and \(r_n > 0\) so that \(F_n = X \cap \bar B(y_n, r_n)\) satisfies, for every \(n\):

(i) \(y_n \in F_n\), so \(F_n \ne \emptyset\); (ii) \(x_n \notin \bar B(y_n, r_n)\), so \(x_n \notin F_n\); (iii) \(\bar B(y_{n+1}, r_{n+1}) \subseteq B(y_n, r_n)\), so \(F_{n+1} \subseteq F_n\).

Take \(y_1 \in X\) with \(y_1 \ne x_1\) and \(r_1 = \tfrac12|y_1 - x_1|\). Given \(y_n, r_n\), the neighborhood \(B(y_n, r_n)\) of \(y_n \in X\) contains infinitely many points of \(X\), at most one equal to \(x_{n+1}\); choose \(y_{n+1} \in X \cap B(y_n, r_n)\) with \(y_{n+1} \ne x_{n+1}\) and then

\begin{equation*} 0 < 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 entries being positive. The second bound gives (ii) at \(n+1\), and the first gives (iii), since \(|z - y_{n+1}| \le r_{n+1}\) forces

\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*}

Each \(F_n\) is nonempty by (i), closed as the intersection of the closed sets \(X\) and \(\bar B(y_n,r_n)\), and bounded, and \(\{F_n\}\) descends by (iii); so the Nested Set Theorem supplies \(z \in \bigcap_{n \ge 1} F_n\). Then \(z \in F_1 \subseteq X\) gives \(z = x_N\) for some \(N\), while \(z \in F_N\) contradicts (ii). Hence \(X\) is uncountable.

The Cantor set \(\mathbf{C}\) is perfect by Problem 41, and every perfect subset of \(\mathbb{R}\) is uncountable by Problem 42; hence \(\mathbf{C}\) is uncountable.

For a nonempty open set \(\mathcal{O}\), the required subset is \(\mathcal{U} = \mathcal{O} \cap (\mathbb{R} \sim \mathbf{C})\). (Since the empty set is always available, the definition is read as demanding a nonempty open subset, and that is what we produce.)

Indeed \(\mathcal{U}\) is open, \(\mathbf{C}\) being closed by Proposition 19; it lies in \(\mathcal{O}\) and misses \(\mathbf{C}\) by construction; and it is nonempty. For choose \(x \in \mathcal{O}\) and \(r > 0\) with \((x-r, x+r) \subseteq \mathcal{O}\). Were \((x - r, x + r) \subseteq \mathbf{C}\), monotonicity of measure and \(m(\mathbf{C}) = 0\) (Proposition 19) would give

\begin{equation*} 2r = m\big((x-r, x+r)\big) \le m(\mathbf{C}) = 0 , \end{equation*}

which is false; so some \(z \in (x-r, x+r) \subseteq \mathcal{O}\) lies outside \(\mathbf{C}\), and \(z \in \mathcal{U}\). Thus \(\mathbf{C}\) is nowhere dense in \(\mathbb{R}\).

Let \(f\) be strictly increasing on the interval \(I\), put \(E = f(I)\), and let \(g = f^{-1} \colon E \to I\), well defined since strict monotonicity makes \(f\) one-to-one; continuity of \(g\) is relative to \(E\), which need not be an interval. Note that \(g\) too is strictly increasing: if \(y_1 < y_2\) in \(E\) had \(g(y_1) \ge g(y_2)\), applying the increasing \(f\) would give \(y_1 \ge y_2\).

Fix \(y_0 \in E\), put \(x_0 = g(y_0)\), and let \(\varepsilon > 0\). Choose \(\delta = \min\{\delta_1, \delta_2\} > 0\) as follows.

(i) If some \(a \in I\) has \(a < x_0\), then \([a, x_0] \subseteq I\); pick \(u \in I\) with \(\max\{a,\, x_0 - \varepsilon\} < u < x_0\) and set \(\delta_1 = y_0 - f(u) > 0\). Otherwise set \(\delta_1 = 1\).

(ii) If some \(b \in I\) has \(b > x_0\), pick \(v \in I\) with \(x_0 < v < \min\{b,\, x_0 + \varepsilon\}\) and set \(\delta_2 = f(v) - y_0 > 0\). Otherwise set \(\delta_2 = 1\).

Let \(y \in E\) with \(|y - y_0| < \delta\). In case (i), \(y > y_0 - \delta_1 = f(u)\), so monotonicity of \(g\) gives

\begin{equation*} g(y) > g(f(u)) = u > x_0 - \varepsilon ; \end{equation*}

in the alternative, \(x_0 = \min I\) and \(g(y) \in I\) forces \(g(y) \ge x_0 > x_0 - \varepsilon\). Symmetrically in case (ii), \(y < y_0 + \delta_2 = f(v)\) gives \(g(y) < v < x_0 + \varepsilon\), while in its alternative \(x_0 = \max I\) gives \(g(y) \le x_0 < x_0 + \varepsilon\). Hence \(|g(y) - g(y_0)| < \varepsilon\), and \(f^{-1}\) is continuous on \(E\).

Following the hint, put

\begin{equation*} \mathcal{A} = \{\, E \subseteq \mathbb{R} \ : \ f^{-1}(E) \in \mathcal{B} \,\}, \end{equation*}

where \(\mathcal{B}\) is the Borel \(\sigma\)-algebra and \(f\) is continuous on a Borel domain \(D\) (some such assumption is forced, since \(f^{-1}(\mathbb{R}) = D\); \(D = \mathbb{R}\) is the intended case and \(D\) an interval the case needed in Problem 47).

\(\mathcal{A}\) is a \(\sigma\)-algebra: \(f^{-1}(\mathbb{R}) = D \in \mathcal{B}\), and for \(E \in \mathcal{A}\) and \(\{E_k\}_{k=1}^{\infty} \subseteq \mathcal{A}\),

\begin{equation*} \begin{aligned} f^{-1}(\mathbb{R} \sim E) &= D \sim f^{-1}(E), \\ f^{-1}\Big(\bigcup\nolimits_{k} E_k\Big) &= \bigcup\nolimits_{k} f^{-1}(E_k), \end{aligned} \end{equation*}

both of which lie in \(\mathcal{B}\), a \(\sigma\)-algebra containing \(D\) and each \(f^{-1}(E_k)\).

\(\mathcal{A}\) contains every open set \(\mathcal{O}\): by continuity of \(f\) on \(D\) and Proposition 22 of Chapter 1 there is an open \(U\) with \(f^{-1}(\mathcal{O}) = D \cap U \in \mathcal{B}\).

Since \(\mathcal{B}\) is the smallest \(\sigma\)-algebra containing the open sets, \(\mathcal{B} \subseteq \mathcal{A}\); that is, \(f^{-1}(B)\) is Borel for every Borel set \(B\).

For every Borel set \(B\), \(f(B \cap I) = g^{-1}(B)\) where \(g = f^{-1}\), and this set is Borel.

The image \(E = f(I)\) is an interval: if \(y_1 = f(x_1) < y < y_2 = f(x_2)\) with \(y_1, y_2 \in E\), then \(x_1 < x_2\) since \(f\) is strictly increasing, so \([x_1, x_2] \subseteq I\) and, \(f\) being continuous there, the Intermediate Value Theorem gives \(x \in (x_1, x_2) \subseteq I\) with \(f(x) = y \in E\). With \(\alpha = \inf E\) and \(\beta = \sup E\) this yields \((\alpha, \beta) \subseteq E \subseteq [\alpha, \beta]\), and every interval is a Borel set.

By Problem 45 the inverse \(g = f^{-1} \colon E \to I\) is continuous, and its domain \(E\) is Borel, so Problem 46 makes \(g^{-1}(B)\) Borel for every Borel set \(B\).

Finally \(g^{-1}(B) = f(B \cap I)\): if \(y \in E\) with \(x = g(y) \in B\), then \(x \in B \cap I\) and \(f(x) = y\); conversely \(y = f(x)\) with \(x \in B \cap I\) gives \(y \in E\) and \(g(y) = x \in B\). Hence \(f\) carries Borel sets to Borel sets.

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.

Yes on \([a,b]\) (with \(a < b\)); no on a general measurable set.

Put \(h = f - g\), continuous on \([a,b]\), and let \(E_0 = \{x \in [a,b] \mid h(x) \ne 0\}\); the hypothesis \(f = g\) a.e. places \(E_0\) inside a set of measure zero, so \(m^*(E_0) = 0\) by monotonicity of outer measure. Suppose \(h(x_0) \ne 0\) for some \(x_0 \in [a,b]\) and set \(\varepsilon = |h(x_0)|/2 > 0\). Continuity of \(h\) at \(x_0\) yields \(\delta > 0\) with \(|h(x) - h(x_0)| < \varepsilon\) for \(x \in [a,b]\), \(|x - x_0| < \delta\), and then

\begin{equation*} |h(x)| \geq |h(x_0)| - |h(x) - h(x_0)| > 2\varepsilon - \varepsilon = \varepsilon > 0 . \end{equation*}

Thus \(I = (x_0 - \delta, x_0 + \delta) \cap [a,b] \subseteq E_0\), and \(I\) is an interval of positive length whether \(x_0\) is interior to \([a,b]\) or an endpoint of it, since \(a < b\) (Check!). As the outer measure of an interval is its length,

\begin{equation*} 0 < m(I) \le m^*(E_0) = 0 , \end{equation*}

a contradiction. Hence \(h \equiv 0\), that is, \(f = g\) on all of \([a,b]\).

For a general measurable set the assertion fails: take

\begin{equation*} E = [0,1] \cup \{2\}, \qquad f \equiv 0 \text{ on } E, \qquad g = \chi_{\{2\}} . \end{equation*}

Both are continuous on \(E\), since \(2\) is an isolated point of \(E\) and \(g\) vanishes on the relative neighborhood \(E \cap (-1,2) = [0,1]\) of each other point; and \(\{x \in E \mid f(x) \ne g(x)\} = \{2\}\) has measure zero, yet \(f(2) \ne g(2)\).

No: only the forward implication survives.

That one is immediate, continuity being relative to the domain: for \(x_0 \in D\) and \(\varepsilon > 0\), a \(\delta\) that works for \(f\) on \(D \cup E\) works for \(f|_D\), since \(D \subseteq D \cup E\); likewise for \(f|_E\).

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, while \(f\) is discontinuous at \(0\): every \(\delta\)-neighborhood of \(0\) contains points \(x > 0\) with \(|f(x) - f(0)| = 1\). So the analogue of Proposition 5(ii) with “continuous” in place of “measurable” is false.

Yes. Let \(E\) be the measurable domain, let \(F = \{x_1, \ldots, x_n\}\) be the finite set of points at which \(f\) fails to be continuous, and let \(D = E \sim F\); then \(F\) is measurable with \(m(F) = 0\), being a finite union of singletons, and hence \(D\) is measurable.

(i) \(f|_D\) is continuous on \(D\): for \(x_0 \in D\) and \(\varepsilon > 0\), the \(\delta\) furnished by continuity of \(f\) at \(x_0\) as a function on \(E\) serves for \(f|_D\), since \(D \subseteq E\). That continuity also forces \(f(x_0)\) to be finite, so \(f|_D\) is a real-valued continuous function on a measurable set and Proposition 3 makes it measurable.

(ii) \(f|_F\) is measurable: for each \(c\) the set \(\{x \in F \mid f(x) > c\}\) is a subset of \(F\), hence of outer measure zero, hence measurable (Chapter 2), and Proposition 1 applies. (This is valid even where \(f\) takes the values \(\pm\infty\).)

Since \(E = D \cup F\) with \(D, F\) measurable and both restrictions measurable, Proposition 5(ii) gives that \(f\) is measurable on \(E\).

No. Fix a nonmeasurable \(A \subseteq (0,1)\) (Theorem 17 of Chapter 2, applicable since \(m^*((0,1)) > 0\)) and define

\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 is measurable: \(f(x) = c\) gives \(x = c\) when \(x \in A\) and \(x = -c\) otherwise, so

\begin{equation*} f^{-1}( c) \subseteq \{c, -c\} \end{equation*}

is finite, hence measurable.

But \(f\) is not measurable, since \(f(x) = x > 0\) for \(x \in A\) while \(f(x) = -x < 0\) for \(x \in (0,1) \sim A\), whence

\begin{equation*} \{x \in \mathbb{R} \mid f(x) > 0\} \cap (0,1) = A ; \end{equation*}

were \(f\) measurable, Proposition 1 and closure of the measurable sets under intersection would make \(A\) measurable.

Yes. Given a real number \(c\), density of \(\mathbb{Q}\) supplies rationals \(c_k \in (c, c + 1/k)\), and

\begin{equation*} \{x \in E \mid f(x) > c\} = \bigcup_{k=1}^{\infty} \{x \in E \mid f(x) > c_k\} . \end{equation*}

Here \(\supseteq\) holds because \(c_k > c\); for \(\subseteq\), a point with \(f(x) = \infty\) satisfies \(f(x) > c_1\), while for finite \(f(x) > c\) the Archimedean property gives \(k\) with \(1/k < f(x) - c\), so that \(c_k < c + 1/k < f(x)\).

Each set on the right is measurable by hypothesis, so the countable union is measurable, the measurable sets being a \(\sigma\)-algebra. As \(E\) is measurable, Proposition 1(i) holds for every real \(c\) and \(f\) is measurable.

Comparing the sets \(\{g > c\}\) with the sets \(\{f > c\}\) settles both directions; both domains, \(\mathbb{R}\) and \(D\), are measurable, so Proposition 1 is all that is needed.

Let \(f\) be measurable and \(c\) real.

(i) \(c \ge 0\): no point of \(\mathbb{R} \sim D\) qualifies, since \(g = 0 \le c\) there, so

\begin{equation*} \{x \in \mathbb{R} \mid g(x) > c\} = \{x \in D \mid f(x) > c\} . \end{equation*}

(ii) \(c < 0\): every point of \(\mathbb{R} \sim D\) qualifies, so

\begin{equation*} \{x \in \mathbb{R} \mid g(x) > c\} = \{x \in D \mid f(x) > c\} \; \cup \; (\mathbb{R} \sim D). \end{equation*}

In both cases the right side is measurable, \(\mathbb{R} \sim D\) being measurable because \(D\) is; hence \(g\) is measurable.

Conversely, if \(g\) is measurable then for each real \(c\)

\begin{equation*} \{x \in D \mid f(x) > c\} = \{x \in \mathbb{R} \mid g(x) > c\} \cap D \end{equation*}

is an intersection of measurable sets, so \(f\) is measurable. (Equally, this direction is Proposition 5(ii) applied to \(g\) on \(\mathbb{R} = D \cup (\mathbb{R} \sim D)\), whose restriction to \(D\) is \(f\).)

Here \(f\) is real-valued and \(f^{-1}(A) = \{x \in E \mid f(x) \in A\}\); for extended real-valued \(f\) the direction from inverse images to measurability requires \(A\) to range over the Borel subsets of the extended reals, the printed hypothesis being vacuous for an \(f\) that is everywhere infinite.

If \(f^{-1}(A)\) is measurable for every Borel \(A\), then for each real \(c\) the open set \((c, \infty)\) gives

\begin{equation*} \{x \in E \mid f(x) > c\} = f^{-1}\big((c, \infty)\big) , \end{equation*}

measurable; since \(E\) is measurable, Proposition 1 makes \(f\) measurable.

Conversely, let \(f\) be measurable and, following the hint, put \(\mathcal{A} = \{A \subseteq \mathbb{R} \mid f^{-1}(A) \text{ is measurable}\}\). Then \(\mathcal{A}\) is a \(\sigma\)-algebra: \(f^{-1}(\mathbb{R}) = E\) is measurable, and for \(A, \{A_k\}_{k=1}^{\infty}\) in \(\mathcal{A}\),

\begin{equation*} \begin{aligned} f^{-1}(\mathbb{R} \sim A) &= E \sim f^{-1}(A), \\ f^{-1}\Big(\bigcup\nolimits_{k} A_k\Big) &= \bigcup\nolimits_{k} f^{-1}(A_k), \end{aligned} \end{equation*}

both measurable because the measurable sets form a \(\sigma\)-algebra. Moreover \(\mathcal{A}\) contains every open set, this being Proposition 2 for the measurable function \(f\). Since the Borel \(\sigma\)-algebra \(\mathcal{B}\) is the smallest \(\sigma\)-algebra containing the open sets, \(\mathcal{B} \subseteq \mathcal{A}\); that is, \(f^{-1}(A)\) is measurable for every Borel \(A\).

The proofs of Section 3.1 use only two properties of the Lebesgue measurable sets, and the Borel sets \(\mathcal{B}\) have both: \(\mathcal{B}\) is a \(\sigma\)-algebra and \(\mathcal{B}\) contains every interval. So both results transfer verbatim. Below \(\{f > c\}\) abbreviates \(\{x \in E \mid f(x) > c\}\), with \(E\) the (Borel) domain.

Proposition 1. Statements (i) and (iv) are complementary in \(E\), as are (ii) and (iii), and relative complements in the Borel set \(E\) of Borel sets are Borel; and

\begin{equation*} \begin{aligned} \{f \ge c\} &= \bigcap\nolimits_{k=1}^{\infty} \{f > c - 1/k\}, \\ \{f > c\} &= \bigcup\nolimits_{k=1}^{\infty} \{f \ge c + 1/k\}, \end{aligned} \end{equation*}

so all four are equivalent. Then \(f^{-1}( c) = \{f \ge c\} \cap \{f \le c\}\) for finite \(c\) and \(f^{-1}(\pm\infty) = \bigcap_{k} \{\pm f > k\}\) are Borel.

Theorem 6. For Borel measurable real-valued \(f, g\) on \(E\) and \(\alpha \ne 0\),

\begin{equation*} \begin{aligned} \{\alpha f > c\} &= \{f > c/\alpha\} \ (\alpha > 0), \qquad \{f < c/\alpha\} \ (\alpha < 0), \\ \{f + g < c\} &= \bigcup\nolimits_{q \in \mathbb{Q}} \big(\{g < c - q\} \cap \{f < q\}\big), \\ \{h^2 > c\} &= \{h > \sqrt{c}\} \cup \{h < -\sqrt{c}\} \ (c \ge 0), \qquad E \ (c < 0), \end{aligned} \end{equation*}

the middle identity by density of \(\mathbb{Q}\); with \(\alpha f \equiv 0\) when \(\alpha = 0\), these give \(\alpha f + \beta g\) Borel measurable, and then so is \(fg = \frac{1}{2}[(f+g)^2 - f^2 - g^2]\). One caveat: Proposition 5 is false for Borel measurability, since by Proposition 22 of Chapter 2 some measure-zero subset \(A\) of the Cantor set is not Borel and \(\chi_A\) is then not Borel measurable; so “finite a.e.” in Theorem 6 must be read as “finite,” or the conclusion asserted on the Borel set where \(f\) and \(g\) are both finite.

(i) By Theorem 9 of Chapter 2 the Lebesgue measurable sets form a \(\sigma\)-algebra containing the open sets, so \(\mathcal{B}\), the smallest such, is contained in it. Hence a Borel measurable \(f\) has Lebesgue measurable domain and level sets, so is Lebesgue measurable.

(ii) This is Problem 7 with \(\mathcal{B}\) in place of the Lebesgue measurable sets: \(\mathcal{A} = \{A \subseteq \mathbb{R} \mid f^{-1}(A) \in \mathcal{B}\}\) is a \(\sigma\)-algebra by the same three identities \(f^{-1}(\mathbb{R}) = E\), \(f^{-1}(\mathbb{R} \sim A) = E \sim f^{-1}(A)\), \(f^{-1}(\bigcup_k A_k) = \bigcup_k f^{-1}(A_k)\), and it contains every ray \((c, \infty)\). The rays generate \(\mathcal{B}\): a \(\sigma\)-algebra containing them contains \((-\infty, c] = \mathbb{R} \sim (c,\infty)\), hence \((-\infty, c) = \bigcup_{k} (-\infty, c - 1/k]\), hence every open interval \((a,b) = (a,\infty) \cap (-\infty,b)\) bounded or not, hence every open set as a countable union of such. So \(\mathcal{B} \subseteq \mathcal{A}\). (If \(f\) takes infinite values, apply this to \(f\) restricted to \(E^{\prime} = E \sim f^{-1}(\{\pm\infty\})\), Borel by the above, noting \(f^{-1}(B) = (f|_{E^{\prime}})^{-1}(B)\) for \(B \subseteq \mathbb{R}\).)

(iii) The natural domain \(g^{-1}(E)\) of \(f \circ g\) is Borel by (ii) applied to \(g\), 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 \(g\)-inverse image of a Borel set, hence Borel by (ii) again.

(iv) The same identity, with Problem 7 supplying the transfer: \(\{A \subseteq \mathbb{R} \mid g^{-1}(A) \text{ Lebesgue measurable}\}\) is a \(\sigma\)-algebra containing every ray, hence contains \(\mathcal{B}\) by the generation argument of (ii). So \(g^{-1}(B)\) is Lebesgue measurable for each Borel \(B\); in particular the domain \(g^{-1}(E)\) and each level set above are measurable, and \(f \circ g\) is Lebesgue measurable.

Yes: reading “converges” as “converges to a finite limit,” \(E_0\) is the Cauchy set

\begin{equation*} F = \bigcap_{k=1}^{\infty} \ \bigcup_{N=1}^{\infty} \ \bigcap_{n \ge N} \ \bigcap_{m \ge N} A_{n,m}^{k}, \qquad A_{n,m}^{k} = \{x \in E_{n,m} \mid |f_n(x) - f_m(x)| < 1/k\}, \end{equation*}

where \(E_{n,m}\) is the set of points of \(E\) at which \(f_n\) and \(f_m\) are both finite.

Each \(E_{n,m}\) is measurable, \(f_n^{-1}(\pm\infty)\) being measurable by Proposition 1; on it \(f_n\) and \(f_m\) are real-valued and measurable by Proposition 5(ii), so \(f_n - f_m\) is measurable there by Theorem 6 and

\begin{equation*} A_{n,m}^{k} = \{f_n - f_m < 1/k\} \cap \{f_n - f_m > -1/k\} \end{equation*}

is measurable. Only countable set operations build \(F\) from these, so \(F\) is measurable.

\(E_0 \subseteq F\): if \(f_n(x) \to L \in \mathbb{R}\), then for fixed \(k\) some \(N\) has \(|f_n(x) - L| < 1/(2k)\) for \(n \ge N\), so \(f_n(x)\) is finite and \(x \in E_{n,m}\) for \(n, m \ge N\), and the triangle inequality gives \(|f_n(x) - f_m(x)| < 1/k\) there.

\(F \subseteq E_0\): taking \(k = 1\) yields \(N_1\) with \(f_n(x)\) finite for \(n \ge N_1\), and given \(\epsilon > 0\), choosing \(k\) with \(1/k < \epsilon\) yields \(N \ge N_1\) with \(|f_n(x) - f_m(x)| < \epsilon\) for \(n, m \ge N\); so the tail is Cauchy in \(\mathbb{R}\), hence convergent, and \(x \in E_0\).

Under the wider reading one adjoins the measurable sets \(\{x \in E \mid f_n(x) \to \pm\infty\} = \bigcap_{k} \bigcup_{N} \bigcap_{n \ge N} \{x \in E \mid \pm f_n(x) > k\}\); the answer is affirmative either way.

No: take \(f = \chi_A\) and \(g = \psi^{-1}\), as in the Example preceding Proposition 7. (Proposition 7 requires the outer function to be continuous; with the roles reversed the conclusion fails.)

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 continuous and strictly increasing from \([0,1]\) onto \([0,2]\) and carries some measurable subset \(A\) of the Cantor set onto a nonmeasurable set \(\psi(A)\). Extend \(\psi\) by

\begin{equation*} \psi(x) = x \ \text{ for } x < 0, \qquad \psi(x) = x + 1 \ \text{ for } x > 1 ; \end{equation*}

since \(\psi(0) = 0\) and \(\psi(1) = 2\), the extension is continuous and strictly increasing from \(\mathbb{R}\) onto \(\mathbb{R}\), so by Problem 45 of Chapter 2 its inverse \(g = \psi^{-1}\) is a continuous real-valued function on \(\mathbb{R}\).

With \(A\) measurable, \(f = \chi_A\) is measurable: \(\{f > c\}\) is \(\emptyset\), \(A\), or \(\mathbb{R}\) according as \(c \ge 1\), \(0 \le c < 1\), or \(c < 0\). Yet, \(\psi\) being a bijection of \(\mathbb{R}\),

\begin{equation*} \{y \in \mathbb{R} \mid (f \circ g)(y) > 1/2\} = \{y \in \mathbb{R} \mid \psi^{-1}(y) \in A\} = \psi(A) \end{equation*}

is nonmeasurable, so \(f \circ g\) is not measurable.

Write \(h = g^{-1}\), Lipschitz by hypothesis. Since \(g\) is a bijection of \(\mathbb{R}\) with inverse \(h\), inverse images under \(g\) are forward images under \(h\):

\begin{equation*} g^{-1}(A) = \{x \in \mathbb{R} \mid g(x) \in A\} = h(A) \qquad (A \subseteq \mathbb{R}), \end{equation*}

since \(g(x) = a \in A\) holds exactly when \(x = h(a)\). And \(h\) maps measurable sets to measurable sets: for measurable \(A\), each \(A \cap [-n,n]\) is a measurable subset of a closed bounded interval on which \(h\) is Lipschitz, so \(h(A \cap [-n,n])\) is measurable by Problem 38 of Chapter 2, and

\begin{equation*} h(A) = \bigcup_{n=1}^{\infty} h(A \cap [-n,n]) \end{equation*}

is measurable as a countable union.

Hence the domain of \(f \circ g\) is \(g^{-1}(E) = h(E)\), measurable; and for each real \(c\), with \(E_c = \{y \in E \mid f(y) > c\}\) measurable by the measurability of \(f\),

\begin{equation*} \{x \in h(E) \mid (f \circ g)(x) > c\} = g^{-1}(E_c) = h(E_c) \end{equation*}

is measurable. Therefore \(f \circ g\) is measurable.

Take \(\varphi_n = \max\{\alpha_1, \ldots, \alpha_n\}\) and \(\psi_n = \min\{\beta_1, \ldots, \beta_n\}\), where the Simple Approximation Lemma, applicable because \(f\) is bounded, measurable and real-valued on \(E\), supplies for each \(n\) simple functions 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*}

Both are simple: measurable by Proposition 8, real-valued, and taking values in the union of the ranges of \(\alpha_1, \ldots, \alpha_n\), resp. \(\beta_1, \ldots, \beta_n\), a finite set. Monotonicity is immediate from \(\varphi_n = \max\{\varphi_{n-1}, \alpha_n\}\) and \(\psi_n = \min\{\psi_{n-1}, \beta_n\}\). Since every \(\alpha_k \le f \le \beta_k\) on \(E\),

\begin{equation*} \alpha_n \le \varphi_n \le f \le \psi_n \le \beta_n \quad \text{on } E, \end{equation*}

so at every \(x \in E\),

\begin{equation*} \begin{aligned} 0 \le f(x) - \varphi_n(x) &\le \beta_n(x) - \alpha_n(x) < 1/n, \\ 0 \le \psi_n(x) - f(x) &\le \beta_n(x) - \alpha_n(x) < 1/n . \end{aligned} \end{equation*}

These bounds are independent of \(x\), so the increasing \(\{\varphi_n\}\) and the decreasing \(\{\psi_n\}\) both converge to \(f\) uniformly on \(E\).

Round \(f\) down to the nearest multiple of \(1/n\): take \(f_n = \frac{1}{n}\lfloor n f\rfloor\), that is, \(f_n = k/n\) on

\begin{equation*} E_{n,k} = \{x \in E \mid k/n \le f(x) < (k+1)/n\}, \qquad k \in \mathbb{Z} . \end{equation*}

(Here \(f\) is real-valued, since semisimple functions are and uniform limits of them are too; no boundedness is needed, countably many values being allowed.)

Each \(E_{n,k}\) is measurable by Proposition 1, and for fixed \(n\) these sets are disjoint with union \(E\), the unique \(k\) containing a given \(x\) being \(k = \lfloor n f(x)\rfloor\); so \(f_n\) is well defined, real-valued, and countably valued. It is measurable, since for each real \(c\)

\begin{equation*} \{x \in E \mid f_n(x) > c\} = \bigcup_{\{k \in \mathbb{Z} \,\mid\, k/n > c\}} E_{n,k} \end{equation*}

is a countable union of measurable sets; hence each \(f_n\) is semisimple. Finally, for \(x \in E_{n,k}\),

\begin{equation*} 0 \le f(x) - f_n(x) = f(x) - k/n < 1/n , \end{equation*}

a bound independent of \(x\), so \(\{f_n\} \to f\) uniformly on \(E\).

Take \(F = E_N\), where \(E_n = \{x \in E \mid |f(x)| \le n\}\) and \(N\) is chosen below.

Each \(E_n = \{f \le n\} \cap \{f \ge -n\}\) is a measurable subset of \(E\) by Proposition 1, the collection ascends, and

\begin{equation*} \bigcup_{n=1}^{\infty} E_n = \{x \in E \mid |f(x)| < \infty\} =: E_{\mathrm{fin}} . \end{equation*}

Since \(f\) is finite a.e., the measurable set \(E \sim E_{\mathrm{fin}} = f^{-1}(\infty) \cup f^{-1}(-\infty)\) has measure zero, so additivity over \(E = E_{\mathrm{fin}} \cup (E \sim E_{\mathrm{fin}})\) gives \(m(E_{\mathrm{fin}}) = m(E)\), and continuity of measure (Theorem 15(i) of Chapter 2, for the ascending \(\{E_n\}\)) gives

\begin{equation*} \lim_{n \to \infty} m(E_n) = m(E_{\mathrm{fin}}) = m(E) < \infty . \end{equation*}

Choose \(N\) with \(m(E_N) > m(E) - \epsilon\). Then \(|f| \le N\) on \(F = E_N\), and since \(F \subseteq E\) with \(m(F) \le m(E) < \infty\), excision gives

\begin{equation*} m(E \sim F) = m(E) - m(E_N) < \epsilon . \end{equation*}

By Problem 14 there is a measurable \(F \subseteq E\) with \(m(E \sim F) < \epsilon\) on which \(f\) is bounded, say \(|f| \le N\); in particular \(f\) is real-valued there, and \(f|_F\) is measurable by Proposition 5(ii).

For each \(n\), the Simple Approximation Lemma applied to the bounded measurable function \(f|_F\) with the number \(1/n\) provides simple functions on \(F\) with \(\phi_n \le f \le \psi_n\) and \(\psi_n - \phi_n < 1/n\), so that

\begin{equation*} 0 \le f - \phi_n \le \psi_n - \phi_n < 1/n \quad \text{on } F. \end{equation*}

Extend to \(E\) 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, takes only the finitely many values of \(\phi_n\) together with \(0\), and is measurable on \(E\) by Proposition 5(ii), its restrictions to \(F\) and to \(E \sim F\) being \(\phi_n\) and \(0\); so \(\varphi_n\) is simple on \(E\). Since \(|f - \varphi_n| < 1/n\) at every point of \(F\), \(\{\varphi_n\} \to f\) uniformly on \(F\), while \(m(E \sim F) < \epsilon\).

Take \(h = \chi_{\mathcal{O}}\) restricted to \(I = [a,b]\) and \(F = I \sim (E \,\Delta\, \mathcal{O})\), where Theorem 12 of Chapter 2, applicable since \(E\) is measurable with \(m^{*}(E) \le b - a < \infty\), supplies a finite disjoint collection of open intervals \(I_k = (a_k, b_k)\), \(1 \le k \le n\), whose union \(\mathcal{O}\) satisfies

\begin{equation*} m^{*}(E \,\Delta\, \mathcal{O}) < \epsilon . \end{equation*}

\(h\) is a step function on \(I\): let \(a = t_0 < \cdots < t_m = b\) be the partition whose points are \(a\), \(b\), and those \(a_k, b_k\) lying in \((a,b)\). No \(a_k\) or \(b_k\) lies in \((t_{j-1}, t_j)\), so \((t_{j-1}, t_j) \cap (a_k, b_k)\) is either empty or all of \((t_{j-1}, t_j)\) - otherwise some point of \(\{a_k, b_k\}\) would lie between two points of \((t_{j-1}, t_j)\), hence in it. So on each \((t_{j-1}, t_j)\) either \(h \equiv 1\) or \(h \equiv 0\).

\(F\) is a measurable subset of \(I\), \(E\) being measurable and \(\mathcal{O}\) open; and \(x \in F\) lies in \(E\) exactly when it lies in \(\mathcal{O}\), so \(\chi_E(x) = \chi_{\mathcal{O}}(x) = h(x)\). Finally

\begin{equation*} I \sim F = I \cap (E \,\Delta\, \mathcal{O}) \subseteq E \,\Delta\, \mathcal{O} , \end{equation*}

so monotonicity of measure gives \(m(I \sim F) < \epsilon\).

Take \(h = \sum_{k=1}^{n} c_k h_k\) and \(F = \bigcap_{k=1}^{n} F_k\), built as follows. Let \(c_1, \ldots, c_n\) be the distinct values of \(\psi\) and \(E_k = \{x \in I \mid \psi(x) = c_k\}\), measurable by Proposition 1, so that

\begin{equation*} \psi = \sum_{k=1}^{n} c_k \cdot \chi_{E_k} \quad \text{on } I ; \end{equation*}

each \(E_k\) is a measurable subset of the closed bounded interval \(I\), so Problem 16 with the number \(\epsilon/n\) yields a step function \(h_k\) on \(I\) and a measurable \(F_k \subseteq I\) with

\begin{equation*} h_k = \chi_{E_k} \text{ on } F_k \quad \text{and} \quad m(I \sim F_k) < \epsilon/n . \end{equation*}

Then \(h\) is a step function: on each open subinterval determined by the common refinement of partitions witnessing \(h_1, \ldots, h_n\), every \(h_k\) is constant, hence so is \(h\). By De Morgan and finite subadditivity,

\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) < \epsilon . \end{equation*}

And for \(x \in F\) we have \(h_k(x) = \chi_{E_k}(x)\) for every \(k\), so \(h(x) = \psi(x)\).

Take the step function \(h\) and set \(F\) that Problem 17 provides for a simple function \(\psi\) lying within \(\epsilon/2\) of \(f\).

Such a \(\psi\) comes from the Simple Approximation Lemma, applicable since \(f\) is bounded, measurable and real-valued on \(I\): with the number \(\epsilon/2\) it gives simple functions \(\phi \le f \le \eta\) on \(I\) with \(\eta - \phi < \epsilon/2\), and \(\psi = \phi\) then satisfies

\begin{equation*} 0 \le f - \psi \le \eta - \phi < \epsilon/2 \quad \text{on all of } I . \end{equation*}

Problem 17, applied to \(\psi\) on \(I\) with the number \(\epsilon\), yields a step function \(h\) and a measurable \(F \subseteq I\) with \(h = \psi\) on \(F\) and \(m(I \sim F) < \epsilon\). Hence for \(x \in F\),

\begin{equation*} |h(x) - f(x)| = |\psi(x) - f(x)| < \epsilon/2 < \epsilon . \end{equation*}

Let \(\varphi, \psi\) be simple on a common measurable domain \(E\), with distinct values \(a_1, \ldots, a_n\) and \(b_1, \ldots, b_m\), and put \(A_i = \{\varphi = a_i\}\), \(B_j = \{\psi = b_j\}\), measurable by Proposition 1. The sets \(C_{ij} = A_i \cap B_j\) are then measurable, pairwise disjoint, and cover \(E\).

Let \(g\) be any one of \(\varphi + \psi\), \(\varphi\psi\), \(\max\{\varphi,\psi\}\), \(\min\{\varphi,\psi\}\), and \(\gamma_{ij}\) correspondingly \(a_i + b_j\), \(a_i b_j\), \(\max\{a_i,b_j\}\), \(\min\{a_i,b_j\}\). Since \(\varphi \equiv a_i\) and \(\psi \equiv b_j\) on \(C_{ij}\), in each case \(g \equiv \gamma_{ij}\) there, so \(g(E) \subseteq \{\gamma_{ij}\}\), a set of at most \(nm\) real numbers (simple functions are real-valued, so no \(\infty - \infty\) or \(0 \cdot \infty\) arises). For each \(c \in \mathbb{R}\),

\begin{equation*} \{x \in E \mid g(x) > c\} = \bigcup_{\gamma_{ij} > c} C_{ij}, \end{equation*}

a finite union of measurable sets, so \(g\) is measurable by Proposition 1. Each of the four functions is thus real-valued, measurable, and finitely valued, that is, simple. \(\blacksquare\)

Fix an ambient set \(X \supseteq A \cup B\), with \(A^{c} = X \sim A\), and evaluate both sides at \(x \in X\), writing \(u = \chi_A(x)\) and \(v = \chi_B(x)\), each in \(\{0,1\}\).

(i) \(\chi_{A \cap B}(x) = 1\) iff \(x \in A\) and \(x \in B\) iff \(u = v = 1\) iff \(uv = 1\), and both sides are \(0\) otherwise; hence \(\chi_{A \cap B} = \chi_A \cdot \chi_B\).

(ii) \(\chi_{A^{c}}(x) = 1\) iff \(x \notin A\) iff \(u = 0\) iff \(1 - u = 1\), and both sides are \(0\) otherwise; hence \(\chi_{A^{c}} = 1 - \chi_A\).

(iii) By (ii), De Morgan’s law, and (i),

\begin{equation*} \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 \cdot \chi_B . \end{equation*}

\(\blacksquare\)

Fix \(c \in \mathbb{R}\) and put \(g = \sup\{f_n\}\), \(h = \inf\{f_n\}\); all sets below are taken inside \(E\). Then

\begin{equation*} \{g > c\} = \bigcup_{n=1}^{\infty} \{f_n > c\}, \qquad \{h < c\} = \bigcup_{n=1}^{\infty} \{f_n < c\}, \end{equation*}

since \(g(x) > c\) says \(c\) is not an upper bound of \(\{f_n(x)\}\), hence \(f_n(x) > c\) for some \(n\), while conversely \(f_n(x) > c\) gives \(g(x) \ge f_n(x) > c\); and dually for \(h\). Each right-hand side is a countable union of measurable sets, hence measurable, so criteria (i) and (iii) of Proposition 1 make \(g\) and \(h\) measurable.

The argument used only the countability of the index set, so for every \(k\) the functions

\begin{equation*} g_k = \inf_{n \ge k} f_n \qquad \text{and} \qquad G_k = \sup_{n \ge k} f_n \end{equation*}

are measurable on \(E\). Applying the supremum case to \(\{g_k\}\) and the infimum case to \(\{G_k\}\), the functions \(\liminf\{f_n\} = \sup_k g_k\) and \(\limsup\{f_n\} = \inf_k G_k\) are measurable. \(\blacksquare\)

Let \(\epsilon > 0\) and, following the hint, set \(E_n = \{x \in [a,b] \mid f(x) - f_n(x) < \epsilon\}\). Letting \(m \to \infty\) in \(f_n \le f_m\) gives \(f_n \le f\) on \([a,b]\) for every \(n\), so \(|f - f_n| = f - f_n\) and it suffices to force \(f - f_n < \epsilon\) uniformly.

Each \(E_n\) is relatively open, being \((f - f_n)^{-1}\big((-\infty,\epsilon)\big)\) for the continuous function \(f - f_n\); write \(E_n = \mathcal{O}_n \cap [a,b]\) with \(\mathcal{O}_n \subseteq \mathbb{R}\) open. From \(f_n \le f_{n+1}\) we get \(f - f_{n+1} \le f - f_n\), hence \(E_n \subseteq E_{n+1}\); and pointwise convergence puts each \(x \in [a,b]\) in some \(E_n \subseteq \mathcal{O}_n\), so \(\{\mathcal{O}_n\}\) is an open cover of \([a,b]\).

Since \([a,b]\) is closed and bounded, the Heine-Borel Theorem yields a finite subcover \(\mathcal{O}_{n_1}, \ldots, \mathcal{O}_{n_k}\). With \(N = \max_i n_i\) the ascending property gives \(\mathcal{O}_{n_i} \cap [a,b] = E_{n_i} \subseteq E_N\), so

\begin{equation*} [a,b] = \bigcup_{i=1}^{k}(\mathcal{O}_{n_i} \cap [a,b]) \subseteq E_N \subseteq [a,b], \end{equation*}

that is, \(E_N = [a,b]\). For \(n \ge N\) we have \(E_N \subseteq E_n\), hence \(E_n = [a,b]\), which says \(|f - f_n| < \epsilon\) at every point of \([a,b]\). As \(N\) does not depend on \(x\), the convergence is uniform. \(\blacksquare\)

Write \(f = f^{+} - f^{-}\) with \(f^{+} = \max\{f, 0\}\) and \(f^{-} = \max\{-f, 0\}\). Both are nonnegative and at each point at most one is nonzero, so no \(\infty - \infty\) arises and \(|f| = f^{+} + f^{-}\). They are measurable whenever \(f\) is, since for \(c \ge 0\)

\begin{equation*} \{f^{+} > c\} = \{f > c\}, \qquad \{f^{-} > c\} = \{f < -c\}, \end{equation*}

while for \(c < 0\) each of these sets is all of \(E\).

Now the general Simple Approximation Theorem. If simple \(\varphi_n \to f\) pointwise on \(E\), then \(f\) is measurable by Proposition 9. Conversely let \(f\) be measurable. The nonnegative case applied to \(f^{+}\) and \(f^{-}\) gives increasing sequences of simple functions \(\widetilde\varphi_n \to f^{+}\) and \(\widetilde\psi_n \to f^{-}\) pointwise with \(|\widetilde\varphi_n| \le f^{+}\) and \(|\widetilde\psi_n| \le f^{-}\). Replace \(\widetilde\varphi_n\) by \(\varphi_n = \max\{\widetilde\varphi_n, 0\}\): this is simple by Problem 19, still increasing and still convergent to \(\max\{f^{+},0\} = f^{+}\) because \(t \mapsto \max\{t,0\}\) is increasing and continuous (and \(\widetilde\varphi_n(x) \to \infty\) forces \(\varphi_n(x) \to \infty\)), and now \(0 \le \varphi_n \le f^{+}\). Do the same for \(\widetilde\psi_n\), giving \(0 \le \psi_n \le f^{-}\).

Put \(h_n = \varphi_n - \psi_n\), simple by Problem 19 (and real-valued, so the difference is defined). At \(x\) with \(f(x) \ge 0\) we have \(f^{-}(x) = 0\), hence \(\psi_n(x) = 0\) and \(|h_n(x)| = \varphi_n(x) \le f^{+}(x) = |f(x)|\); at \(x\) with \(f(x) < 0\), symmetrically \(\varphi_n(x) = 0\) and \(|h_n(x)| = \psi_n(x) \le |f(x)|\). The same two computations give

\begin{equation*} h_n(x) \longrightarrow f^{+}(x) - f^{-}(x) = f(x) \end{equation*}

at every \(x \in E\), the cases \(f(x) = \pm\infty\) included, since there one of \(\varphi_n(x), \psi_n(x)\) vanishes identically while the other tends to \(\infty\). Finally, if \(f \ge 0\) then \(f^{-} \equiv 0\), so \(\psi_n \equiv 0\) and \(h_n = \varphi_n\) is increasing. \(\blacksquare\)

Every interval is measurable, so \(I\) is, and it suffices to show \(\{x \in I \mid f(x) > c\}\) is measurable for each \(c \in \mathbb{R}\).

Fix a natural number \(n\) and set \(g_n(x) = f(x) + x/n\); this is strictly increasing on \(I\), since \(x < y\) gives \(f(x) \le f(y)\) and \(x/n < y/n\). Any increasing real-valued \(g\) on the interval \(I\) is measurable: the set \(S = \{x \in I \mid g(x) > c\}\) is upward closed in \(I\) (if \(x \in S\), \(y \in I\), \(y > x\), then \(g(y) \ge g(x) > c\)), hence convex, because for \(x, y \in S\) and \(x < z < y\) we get \(z \in I\) as \(I\) is an interval and \(z \in S\) as \(z > x\); so \(S\) is an interval, and therefore measurable. In particular each \(g_n\) is measurable on \(I\).

For each fixed \(x \in I\) we have \(x/n \to 0\), so

\begin{equation*} g_n(x) = f(x) + \frac{x}{n} \longrightarrow f(x), \end{equation*}

and \(f\), the pointwise limit on the measurable set \(I\) of the measurable \(g_n\), is measurable by Proposition 9. \(\blacksquare\)

If \(F = \emptyset\) take \(g \equiv 0\), and if \(F = \mathbb{R}\) take \(g = f\). Otherwise put \(\mathcal{O} = \mathbb{R} \sim F\) and write \(\mathcal{O} = \bigcup_k I_k\) as a countable disjoint union of open intervals (Proposition 9 of Chapter 1), whose proof also shows each finite endpoint of each \(I_k\) lies in \(F\); since \(F \ne \emptyset\), no \(I_k\) is all of \(\mathbb{R}\). 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), \\[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*}

This is well defined and real-valued, the \(I_k\) being disjoint from each other and from \(F\), and \(g = f\) on \(F\). On a bounded \(I_k\), with \(t = (x - a_k)/(b_k - a_k) \in (0,1)\),

\begin{equation*} g(x) = (1 - t) f(a_k) + t f(b_k). \tag{C} \end{equation*}

Continuity at \(x_0 \in \mathcal{O}\): the component \(I_k\) containing \(x_0\) is an open neighbourhood on which \(g\) is affine or constant.

Continuity at \(x_0 \in F\): given \(\epsilon > 0\), continuity of \(f\) on \(F\) supplies \(\delta > 0\) with \(|f(y) - f(x_0)| < \epsilon/2\) for all \(y \in F\) with \(|y - x_0| < \delta\), which we call (D). We produce \(\delta_{+} > 0\) working on \([x_0, x_0 + \delta_{+})\); the left-hand estimate is the mirror image and gives \(\delta_{-}\), and \(\delta_0 = \min\{\delta_{+}, \delta_{-}\}\) then works.

(i) \(x_0\) is the left endpoint of a component \(I_k\). If \(I_k = (x_0, \infty)\), then \(g \equiv f(x_0) = g(x_0)\) on \([x_0,\infty)\) and any \(\delta_{+}\) serves. If \(I_k = (x_0, b_k)\), then \(g\) is the continuous affine function on \([x_0, b_k]\) with \(g(x_0) = f(x_0)\), so \(g(x) \to g(x_0)\) as \(x \to x_0^{+}\); choose \(\delta_{+} \le b_k - x_0\) accordingly.

(ii) \(x_0\) is the left endpoint of no component. Then \(F \cap (x_0, x_0 + \eta) \ne \emptyset\) for every \(\eta > 0\): otherwise \((x_0, x_0 + \eta) \subseteq \mathcal{O}\), and the component \(I_k\) containing \(x_0 + \eta/2\), being by the construction of Proposition 9 of Chapter 1 the largest open interval about that point inside \(\mathcal{O}\), would contain \((x_0, x_0+\eta)\), forcing \(a_k \le x_0\); and \(a_k < x_0\) would put \(x_0 \in \mathcal{O}\), so \(a_k = x_0\), contrary to the case. Choose \(c \in F\) with \(x_0 < c < x_0 + \delta\) and set \(\delta_{+} = c - x_0\). Let \(x_0 \le x < c\). If \(x \in F\), (D) gives \(|g(x) - g(x_0)| = |f(x) - f(x_0)| < \epsilon/2\). If \(x \in \mathcal{O}\) with component \(I_k\), then \(a_k \ge x_0\) (as \(x_0 \notin \mathcal{O}\)) and \(b_k \le c\) (as \(c \in F\) and \(x < c\)), so \(x_0 \le a_k < b_k \le c < x_0 + \delta\) with \(a_k, b_k \in F\), and (C) with (D) gives

\begin{equation*} |g(x) - g(x_0)| \le (1-t)\frac{\epsilon}{2} + t\frac{\epsilon}{2} = \frac{\epsilon}{2} < \epsilon . \end{equation*}

Hence \(g\) is continuous at every point of \(\mathbb{R}\) and extends \(f\). \(\blacksquare\)

The restriction \(f|_F\) is continuous, but \(f\) itself need be continuous at no point of \(E\).

Lusin’s Theorem supplies a continuous \(g\) on \(\mathbb{R}\) with \(f = g\) on \(F\), and the restriction of a continuous function on \(\mathbb{R}\) to any subset of \(\mathbb{R}\) is continuous on that subset; since \(f|_F = g|_F\), the restriction \(f|_F\) is continuous.

For the second question the answer is no: continuity relative to \(F\) says nothing about points of \(E \sim F\) near \(x_0\). Take \(E = [0,1]\) and \(f = \chi_{\mathbb{Q} \cap [0,1]}\), measurable because \(\mathbb{Q} \cap [0,1]\) is countable, hence of measure zero, hence measurable, and \(m(E) = 1 < \infty\) so Lusin’s Theorem applies. Given \(x_0 \in [0,1]\) and \(\delta > 0\), density of the rationals and of the irrationals puts a rational \(x_1\) and an irrational \(x_2\) of \([0,1]\) in \((x_0 - \delta, x_0 + \delta)\), and

\begin{equation*} |f(x_1) - f(x_2)| = 1, \end{equation*}

so at least one of \(|f(x_1) - f(x_0)|\), \(|f(x_2) - f(x_0)|\) is \(\ge 1/2\). No \(\delta\) works, so \(f\) is continuous at no point of \([0,1]\). \(\blacksquare\)

Take \(E = \mathbb{R}\), so \(m(E) = \infty\), and \(f_n = \chi_{[n,\infty)}\), each measurable, converging pointwise to \(f \equiv 0\): given \(x\), the Archimedean property supplies \(N > x\), and \(f_n(x) = 0\) for all \(n \ge N\). Every hypothesis of Egoroff’s Theorem but the finiteness of \(m(E)\) holds.

Let \(\epsilon > 0\) and let \(F\) be any measurable subset of \(\mathbb{R}\) with \(m(\mathbb{R} \sim F) < \epsilon\); closed sets are in particular measurable. Then \(F \cap [n,\infty) \ne \emptyset\) for every \(n\), since otherwise \([n,\infty) \subseteq \mathbb{R} \sim F\) and monotonicity of measure gives

\begin{equation*} \infty = m\big([n,\infty)\big) \le m(\mathbb{R} \sim F) < \epsilon < \infty . \end{equation*}

Choosing \(x_n \in F \cap [n,\infty)\) gives \(f_n(x_n) = 1\) and \(f(x_n) = 0\), so

\begin{equation*} \sup_{x \in F} |f_n(x) - f(x)| \ge 1 \qquad \text{for every } n, \end{equation*}

and \(\{f_n\}\) does not converge uniformly to \(f\) on \(F\) (take \(\eta = 1/2\)). Hence the conclusion of Egoroff’s Theorem fails here for every \(\epsilon > 0\). \(\blacksquare\)

Put \(A = \{x \in E \mid |f(x)| = \infty\}\), \(B = \{x \in E \mid f_n(x) \not\to f(x)\}\), and \(E_0 = A \cup B\). By hypothesis each of \(A, B\) is contained in a set of measure zero, so \(m^{*}(E_0) = 0\) by monotonicity and countable subadditivity of outer measure; a set of outer measure zero is measurable by the completeness of Lebesgue measure (Chapter 2), so \(E_0\) is measurable with \(m(E_0) = 0\).

Set \(\widetilde{E} = E \sim E_0\), measurable with \(m(\widetilde{E}) \le m(E) < \infty\). Each \(f_n\) is measurable on \(\widetilde{E}\), since \(\{x \in \widetilde{E} \mid f_n(x) > c\} = \widetilde{E} \cap \{x \in E \mid f_n(x) > c\}\); and on \(\widetilde{E}\) the function \(f\) is real-valued with \(f_n \to f\) at every point. These are exactly the hypotheses of Egoroff’s Theorem, which for a given \(\epsilon > 0\) yields a closed \(F \subseteq \widetilde{E}\) with \(\{f_n\} \to f\) uniformly on \(F\) and \(m(\widetilde{E} \sim F) < \epsilon\).

Then \(F\) is a closed subset of \(E\) carrying uniform convergence, and since \(F \subseteq \widetilde{E}\) gives \(E \sim F = E_0 \cup (\widetilde{E} \sim F)\), subadditivity yields

\begin{equation*} m(E \sim F) \le m(E_0) + m(\widetilde{E} \sim F) = m(\widetilde{E} \sim F) < \epsilon . \end{equation*}

\(\blacksquare\)

Let \(\epsilon > 0\). For \(n \in \mathbb{Z}\) put \(A_n = E \cap (n, n+1)\) and \(Z = E \cap \mathbb{Z}\); each \(A_n\) is measurable with \(m(A_n) \le 1 < \infty\), the set \(Z\) is countable so \(m(Z) = 0\), and \[ E = Z \cup \bigcup_{n \in \mathbb{Z}} A_n \] disjointly. Fix a bijection \(k \mapsto n_k\) of \(\mathbb{N}\) onto \(\mathbb{Z}\). On \(A_{n_k}\) the function \(f\) is real-valued and measurable, since \(\{x \in A_{n_k} \mid f(x) > c\} = \{x \in E \mid f(x) > c\} \cap A_{n_k}\), and \(m(A_{n_k}) < \infty\), so Lusin’s Theorem as proved in the text gives a continuous \(g_k\) on \(\mathbb{R}\) and a closed \(F_{n_k} \subseteq A_{n_k}\) with

\begin{equation*} f = g_k \text{ on } F_{n_k}, \qquad m(A_{n_k} \sim F_{n_k}) < \epsilon/2^{k+1} . \end{equation*}

Put \(F = \bigcup_{n \in \mathbb{Z}} F_n \subseteq E\). Since \(F_n \subseteq A_n \subseteq (n, n+1)\), the \(F_n\) are pairwise disjoint and \(F \cap (n, n+1) = F_n\) for every \(n\); call this the localization identity.

\(F\) is closed: if \(x_j \to x\) with \(x_j \in F\), then \(\{x_j\}\) is bounded, say \(|x_j| \le M\), and \(x_j \in F_n\) forces \(-M-1 \le n \le M\), so the \(x_j\) lie in finitely many \(F_n\); some single \(F_n\) then contains a subsequence converging to \(x\), and \(F_n\) is closed, so \(x \in F_n \subseteq F\).

\(m(E \sim F) < \epsilon\): the localization identity gives \(A_n \sim F = A_n \sim F_n\), so \(E \sim F \subseteq Z \cup \bigcup_k (A_{n_k} \sim F_{n_k})\), and monotonicity with countable subadditivity gives \[ m(E \sim F) \le \sum_{k \ge 1} \epsilon/2^{k+1} = \epsilon/2 < \epsilon . \]

\(f|_F\) is continuous: if \(x \in F_n\) and \(x_j \to x\) in \(F\), then \(x_j \in (n,n+1)\) for large \(j\), hence \(x_j \in F_n\) by the localization identity, and with \(n = n_k\) continuity of \(g_k\) gives \(f(x_j) = g_k(x_j) \to g_k(x) = f(x)\). By Problem 25 this continuous function on the closed set \(F\) extends to a continuous \(g\) on \(\mathbb{R}\); then \(F \subseteq E\) is closed, \(f = g\) on \(F\), and \(m(E \sim F) < \epsilon\). \(\blacksquare\)

Let \(\epsilon > 0\) and put \(E_0 = \{f = \infty\} \cup \{f = -\infty\}\), measurable because

\begin{equation*} \{f = \infty\} = \bigcap_{n=1}^{\infty}\{f > n\}, \qquad \{f = -\infty\} = \bigcap_{n=1}^{\infty}\{f < -n\}, \end{equation*}

with \(m(E_0) = 0\), which is precisely the hypothesis that \(f\) is finite a.e. On \(E^{\prime} = E \sim E_0\) the function \(f\) is real-valued and measurable, since \(\{x \in E^{\prime} \mid f(x) > c\} = \{x \in E \mid f(x) > c\} \cap E^{\prime}\).

Apply Lusin’s Theorem to \(f\) on \(E^{\prime}\), in the text’s form if \(m(E^{\prime}) < \infty\) and in the form of Problem 29 otherwise, obtaining a continuous \(g\) on \(\mathbb{R}\) and a closed \(F \subseteq E^{\prime}\) with \(f = g\) on \(F\) and \(m(E^{\prime} \sim F) < \epsilon\). Then \(F\) is a closed subset of \(E\), \(f = g\) on \(F\), and since \(E \sim F \subseteq E_0 \cup (E^{\prime} \sim F)\),

\begin{equation*} m(E \sim F) \le m(E_0) + m(E^{\prime} \sim F) = m(E^{\prime} \sim F) < \epsilon . \end{equation*}

\(\blacksquare\)

(i) The case \(m(E) < \infty\). For each integer \(k \ge 2\), Egoroff’s Theorem applied with \(\epsilon = 1/k\) gives a closed \(A_k \subseteq E\) with \(\{f_n\} \to f\) uniformly on \(A_k\) and \(m(E \sim A_k) < 1/k\). Put \[ N = E \sim \bigcup_{k \ge 2} A_k = \bigcap_{k \ge 2}(E \sim A_k), \] measurable, with \(m(N) \le m(E \sim A_k) < 1/k\) for every \(k\), hence \(m(N) = 0\); and \(E = N \cup \bigcup_{k \ge 2} A_k\) by the definition of \(N\).

(ii) The general case. Put \(C_1 = E \cap [-1,1]\) and \(C_j = E \cap \{x \mid j-1 < |x| \le j\}\) for \(j \ge 2\); these are measurable with \(m(C_j) \le 2 < \infty\), and \(E = \bigcup_{j \ge 1} C_j\). On each \(C_j\) the \(f_n\) are measurable, since \(\{x \in C_j \mid f_n(x) > c\} = \{x \in E \mid f_n(x) > c\} \cap C_j\), and they converge pointwise to the real-valued \(f\); so (i) gives a measurable \(N_j \subseteq C_j\) with \(m(N_j) = 0\) and measurable \(A_{j,2}, A_{j,3}, \ldots \subseteq C_j\) carrying uniform convergence, with \(C_j = N_j \cup \bigcup_{k \ge 2} A_{j,k}\).

Set \(E_1 = \bigcup_{j \ge 1} N_j\), measurable with \(m(E_1) \le \sum_j m(N_j) = 0\) by countable subadditivity. The index set \(\{(j,k) \mid j \ge 1,\ k \ge 2\}\) is countably infinite, so fix a bijection \(i \mapsto (j(i), k(i))\) of \(\{2,3,\ldots\}\) onto it and put \(E_i = A_{j(i),k(i)}\) for \(i > 1\). Each such \(E_i\) is measurable with \(\{f_n\} \to f\) uniformly on it (vacuously so if empty), and

\begin{equation*} \begin{aligned} \bigcup_{k=1}^{\infty} E_k &= \bigcup_{j=1}^{\infty}\Big(N_j \cup \bigcup_{k \ge 2} A_{j,k}\Big) \\ &= \bigcup_{j=1}^{\infty} C_j = E . \end{aligned} \end{equation*}

\(\blacksquare\)

The convergence is not uniform, because \(\sup_{x \in [0,1]}|f(x) - f_n(x)| = 1\) for every \(n\).

Fix \(n\). The set \(\{q_1, \ldots, q_n\}\) is finite while \(\mathbb{Q} \cap [0,1]\) is infinite (it contains \(1/m\) for every natural number \(m\)), so there is a rational \(r \in [0,1]\) outside it; then \(f( r) = 1\) and \(f_n( r) = 0\), whence

\begin{equation*} \sup_{x \in [0,1]}|f(x) - f_n(x)| \ge |f( r) - f_n( r)| = 1, \end{equation*}

with equality since \(0 \le f_n \le f \le 1\) on \([0,1]\). Were the convergence uniform, taking \(\epsilon = 1/2\) would give an index \(N\) with \(|f - f_N| < 1/2\) throughout \([0,1]\), hence \(1 \le 1/2\) on passing to the supremum. \(\blacksquare\)

Both claims reduce to one monotonicity fact: if \(\emptyset \ne A \subseteq B\) and \(f\) is bounded, then \(\inf_B f \le \inf_A f\) and \(\sup_A f \le \sup_B f\), since \(\inf_B f\) is a lower bound and \(\sup_B f\) an upper bound for \(f\) on \(A\).

(i) One adjoined point. Let \(P^{\prime} = P \cup \{c\}\) with \(x_{k-1} < c < x_k\), and put \(m^{\prime} = \inf_{(x_{k-1},c)} f\), \(m^{\prime\prime} = \inf_{(c,x_k)} f\). All terms with \(i \ne k\) agree, and \(m^{\prime}, m^{\prime\prime} \ge m_k\) by the monotonicity fact, so

\begin{equation*} \begin{aligned} L(f,P^{\prime}) - L(f,P) &= m^{\prime}(c - x_{k-1}) + m^{\prime\prime}(x_k - c) - m_k(x_k - x_{k-1}) \\ &\ge m_k\big[(c - x_{k-1}) + (x_k - c) - (x_k - x_{k-1})\big] = 0 . \end{aligned} \end{equation*}

Symmetrically \(M^{\prime}, M^{\prime\prime} \le M_k\) gives \(U(f,P^{\prime}) - U(f,P) \le 0\).

(ii) The general case. If \(P^{\prime}\) refines \(P\), then \(P^{\prime} \sim P\) is finite; induct on its cardinality \(p\), the case \(p = 0\) being \(P^{\prime} = P\). For the step, pick \(c \in P^{\prime} \sim P\) and put \(Q = P \cup \{c\}\), so that \(P^{\prime}\) refines \(Q\) by \(p\) points; by (i) and the inductive hypothesis,

\begin{equation*} L(f,P) \le L(f,Q) \le L(f,P^{\prime}), \qquad U(f,P^{\prime}) \le U(f,Q) \le U(f,P). \end{equation*}

\(\blacksquare\)

For any two partitions \(P_1, P_2\) of \([a,b]\), the common refinement \(P = P_1 \cup P_2\) (a finite subset of \([a,b]\) containing \(a\) and \(b\), hence a partition, refining both) gives, using Problem 2 together with \(m_i \le M_i\) on each nonempty \((x_{i-1},x_i)\),

\begin{equation*} L(f,P_1) \le L(f,P) \le U(f,P) \le U(f,P_2). \end{equation*}

So each lower Darboux sum is no greater than each upper Darboux sum.

Now fix \(P_2\). The number \(U(f,P_2)\) is an upper bound for the set of lower Darboux sums, which is nonempty and bounded above since \(f\) is bounded, so its supremum satisfies \(( R)\underline{\int_a^b} f \le U(f,P_2)\). As \(P_2\) was arbitrary, \(( R)\underline{\int_a^b} f\) is a lower bound for the set of upper Darboux sums, and since the infimum is the greatest lower bound,

\begin{equation*} ( R)\underline{\int_a^b} f \le \inf_{P_2} U(f,P_2) = ( R)\overline{\int_a^b} f . \end{equation*}

\(\blacksquare\)

Let \(I\) be the common value of the lower and upper Riemann integrals, a real number since \(f\) is bounded. Fix a natural number \(n\). As \(I = \sup_P L(f,P)\), some partition \(P_n^{\prime}\) has \(L(f,P_n^{\prime}) > I - 1/(2n)\); as \(I = \inf_P U(f,P)\), some partition \(P_n^{\prime\prime}\) has \(U(f,P_n^{\prime\prime}) < I + 1/(2n)\). Put \(P_n = P_n^{\prime} \cup P_n^{\prime\prime}\), a partition of \([a,b]\) refining both, so that by Problem 2

\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*}

Hence, the left inequality below being \(L(f,P) \le U(f,P)\) from Problem 3,

\begin{equation*} 0 \le U(f,P_n) - L(f,P_n) < \frac{1}{n} \longrightarrow 0 . \end{equation*}

\(\blacksquare\)

Both \(( R)\underline{\int_a^b} f\) and \(( R)\overline{\int_a^b} f\) are real numbers, since \(|f| \le M\) puts every Darboux sum in \([-M(b-a), M(b-a)]\), so the relevant sets are nonempty and bounded. For each \(n\) the supremum defining the lower integral dominates \(L(f,P_n)\), the infimum defining the upper integral is dominated by \(U(f,P_n)\), and Problem 3 orders the two integrals, giving

\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*}

Hence

\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) \longrightarrow 0, \end{equation*}

and the left-hand quantity, independent of \(n\), is \(0\). So the lower and upper Riemann integrals coincide and \(f\), being bounded, is Riemann integrable over \([a,b]\). \(\blacksquare\)

Assume \(a < b\). By the Extreme Value Theorem of Chapter 1, \(f\) attains a minimum and a maximum on the nonempty closed, bounded set \([a,b]\), so \(f\) is bounded there and its Darboux sums are defined; and by Theorem 23 of Chapter 1, \(f\) is uniformly continuous on \([a,b]\).

Fix \(n\) and choose \(\delta_n > 0\) with \(|f(u) - f(v)| < 1/n\) whenever \(u, v \in [a,b]\) and \(|u - v| < \delta_n\). Choose \(k_n \in \mathbb{N}\) with \((b-a)/k_n < \delta_n\), possible by the Archimedean property, and let \(P_n\) be the regular partition of \([a,b]\) into \(k_n\) subintervals of length \((b-a)/k_n\). For \(u, v \in (x_{i-1}, x_i)\) we have \(|u - v| < (b-a)/k_n < \delta_n\), so \(f(u) < f(v) + 1/n\); the supremum over \(u\) gives \(M_i \le f(v) + 1/n\), so \(M_i - 1/n\) is a lower bound for \(f\) on \((x_{i-1},x_i)\), and the infimum over \(v\) gives \(M_i - m_i \le 1/n\). Therefore

\begin{equation*} \begin{aligned} 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} \longrightarrow 0 . \end{aligned} \end{equation*}

Since also \(U(f,P_n) - L(f,P_n) \ge 0\) by Problem 3, Problem 5 makes \(f\) Riemann integrable over \([a,b]\). \(\blacksquare\)

Since \(f\) is increasing, \(f(0) \le f \le f(1)\) on \([0,1]\), so \(f\) is bounded and its Darboux sums are defined; no continuity is assumed. With \(x_i = i/n\), each subinterval has length \(1/n\), and for \(x_{i-1} < x < x_i\) monotonicity gives \(f(x_{i-1}) \le f(x) \le f(x_i)\), so

\begin{equation*} f(x_{i-1}) \le m_i \le M_i \le f(x_i), \quad\text{hence}\quad 0 \le M_i - m_i \le f(x_i) - f(x_{i-1}), \end{equation*}

the middle inequality because \((x_{i-1},x_i)\) is nonempty. Therefore

\begin{equation*} \begin{aligned} U(f,P_n) - L(f,P_n) &= \frac{1}{n}\sum_{i=1}^{n}(M_i - m_i) \\ &\le \frac{1}{n}\sum_{i=1}^{n}\big[f(x_i) - f(x_{i-1})\big] \\ &= \frac{1}{n}\big[f(1) - f(0)\big], \end{aligned} \end{equation*}

the last step by telescoping. Since \(f(1) - f(0)\) is a constant and \(U(f,P_n) - L(f,P_n) \ge 0\) by Problem 3, the difference tends to \(0\), and Problem 5 makes \(f\) Riemann integrable over \([0,1]\). \(\blacksquare\)

Yes to both: \(f\) is Riemann integrable over \([a,b]\) and \(\lim_n \int_a^b f_n = \int_a^b f\). Assume \(a < b\), the case \(a = b\) being trivial, and put \(\eta_n = \sup_{[a,b]}|f_n - f|\); uniform convergence says exactly that \(\eta_n \le \epsilon\) for all large \(n\), so \(\eta_n\) is finite for all large \(n\) and \(\eta_n \to 0\).

\(f\) is bounded: choose \(N\) with \(\eta_N \le 1\); then \(|f| \le |f_N| + |f - f_N| \le M + 1\), where \(|f_N| \le M\).

Comparison of Darboux sums: if \(g, h\) are bounded with \(|g - h| \le \eta\) on \([a,b]\), then \(M_i(g) \le M_i(h) + \eta\) and \(m_i(g) \ge m_i(h) - \eta\) on each subinterval, so multiplying by the lengths, which total \(b - a\), and summing,

\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*}

Integrability. Fix \(k \in \mathbb{N}\) and choose \(n_k\) with \(\eta_{n_k} < 1/[4k(b-a)]\). Since \(f_{n_k}\) is Riemann integrable, Problem 4 supplies a partition \(P_k\) of \([a,b]\) with \(U(f_{n_k},P_k) - L(f_{n_k},P_k) < 1/(2k)\), and \((\ast)\) with \(g = f\), \(h = f_{n_k}\) gives

\begin{equation*} U(f,P_k) - L(f,P_k) < \frac{1}{2k} + 2 \cdot \frac{b-a}{4k(b-a)} = \frac{1}{k} . \end{equation*}

So \(U(f,P_k) - L(f,P_k) \to 0\), and \(f\) is Riemann integrable by Problem 5.

Convergence of the integrals. For \(n\) with \(\eta_n < \infty\), applying \((\ast)\) with \(g = f\), \(h = f_n\), \(\eta = \eta_n\) and taking the infimum over all partitions gives \(( R)\overline{\int_a^b} f \le ( R)\overline{\int_a^b} f_n + \eta_n(b-a)\); both functions being Riemann integrable, their upper integrals are their integrals. Exchanging \(f\) and \(f_n\), the bound \(|f - f_n| \le \eta_n\) being symmetric, we get

\begin{equation*} \left|\int_a^b f_n - \int_a^b f\right| \le \eta_n (b-a) \longrightarrow 0 . \end{equation*}

\(\blacksquare\)

\(f\) is measurable and \(\int_E f = 0\).

Measurability: \(E\) has outer measure zero, hence is measurable by Proposition 4 of Chapter 2, and for each \(c \in \mathbb{R}\) the set \(\{x \in E \mid f(x) > c\}\) is a subset of \(E\), so has outer measure zero by monotonicity and is measurable by the same proposition.

Every simple \(\varphi\) on \(E\) has \(\int_E \varphi = 0\): in the canonical representation \(\varphi = \sum_{i=1}^{n} a_i \cdot \chi_{E_i}\) each \(E_i\) is a measurable subset of \(E\), so \(m(E_i) = 0\) by monotonicity and

\begin{equation*} \int_E \varphi = \sum_{i=1}^{n} a_i \cdot m(E_i) = 0 . \end{equation*}

Since \(f\) is bounded, say \(|f| \le M\), the constant simple functions \(-M\) and \(M\) lie below and above \(f\), so the competing families in the definitions of the lower and upper Lebesgue integrals are nonempty; Theorem 4 applies (\(f\) bounded measurable on \(E\) with \(m(E) = 0 < \infty\)), and both integrals equal \(\sup\{0\} = \inf\{0\} = 0\). Hence \(\int_E f = 0\). \(\blacksquare\)

The device is to extend simple functions on \(A\) by zero. Let \(\sigma = \sum_{i=1}^{n} a_i \cdot \chi_{A_i}\) be the canonical representation of a simple function on \(A\), so the \(A_i = \sigma^{-1}(a_i)\) are disjoint measurable subsets of \(A\), hence of \(E\), with union \(A\); put \(\tilde\sigma = \sigma\) on \(A\) and \(\tilde\sigma = 0\) on \(E \sim A\). Then \(\tilde\sigma\) is simple on \(E\), represented over the finite disjoint collection \(\{A_1, \ldots, A_n, E \sim A\}\) of measurable subsets of \(E\), so Lemma 1 gives

\begin{equation*} \int_E \tilde\sigma = \sum_{i=1}^{n} a_i \cdot m(A_i) + 0 \cdot m(E \sim A) = \int_A \sigma . \tag{\(\dagger\)} \end{equation*}

Both integrals in question exist by Theorem 4: \(m(A) \le m(E) < \infty\) and \(f|_A\) is bounded measurable, while \(f \cdot \chi_A\) is bounded and measurable on \(E\); in particular each of \(\int_A f\) and \(\int_E f \cdot \chi_A\) equals both the corresponding upper and the corresponding lower Lebesgue integral. If \(A = \emptyset\) both sides are \(0\), so assume \(A \ne \emptyset\).

(i) Let \(\psi\) be simple on \(A\) with \(\psi \ge f\). Then \(f \cdot \chi_A \le \tilde\psi\) on \(E\), the two being equal to \(0\) off \(A\), so by \((\dagger)\) and the upper-integral description, \(\int_E f \cdot \chi_A \le \int_E \tilde\psi = \int_A \psi\); the infimum over such \(\psi\) gives \(\int_E f \cdot \chi_A \le \int_A f\).

(ii) Let \(\varphi\) be simple on \(A\) with \(\varphi \le f\). Then \(\tilde\varphi \le f \cdot \chi_A\) on \(E\), so by \((\dagger)\) and the lower-integral description, \(\int_A \varphi = \int_E \tilde\varphi \le \int_E f \cdot \chi_A\); the supremum over such \(\varphi\) gives \(\int_A f \le \int_E f \cdot \chi_A\).

Hence \(\int_A f = \int_E f \cdot \chi_A\). \(\blacksquare\)

No. The Dirichlet example of Section 4.1 satisfies every hypothesis while the limit function is not Riemann integrable at all, so the asserted conclusion is meaningless.

Enumerate \(\mathbb{Q} \cap [0,1] = \{q_k\}_{k=1}^{\infty}\) and define on \([0,1]\)

\begin{equation*} f_n(x) = \begin{cases} 1 & \text{if } x = q_k \text{ for some } k \le n,\\[2pt] 0 & \text{otherwise.} \end{cases} \end{equation*}

Each \(f_n\) is a step function, hence Riemann integrable with \(( R)\int_0^1 f_n = 0\); the sequence is uniformly bounded by \(1\) and increases pointwise to the Dirichlet function \(f = \chi_{\mathbb{Q} \cap [0,1]}\). But 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 \([0,1]\). Thus \(\lim_n ( R)\int_0^1 f_n\) exists and equals \(0\), but there is no number \(( R)\int_0^1 f\) for it to equal. \(\blacksquare\)

Let \(E_0 \subseteq E\) have \(m(E_0) = 0\) with \(f = g\) on \(E \sim E_0\); we may take \(E_0\) measurable, since \(\{x \in E \mid f(x) \ne g(x)\}\) is a subset of a null set, hence of outer measure zero, hence measurable by Proposition 4 of Chapter 2.

By Proposition 5 of Chapter 3, \(g\) is measurable on \(E\), agreeing a.e. with the measurable \(f\). Both \(f\) and \(g\) are then bounded measurable functions on the set \(E\) of finite measure, so both integrals exist by Theorem 4.

Put \(h = g - f\), bounded and measurable on \(E\), with \(h = 0\) on \(E \sim E_0\); choose \(M \ge 0\) with \(|h| \le M\). Then \(0 \le |h| \le M \cdot \chi_{E_0}\) on \(E\), and \(M \cdot \chi_{E_0}\) is simple over the finite disjoint collection \(\{E_0, E \sim E_0\}\), so Lemma 1 gives \(\int_E M \cdot \chi_{E_0} = M \cdot m(E_0) = 0\). Monotonicity of integration (Theorem 5) therefore forces \(\int_E |h| = 0\), and by Corollary 7 together with linearity (Theorem 5),

\begin{equation*} \left|\int_E g - \int_E f\right| = \left|\int_E h\right| \le \int_E |h| = 0 . \end{equation*}

\(\blacksquare\)

No. On \(E=[0,1]\), with \(m(E)=1<\infty\), take \(f_n=n\cdot\chi_{(0,\,1/n]}\). Each \(f_n\) is simple, so by Lemma 1

\begin{equation*} \int_{[0,1]} f_n = n\cdot m\bigl((0,1/n]\bigr) = 1 \quad\text{for every } n, \end{equation*}

while \(\{f_n\}\to f\equiv 0\) pointwise on \([0,1]\) (given \(x>0\), \(f_n(x)=0\) once \(1/n<x\)), so \(\int_{[0,1]} f=0\). Every hypothesis of the Bounded Convergence Theorem holds except uniform boundedness, which fails since \(\sup_{[0,1]}|f_n|=n\to\infty\), and the conclusion fails with it:

\begin{equation*} \lim_{n\to\infty}\int_{[0,1]} f_n = 1 \ne 0 = \int_{[0,1]} f . \end{equation*}

The hypotheses of Proposition 8 imply those of the Bounded Convergence Theorem, whose conclusion is that of Proposition 8. Measurability of the \(f_n\) and \(m(E)<\infty\) are common to both, and uniform convergence on \(E\) forces pointwise convergence on \(E\); only the uniform bound must be manufactured.

Apply uniform convergence with \(\epsilon=1\) to get \(N\) with \(|f_n-f|<1\) on \(E\) for all \(n\ge N\), and choose \(M_j\ge 0\) with \(|f_j|\le M_j\) on \(E\) for \(1\le j\le N\) (each \(f_j\) is bounded by hypothesis). Then \(|f|\le|f-f_N|+|f_N|<1+M_N\) on \(E\), so for \(n\ge N\) we get \(|f_n|\le|f_n-f|+|f|<M_N+2\) on \(E\), and hence

\begin{equation*} |f_n|\le M:=\max\{M_1,\dots,M_{N-1},\,M_N+2\}\quad\text{on }E\text{ for every }n . \end{equation*}

So \(\{f_n\}\) is uniformly pointwise bounded on \(E\) and the Bounded Convergence Theorem gives \(\lim_{n\to\infty}\int_E f_n=\int_E f\). No step in the text’s proof of the Bounded Convergence Theorem invokes Proposition 8, so the deduction is not circular.

All three assertions hold. Throughout, additivity means additivity of \(m\) over disjoint measurable sets (finite additivity is Proposition 6 of Chapter 2, countable additivity Proposition 13); the countable subadditivity of outer measure \(m^*\), which builds the \(\sigma\)-algebra and the measurability machinery of Chapter 3, is a different statement and is freely available throughout.

The audit of Section 4.2. (a) The definition of \(\int_E\psi\) for simple \(\psi\) needs only that the level sets are measurable of finite measure: no additivity. (b) Lemma 1 groups the indices by the value \(\lambda_j\) of \(\varphi\) and asserts \(m(A_j)=\sum_{i\in I_j} m(E_i)\) for the finite disjoint union \(A_j=\bigcup_{i\in I_j}E_i\); this is finite additivity, and it is the only appearance of additivity of \(m\) in the section. (c) Proposition 2 uses set algebra (closure of the measurable sets under finite intersection), Lemma 1, and arithmetic; the following remark that disjointness in Lemma 1 is superfluous uses only Proposition 2. (d) The step-function discussion uses that the measure of an interval is its length (Proposition 1 of Chapter 2, a statement about \(m^*\)) and that a singleton has measure zero, then Proposition 2. (e) The upper and lower integrals and integrability are definitions; their comparison is the monotonicity half of Proposition 2. (f) Theorem 3 uses only that every step function is simple, together with (d). (g) Theorem 4 uses the Simple Approximation Lemma of Chapter 3 (which invokes no property of \(m\)), Proposition 2, and \(m(E)<\infty\); the Dirichlet example uses only that a countable set has outer measure zero. (h) Theorem 5 passes between upper and lower integrals using Theorem 4 and Proposition 2 alone. (i) Corollary 6 comes from the identity \(\chi_{A\cup B}=\chi_A+\chi_B\) between functions, valid for disjoint \(A,B\), plus Problem 10 and the linearity of Theorem 5; despite its name it is additivity of the integral, not of \(m\). (j) Corollary 7 follows from \(-|f|\le f\le|f|\) and Theorem 5, and (k) Proposition 8 from Theorem 5, Corollary 7 and \(m(E)<\infty\). So countable additivity is never invoked and finite additivity exactly once.

The Bounded Convergence Theorem uses Egoroff’s Theorem. Its proof applies Egoroff’s Theorem to produce measurable \(A\subseteq E\) with \(\{f_n\}\to f\) uniformly on \(A\) and \(m(E\sim A)<\epsilon/4M\), then finishes from

\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*}

Egoroff’s Theorem rests on Lemma 10 of Chapter 3, whose proof takes the ascending measurable sets \(E_n=\{x\in E\mid |f-f_k|<\eta \text{ for all } k\ge n\}\), notes \(\bigcup_n E_n=E\) by pointwise convergence, and concludes \(m(E)=\lim_n m(E_n)\) by the continuity of measure, Theorem 15(i) of Chapter 2; since \(m(E)<\infty\), excision then gives \(m(E\sim E_N)<\delta\) for some \(N\).

Continuity of measure follows from countable additivity. For an ascending \(\{A_k\}\) put \(A_0=\emptyset\). The sets \(A_k\sim A_{k-1}\) are measurable and pairwise disjoint, and sending each \(x\) to the least \(k\) with \(x\in A_k\) shows

\begin{equation*} \begin{aligned} \bigcup_{k=1}^{\infty}A_k &= \bigcup_{k=1}^{\infty}\left[A_k\sim A_{k-1}\right], \\ A_n &= \bigcup_{k=1}^{n}\left[A_k\sim A_{k-1}\right]. \end{aligned} \end{equation*}

So Proposition 13 of Chapter 2 on the left and finite additivity on the right give

\begin{equation*} \begin{aligned} m\left(\bigcup_{k=1}^{\infty}A_k\right) &= \sum_{k=1}^{\infty}m(A_k\sim A_{k-1}) \\ &= \lim_{n\to\infty}m(A_n), \end{aligned} \end{equation*}

which is Theorem 15(i); part (ii) follows by applying (i) to \(\{B_1\sim B_k\}\) and excising, legitimate since \(m(B_1)<\infty\).

Set \(E_n=\{x\in E\mid f(x)\ge 1/n\}\), a measurable subset of \(E\). Then \(\tfrac{1}{n}\cdot\chi_{E_n}\le f\) on \(E\) (on \(E_n\) by definition, off \(E_n\) since \(f\ge 0\)), and both sides are bounded and measurable on the set \(E\) of finite measure, so the monotonicity of Theorem 5 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*}

whence \(m(E_n)=0\) for every \(n\). 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*}

using the Archimedean property for the second equality, and countable subadditivity of Lebesgue measure makes this union a set of measure zero. Hence \(f=0\) a.e. on \(E\).

Zero, because every competitor in the defining supremum has integral zero. By (8),

\begin{equation*} \int_E f=\sup\left\{\int_E h \;\middle|\; \begin{gathered} h \text{ bounded, measurable,} \\ \text{of finite support}, \ 0\le h\le f \end{gathered}\right\}, \end{equation*}

a nonempty set of numbers since \(h\equiv 0\) qualifies. Let \(h\) be any competitor; as \(m(E)=0<\infty\), \(\int_E h\) is just the Section 4.2 integral of a bounded measurable function on \(E\). For every simple \(\psi\ge h\) on \(E\) with canonical representation \(\psi=\sum_{i=1}^{n}a_i\cdot\chi_{A_i}\) we have \(m(A_i)\le m(E)=0\) by monotonicity of measure, so

\begin{equation*} \int_E\psi=\sum_{i=1}^{n}a_i\cdot m(A_i)=0 . \end{equation*}

Thus the upper Lebesgue integral of \(h\) over \(E\) is \(0\), and \(\int_E h=0\) by Theorem 4. Therefore \(\int_E f=\sup\{0\}=0\).

Both requirements hold. If \(E_0\subseteq E\) is measurable with \(m(E_0)<\infty\) and \(f=0\) on \(E\sim E_0\), then \(f\) restricted to \(E_0\) is bounded and measurable on a set of finite measure, so \(\int_{E_0} f\) is a well-defined real number by Theorem 4.

For independence, let \(E_0\) and \(E_1\) both be admissible. Corollary 6 applied on \(E_0\) to the disjoint decomposition \(E_0=(E_0\cap E_1)\cup(E_0\sim E_1)\) gives

\begin{equation*} \int_{E_0} f=\int_{E_0\cap E_1} f+\int_{E_0\sim E_1} f=\int_{E_0\cap E_1} f , \end{equation*}

since \(E_0\sim E_1\subseteq E\sim E_1\), where \(f\) vanishes, so that the second integral is \(0\cdot m(E_0\sim E_1)=0\). Interchanging the roles of \(E_0\) and \(E_1\) gives \(\int_{E_1} f=\int_{E_0\cap E_1} f\) likewise, so \(\int_{E_0} f=\int_{E_1} f\) and the definition is proper.

\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*}

Here \(f\) is nonnegative and measurable on \([0,1]\), being continuous on \((0,1]\) and \(0\) at the origin. Put \(f_n=f\cdot\chi_{[1/n,\,1]}\) for \(n\ge 2\): the sets \([1/n,1]\) ascend and \(f\ge 0\), so \(\{f_n\}\) is an increasing sequence of nonnegative measurable functions, and it converges pointwise on \([0,1]\) to \(f\) (for \(x>0\), \(f_n(x)=f(x)\) once \(1/n<x\)). The Monotone Convergence Theorem gives \(\int_{[0,1]} f=\lim_n\int_{[0,1]} f_n\). Since \(f_n\) vanishes on \([0,1/n)\), Theorem 11 (additivity over domains) reduces \(\int_{[0,1]}f_n\) to \(\int_{[1/n,1]}x^{\alpha}\,dx\), and \(x\mapsto x^{\alpha}\) is continuous, hence Riemann integrable, on \([1/n,1]\), so Theorem 3 evaluates it:

\begin{equation*} \int_{[0,1]} f_n=\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*}

Letting \(n\to\infty\): (i) if \(\alpha>-1\) then \(\alpha+1>0\), so \((1/n)^{\alpha+1}\to 0\) and the limit is \(1/(\alpha+1)\); (ii) if \(\alpha=-1\) then \(\ln n\to\infty\); (iii) if \(\alpha<-1\) then \((1/n)^{\alpha+1}=n^{-(\alpha+1)}\to\infty\) while \(\alpha+1<0\), so the quotient tends to \(+\infty\).

Both implications. If \(\{f_n\}\) is nonnegative and measurable with \(\{f_n\}\to f\) pointwise on \(E\) and \(\int_E f_n\le M\) for all \(n\), then Fatou’s Lemma gives

\begin{equation*} \int_E f\le\liminf_{n\to\infty}\int_E f_n\le M , \end{equation*}

which is the asserted property; call it \((P)\).

Conversely, assume \((P)\), let \(\{g_n\}\) be nonnegative and measurable on \(E\) with \(\{g_n\}\to g\) pointwise a.e. on \(E\), and put \(L=\liminf_n\int_E g_n\in[0,\infty]\). If \(L=\infty\) there is nothing to prove, so suppose \(L<\infty\) and fix \(\epsilon>0\). Choose a subsequence with \(\int_E g_{n_k}\to L\) and then \(K\) with \(\int_E g_{n_k}\le L+\epsilon\) for \(k\ge K\). Its tail \(h_k=g_{n_{K+k-1}}\) consists of nonnegative measurable functions converging to \(g\) pointwise a.e. on \(E\), with \(\int_E h_k\le L+\epsilon\) for every \(k\), so \((P)\) with \(M=L+\epsilon\) gives \(\int_E g\le L+\epsilon\). As \(\epsilon>0\) was arbitrary, \(\int_E g\le L\), which is Fatou’s Lemma.

Here \((P)\) has been applied under a.e. rather than everywhere convergence; this is legitimate because excising the null set off which convergence holds changes no integral, by the excision formula (15) of Theorem 11, exactly the reduction performed at the start of the text’s proof of Fatou’s Lemma.

Part 1. Since \(f\) is integrable, \(\int_E f<\infty\), so the definition (8) provides a bounded measurable \(h_1\) of finite support with \(0\le h_1\le f\) on \(E\) and \(\int_E h_1>\int_E f-\epsilon/2\). Pick measurable \(E_0\subseteq E\) with \(m(E_0)<\infty\) and \(h_1=0\) on \(E\sim E_0\); by Problem 18 the value \(\int_E h_1=\int_{E_0}h_1\) is independent of this choice. The restriction of \(h_1\) to \(E_0\) is bounded, real-valued and measurable, so the Simple Approximation Lemma of Chapter 3 applied with \(\epsilon’=\epsilon/\bigl(2(m(E_0)+1)\bigr)\) gives simple \(\varphi\le h_1\le\psi\) on \(E_0\) with \(0\le\psi-\varphi<\epsilon’\). Set \(\eta_0=\max\{\varphi,0\}\), which is simple since each of its level sets is a finite union of level sets of \(\varphi\), and extend it by \(0\) to a function \(\eta\) on \(E\). Then \(\eta\) is simple with finite support, \(0\le\eta\le h_1\le f\) on \(E\), and \(0\le h_1-\eta\le\psi-\varphi<\epsilon’\) on \(E_0\). All of \(h_1,\eta,h_1-\eta\) vanish off \(E_0\), so Theorem 5 on \(E_0\) applies and yields

\begin{equation*} \int_E (h_1-\eta)\le\epsilon’\cdot m(E_0)<\frac{\epsilon}{2}, \qquad \int_E\eta=\int_E h_1-\int_E (h_1-\eta)>\int_E f-\epsilon . \end{equation*}

Since \(0\le\eta\le f\), the function \(f-\eta\) is nonnegative and measurable with \((f-\eta)+\eta=f\) on \(E\), so the linearity of Theorem 10 gives \(\int_E f=\int_E(f-\eta)+\int_E\eta\); both \(\int_E f\) and \(\int_E\eta\) are finite, so

\begin{equation*} \int_E|f-\eta|=\int_E (f-\eta)=\int_E f-\int_E\eta<\epsilon . \end{equation*}

Part 2. Let \(E=[a,b]\); every function on \(E\) has finite support since \(m(E)<\infty\), and \(h\) is not required to satisfy \(0\le h\le f\). Take \(\eta\) from Part 1 with \(\int_E|f-\eta|<\epsilon/2\) and write its canonical representation \(\eta=\sum_{k=1}^{n}a_k\cdot\chi_{E_k}\) over the distinct nonzero values \(a_k\) (if \(n=0\), take \(h\equiv 0\)). Put \(\delta=\epsilon/\bigl(2(1+\sum_k|a_k|)\bigr)\). Each \(E_k\) is measurable of finite measure, so Theorem 12 of Chapter 2 gives a finite disjoint union of open intervals \(O_k\) with \(m(E_k\,\Delta\,O_k)<\delta\); set

\begin{equation*} h=\sum_{k=1}^{n}a_k\cdot\chi_{O_k\cap[a,b]} , \end{equation*}

a step function on \([a,b]\). Since \(|\chi_A-\chi_B|=\chi_{A\Delta B}\) and, as \(E_k\subseteq[a,b]\), \(E_k\,\Delta\,(O_k\cap[a,b])\subseteq E_k\,\Delta\,O_k\), the monotonicity and linearity of Theorem 5 on \([a,b]\) give

\begin{equation*} \begin{aligned} \int_{[a,b]}|\eta-h| &\le \sum_{k=1}^{n}|a_k|\cdot m\bigl(E_k\,\Delta\,O_k\bigr) \\ &\le \delta\sum_{k=1}^{n}|a_k|<\frac{\epsilon}{2} . \end{aligned} \end{equation*}

Finally \(f\) is finite off the null set \(Z=\{x\in E\mid f(x)=\infty\}\) (Proposition 13), so \(|f-h|\le|f-\eta|+|\eta-h|\) on \(E\sim Z\), and excision (15) with Theorem 10 gives

\begin{equation*} \int_E|f-h|\le\int_E|f-\eta|+\int_E|\eta-h|<\epsilon . \end{equation*}

Let \(E\) be measurable, \(E^c=\mathbb{R}\setminus E\), and \(L=\int_{\mathbb{R}}f=\lim_n\int_{\mathbb{R}}f_n\), finite since \(f\) is integrable over \(\mathbb{R}\). By Theorem 11 of Chapter 4 (Additivity Over Domains of Integration), applied to the nonnegative measurable functions \(f\) and \(f_n\) on the disjoint pair \(E,E^c\),

\begin{equation*} \begin{aligned} \int_E f+\int_{E^c} f &= L<\infty, \\ \int_E f_n+\int_{E^c} f_n &= \int_{\mathbb{R}} f_n , \end{aligned} \end{equation*}

so \(\int_E f\) and \(\int_{E^c}f\) are finite, and since \(\int_{\mathbb{R}}f_n\le L+1\) for all large \(n\), so are \(\int_E f_n\) and \(\int_{E^c}f_n\) for such \(n\). Fatou’s Lemma on \(E\) and on \(E^c\) gives

\begin{equation*} \int_E f\le\liminf_{n\to\infty}\int_E f_n, \qquad \int_{E^c} f\le\liminf_{n\to\infty}\int_{E^c} f_n . \end{equation*}

Because \(\left\{\int_{\mathbb{R}}f_n\right\}\) converges to the finite \(L\), we have \(\limsup_n\bigl(\int_{\mathbb{R}}f_n-\int_{E^c}f_n\bigr)=L-\liminf_n\int_{E^c}f_n\), so the second inequality yields

\begin{equation*} \begin{aligned} \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_E f , \end{aligned} \end{equation*}

the last equality being the additivity relation above, a legitimate subtraction since \(\int_{E^c}f\) is finite. Together with the Fatou bound on \(E\) this forces \(\lim_{n\to\infty}\int_E f_n=\int_E f\).

Put \(I_n=[n,n+1)\) and \(u_n=a_n\cdot\chi_{I_n}\), a nonnegative simple function on \(E=[1,\infty)\). The \(I_n\) are pairwise disjoint with union \(E\), so at each point of \(E\) at most one \(u_n\) is nonzero and \(f=\sum_{n=1}^{\infty}u_n\) pointwise on \(E\), while

\begin{equation*} \int_E u_n=a_n\cdot m(I_n)=a_n . \end{equation*}

By Corollary 12 of Chapter 4,

\begin{equation*} \int_E f=\sum_{n=1}^{\infty}\int_E u_n=\sum_{n=1}^{\infty}a_n , \end{equation*}

an equality in \([0,\infty]\).

Part (i). Take

\begin{equation*} \varphi_n=g_n\cdot\chi_{E_n}, \qquad g_n=\min\left\{n, \ \frac{\lfloor 2^n f\rfloor}{2^n}\right\}, \qquad E_n=E\cap[-n,n], \end{equation*}

with the convention \(g_n=n\) where \(f=\infty\). Each \(g_n\) takes values in the finite set \(\{k/2^n\mid 0\le k\le n2^n\}\), and each level set is described by inequalities on the measurable \(f\), so \(\varphi_n\) is simple; it vanishes off \(E_n\), a set of measure at most \(2n\), so it has finite support; and \(0\le\varphi_n\le g_n\le f\) since \(\lfloor 2^n t\rfloor/2^n\le t\) for \(t\ge 0\).

The sequence increases: \(2^{n+1}t\ge 2\lfloor 2^n t\rfloor\) with the right-hand side an integer, so \(\lfloor 2^{n+1}t\rfloor\ge 2\lfloor 2^n t\rfloor\) and hence \(g_n\le g_{n+1}\) (trivially so where \(f=\infty\)); also \(E_n\subseteq E_{n+1}\), and all quantities are nonnegative, so \(\varphi_n\le\varphi_{n+1}\) on \(E\).

It converges pointwise to \(f\): fix \(x\in E\), so that \(\varphi_n(x)=g_n(x)\) for all \(n\ge|x|\). (i) If \(f(x)<\infty\), then for \(n>f(x)\) we have \(g_n(x)=\lfloor 2^n f(x)\rfloor/2^n\) and \(0\le f(x)-g_n(x)<2^{-n}\). (ii) If \(f(x)=\infty\), then \(\varphi_n(x)=n\to\infty=f(x)\).

Part (ii). Write \(S\) for the supremum in question. Each competitor \(\varphi\) takes finitely many real values, so it is a bounded measurable function of finite support with \(0\le\varphi\le f\), hence one of the competitors in the definition (8) of \(\int_E f\); thus \(S\le\int_E f\). Conversely the \(\varphi_n\) of part (i) are themselves competitors, so \(\int_E\varphi_n\le S\), and the Monotone Convergence Theorem applied to \(\{\varphi_n\}\uparrow f\) gives

\begin{equation*} \int_E f=\lim_{n\to\infty}\int_E\varphi_n\le S . \end{equation*}

Hence \(\int_E f=S\).

The monotonicity of integration for nonnegative measurable functions applied to \(0\le f_n\le f\) gives \(\int_E f_n\le\int_E f\) for every \(n\), while Fatou’s Lemma, applicable since the \(f_n\) are nonnegative and measurable with \(\{f_n\}\to f\) pointwise on \(E\), bounds the lower limit from below:

\begin{equation*} \begin{aligned} \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{aligned} \end{equation*}

Hence \(\lim_{n\to\infty}\int_E f_n\) exists in \([0,\infty]\) and equals \(\int_E f\).

Take \(f_n=\chi_{[n,\infty)}\) on \(E=\mathbb{R}\). Since \([n+1,\infty)\subseteq[n,\infty)\) the sequence of nonnegative measurable functions \(\{f_n\}\) is decreasing, and it converges pointwise on \(\mathbb{R}\) to \(f\equiv 0\) because \(f_n(x)=0\) once \(n>x\). But for every \(k\) the function \(\chi_{[n,n+k]}\) is a bounded measurable competitor of finite support with \(0\le\chi_{[n,n+k]}\le f_n\) and integral \(k\), so

\begin{equation*} \int_{\mathbb{R}} f_n=\infty \ \text{ for every } n, \qquad \int_{\mathbb{R}} f=0 , \end{equation*}

and the conclusion of the Monotone Convergence Theorem fails.

Method (2): on a set of finite measure. On \(E=(0,1)\) take \(f_n(x)=\tfrac{1}{x}\cdot\chi_{(0,1/n)}(x)\), again a decreasing sequence of nonnegative measurable functions with pointwise limit \(f\equiv 0\). Fix \(n\), choose \(N\) with \(2^{-N}<1/n\), and put \(A_j=\bigl(2^{-(j+1)},2^{-j}\bigr)\) for \(j\ge N\); these are disjoint subsets of \((0,1/n)\) of measure \(2^{-(j+1)}\) on which \(1/x>2^{j}\), so \(h_M=\sum_{j=N}^{M}2^{j}\chi_{A_j}\) is a competitor with \(0\le h_M\le f_n\) and

\begin{equation*} \int_{(0,1)} h_M=\sum_{j=N}^{M}2^{j}\cdot 2^{-(j+1)}=\frac{M-N+1}{2} , \end{equation*}

unbounded in \(M\). Hence \(\int_{(0,1)} f_n=\infty\ne 0=\int_{(0,1)} f\).

Put \(g_n=\inf_{k\ge n}f_k\) and \(g=\liminf_{n\to\infty}f_n\), so that \(g=\lim_n g_n\) pointwise on \(E\). Each \(g_n\) is nonnegative and measurable, since 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*}

is a countable intersection of measurable sets; and \(\{g_n\}\) increases to \(g\), the infimum over the smaller index set \(\{k\ge n+1\}\) being at least that over \(\{k\ge n\}\). For every \(k\ge n\) we have \(0\le g_n\le f_k\) on \(E\), so monotonicity of integration gives \(\int_E g_n\le\inf_{k\ge n}\int_E f_k\); and the Monotone Convergence Theorem applies to \(\{g_n\}\uparrow g\), whence

\begin{equation*} \begin{aligned} \int_E\liminf_{n\to\infty}f_n &= \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{aligned} \end{equation*}

For any nonnegative measurable \(h\) on \(E\), Theorem 11 of Chapter 4 (Additivity Over Domains of Integration) applied to the disjoint decomposition \(E=C\cup(E\setminus C)\), together with \(\chi_C\equiv 1\) on \(C\) and \(h\cdot\chi_C\equiv 0\) on \(E\setminus C\), gives

\begin{equation*} \int_E h\cdot\chi_C=\int_C h\cdot\chi_C+\int_{E\setminus C} h\cdot\chi_C=\int_C h ; \end{equation*}

call this \((\ast)\). Since \(|f\cdot\chi_C|\le|f|\) with \(|f|\) integrable over \(E\), the Integral Comparison Test (Proposition 16 of Chapter 4) makes \(f\cdot\chi_C\) integrable over \(E\); and \((\ast)\) with \(h=|f|\) gives \(\int_C|f|=\int_E|f|\cdot\chi_C\le\int_E|f|<\infty\), so \(f\) is integrable over \(C\) too. Because \(\chi_C\) takes only the values \(0\) and \(1\),

\begin{equation*} (f\cdot\chi_C)^{+}=f^{+}\cdot\chi_C, \qquad (f\cdot\chi_C)^{-}=f^{-}\cdot\chi_C \ \text{ on } E , \end{equation*}

so applying \((\ast)\) to the nonnegative measurable functions \(f^{+}\) and \(f^{-}\), all four integrals being finite,

\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*}

No to both questions: integrability implies absolute convergence of \(\sum_n a_n\), but neither converse holds.

Each \(a_n\) is a real number, since \(f\) is measurable and bounded on the finite-measure set \([n,n+1]\). Suppose \(f\) is integrable over \([1,\infty)\). With \(u_n=|f|\cdot\chi_{[n,n+1)}\) the disjoint decomposition \([1,\infty)=\bigcup_n[n,n+1)\) gives \(|f|=\sum_n u_n\) pointwise, so Corollary 12 and Theorem 11 of Chapter 4 give

\begin{equation*} \int_{[1,\infty)}|f|=\sum_{n=1}^{\infty}\int_{[n,n+1)}|f| , \end{equation*}

while the Integral Comparison Test (Proposition 16) together with (18) of Proposition 15, which lets the null set \(\{n+1\}\) be excised so that \(a_n=\int_{[n,n+1)}f\), gives \(|a_n|\le\int_{[n,n+1)}|f|\). Hence \(\sum_n|a_n|\le\int_{[1,\infty)}|f|<\infty\).

(i) Convergence of \(\sum_n a_n\) does not imply integrability. Take \(f=(-1)^n/n\) on \([n,n+1)\), so \(|f|\le 1\) and \(a_n=(-1)^n/n\); the alternating series \(\sum_n(-1)^n/n\) converges, yet by the display above \(\int_{[1,\infty)}|f|=\sum_n 1/n=\infty\).

(ii) Absolute convergence does not imply integrability either. Take \(f=1\) on \([n,n+1/2)\) and \(f=-1\) on \([n+1/2,n+1)\), so \(|f|\equiv 1\); by Corollary 18 of Chapter 4, \(a_n=\tfrac12-\tfrac12=0\) for every \(n\), so \(\sum_n|a_n|=0\), yet \(\int_{[1,\infty)}|f|=\sum_n 1=\infty\).

Everything follows from the generalized Fatou inequality of Problem 27, \(\int_E\liminf h_n\le\liminf\int_E h_n\) for nonnegative measurable \(\{h_n\}\), applied to \(\{g+f_n\}\) and to \(\{g-f_n\}\).

First excise the null set \(Z\) formed by the points where some \(|f_n|>g\) together with the points where \(g=\infty\); it is null by Proposition 13 of Chapter 4 and countable subadditivity, and by (15) and (18) of Proposition 15 no integral below is changed, so we may assume \(|f_n|\le g\) on \(E\) for all \(n\) and \(g\) finite on \(E\). Then each \(f_n\), and likewise \(\liminf f_n\) and \(\limsup f_n\), is measurable and dominated in absolute value by the integrable \(g\), hence integrable over \(E\) by the Integral Comparison Test (Proposition 16); so all four quantities in the assertion are finite and the middle inequality is automatic. Also \(g\pm f_n\) is properly defined (\(g\) being finite), measurable, nonnegative and integrable.

Since \(g\) is finite, \(\liminf(g+f_n)=g+\liminf f_n\) pointwise on \(E\), so Problem 27 and the linearity of Theorem 17 of Chapter 4 give

\begin{equation*} \int_E g+\int_E\liminf f_n\le\liminf\int_E (g+f_n)=\int_E g+\liminf\int_E f_n , \end{equation*}

and cancelling the finite \(\int_E g\) leaves \(\int_E\liminf f_n\le\liminf\int_E f_n\). Likewise \(\liminf(g-f_n)=g-\limsup f_n\), so

\begin{equation*} \int_E g-\int_E\limsup f_n\le\liminf\int_E (g-f_n)=\int_E g-\limsup\int_E f_n , \end{equation*}

using \(\liminf(-\int_E f_n)=-\limsup\int_E f_n\); cancelling \(\int_E g\) and multiplying by \(-1\) gives \(\limsup\int_E f_n\le\int_E\limsup f_n\).

Any two admissible decompositions give the same value. Note first that for a single decomposition \(f=g+h\) the proposed value is unambiguous: \(h=f-g\) involves no \(\infty-\infty\) since \(g\) is finite and is measurable, so \(\int_E h\in[0,\infty]\) while \(\int_E g\in\mathbb{R}\), and the sum lies in \((-\infty,\infty]\).

Let \(f=g_1+h_1=g_2+h_2\) on \(E\) with \(g_i\) finite and integrable and \(h_i\ge 0\) measurable. Adding the finite nonnegative function \(g_1^{-}+g_2^{-}\) to both sides and using \(g_i+g_i^{-}=g_i^{+}\) gives

\begin{equation*} h_1+g_1^{+}+g_2^{-}=h_2+g_2^{+}+g_1^{-} \quad\text{on } E , \end{equation*}

an identity between nonnegative measurable functions: where \(h_1\) and \(h_2\) are finite it is that elementary rearrangement, and if \(h_1(x)=\infty\) then \(f(x)=\infty\) and hence \(h_2(x)=\infty\) as well (both \(g_i(x)\) being finite), so both sides equal \(\infty\) there. Integrating, 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 in \([0,\infty]\). The four integrals of \(g_i^{\pm}\) are finite by Proposition 14 of Chapter 4, so subtracting the finite number \(\int_E g_1^{-}+\int_E g_2^{-}\) is legitimate and yields

\begin{equation*} \int_E h_1+\int_E g_1=\int_E h_2+\int_E g_2 , \end{equation*}

which is the asserted independence.

Since \(\int_E g_n\to\int_E g<\infty\), all but finitely many \(g_n\) are integrable over \(E\); discarding those finitely many terms, which affects neither hypothesis nor conclusion, we may assume every \(g_n\) is integrable. Then each \(f_n\) is integrable over \(E\) by the Integral Comparison Test (Proposition 16 of Chapter 4), as is \(g\), and passing to the limit in \(|f_n|\le g_n\) gives \(|f|\le g\) a.e., so the measurable \(f\) is integrable too. Excising the null set on which any of the following fails (a countable union of null sets, so null, and no integral changes, by (15) applied to positive and negative parts), we may assume that on \(E\): \(f_n\to f\) and \(g_n\to g\) pointwise, every \(g_n\) and \(g\) is finite, \(|f_n|\le g_n\) for all \(n\), and \(|f|\le g\).

Then \(g_n-f_n\) and \(g_n+f_n\) are nonnegative measurable functions on \(E\) converging pointwise to \(g-f\) and \(g+f\). Fatou’s Lemma applied to \(\{g_n-f_n\}\), with the linearity of Theorem 17 of Chapter 4 and the finiteness of \(\lim_n\int_E g_n=\int_E g\), gives

\begin{equation*} \begin{aligned} \int_E g-\int_E f &\le \liminf\left(\int_E g_n-\int_E f_n\right) \\ &= \int_E g-\limsup\int_E f_n , \end{aligned} \end{equation*}

so cancelling \(\int_E g\) leaves \(\limsup\int_E f_n\le\int_E f\). Fatou’s Lemma applied to \(\{g_n+f_n\}\) gives, in the same way,

\begin{equation*} \int_E g+\int_E f\le\int_E g+\liminf\int_E f_n , \end{equation*}

so \(\int_E f\le\liminf\int_E f_n\). The two bounds force \(\lim_{n\to\infty}\int_E f_n=\int_E f\).

Both directions follow from the estimates below.

Necessity. At a.e. \(x\in E\) both \(f\) and \(f_n\) are finite, by Proposition 15 of Chapter 4, so the reverse triangle inequality gives \(\bigl||f_n|-|f|\bigr|\le|f-f_n|\) a.e. on \(E\); hence, by the linearity of Theorem 17 and the Integral Comparison Test (Proposition 16),

\begin{equation*} \begin{aligned} \left|\int_E|f_n|-\int_E|f|\right| &\le \int_E\bigl||f_n|-|f|\bigr| \\ &\le \int_E|f-f_n|\longrightarrow 0 . \end{aligned} \end{equation*}

Sufficiency. Put \(h_n=|f-f_n|\) and \(g_n=|f|+|f_n|\) on \(E\). Each \(h_n\) is measurable with \(h_n\to 0\) a.e. on \(E\) (at a.e. point \(f_n\to f\) with \(f\) finite), each \(g_n\) is nonnegative and measurable with \(g_n\to 2|f|\) a.e., and \(|h_n|\le g_n\) at every point where \(f\) and \(f_n\) are finite, hence a.e.; excising a null set, which by (15) changes no integral, we may assume the domination holds on \(E\). The hypothesis supplies the remaining condition:

\begin{equation*} \lim_{n\to\infty}\int_E g_n=\int_E|f|+\int_E|f|=\int_E 2|f|<\infty , \end{equation*}

the right-hand side being the integral of the a.e. limit of \(\{g_n\}\), finite since \(f\) is integrable. So the General Lebesgue Dominated Convergence Theorem (Theorem 19 of Chapter 4) applies to \(\{h_n\}\) dominated by \(\{g_n\}\) and gives \(\int_E|f-f_n|\to\int_E 0=0\).

Put \(f_n=f\cdot\chi_{[-n,n]}\) on \(\mathbb{R}\). This is an increasing sequence of nonnegative measurable functions, since the sets \([-n,n]\) ascend and \(f\ge 0\), and it converges pointwise on \(\mathbb{R}\) to \(f\), since \(f_n(x)=f(x)\) as soon as \(n>|x|\). Additivity over domains for nonnegative measurable functions (Theorem 11 of Chapter 4), applied to \(\mathbb{R}=[-n,n]\cup(\mathbb{R}\sim[-n,n])\), gives \(\int_{\mathbb{R}}f_n=\int_{-n}^{n}f\), because \(f_n=f\) on \([-n,n]\) and \(f_n\equiv 0\) off it. So the Monotone Convergence Theorem yields

\begin{equation*} \lim_{n\to\infty}\int_{-n}^{n} f =\lim_{n\to\infty}\int_{\mathbb{R}} f_n=\int_{\mathbb{R}} f , \end{equation*}

an equality in \([0,\infty]\).

Write \(E=[0,1]\) and \(f_y\) for the function \(x\mapsto f(x,y)\) on \(E\). Each \(f_y\) is measurable with \(|f_y|\le g\), and \(g\) is integrable over \(E\) (and nonnegative, since \(g\ge|f(\cdot,0)|\)), so each \(f_y\) is integrable over \(E\) by the Integral Comparison Test (Proposition 16 of Chapter 4) and \(h(y)=\int_0^1 f(x,y)\,dx\) is a real number for every \(y\in[0,1]\).

First assertion. Let \(\{y_n\}\) be any sequence in \((0,1]\) with \(y_n\to 0\). Then \(f_{y_n}\to f\) pointwise on \(E\), so \(f\) is measurable with \(|f|\le g\), hence integrable over \(E\); and \(\{f_{y_n}\}\) is dominated by the single integrable function \(g\), so the Lebesgue Dominated Convergence Theorem gives \(h(y_n)\to\int_0^1 f(x)\,dx\). As this holds for every such sequence, the sequential criterion for limits gives

\begin{equation*} \lim_{y\to 0}\int_0^1 f(x,y)\,dx=\int_0^1 f(x)\,dx . \end{equation*}

Second assertion. Fix \(y_0\in[0,1]\) and let \(\{y_n\}\) be any sequence in \([0,1]\) with \(y_n\to y_0\). Continuity of \(f(x,\cdot)\) at \(y_0\) gives \(f_{y_n}\to f_{y_0}\) pointwise on \(E\), with \(|f_{y_n}|\le g\) for every \(n\), so the Lebesgue Dominated Convergence Theorem gives \(h(y_n)\to\int_0^1 f(x,y_0)\,dx=h(y_0)\). By the sequential criterion for continuity, \(h\) is continuous at \(y_0\), hence on all of \([0,1]\).

The printed hypotheses do not force the inner integral to be finite: \(f(x,y)=1/x\) on \(0<x\le 1\) with \(f(0,y)=0\) satisfies them all with \(g\equiv 0\), yet \(\int_0^1 f(x,y)\,dx=\infty\). We therefore assume, as intended, that \(f(\cdot,y_0)\) is integrable over \([0,1]\) for some \(y_0\in[0,1]\).

Integrability then holds for every \(y\). For fixed \(x\), the function \(t\mapsto f(x,t)\) is differentiable, hence continuous, on \([0,1]\), so the Mean Value Theorem gives a \(\xi\) between \(y_0\) and \(y\) with \(f(x,y)-f(x,y_0)=\tfrac{\partial f}{\partial y}(x,\xi)\,(y-y_0)\); since \(|y-y_0|\le 1\) and \(|\partial f/\partial y|\le g\) on \(Q\),

\begin{equation*} |f(x,y)|\le|f(x,y_0)|+g(x) \quad\text{for all } x\in[0,1] , \end{equation*}

so \(f(\cdot,y)\) is integrable over \([0,1]\) by the Integral Comparison Test (Proposition 16 of Chapter 4) and \(F(y)=\int_0^1 f(x,y)\,dx\) is a real number.

Now fix \(y\in[0,1]\) and let \(\{y_n\}\) be any sequence in \([0,1]\setminus\{y\}\) with \(y_n\to y\). The difference quotients \(h_n(x)=[f(x,y_n)-f(x,y)]/(y_n-y)\) are measurable, and the same Mean Value Theorem argument gives \(h_n(x)=\tfrac{\partial f}{\partial y}(x,\xi_{n,x})\) for some \(\xi_{n,x}\in[0,1]\), hence \(|h_n|\le g\) on \([0,1]\) for every \(n\). By the definition of the partial derivative, which exists by hypothesis, \(h_n\to\partial f/\partial y(\cdot,y)\) pointwise on \([0,1]\), so that function is measurable with \(|\partial f/\partial y(\cdot,y)|\le g\), hence integrable. The Lebesgue Dominated Convergence Theorem and the linearity of Theorem 17 of Chapter 4 therefore give

\begin{equation*} \begin{aligned} \lim_{n\to\infty}\frac{F(y_n)-F(y)}{y_n-y} &= \lim_{n\to\infty}\int_0^1 h_n(x)\,dx \\ &= \int_0^1\frac{\partial f}{\partial y}(x,y)\,dx . \end{aligned} \end{equation*}

As \(\{y_n\}\) was arbitrary and the limit does not depend on it, the sequential characterization of limits makes \(F\) differentiable at \(y\) (one-sidedly at \(y=0\) and \(y=1\)) with

\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*}

Immediate from the Continuity of Integration, Theorem 21 of Chapter 4. Each \(E_n=E\cap\bigl((-\infty,-n]\cup[n,\infty)\bigr)\) is measurable, \(\{E_n\}\) is descending, \(f\) is integrable over \(E\), and \(\bigcap_{n=1}^{\infty}E_n=\varnothing\) since every \(x\in E\) fails \(|x|\ge n\) once \(n>|x|\); so formula (25) gives

\begin{equation*} \lim_{n\to\infty}\int_{E_n} f=\int_{\varnothing} f=0 , \end{equation*}

and the definition of this limit supplies, for each \(\epsilon>0\), a natural number \(N\) with \(\left|\int_{E_n} f\right|<\epsilon\) whenever \(n\ge N\).

Method (2): put \(g_n=|f|\cdot\chi_{E_n}\) on \(E\). Then \(g_n\) is measurable, \(0\le g_n\le|f|\) with \(|f|\) integrable over \(E\), and \(g_n\to 0\) pointwise on \(E\), so the Lebesgue Dominated Convergence Theorem gives \(\int_{E_n}|f|=\int_E g_n\to 0\); the Integral Comparison Test (Proposition 16) then gives \(\left|\int_{E_n} f\right|\le\int_{E_n}|f|<\epsilon\) for \(n\) large.

In each case the limit exists while \(\int_{[1,\infty)} |f| = \infty\); there is no contradiction, since Theorem 21 of Chapter 4 presupposes integrability.

(i) By additivity over domains, \(\int_1^N f = \sum_{n=1}^{N-1} (-1)^n/n\), so by the alternating series test the limit exists:

\begin{equation*} \lim_{N \to \infty} \int_1^N f = \sum_{n=1}^\infty \frac{(-1)^n}{n} = -\ln 2 . \end{equation*}

But \(|f| = \sum_{n \ge 1} n^{-1} \chi_{[n,n+1)}\) pointwise with each summand nonnegative, so Corollary 12 of Chapter 4 (countable additivity for nonnegative functions, which presupposes no integrability, unlike Theorem 20) gives

\begin{equation*} \int_{[1, \infty)} |f| = \sum_{n=1}^\infty \frac{1}{n}\, m\big([n, n+1)\big) = \sum_{n=1}^\infty \frac{1}{n} = \infty . \end{equation*}

(ii) \(f\) is continuous, hence Riemann integrable on each \([1,b]\), where by Theorem 3 of Chapter 4 the Riemann and Lebesgue integrals agree; integrating by parts,

\begin{equation*} \int_1^b \frac{\sin x}{x}\, dx = \cos 1 - \frac{\cos b}{b} - \int_1^b \frac{\cos x}{x^2}\, dx . \end{equation*}

Here \(|\cos b/b| \le 1/b \to 0\), while \(|\cos x/x^2| \le 1/x^2\) with \(\int_{[1,\infty)} x^{-2}\,dx = 1 < \infty\) (Monotone Convergence Theorem applied to \(x^{-2}\chi_{[1,n]}\), each integral evaluated by Theorem 3 of Chapter 4), so the Dominated Convergence Theorem applied to \((\cos x)x^{-2}\chi_{[1,n]}\), dominated by \(1/x^2\), gives

\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*}

For non-integrability, \(\int_{k\pi}^{(k+1)\pi} |\sin x|\,dx = 2\) and \(x \le (k+1)\pi\) on that interval, so with the nonnegative \(v_k = |f| \cdot \chi_{[k\pi, (k+1)\pi)}\), whose sum is \(\le |f|\) pointwise on \([1,\infty)\), Corollary 12 of Chapter 4 and monotonicity give

\begin{equation*} \int_{[1, \infty)} \frac{|\sin x|}{x}\, dx \ge \sum_{k=1}^\infty \frac{2}{(k+1)\pi} = \frac{2}{\pi} \sum_{k=2}^\infty \frac{1}{k} = \infty . \end{equation*}

No contradiction: Theorem 21 of Chapter 4 requires \(f\) integrable over \(E = [1,\infty)\), and here \(\int_{[1,\infty)} f^+ = \int_{[1,\infty)} f^- = \infty\), so \(\int_{[1,\infty)} f\) is not even defined.

Disjointify and apply the Countable Additivity of Integration (Theorem 20 of Chapter 4). Every integral below is a finite real number, since \(\int_A |f| \le \int_E |f| < \infty\) for each measurable \(A \subseteq E\).

(i) Put \(E_0 = \varnothing\), \(F_k = E_k \sim E_{k-1}\), \(E_\infty = \bigcup_{n} E_n\). The \(F_k\) are disjoint (for \(j < k\), \(F_j \subseteq E_j \subseteq E_{k-1}\) while \(F_k \cap E_{k-1} = \varnothing\)) measurable subsets of \(E_\infty\) with \(\bigcup_{k=1}^n F_k = E_n\) for each \(n\) (induction, using \(E_n \subseteq E_{n+1}\)), hence with union \(E_\infty\). So Theorem 20 of Chapter 4, and then additivity over domains (Corollary 18 of Chapter 4) on each partial sum, give

\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 = \lim_{n \to \infty} \int_{E_n} f , \end{equation*}

which is (24).

(ii) Put \(E_\infty = \bigcap_n E_n\) and \(G_n = E_1 \sim E_n\); the \(G_n\) ascend and, by De Morgan, \(\bigcup_n G_n = E_1 \sim E_\infty\), so part (i) gives \(\int_{E_1 \sim E_\infty} f = \lim_n \int_{E_1 \sim E_n} f\). Additivity over domains applied to the disjoint decompositions \(E_1 = E_\infty \cup (E_1 \sim E_\infty)\) and \(E_1 = E_n \cup (E_1 \sim E_n)\) then yields

\begin{equation*} \int_{E_n} f = \int_{E_1} f - \int_{E_1 \sim E_n} f \longrightarrow \int_{E_1} f - \int_{E_1 \sim E_\infty} f = \int_{E_\infty} f , \end{equation*}

the subtraction being licensed by the finiteness of \(\int_{E_1} f\) — exactly where integrability of \(f\) enters, as \(m(E_1) < \infty\) does in the continuity of measure. This is (25).

\(F\) is properly defined and uniformly continuous, but need not be Lipschitz.

Properly defined: monotonicity gives \(\int_{(-\infty, x]} |f| \le \int_{\mathbb{R}} |f| < \infty\), so \(f\) is integrable over \((-\infty, x]\) and \(F(x) \in \mathbb{R}\).

Continuity: let \(\epsilon > 0\) and let \(\delta > 0\) be supplied by Proposition 23 of Chapter 4 (applicable, \(f\) being integrable over \(\mathbb{R}\)), so that \(\int_A |f| < \epsilon\) whenever \(A \subseteq \mathbb{R}\) is measurable with \(m(A) < \delta\). For \(x < y\) with \(y - x < \delta\), additivity over domains (Corollary 18 of Chapter 4) gives \(F(y) - F(x) = \int_{(x,y]} f\), all terms finite, whence 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 \(m((x, y]) = y - x < \delta\). As \(\delta\) depends only on \(\epsilon\), \(F\) is uniformly continuous.

Not Lipschitz: take \(f = 1/\sqrt{x}\) on \((0,1]\) and \(f = 0\) elsewhere. The functions \(f \cdot \chi_{[1/n,1]}\) increase to \(f\), so by the Monotone Convergence Theorem with Theorem 3 of Chapter 4 evaluating each integral,

\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*}

so \(f\) is integrable over \(\mathbb{R}\), and the same computation on \((0,x]\) gives \(F(x) = 2\sqrt{x}\) for \(0 \le x \le 1\) while \(F(x) = 0\) for \(x \le 0\). A Lipschitz constant \(C\) would force

\begin{equation*} 2\sqrt{x} = |F(x) - F(0)| \le Cx, \quad \text{i.e.} \quad \frac{2}{\sqrt{x}} \le C \quad \text{for all } x \in (0,1] , \end{equation*}

which fails as \(x \to 0^+\).

Take \(f_n = \chi_{[-n,n]}\) on \(E = \mathbb{R}\).

Each \(f_n\) is integrable over \(\mathbb{R}\), with \(\int_{\mathbb{R}} |f_n| = 2n < \infty\), and \(\{f_n\} \to f \equiv 1\) pointwise on all of \(\mathbb{R}\), since \(f_n(x) = 1\) as soon as \(n > |x|\). The family is uniformly integrable over \(\mathbb{R}\): given \(\epsilon > 0\) take \(\delta = \epsilon\), for if \(A\) is measurable with \(m(A) < \delta\) then \(0 \le f_n \le 1\) gives

\begin{equation*} \int_A |f_n| \le \int_A 1 = m(A) < \epsilon \qquad \text{for every } n , \end{equation*}

which is criterion (27). Yet by monotonicity \(\int_{\mathbb{R}} 1 \ge \int_{\mathbb{R}} \chi_{[-n,n]} = 2n\) for every \(n\), so

\begin{equation*} \int_{\mathbb{R}} |f| = \int_{\mathbb{R}} 1 = \infty \end{equation*}

and the limit is not integrable over \(\mathbb{R}\). Every hypothesis of Proposition 25 save \(m(E) < \infty\) holds, and the conclusion fails.

On \(E = [0,1]\) take

\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*}

It is the converse implication that fails; the forward one is the Vitali Convergence Theorem applied to the limit \(h \equiv 0\), and makes no sign assumption.

Each \(h_n\) is simple with \(\int_E |h_n| = n \cdot \frac{1}{2n} + n \cdot \frac{1}{2n} = 1 < \infty\), hence integrable over \(E\), and \(\{h_n\} \to 0\) pointwise on \([0,1]\), since \(h_n(0) = 0\) and \(h_n(x) = 0\) as soon as \(1/n < x\). By linearity the positive and negative parts cancel exactly:

\begin{equation*} \int_E h_n = n \cdot \frac{1}{2n} - n \cdot \frac{1}{2n} = 0 \qquad \text{for every } n , \end{equation*}

so \(\lim_{n \to \infty} \int_E h_n = 0\). But \(\{h_n\}\) is not uniformly integrable over \(E\): take \(\epsilon = 1/2\), let \(\delta > 0\) be arbitrary, choose \(n\) with \(1/(2n) < \delta\), and put \(A_n = (0, 1/(2n))\), so that \(m(A_n) < \delta\) while

\begin{equation*} \int_{A_n} |h_n| = n \cdot \frac{1}{2n} = \frac{1}{2} \not< \epsilon . \end{equation*}

Thus no \(\delta\) responds to \(\epsilon = 1/2\) in criterion (27), and Theorem 26 fails without nonnegativity.

Put \(M = |\alpha| + |\beta| + 1 > 0\) and take \(\delta = \min\{\delta_1, \delta_2\}\), where \(\delta_i\) responds to the challenge \(\epsilon/(2M)\) for \(\{f_n\}\), resp. \(\{g_n\}\).

First, every member \(h\) of a uniformly integrable family is finite a.e.: were \(S = \{x \in E \mid |h(x)| = \infty\}\) of positive measure, the continuity of measure (Theorem 15 of Chapter 2) would give an \(n\) with \(m(S \cap [-n,n]) > 0\), and since \(t \mapsto m(S \cap [-n,t])\) is nondecreasing and (increasing by at most \(t^{\prime} - t\) on \([t,t^{\prime}]\)) continuous, the Intermediate Value Theorem would supply \(A = S \cap [-n,t]\) with \(0 < m(A) < \delta\) for the \(\delta\) responding to \(\epsilon = 1\), forcing \(\int_A |h| = \infty\). So \(\alpha f_n + \beta g_n\) is defined a.e. on \(E\), and since null sets carry no integral, all integrals below are unambiguous.

Now let \(\epsilon > 0\), let \(A \subseteq E\) be measurable with \(m(A) < \delta\), and let \(n\) be any index. By the pointwise bound \(|\alpha f_n + \beta g_n| \le |\alpha|\,|f_n| + |\beta|\,|g_n|\) and the monotonicity and linearity of integration for nonnegative measurable functions (Theorem 10 of Chapter 4),

\begin{equation*} \int_A |\alpha f_n + \beta g_n| \le |\alpha| \int_A |f_n| + |\beta| \int_A |g_n| < M \cdot \frac{\epsilon}{2M} + M \cdot \frac{\epsilon}{2M} = \epsilon , \end{equation*}

using \(|\alpha| \le M\), \(|\beta| \le M\) and \(M > 0\). As \(\delta\) is independent of \(n\) and \(A\), the sequence \(\{\alpha f_n + \beta g_n\}\) is uniformly integrable over \(E\).

Each approximant is obtained from the previous one, “finite support” meaning vanishing off a set of finite measure.

(i) Assume first \(f \ge 0\). Since \(\int_{\mathbb{R}} f\) is by definition the supremum of \(\int_{\mathbb{R}} h\) over bounded measurable \(h\) of finite support with \(0 \le h \le f\), and is finite, there is such an \(h\) with \(\int_{\mathbb{R}} f - \int_{\mathbb{R}} h < \epsilon/2\); as \(\int_{\mathbb{R}} h\) is finite, linearity for nonnegative functions (Theorem 10 of Chapter 4) gives

\begin{equation*} \int_{\mathbb{R}} |f - h| = \int_{\mathbb{R}} f - \int_{\mathbb{R}} h < \frac{\epsilon}{2} . \end{equation*}

Let \(h\) vanish off the finite-measure set \(E_0\) (if \(m(E_0) = 0\) take \(\eta = 0\)). The Simple Approximation Lemma, applied to the bounded measurable \(h\) on \(E_0\) with \(\epsilon^{\prime} = \epsilon/(2m(E_0)+1)\), gives simple \(\varphi \le h \le \psi\) on \(E_0\) with \(\psi - \varphi < \epsilon^{\prime}\), so \(|h - \varphi| < \epsilon^{\prime}\) there; replacing \(\varphi\) by \(\max\{\varphi,0\}\) preserves this, since \(h \ge 0\). Let \(\eta = \varphi\) on \(E_0\) and \(\eta = 0\) elsewhere, a simple function of finite support with

\begin{equation*} \int_{\mathbb{R}} |h - \eta| = \int_{E_0} |h - \varphi| \le \epsilon^{\prime} \cdot m(E_0) < \frac{\epsilon}{2} , \end{equation*}

whence \(\int_{\mathbb{R}} |f - \eta| < \epsilon\). In general \(f^{\pm}\) are nonnegative and integrable (Proposition 14 of Chapter 4), so the above supplies simple \(\eta_1, \eta_2\) of finite support within \(\epsilon/2\) of \(f^+\), \(f^-\); put \(\eta = \eta_1 - \eta_2\), simple (Problem 19 of Chapter 3) of finite support. Since \(f\) is finite a.e. (Proposition 15 of Chapter 4), \(|f - \eta| \le |f^+ - \eta_1| + |f^- - \eta_2|\) a.e., and integrating gives \(\int_{\mathbb{R}} |f - \eta| < \epsilon\).

(ii) Take \(\eta = \sum_{k=1}^p a_k \chi_{E_k}\) from (i) with \(\int_{\mathbb{R}} |f - \eta| < \epsilon/2\), the \(E_k\) disjoint with \(m(E_k) < \infty\). The sets \(E_k \sim [-N,N]\) descend to \(\emptyset\) with \(m(E_k) < \infty\), so by the continuity of measure we may fix \(I = [-N,N]\) with \(\sum_k |a_k| \, m(E_k \sim I) < \epsilon/4\); then \(\eta^{\prime} = \eta \chi_I\) is simple, vanishes off \(I\), and

\begin{equation*} \int_{\mathbb{R}} |\eta - \eta^{\prime}| = \int_{\mathbb{R} \sim I} |\eta| \le \sum_{k=1}^{p} |a_k| \cdot m(E_k \sim I) < \frac{\epsilon}{4} . \end{equation*}

Let \(M = \max_k |a_k|\), so \(|\eta^{\prime}| \le M\). By Problem 18 of Chapter 3 (applicable: \(\eta^{\prime}\) is bounded and measurable on the closed bounded interval \(I\)) there are a step function \(h\) on \(I\) and measurable \(F \subseteq I\) with \(|h - \eta^{\prime}| < \epsilon^{\prime}\) on \(F\) and \(m(I \sim F) < \epsilon^{\prime}\). Its truncation \(\tilde h = \min\{\max\{h,-M\},M\}\) is again a step function, and satisfies \(|\tilde h - \eta^{\prime}| \le |h - \eta^{\prime}|\) pointwise (since \(|\eta^{\prime}| \le M\)) and \(|\tilde h - \eta^{\prime}| \le 2M\) on \(I\), so by additivity over domains

\begin{equation*} \int_I |\eta^{\prime} - \tilde h| \le \epsilon^{\prime} \cdot m(I) + 2M \cdot \epsilon^{\prime} = \epsilon^{\prime}(2N + 2M) < \frac{\epsilon}{4} \end{equation*}

for \(\epsilon^{\prime}\) chosen small enough. Setting \(s = \tilde h\) on \(I\) and \(s = 0\) off \(I\) gives a step function on \(\mathbb{R}\) vanishing off \(I\), and adding the three estimates gives \(\int_{\mathbb{R}} |f - s| < \epsilon/2 + \epsilon/4 + \epsilon/4 = \epsilon\).

(iii) Take \(s\) from (ii) vanishing off \([a,b]\) with \(\int_{\mathbb{R}} |f - s| < \epsilon/2\), say \(s \equiv c_i\) on \((x_{i-1}, x_i)\) for a partition \(a = x_0 < \cdots < x_n = b\), and let \(M = \max_i |c_i|\) (if \(M = 0\) take \(g \equiv 0\)). Fix \(\sigma > 0\) with \(\sigma < \min_i (x_i - x_{i-1})\) and \(2M(n+1)\sigma < \epsilon/2\), so the intervals \(J_i = (x_i - \sigma/2, x_i + \sigma/2)\), \(0 \le i \le n\), are disjoint. Define \(g = s\) off \(\bigcup_i J_i\) and \(g = 0\) off \([a,b]\), and on each \(J_i\) let \(g\) be affine between the two neighbouring constant values (for \(i = 0\) the left value is \(0\), for \(i = n\) the right value is \(0\)). Then \(g\) is continuous (locally constant off the \(\overline{J_i}\), affine on each \(J_i\), the definitions agreeing at their endpoints), vanishes off \([a-\sigma, b+\sigma]\), and \(|g| \le M\), being on each \(J_i\) a convex combination of two values of modulus at most \(M\). Hence \(g = s\) off a set of measure \((n+1)\sigma\) with \(|g - s| \le 2M\), so

\begin{equation*} \int_{\mathbb{R}} |f - g| \le \int_{\mathbb{R}} |f - s| + 2M(n+1)\sigma < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

\(\int_{\mathbb{R}} \hat f = \int_E f\), since extension by \(0\) is an integral-preserving bijection between the test functions competing in the two suprema.

Assume first \(f \ge 0\). If \(h\) is bounded, measurable, of finite support on \(\mathbb{R}\) with \(0 \le h \le \hat f\), then \(h = 0\) off \(E\), so additivity over domains (Theorem 11 of Chapter 4) and the definition of \(\int_E f\) as a supremum give \(\int_{\mathbb{R}} h = \int_E h \le \int_E f\); taking suprema, \(\int_{\mathbb{R}} \hat f \le \int_E f\). Conversely, each such \(h\) on \(E\) extends by \(0\) to a bounded measurable \(\hat h\) of finite support with \(0 \le \hat h \le \hat f\), and \(\int_E h = \int_{\mathbb{R}} \hat h \le \int_{\mathbb{R}} \hat f\), so the reverse inequality holds too.

In general \((\hat f)^{\pm} = \widehat{f^{\pm}}\) and \(|\hat f| = \widehat{|f|}\), so the nonnegative case applied to \(|f|\) gives \(\int_{\mathbb{R}} |\hat f| = \int_E |f| < \infty\), that is, \(\hat f\) is integrable over \(\mathbb{R}\), and applied to \(f^+\) and \(f^-\) it gives

\begin{equation*} \int_{\mathbb{R}} \hat f = \int_E f^+ - \int_E f^- = \int_E f . \end{equation*}

Now let \(\epsilon > 0\). Problem 44(i), applied to the integrable \(\hat f\), gives a simple \(\tilde\eta\) of finite support on \(\mathbb{R}\) with \(\int_{\mathbb{R}} |\hat f - \tilde\eta| < \epsilon\); its restriction \(\eta\) to \(E\) is simple on \(E\), and since \(\hat f = f\) on \(E\) and the integrand is nonnegative, additivity over domains gives

\begin{equation*} \int_E |f - \eta| = \int_E |\hat f - \tilde\eta| \le \int_{\mathbb{R}} |\hat f - \tilde\eta| < \epsilon . \end{equation*}

Problem 44(iii) in place of 44(i) supplies, in exactly the same way, a continuous \(g\) on \(E\) with \(\int_E |f - g| < \epsilon\).

By Problem 44(ii) it suffices to treat a step function vanishing outside a closed, bounded interval. Note throughout that \(|f(x)\cos nx| \le |f(x)|\), so by the Integral Comparison Test (Proposition 16 of Chapter 4) each \(f(x)\cos nx\) is integrable over \(\mathbb{R}\) and every integral below is finite.

(i) Let \(s \equiv c_i\) on \((x_{i-1}, x_i)\), \(1 \le i \le p\), for a partition \(a = x_0 < \cdots < x_p = b\), with \(s = 0\) off \([a,b]\). The partition points form a null set, so by additivity over domains, linearity, and Theorem 3 of Chapter 4 (the integrand being continuous on each bounded interval),

\begin{equation*} \int_{-\infty}^{\infty} s(x) \cos nx \, dx = \sum_{i=1}^{p} c_i \cdot \frac{\sin n x_i - \sin n x_{i-1}}{n} , \end{equation*}

so that \(\left| \int_{\mathbb{R}} s(x)\cos nx\,dx \right| \le \frac{2}{n}\sum_{i=1}^p |c_i| \to 0\).

(ii) In general, let \(\epsilon > 0\) and take by Problem 44(ii) such a step function \(s\) with \(\int_{\mathbb{R}} |f - s| < \epsilon/2\); being bounded of finite support, \(s\) is integrable, so \(f - s\) is integrable (Theorem 17 of Chapter 4) and, by linearity, the Integral Comparison Test and \(|\cos nx| \le 1\),

\begin{equation*} \left| \int_{\mathbb{R}} f(x)\cos nx\,dx - \int_{\mathbb{R}} s(x)\cos nx\,dx \right| \le \int_{\mathbb{R}} |f - s| < \frac{\epsilon}{2} . \end{equation*}

Choosing \(N\) by (i) with \(\left| \int_{\mathbb{R}} s(x)\cos nx\,dx \right| < \epsilon/2\) for \(n \ge N\) gives \(\left| \int_{-\infty}^{\infty} f(x)\cos nx\,dx \right| < \epsilon\) for all \(n \ge N\), and the limit is \(0\).

Write \(h_t(x) = h(x+t)\).

(i) For \(f \ge 0\), the map \(h \mapsto h_t\) is a bijection, with inverse \(k \mapsto k_{-t}\), of the bounded measurable functions of finite support with \(0 \le h \le f\) onto the corresponding family for \(f_t\): translates of measurable sets are measurable (Proposition 10 of Chapter 2) and \(m(E_0 - t) = m(E_0)\) (Theorem 14 of Chapter 2). It preserves integrals, since for simple \(\varphi = \sum_k a_k \chi_{E_k}\) of finite support \(\varphi_t = \sum_k a_k \chi_{E_k - t}\) has the same integral by that translation invariance, while \(\varphi \le h\) if and only if \(\varphi_t \le h_t\), so the sup and inf over simple functions defining the integral of a bounded function of finite support are unchanged. Hence the two suprema agree, \(\int_{\mathbb{R}} f = \int_{\mathbb{R}} f_t\). For general integrable \(f\) we have \((f_t)^{\pm} = (f^{\pm})_t\) and \(|f_t| = |f|_t\), so \(f_t\) is integrable and

\begin{equation*} \int_{-\infty}^{\infty} f(x+t)\,dx = \int_{\mathbb{R}} f^+ - \int_{\mathbb{R}} f^- = \int_{-\infty}^{\infty} f(x)\,dx . \end{equation*}

(ii) Let \(|g| \le M\), \(M > 0\). By (i) each \(f_t\) is integrable, hence so is \(f - f_t\) (Theorem 17 of Chapter 4) and, \(g\) being bounded and measurable, so is \(g \cdot (f - f_t)\) by Problem 48; the Integral Comparison Test gives

\begin{equation*} \left| \int_{\mathbb{R}} g \cdot u \right| \le M \int_{\mathbb{R}} |u| \qquad (\ast) \end{equation*}

for every \(u\) integrable over \(\mathbb{R}\).

(a) \(f\) continuous vanishing off \([-N,N]\). Such an \(f\) is uniformly continuous on all of \(\mathbb{R}\): a \(\delta^{\prime} \in (0,1)\) witnessing uniform continuity on the compact \([-N-1,N+1]\) serves globally, since if \(|x-y| < \delta^{\prime}\) and not both lie in \([-N-1,N+1]\) then both lie off \([-N,N]\), where \(f = 0\). Given \(\epsilon > 0\), pick \(\delta \in (0,1)\) with \(|f(x)-f(y)| < \epsilon/(M(2N+2)+1)\) whenever \(|x-y| < \delta\). For \(|t| < \delta\), \(f - f_t\) vanishes off \([-N-1,N+1]\) (if \(|x| > N+1\) then \(|x+t| > N\)), a set of measure \(2N+2\), so by monotonicity and \((\ast)\),

\begin{equation*} \left| \int_{\mathbb{R}} g \cdot [f - f_t] \right| \le M (2N+2) \cdot \frac{\epsilon}{M(2N+2)+1} < \epsilon . \end{equation*}

(b) General \(f\). Given \(\epsilon > 0\), Problem 44(iii) supplies \(h\) continuous vanishing off a bounded set with \(\int_{\mathbb{R}} |f - h| < \epsilon/(3(M+1))\), and all four functions below are integrable by (i), Theorem 17 of Chapter 4 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 less than \(\epsilon/3\) in absolute value by \((\ast)\); so is the third, since \(|h_t - f_t| = |h-f|_t\) and (i) applies to the nonnegative integrable \(|h-f|\); and so is the second whenever \(|t| < \delta\), by (a) applied to \(h\). Hence for \(0 < |t| < \delta\) the left side has modulus less than \(\epsilon\), and the limit is \(0\).

The dominating function is \(M|f|\), where \(|g| \le M\) on \(E\).

Indeed \(|f \cdot g| = |f|\,|g| \le M|f|\) at every point of \(E\), and \(M|f|\) is integrable over \(E\), with \(\int_E M|f| = M \int_E |f| < \infty\) by linearity (Theorem 10 of Chapter 4). Moreover \(f \cdot g\) is measurable on \(E\): \(f\) is finite off a null set \(E_0\) (Proposition 15 of Chapter 4), the product is measurable on \(E \sim E_0\) by Theorem 6 of Chapter 3, and completeness of Lebesgue measure extends this to \(E\) (Proposition 5 of Chapter 3). So the Integral Comparison Test (Proposition 16 of Chapter 4) gives that \(f \cdot g\) is integrable over \(E\), with

\begin{equation*} \left| \int_E f \cdot g \right| \le \int_E |f \cdot g| \le M \int_E |f| . \end{equation*}

We close the cycle (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (iv) \(\Rightarrow\) (i), using throughout that \(f\) is integrable over every measurable \(B\) (since \(\int_B |f| \le \int_{\mathbb{R}} |f| < \infty\)) and that an integrable function vanishing a.e. on \(B\) has \(\int_B = 0\) (Proposition 15 of Chapter 4).

(i) \(\Rightarrow\) (ii). \(fg\) is integrable by Problem 48 and vanishes a.e., so \(\int_{\mathbb{R}} fg = 0\).

(ii) \(\Rightarrow\) (iii). \(\chi_A\) is measurable and bounded by \(1\), so \(\int_{\mathbb{R}} f\chi_A = 0\); and since \(f\chi_A = f\) on \(A\) and \(0\) off \(A\), additivity over domains (Corollary 18 of Chapter 4) gives \(\int_{\mathbb{R}} f\chi_A = \int_A f\).

(iii) \(\Rightarrow\) (iv). Open sets are measurable.

(iv) \(\Rightarrow\) (i). First, \(\int_A f = 0\) whenever \(m(A) < \infty\): given \(\epsilon > 0\), Proposition 23 of Chapter 4 (applicable, \(f\) being integrable over \(\mathbb{R}\)) gives \(\delta > 0\) with \(\int_B |f| < \epsilon\) whenever \(m(B) < \delta\), and Theorem 11(i) of Chapter 2 gives an open \(\mathcal{O} \supseteq A\) with \(m(\mathcal{O} \sim A) < \delta\), so additivity over domains and the Integral Comparison Test yield

\begin{equation*} \left| \int_A f \right| = \left| \int_{\mathcal{O}} f - \int_{\mathcal{O} \sim A} f \right| = \left| \int_{\mathcal{O} \sim A} f \right| \le \int_{\mathcal{O} \sim A} |f| < \epsilon . \end{equation*}

Now put \(A_n^+ = \{x \in (-n,n) \mid f(x) > 1/n\}\), of measure at most \(2n\); the above and monotonicity give

\begin{equation*} 0 = \int_{A_n^+} f \ge \frac{1}{n}\, m(A_n^+) \ge 0 , \end{equation*}

so \(m(A_n^+) = 0\). Since \(\{f > 0\} = \bigcup_{n=1}^\infty A_n^+\) (choose \(n\) with \(|x| < n\) and \(f(x) > 1/n\), possible also when \(f(x) = \infty\)), countable subadditivity gives \(m\{f > 0\} = 0\); applying this to \(-f\), which satisfies (iv) as well, gives \(m\{f < 0\} = 0\). Hence \(f = 0\) a.e. on \(\mathbb{R}\).

Necessity is the inequality \(\left|\int_A f\right| \le \int_A |f|\); sufficiency comes from splitting \(A\) at the sign of \(f\). Each \(f \in \mathcal{F}\) is integrable over every measurable \(A \subseteq E\), since \(|f|\chi_A \le |f|\), so all integrals below are finite.

Necessary. Let \(\delta\) respond to the \(\epsilon\) challenge in the definition of uniform integrability, and let \(m(A) < \delta\). From \(-|f| \le f \le |f|\) on \(A\), monotonicity of integration (Theorem 17 of Chapter 4) gives \(\left|\int_A f\right| \le \int_A |f| < \epsilon\), so the same \(\delta\) works.

Sufficient. Let \(\delta\) respond to the challenge \(\epsilon/2\) in the stated criterion, let \(f \in \mathcal{F}\), and let \(A \subseteq E\) be measurable with \(m(A) < \delta\). The sets \(A^{+} = \{x \in A \mid f(x) \ge 0\}\) and \(A^{-} = A \sim A^{+}\) are disjoint measurable subsets of \(E\) of measure at most \(m(A) < \delta\), so \(\left|\int_{A^{+}} f\right|\) and \(\left|\int_{A^{-}} f\right|\) are each less than \(\epsilon/2\). Since \(|f| = f\) on \(A^{+}\) with \(\int_{A^{+}} f \ge 0\), and \(|f| = -f\) on \(A^{-}\) with \(\int_{A^{-}} f \le 0\), additivity over domains (Corollary 18 of Chapter 4) gives

\begin{equation*} \int_A |f| = \int_{A^{+}} f - \int_{A^{-}} f = \left| \int_{A^{+}} f \right| + \left| \int_{A^{-}} f \right| < \epsilon . \end{equation*}

As \(\delta\) depends only on \(\epsilon\), \(\mathcal{F}\) is uniformly integrable over \(E\).

Necessity is immediate, \(E \cap \mathcal{U}\) being a measurable subset of \(E\); for sufficiency take \(\delta = \delta_0/2\) and enclose \(A\) in an open set. As in Problem 50, each \(f \in \mathcal{F}\) is integrable over every measurable subset of \(E\), and additivity over domains (Corollary 18 of Chapter 4) applied to the nonnegative \(|f|\) gives \(\int_A |f| \le \int_B |f|\) whenever \(A \subseteq B\) are measurable subsets of \(E\).

Necessary. Let \(\delta\) respond to the \(\epsilon\) challenge in the definition. If \(\mathcal{U}\) is open with \(m(E \cap \mathcal{U}) < \delta\), then \(E \cap \mathcal{U}\) is a measurable subset of \(E\) of measure less than \(\delta\), so \(\int_{E \cap \mathcal{U}} |f| < \epsilon\) for every \(f \in \mathcal{F}\).

Sufficient. Let \(\delta_0\) respond to the \(\epsilon\) challenge in the stated criterion, put \(\delta = \delta_0/2\), and let \(A \subseteq E\) be measurable with \(m(A) < \delta\). By Theorem 11(i) of Chapter 2 there is an open \(\mathcal{U} \supseteq A\) with \(m^{*}(\mathcal{U} \sim A) < \delta_0/2\); since \(E \cap \mathcal{U} \subseteq A \cup (\mathcal{U} \sim A)\), monotonicity and finite subadditivity of outer measure give

\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*}

Hence \(\int_{E \cap \mathcal{U}} |f| < \epsilon\), and as \(A \subseteq E \cap \mathcal{U}\), the monotonicity over domains noted above gives \(\int_A |f| \le \int_{E \cap \mathcal{U}} |f| < \epsilon\) for every \(f \in \mathcal{F}\).

(a) No. (b) Yes. (c) Yes.

(a) Take \(f_n = n \cdot \chi_{(0,\,1/n]}\), so \(\int_0^1 |f_n| = 1\) and \(f_n \in \mathcal{F}\). For the challenge \(\epsilon = 1/2\) and any \(\delta > 0\), choose \(n\) with \(1/n < \delta\) and put \(A_n = (0,1/n]\); then \(m(A_n) < \delta\) while

\begin{equation*} \int_{A_n} |f_n| = n \cdot \frac{1}{n} = 1 > \frac{1}{2} = \epsilon . \end{equation*}

(b) The single integrable function \(1\) dominates the whole family. Each \(f \in \mathcal{F}\) is integrable over \([0,1]\), being measurable (continuity, Proposition 3 of Chapter 3) and bounded on a set of finite measure (Theorem 4 of Chapter 4). Take \(\delta = \epsilon\); if \(A \subseteq [0,1]\) is measurable with \(m(A) < \delta\), then \(|f| \le 1\) on \(A\) and monotonicity for nonnegative functions (Theorem 10 of Chapter 4) gives

\begin{equation*} \int_A |f| \;\le\; \int_A 1 \;=\; m(A) \;<\; \epsilon , \end{equation*}

uniformly over \(\mathcal{F}\).

(c) By Problem 51 it suffices to test the criterion on open sets, and there the hypothesis passes from intervals to open sets. Claim: \(\int_{[0,1] \cap \mathcal{U}} |f| \le m([0,1] \cap \mathcal{U})\) for every open \(\mathcal{U}\) and \(f \in \mathcal{F}\). Put \(\mathcal{V} = (0,1) \cap \mathcal{U}\), so that \(([0,1] \cap \mathcal{U}) \sim \mathcal{V} \subseteq \{0,1\}\) is null and both the measures and (by Theorem 11 of Chapter 4) the integrals over the two sets agree. If \(\mathcal{V} = \emptyset\) the claim is trivial. Otherwise \(\mathcal{V} = \bigcup_k I_k\) is a countable disjoint union of open intervals \(I_k = (a_k,b_k)\) with \([a_k,b_k] \subseteq [0,1]\) (Proposition 9 of Chapter 1), and discarding the null endpoint pairs the defining hypothesis gives \(\int_{I_k} |f| = \int_{a_k}^{b_k} |f| \le b_k - a_k = m(I_k)\). Since \(|f| \chi_{\mathcal{V}} \le |f|\) makes \(|f|\) integrable over \(\mathcal{V}\), the Countable Additivity of Integration (Theorem 20 of Chapter 4) and countable additivity of measure give

\begin{equation*} \int_{[0,1] \cap \mathcal{U}} |f| = \int_{\mathcal{V}} |f| = \sum_k \int_{I_k} |f| \le \sum_k m(I_k) = m(\mathcal{V}) , \end{equation*}

which is the claim. Now take \(\delta = \epsilon\): if \(\mathcal{U}\) is open with \(m([0,1] \cap \mathcal{U}) < \delta\) then \(\int_{[0,1] \cap \mathcal{U}} |f| < \epsilon\) for every \(f \in \mathcal{F}\), so \(\mathcal{F}\) is uniformly integrable over \([0,1]\) by Problem 51.

Forward: the Vitali Convergence Theorem of this section, in its version for sets of possibly infinite measure, applies to the a.e. pointwise limit \(h \equiv 0\) and gives \(\lim_{n \to \infty} \int_E h_n = 0\). Converse: the tail is controlled by \(\int_E h_n\) itself and the finitely many initial terms by Proposition 24 of Chapter 4 and Proposition 1 of this section.

So assume \(\lim_{n \to \infty} \int_E h_n = 0\), let \(\epsilon > 0\), and choose \(N\) with \(\int_E h_n < \epsilon\) for \(n \ge N\). Since \(h_n \ge 0\), monotonicity and additivity over domains give, for every measurable \(A \subseteq E\) 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*}

For the initial indices \(1 \le n \le N-1\) (there are none if \(N = 1\); then take \(\delta = 1\), \(E_0 = \emptyset\)), Proposition 23 of Chapter 4 gives \(\delta_n > 0\) with \(\int_A |h_n| < \epsilon\) whenever \(m(A) < \delta_n\), and Proposition 1 of this section gives \(E_n\) of finite measure with \(\int_{E \sim E_n} h_n < \epsilon\); these are applicable since each \(h_n\) is integrable over \(E\). Put

\begin{equation*} \delta = \min\{\delta_1, \dots, \delta_{N-1}\} > 0 , \qquad E_0 = \bigcup_{n=1}^{N-1} E_n , \end{equation*}

the latter of finite measure. Then \(m(A) < \delta\) forces \(\int_A |h_n| < \epsilon\) for every \(n\), by (1) when \(n \ge N\); and \(E \sim E_0 \subseteq E \sim E_n\) gives \(\int_{E \sim E_0} h_n \le \int_{E \sim E_n} h_n < \epsilon\) for \(n < N\), while (1) with \(A = E \sim E_0\) covers \(n \ge N\). Hence \(\{h_n\}\) is uniformly integrable and tight over \(E\).

Take \(\delta = \min\{\delta_1, \dots, \delta_n\}\) and \(E_0 = \bigcup_{k=1}^{n} E_k\), where \(\delta_k\) and \(E_k\) respond to the \(\epsilon\) challenge for \(f_k\) alone.

Each \(f_k\) is integrable over \(E\), so Proposition 23 of Chapter 4 supplies \(\delta_k > 0\) with \(\int_A |f_k| < \epsilon\) whenever \(A \subseteq E\) is measurable with \(m(A) < \delta_k\), and Proposition 1 of this section supplies \(E_k \subseteq E\) of finite measure with \(\int_{E \sim E_k} |f_k| < \epsilon\). Then \(\delta > 0\), being a minimum of finitely many positive numbers, and \(m(E_0) \le \sum_{k=1}^n m(E_k) < \infty\) by finite subadditivity; and for every \(k\),

\begin{equation*} m(A) < \delta \;\Longrightarrow\; \int_A |f_k| < \epsilon , \qquad \int_{E \sim E_0} |f_k| \le \int_{E \sim E_k} |f_k| < \epsilon , \end{equation*}

the second inequality by monotonicity over domains, since \(E \sim E_0 \subseteq E \sim E_k\). These are precisely uniform integrability (Proposition 24 of Chapter 4) and tightness.

Take \(\delta = \min\{\delta_1, \delta_2\}\) and \(E_0 = E_1 \cup E_2\), where \(\delta_i, E_i\) respond to the challenge \(\epsilon/M\) for \(\{f_n\}\), resp. \(\{g_n\}\), with \(M = |\alpha| + |\beta| + 1\).

A family that is uniformly integrable and tight over \(E\) consists of functions integrable over \(E\), hence finite a.e. (Proposition 15 of Chapter 4); off the null set \(Z\) where some \(f_n\) or \(g_n\) is infinite, a countable union of null sets, each \(\alpha f_n + \beta g_n\) is defined, measurable and finite, and no integral is affected. There the triangle inequality and the monotonicity and linearity of integration for nonnegative functions give, for every measurable \(A \subseteq E\),

\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: if \(m(A) < \delta\), then (2) gives, for every index \(n\),

\begin{equation*} \int_A |\alpha f_n + \beta g_n| \le \frac{|\alpha| + |\beta|}{M} \, \epsilon < \epsilon , \end{equation*}

since \(|\alpha| + |\beta| < M\).

Tightness: \(m(E_0) \le m(E_1) + m(E_2) < \infty\), and \(E \sim E_0\) lies in both \(E \sim E_1\) and \(E \sim E_2\), so (2) with \(A = E \sim E_0\) together with monotonicity over domains gives the same bound \(< \epsilon\) for every \(n\).

Necessity: split \(A\) into \(A \cap E_0\) and \(A \sim E_0\). Sufficiency: use the same \(E_0\) and \(\delta\), testing with \(A = E \sim E_0\) for tightness.

Necessary. Given \(\epsilon > 0\), tightness supplies \(E_0 \subseteq E\) of finite measure with \(\int_{E \sim E_0} |f_n| < \epsilon/2\) for all \(n\), and uniform integrability supplies \(\delta > 0\) with \(\int_B |f_n| < \epsilon/2\) for all \(n\) whenever \(B \subseteq E\) is measurable with \(m(B) < \delta\). If \(m(A \cap E_0) < \delta\), then additivity over domains for the nonnegative \(|f_n|\) gives, the first term by the choice of \(\delta\) and the second by \(A \sim E_0 \subseteq E \sim E_0\) and monotonicity,

\begin{equation*} \int_A |f_n| = \int_{A \cap E_0} |f_n| + \int_{A \sim E_0} |f_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

Sufficient. Let \(E_0\) and \(\delta\) be as supplied for \(\epsilon\). If \(A \subseteq E\) is measurable with \(m(A) < \delta\), then \(m(A \cap E_0) \le m(A) < \delta\), so \(\int_A |f_n| < \epsilon\) for every \(n\): uniform integrability. Taking \(A = E \sim E_0\), for which \(m(A \cap E_0) = 0 < \delta\), gives \(\int_{E \sim E_0} |f_n| < \epsilon\) for every \(n\) with \(m(E_0) < \infty\): tightness.

Necessity: take \(\delta\) from uniform integrability at level \(\epsilon/3\) and \(r\) so large that \(m(E_0 \sim (-r,r)) < \delta\). Sufficiency: take \(E_0 = (-r,r)\) and halve \(\delta\).

Necessary. Given \(\epsilon > 0\), let \(\delta > 0\) satisfy \(\int_A |f_n| < \epsilon/3\) for all \(n\) whenever \(A\) is measurable with \(m(A) < \delta\), and let \(E_0\) of finite measure satisfy \(\int_{\mathbb{R} \sim E_0} |f_n| < \epsilon/3\) for all \(n\). The sets \(E_0 \sim (-k,k)\) descend, have finite measure and empty intersection, so the Continuity of Measure (Theorem 15 of Chapter 2) lets us fix \(r > 0\) with \(m(E_0 \sim (-r,r)) < \delta\). Since \(\mathbb{R} \sim (-r,r) \subseteq (\mathbb{R} \sim E_0) \cup (E_0 \sim (-r,r))\), monotonicity and subadditivity over domains give, for all \(n\),

\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| < \frac{2\epsilon}{3} . \end{equation*}

Hence, for \(\mathcal{O}\) open with \(m(\mathcal{O} \cap (-r,r)) < \delta\), splitting \(\mathcal{O}\) at \((-r,r)\) and using additivity over domains, the choice of \(\delta\), and the display above,

\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} + \frac{2\epsilon}{3} = \epsilon . \end{equation*}

Sufficient. Let \(r, \delta\) respond to \(\epsilon\). Tightness: \(E_0 = (-r,r)\) has measure \(2r < \infty\), and \(\mathcal{O} = \mathbb{R} \sim [-r,r]\) is open with \(\mathcal{O} \cap (-r,r) = \emptyset\), so \(\int_{\mathbb{R} \sim E_0} |f_n| = \int_{\mathcal{O}} |f_n| < \epsilon\) for all \(n\), the two domains differing by the null set \(\{-r,r\}\). Uniform integrability: if \(m(A) < \delta/2\), Theorem 11(i) of Chapter 2 gives an open \(\mathcal{O} \supseteq A\) with \(m(\mathcal{O} \sim A) < \delta/2\), whence \(m(\mathcal{O} \cap (-r,r)) \le m(\mathcal{O}) = m(A) + m(\mathcal{O} \sim A) < \delta\) and so, by monotonicity over domains, \(\int_A |f_n| \le \int_{\mathcal{O}} |f_n| < \epsilon\) for all \(n\).

Each direction rests on one inclusion of bad sets. Let \(Z\) be the null set of points of \(E\) at which \(f\) or \(g\) fails to be finite; \(f\) is measurable and finite a.e. by the definition of convergence in measure, and \(g\) by hypothesis.

Sufficiency. Assume \(f = g\) a.e. and put \(W = Z \cup \{x \in E \mid f(x) \ne g(x)\}\), a null set. Off \(W\) the two functions take the same real value, so \(|f_n - g| > \eta\) exactly when \(|f_n - f| > \eta\), whence for each \(\eta > 0\) and each \(n\)

\begin{equation*} \{x \in E \mid |f_n - g| > \eta\} \subseteq W \cup \{x \in E \mid |f_n - f| > \eta\} , \end{equation*}

and monotonicity with countable subadditivity of measure gives \(m\{|f_n - g| > \eta\} \le m\{|f_n - f| > \eta\} \to 0\). As \(g\) is measurable and finite a.e., \(\{f_n\} \to g\) in measure on \(E\).

Necessity. Assume also \(\{f_n\} \to g\) in measure and fix \(\eta > 0\). At a point of \(E \sim Z\) where \(|f - g| > \eta\) the values \(f, g\) are finite, so the triangle inequality forces \(|f - f_n| > \eta/2\) or \(|f_n - g| > \eta/2\) (trivially so if \(f_n\) is infinite there); hence for every \(n\)

\begin{equation*} \{x \in E \mid |f - g| > \eta\} \subseteq Z \cup \{|f_n - f| > \eta/2\} \cup \{|f_n - g| > \eta/2\} . \end{equation*}

The left side is independent of \(n\) while the measures of the last two sets tend to \(0\), so \(m\{|f - g| > \eta\} = 0\) for every \(\eta > 0\). Since \(f\) and \(g\) are finite off \(Z\),

\begin{equation*} \{x \in E \sim Z \mid f(x) \ne g(x)\} = \bigcup_{k=1}^{\infty} \{x \in E \sim Z \mid |f - g| > 1/k\} \end{equation*}

is null by countable subadditivity, and adjoining \(Z\) gives \(f = g\) a.e. on \(E\).

Truncate \(g\) at a level \(k\) chosen by the Continuity of Measure, which is where \(m(E) < \infty\) enters; squares and general products then follow by polarization.

Lemma. If \(\{u_n\} \to u\) and \(\{v_n\} \to v\) in measure on \(E\), then \(\{\alpha u_n + \beta v_n\} \to \alpha u + \beta v\) in measure on \(E\). Indeed, put \(M = |\alpha| + |\beta| + 1\) and let \(Z\) be the null set (a countable union of null sets) off which all of \(u, v, u_n, v_n\) are finite. Off \(Z\) the triangle inequality forces \(|u_n - u| > \eta/(2M)\) or \(|v_n - v| > \eta/(2M)\) wherever the left side below exceeds \(\eta\), since otherwise it 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| > \tfrac{\eta}{2M}\} \cup \{|v_n - v| > \tfrac{\eta}{2M}\} , \end{equation*}

whose measure tends to \(0\) by monotonicity and countable subadditivity; and \(\alpha u + \beta v\) is measurable and finite a.e.

Part 1: \(\{f_n \cdot g\} \to f \cdot g\). Fix \(\eta, \epsilon > 0\). The measurable sets \(E_k = \{x \in E \mid |g(x)| > k\}\) descend with \(\bigcap_k E_k = \{|g| = \infty\}\) null and \(m(E_1) \le m(E) < \infty\), so the Continuity of Measure (Theorem 15 of Chapter 2) supplies \(k\) with \(m(E_k) < \epsilon\). Off \(E_k\) and off the null set \(Z\) where one of \(f, g, f_1, f_2, \dots\) is infinite, the factorization \(f_n g - fg = g[f_n - f]\) is legitimate and \(|g| \le k\), so

\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*}

giving \(m\{|f_n g - fg| > \eta\} \le \epsilon + m\{|f_n - f| > \eta/k\}\) and hence \(\limsup_n m\{|f_n g - fg| > \eta\} \le \epsilon\). As \(\epsilon\) and \(\eta\) were arbitrary and \(f \cdot g\) is measurable and finite a.e., \(\{f_n \cdot g\} \to f \cdot g\) in measure.

Part 2: \(\{f_n^2\} \to f^2\). Off the null set \(Z\) where \(f\) or some \(f_n\) is infinite, \((f_n - f)^2 > \eta\) precisely when \(|f_n - f| > \sqrt{\eta}\), and \(\{f_n - f\} \to 0\) in measure by the Lemma, so \(\{(f_n - f)^2\} \to 0\) in measure; and \(\{f_n f\} \to f^2\) in measure by Part 1 with \(g = f\), \(f\) being measurable and finite a.e. Since \(f_n^2 = (f_n - f)^2 + [2 f f_n - f^2]\) on \(E \sim Z\), the Lemma applied first to \(2 f f_n - f^2\) (limit \(f^2\)) and then to the sum gives \(\{f_n^2\} \to f^2\) in measure.

Part 3: \(\{f_n \cdot g_n\} \to f \cdot g\). By the Lemma \(\{f_n + g_n\} \to f + g\) in measure, and Part 2 used only \(m(E) < \infty\) and convergence in measure to an a.e. finite measurable limit, so it applies alike to \(\{f_n + g_n\}\), \(\{f_n\}\) and \(\{g_n\}\). Two further applications of the Lemma to

\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*}

give \(\{k_n\} \to \tfrac{1}{2}[(f+g)^2 - f^2 - g^2]\) in measure. Off the null set \(W\) where any of \(f, g, f_n, g_n\) is infinite, the identity \((a+b)^2 - a^2 - b^2 = 2ab\) makes \(k_n = f_n g_n\) and the limit \(f \cdot g\), so

\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., \(\{f_n \cdot g_n\} \to f \cdot g\) in measure on \(E\).

Everything follows from the Riesz Theorem (Theorem 4 of this chapter) and the fact that subsequences inherit convergence in measure: \((\ast)\) every subsequence of \(\{f_n\}\) has a further subsequence converging to \(f\) pointwise a.e. on \(E\). We also use the subsequence criterion for real sequences: if every subsequence of \(\{a_n\}\) has a further subsequence converging to \(a\), then \(a_n \to a\) (else \(|a_{n_k} - a| \ge \epsilon_0\) along some subsequence, which then has no further subsequence tending to \(a\)).

(i) Fatou’s Lemma. Some subsequence tends to \(f\) pointwise a.e. (Riesz), so \(f \ge 0\) a.e. and \(\int_E f\) is defined. Put \(L = \liminf_n \int_E f_n\), pick \(\{f_{n_k}\}\) with \(\int_E f_{n_k} \to L\), and by \((\ast)\) a further subsequence converging to \(f\) pointwise a.e.; Fatou’s Lemma of Chapter 4 applied to it 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*}

(ii) Monotone Convergence. \(\{\int_E f_n\}\) increases, say to \(L \in [0,\infty]\), and \(\int_E f \le L\) by (i). Conversely fix \(p\) and take by Riesz a subsequence converging to \(f\) a.e.; at a.e. \(x\), monotonicity gives \(f_p(x) \le f_{n_k}(x)\) once \(n_k \ge p\), so \(f_p \le f\) a.e. and \(\int_E f_p \le \int_E f\). Let \(p \to \infty\): \(L \le \int_E f\).

(iii) Dominated Convergence. A subsequence converging to \(f\) a.e. (Riesz) turns \(|f_{n_k}| \le g\) into \(|f| \le g\) a.e., so \(f\) is integrable by the Integral Comparison Test (Proposition 16 of Chapter 4) and every \(\int_E f_n\), \(\int_E f\) is real. Given any subsequence of \(\{\int_E f_n\}\), \((\ast)\) supplies a further subsequence converging to \(f\) a.e. and dominated by \(g\), whose integrals tend to \(\int_E f\) by the Dominated Convergence Theorem of Chapter 4; the subsequence criterion gives \(\lim_n \int_E f_n = \int_E f\).

(iv) Vitali. Verbatim (iii), with uniform integrability and tightness in place of domination. Each \(f_n\) is integrable over \(E\), this being exactly what uniform integrability together with tightness yields by the remark preceding the general Vitali Convergence Theorem of Section 5.1. Any subsequence has, by \((\ast)\), a further subsequence converging to \(f\) a.e., itself uniformly integrable and tight (the \(\delta\) and the set of finite measure for the whole family serve any subfamily), so the Vitali Convergence Theorem gives that \(f\) is integrable over \(E\) and that its integrals tend to \(\int_E f\); the subsequence criterion then gives \(\lim_n \int_E f_n = \int_E f\).

Take \(E = \mathbb{R}\) and \(f_n = \chi_{[n,\,n+1)}\).

Each \(f_n\) is measurable and real-valued, and \(\{f_n\} \to f \equiv 0\) pointwise on all of \(\mathbb{R}\), since \(f_n(x) = 0\) for every \(n > x\); so every hypothesis of Proposition 3 holds save \(m(E) < \infty\), the limit \(f\) being measurable and finite everywhere. Yet with \(\eta = 1/2\),

\begin{equation*} \{x \in \mathbb{R} \mid |f_n(x) - f(x)| > 1/2\} = [n,\, n+1) , \end{equation*}

of measure \(1\) for every \(n\), so this measure does not tend to \(0\) and \(\{f_n\}\) fails to converge to \(f\) in measure on \(\mathbb{R}\).

If \(\{f_n\} \to f\) and \(\{g_n\} \to g\) in measure on \(E\) and \(\alpha, \beta \in \mathbb{R}\), then \(\{\alpha f_n + \beta g_n\} \to \alpha f + \beta g\) in measure on \(E\); induction extends this to any finite linear combination.

Let \(E_0\) be the set where at least one of \(f, g, f_1, g_1, f_2, g_2, \dots\) is infinite, a countable union of null sets and hence null. Off \(E_0\) all these functions are real-valued, so the combinations are defined, measurable (Theorem 6 of Chapter 3) and finite there, and null sets do not affect the measures occurring in the definition of convergence in measure; so we argue on \(E \sim E_0\). If \(\alpha = \beta = 0\) the claim is trivial.

Fix \(\eta, \epsilon > 0\). By the triangle inequality, at a point of \(E \sim E_0\) where \(|(\alpha f_n + \beta g_n) - (\alpha f + \beta g)| > \eta\) at least one of \(|\alpha| \, |f_n - f|\) and \(|\beta| \, |g_n - g|\) exceeds \(\eta/2\), since otherwise the sum would be at most \(\eta\). So, with \(A_n\) and \(B_n\) the two sets in \(E \sim E_0\) where those quantities exceed \(\eta/2\),

\begin{equation*} \{x \in E \sim E_0 \mid |(\alpha f_n + \beta g_n) - (\alpha f + \beta g)| > \eta\} \subseteq A_n \cup B_n . \end{equation*}

If \(\alpha = 0\) then \(A_n = \emptyset\); otherwise \(A_n = \{|f_n - f| > \eta/(2|\alpha|)\}\), so convergence in measure gives \(N_1\) with \(m(A_n) < \epsilon/2\) for \(n \ge N_1\). Symmetrically \(m(B_n) < \epsilon/2\) for \(n \ge N_2\). Hence for \(n \ge \max\{N_1, N_2\}\), by monotonicity and finite subadditivity of measure,

\begin{equation*} m\{x \in E \mid |(\alpha f_n + \beta g_n) - (\alpha f + \beta g)| > \eta\} \le m(E_0) + m(A_n) + m(B_n) < \epsilon . \end{equation*}

(\(\Rightarrow\)) If \(\{f_n\} \to f\) in measure, so does every subsequence \(\{f_{n_k}\}\), since \(n_k \ge k\) makes the same \(N\) work; the Riesz Theorem (Theorem 4) then extracts from \(\{f_{n_k}\}\) a further subsequence converging to \(f\) pointwise a.e. on \(E\).

(\(\Leftarrow\)) Suppose not: negating the definition, there are \(\eta_0, \epsilon_0 > 0\) such that for every \(N\) some \(n \ge N\) has \(m\{x \in E \mid |f_n(x) - f(x)| > \eta_0\} \ge \epsilon_0\). Applying this with \(N = n_k + 1\) at each stage produces a strictly increasing \(\{n_k\}\) 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 some further subsequence \(\{f_{n_{k_j}}\}\) converges to \(f\) pointwise a.e.; since \(m(E) < \infty\) and \(f\) is finite a.e., Proposition 3 upgrades this to \(\{f_{n_{k_j}}\} \to f\) in measure, so for large \(j\)

\begin{equation*} m\{x \in E \mid |f_{n_{k_j}}(x) - f(x)| > \eta_0\} < \epsilon_0 , \end{equation*}

contradicting \((\dagger)\) at that index. Hence \(\{f_n\} \to f\) in measure on \(E\).

Telescoping is the whole proof: for \(n > m \ge 1\),

\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} < \frac{1}{2^{m-1}} . \tag{\(\ast\)} \end{equation*}

Given \(\epsilon > 0\), choose \(N\) with \(1/2^{N-1} < \epsilon\) (possible since \(2^k > k\)). For \(m, n \ge N\), either \(m = n\) and \(|a_n - a_m| = 0\), or, relabelling so \(n > m \ge N\), \((\ast)\) gives \(|a_n - a_m| < 1/2^{m-1} \le 1/2^{N-1} < \epsilon\). So \(\{a_j\}\) is Cauchy, and by the Cauchy Convergence Criterion (Theorem 17 of Chapter 1) it converges to some \(a \in \mathbb{R}\). Letting \(n \to \infty\) in \((\ast)\) records the rate \(|a - a_m| \le 1/2^{m-1}\), which is the form used in Problem 13.

Take \(f = \lim_j f_{n_j}\) along the rapidly Cauchy subsequence of the hint (and \(f = 0\) where that limit fails to exist).

Applying the Cauchy-in-measure hypothesis with \(\eta = \epsilon = 1/2^j\) gives an index \(N_j\), and \(n_1 = N_1\), \(n_{j+1} = \max\{N_{j+1}, n_j + 1\}\) is strictly increasing with \(n_{j+1} > n_j \ge N_j\); hence with \(E_j = \{x \in E \mid |f_{n_{j+1}}(x) - f_{n_j}(x)| > 1/2^j\}\),

\begin{equation*} m(E_j) < \frac{1}{2^j} , \qquad \sum_{j=1}^{\infty} m(E_j) < 1 < \infty . \end{equation*}

By the Borel-Cantelli Lemma (Chapter 2), \(m(\limsup_j E_j) = 0\). Let \(Z\) be the null set off which every \(f_n\) is real-valued and put \(E^{\prime} = (E \sim Z) \sim \limsup_j E_j\), so \(m(E \sim E^{\prime}) = 0\). For \(x \in E^{\prime}\) there is \(J(x)\) with \(|f_{n_{j+1}}(x) - f_{n_j}(x)| \le 1/2^j\) for all \(j \ge J\); Problem 12, applied to \(b_i = f_{n_{J+i-1}}(x)\) (whose increments are at most \(2^{-(J+i-1)} \le 2^{-i}\)), makes \(\lim_j f_{n_j}(x)\) exist and be real. Define \(f\) as above; it is the pointwise limit on \(E\) of the measurable functions \(f_{n_j} \cdot \chi_{E^{\prime}}\), hence measurable by Proposition 9 of Chapter 3.

With \(F_k = \bigcup_{j \ge k} E_j\), countable subadditivity gives \(m(F_k) < 1/2^{k-1}\), and for \(x \in E^{\prime} \sim F_k\) the telescoping estimate of Problem 12 gives \(|f_{n_i}(x) - f_{n_k}(x)| < 1/2^{k-1}\) for \(i > k\), so letting \(i \to \infty\),

\begin{equation*} \bigl\{ x \in E \bigm| |f_{n_k}(x) - f(x)| > 1/2^{k-1} \bigr\} \subseteq F_k \cup (E \sim E^{\prime}) , \end{equation*}

of measure less than \(1/2^{k-1}\). Given \(\eta, \epsilon > 0\), taking \(1/2^{k-1} < \min\{\eta, \epsilon\}\) shows \(\{f_{n_k}\} \to f\) in measure.

Finally, given \(\eta, \epsilon > 0\), pick \(N\) with \(m\{|f_n - f_p| \ge \eta/2\} < \epsilon/2\) for \(n, p \ge N\) and then \(k\) with \(n_k \ge N\) and \(m\{|f_{n_k} - f| > \eta/2\} < \epsilon/2\). The triangle inequality gives, for \(n \ge N\),

\begin{equation*} \{|f_n - f| > \eta\} \subseteq \{|f_n - f_{n_k}| \ge \eta/2\} \cup \{|f_{n_k} - f| > \eta/2\} \cup Z , \end{equation*}

so \(m\{x \in E \mid |f_n(x) - f(x)| > \eta\} < \epsilon\) for all \(n \ge N\). Hence \(\{f_n\} \to f\) in measure on \(E\).

Everything follows from two properties of \(\varphi(t) = t/(1+t)\) on \([0,\infty)\): it is increasing (\(\varphi^{\prime}(t) = (1+t)^{-2} > 0\)) and \(0 \le \varphi < 1\). In particular \(0 \le \rho(f_n, f) \le m(E) < \infty\). Write \(A_n = \{x \in E \mid |f_n(x) - f(x)| > \eta\}\).

(\(\Rightarrow\)) Assume \(\{f_n\} \to f\) in measure; we may assume \(m(E) > 0\). Given \(\epsilon > 0\), set \(\eta = \epsilon / (2m(E))\). Splitting the integral over \(A_n\) and \(E \sim A_n\) (additivity over domains, Theorem 11 of Chapter 4), bounding \(\varphi \le 1\) on \(A_n\) and \(\varphi(|f_n - f|) \le \varphi(\eta) \le \eta\) a.e. off \(A_n\) by monotonicity,

\begin{equation*} \rho(f_n, f) \le m(A_n) + \eta \cdot m(E) = m(A_n) + \frac{\epsilon}{2} . \end{equation*}

As \(m(A_n) < \epsilon/2\) for \(n\) large, \(\rho(f_n, f) \to 0\).

(\(\Leftarrow\)) Given \(\eta > 0\), monotonicity of \(\varphi\) gives \(\varphi(|f_n - f|) \ge \varphi(\eta)\) a.e. on \(A_n\), whence the Chebychev-type bound

\begin{equation*} \rho(f_n, f) \ge \int_{A_n} \varphi(|f_n - f|) \ge \frac{\eta}{1+\eta} \, m(A_n) , \qquad m(A_n) \le \frac{1+\eta}{\eta} \rho(f_n, f) . \end{equation*}

Since \(\rho(f_n, f) \to 0\) and \((1+\eta)/\eta\) is a fixed constant, \(m(A_n) < \epsilon\) for \(n\) large. Hence \(\{f_n\} \to f\) in measure on \(E\).

The discontinuity sets satisfy \(D_{fg} \subseteq D_f \cup D_g\), since the product of two functions continuous at \(x_0\) is continuous at \(x_0\). Let \(M\) bound both \(|f|\) and \(|g|\); then \(fg\) is bounded by \(M^2\). By Theorem 8 (Lebesgue) of this chapter, applicable because \(f\) and \(g\) are bounded and Riemann integrable, \(m(D_f) = m(D_g) = 0\); subadditivity of outer measure gives \(m^*(D_{fg}) = 0\), so \(D_{fg}\) is measurable of measure zero. The converse half of Theorem 8 now makes \(fg\) Riemann integrable over \([a,b]\).

Method (2): from the Darboux sums. With \(\operatorname{osc}(h,S) = \sup_{x,y \in S}|h(x) - h(y)|\), the estimate \(|f(x)g(x) - f(y)g(y)| \le |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)|\) gives, on taking suprema over \(x, y \in S\),

\begin{equation*} \operatorname{osc}(fg, S) \le M\bigl[\operatorname{osc}(f,S) + \operatorname{osc}(g,S)\bigr] . \end{equation*}

Since \(U(h,P) - L(h,P) = \sum_i \operatorname{osc}(h, I_i)(x_i - x_{i-1})\) over the subintervals of a partition \(P\),

\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*}

Given \(\epsilon > 0\), put \(\eta = \epsilon/(2M+1)\). Riemann integrability of \(f\) supplies partitions whose upper and lower sums both lie within \(\eta/2\) of \(\int_a^b f\), and their common refinement \(P_1\) has \(U(f,P_1) - L(f,P_1) < \eta\) since refinement decreases upper sums and increases lower ones (Problem 2 of Chapter 4); similarly for \(g\) with some \(P_2\). Taking \(P = P_1 \cup P_2\) and refining once more gives \(U(fg,P) - L(fg,P) < 2M\eta < \epsilon\), so the upper and lower Riemann integrals of \(fg\) coincide.

Let \(D\) be the discontinuity set, \(m(D) = 0\). For \(f\) bounded: Theorem 8 (Lebesgue) of this chapter makes \(f\) Riemann integrable over \([a,b]\); Theorem 3 of Chapter 4 then makes it Lebesgue integrable there; and by Theorem 7 of this chapter a bounded function Lebesgue integrable over the finite-measure set \([a,b]\) is measurable.

Method (2), which covers the unbounded case at once. Put \(E = [a,b] \sim D\), measurable since \(D\) is. The restriction \(f|_E\) is continuous on \(E\), so for each \(c \in \mathbb{R}\) the set \(\{x \in E : f(x) > c\}\) is relatively open in \(E\), of the form \(E \cap U\) with \(U\) open, hence measurable; while \(\{x \in D : f(x) > c\} \subseteq D\) is measurable of measure zero by completeness of Lebesgue measure. Hence

\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 \(c\), and boundedness was never used.

Take \(f(x) = x^{-1}\sin(1/x)\) on \((0,1]\), \(f(0) = 0\). On each \([1/n,1]\) the function \(f\) is continuous, hence Riemann integrable, so its Lebesgue integral agrees with its Riemann integral (Theorem 3 of Chapter 4), and \(u = 1/x\) gives

\begin{equation*} a_n := \int_{[1/n,\,1]} f = \int_{1}^{n} \frac{\sin u}{u}\,du . \end{equation*}

Integrating by parts, for \(2 \le n < m\),

\begin{equation*} \Bigl| \int_{n}^{m} \frac{\sin u}{u}\,du \Bigr| = \Bigl| \frac{\cos n}{n} - \frac{\cos m}{m} - \int_{n}^{m}\frac{\cos u}{u^{2}}\,du \Bigr| \le \frac{2}{n} , \end{equation*}

so \(\{a_n\}\) is Cauchy and converges. But the same substitution applied to \(|f|\), together with \(\int_{k\pi}^{(k+1)\pi} |\sin u| u^{-1} du \ge 2/((k+1)\pi)\) summed over the nonoverlapping intervals inside \([1,n]\), gives

\begin{equation*} \int_{[1/n,\,1]} |f| = \int_{1}^{n} \frac{|\sin u|}{u}\,du \ge \frac{2}{\pi} \sum_{k=1}^{K} \frac{1}{k+1} \longrightarrow \infty , \end{equation*}

where \(K\) is the largest integer with \((K+1)\pi \le n\). Since \(|f|\chi_{[1/n,1]}\) increases pointwise to \(|f|\chi_{(0,1]} = |f|\) a.e., the Monotone Convergence Theorem gives \(\int_{[0,1]}|f| = \infty\), so \(f\) is not Lebesgue integrable over \([0,1]\).

No, this cannot happen for \(f \ge 0\): then \(f\chi_{[1/n,1]}\) increases pointwise to \(f\chi_{(0,1]} = f\) a.e., so the Monotone Convergence Theorem gives

\begin{equation*} \int_{[0,1]} f = \lim_{n \to \infty} \int_{[1/n,\,1]} f < \infty , \end{equation*}

and \(|f| = f\) makes \(f\) Lebesgue integrable over \([0,1]\).

Take \(f = \sum_n 2^{-n}\varphi_n\), the jump function of Proposition 2 modified at the left endpoint: since continuity at \(a\) of a function on \([a,b]\) is continuity from the right, a jump placed at \(a\) must jump to the right.

If \(C = \emptyset\), take \(f(x) = x\). Otherwise let \(\{q_n\}_{n \in \Lambda}\) enumerate \(C\), \(\Lambda = \{1,\dots,N\}\) or \(\{1,2,\dots\}\), and for \(n \in \Lambda\) put

\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*}

Each \(\varphi_n\) is \(\{0,1\}\)-valued, increasing on \([a,b]\), and continuous except at \(q_n\). The series converges (dominated by \(\sum 2^{-n} \le 1\)) and \(f\) is increasing, since for \(a \le u < v \le b\) every term of

\begin{equation*} f(v) - f(u) = \sum_{n \in \Lambda} \frac{1}{2^n} \bigl[ \varphi_n(v) - \varphi_n(u) \bigr] \tag{1} \end{equation*}

is nonnegative.

Discontinuity on \(C\): fix \(k \in \Lambda\) and keep only the \(k\)-th term of (1). (i) If \(q_k > a\), then \(f(q_k) - f(x) \ge 2^{-k}\) for every \(x < q_k\) in \([a,b]\), and such \(x\) come arbitrarily close to \(q_k\). (ii) If \(q_k = a\), then \(f(x) - f(a) \ge 2^{-k}\) for every \(x \in (a,b]\). Either way \(f\) is discontinuous at \(q_k\).

Continuity off \(C\): let \(x_0 \notin C\) and \(\epsilon > 0\). Pick \(M\) with \(2^{-M} < \epsilon\) and set \(\delta = \min\{|x_0 - q_n| : n \in \Lambda, n \le M\} > 0\) (take \(\delta = b - a\) if that set is empty), which is positive because \(x_0 \ne q_n\) for all \(n\). Each \(\varphi_n\) with \(n \le M\) is constant on either side of \(q_n\), so it is constant on \((x_0 - \delta, x_0 + \delta)\); those terms of (1) cancel, leaving

\begin{equation*} |f(x) - f(x_0)| \le \sum_{n > M} \frac{1}{2^n} = \frac{1}{2^M} < \epsilon \end{equation*}

for \(|x - x_0| < \delta\).

Take \(f(x) = g(x) + x\), where \(g\) is the increasing function supplied by Problem 6.1 for the countable set \(C = \mathbb{Q} \cap [0,1]\), so \(g\) is continuous exactly at the irrationals of \([0,1]\).

Then \(f\) is strictly increasing, since for \(0 \le u < v \le 1\),

\begin{equation*} f(v) - f(u) = \bigl[g(v) - g(u)\bigr] + (v - u) \ge v - u > 0 . \end{equation*}

And \(f\) has the same continuity points as \(g\): the identity is continuous, so \(f = g + \mathrm{id}\) is continuous wherever \(g\) is, and \(g = f - \mathrm{id}\) is continuous wherever \(f\) is.

The discontinuity set \(D\) injects into \(\mathbb{Q}\) via a disjoint family of jump intervals. Replacing \(f\) by \(-f\) if necessary, assume \(f\) increasing, and 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*}

Monotonicity makes each supremum and infimum a real number squeezed as \(L(x_0) \le f(x_0) \le R(x_0)\).

If \(L(x_0) = f(x_0) = R(x_0)\), then \(f\) is continuous at \(x_0\): given \(\epsilon > 0\), on the left either \(E \cap (x_0 - \delta_1, x_0) = \emptyset\) for some \(\delta_1 > 0\), or some \(z \in E\), \(z < x_0\), has \(f(z) > f(x_0) - \epsilon\), and \(\delta_1 = x_0 - z\) forces \(f(x_0) - \epsilon < f(z) \le f(x) \le f(x_0)\) for \(x \in E \cap (x_0 - \delta_1, x_0)\); symmetrically on the right, and \(\delta = \min\{\delta_1, \delta_2\}\) serves. Hence every \(x_0 \in D\) has \(L(x_0) < R(x_0)\), so \(J(x_0) = (L(x_0), R(x_0))\) is a nonempty open interval.

These intervals are disjoint: for \(x_0 < x_1\) in \(D\) it suffices that \(R(x_0) \le L(x_1)\). (i) If some \(z \in E\) satisfies \(x_0 < z < x_1\), then \(R(x_0) \le f(z)\) (an infimum over a set containing \(f(z)\), or else \(R(x_0) = f(x_0) \le f(z)\)) and symmetrically \(f(z) \le L(x_1)\). (ii) If no point of \(E\) lies strictly between them, then \(R(x_0) = f(x_0) \le f(x_1) = L(x_1)\).

Choosing a rational in each \(J(x_0)\) gives a one-to-one map \(D \to \mathbb{Q}\), so \(D\) is countable.

No in general: with \(E = \{0\} \cup [1,2]\) and \(C = \{0\}\), the point \(0\) is isolated in \(E\) (take \(\delta = 1\)), so every function on \(E\) is continuous at \(0\) and none can be discontinuous exactly on \(C\).

Precisely, such an increasing function exists if and only if no point of \(C\) is isolated in \(E\). Necessity is the argument just given. For sufficiency, assume no \(q \in C\) is isolated; if \(C = \emptyset\) take \(f(x) = x\), and otherwise enumerate \(C = \{q_n\}_{n \in \Lambda}\). Each \(q_n\), not being isolated, is a limit point of \(E \cap (-\infty, q_n)\) or of \(E \cap (q_n, \infty)\); set, 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; each \(\varphi_n\) is \(\{0,1\}\)-valued, increasing on \(E\), and constant on every interval omitting \(q_n\). Put \(f = \sum_{n \in \Lambda} 2^{-n}\varphi_n\), convergent since dominated by \(\sum 2^{-n}\) and increasing since every term is, with

\begin{equation*} f(v) - f(u) \ge \frac{1}{2^k}\bigl[\varphi_k(v) - \varphi_k(u)\bigr] \quad \text{for } u < v \text{ in } E . \tag{2} \end{equation*}

Discontinuity at \(q_k\): (i) in the first case \(\varphi_k(q_k) - \varphi_k(x) = 1\) for \(x \in E\), \(x < q_k\), so (2) gives \(f(q_k) - f(x) \ge 2^{-k}\), and such \(x\) accumulate at \(q_k\); (ii) in the second case \(f(x) - f(q_k) \ge 2^{-k}\) for \(x \in E\), \(x > q_k\), and again such \(x\) accumulate at \(q_k\). So continuity fails at \(q_k\) for \(\epsilon = 2^{-k}\).

Continuity at \(x_0 \in E \sim C\): given \(\epsilon > 0\), take \(M\) with \(2^{-M} < \epsilon\) and \(\delta = \min\{|x_0 - q_n| : n \in \Lambda, n \le M\} > 0\) (or \(\delta = 1\) if empty). For \(|x - x_0| < \delta\) each \(\varphi_n\) with \(n \le M\) is constant across \(x_0\), so

\begin{equation*} |f(x) - f(x_0)| \le \sum_{n > M} \frac{1}{2^n} = \frac{1}{2^M} < \epsilon . \end{equation*}

Take \(E = [0,1]\) and \(\mathcal{F} = \{[x,x] : x \in [0,1]\}\). With nondegeneracy dropped, \(\mathcal{F}\) covers \(E\) in the sense of Vitali, since \([x,x] \ni x\) has length \(0 < \epsilon\).

But for any finite (indeed any countable) subcollection \(\{I_k\}\), the union \(A = \bigcup_k I_k\) is a countable set of points, so \(m(A) = 0\) and

\begin{equation*} m^*\Bigl[E \sim \bigcup_{k=1}^{n} I_k\Bigr] = m(E) - m(E \cap A) = 1 . \end{equation*}

Hence conclusion (2) of the Vitali Covering Lemma fails for every \(\epsilon \le 1\).

Pass to closures. Let \(\mathcal{F}\) be a collection of nondegenerate general intervals covering \(E\) in the sense of Vitali, \(m^*(E) < \infty\), and put \(\overline{\mathcal{F}} = \{\overline{I} : I \in \mathcal{F}, \ell(I) < \infty\}\); each such \(\overline{I} = [c,d]\) is closed, bounded and nondegenerate with \(\ell(\overline{I}) = \ell(I)\). Since the covering condition only ever calls on intervals of length \(< \epsilon\), hence of finite length, \(\overline{\mathcal{F}}\) covers \(E\) in the sense of Vitali.

The Vitali Covering Lemma, applicable as \(m^*(E) < \infty\), gives a finite disjoint \(\{\overline{I_k}\}_{k=1}^n\) with \(m^*[E \sim \bigcup_k \overline{I_k}] < \epsilon\). The corresponding \(\{I_k\} \subseteq \mathcal{F}\) is disjoint, since \(I_j \cap I_k \subseteq \overline{I_j} \cap \overline{I_k} = \emptyset\), and \(\Delta = \bigcup_k (\overline{I_k} \sim I_k)\) is finite, hence null. As a point of \(E\) in no \(I_k\) lies either in no \(\overline{I_k}\) or in \(\Delta\), monotonicity and subadditivity give

\begin{equation*} m^*\Bigl[E \sim \bigcup_{k=1}^n I_k\Bigr] \le m^*\Bigl[E \sim \bigcup_{k=1}^n \overline{I_k}\Bigr] + m^*(\Delta) < \epsilon . \end{equation*}

No. Take the Takagi function \(T(x) = \sum_{m \ge 0} 2^{-m}\varphi(2^m x)\), where \(\varphi(t) = \operatorname{dist}(t, \mathbb{Z})\); it is continuous on \(\mathbb{R}\) and monotone on no open interval.

Continuity: \(\varphi\) is continuous and \(0 \le 2^{-m}\varphi(2^m x) \le 2^{-m-1}\), so the series converges uniformly by the Weierstrass M-test.

Write \(I_{n,k} = [k/2^n, (k+1)/2^n]\) and let \(x \in I_{n,k}\). For \(m < n\), the point \(2^m x\) ranges over an interval of length \(2^{m-n} \le 1/2\) between consecutive multiples of \(2^{m-n}\), hence inside one \([j/2, (j+1)/2]\) on which \(\varphi\) is affine of slope \(\pm 1\); so each such term is affine on \(I_{n,k}\) of slope \(\pm 1\). For \(m \ge n\), with \(u = 2^n x - k \in [0,1]\), periodicity of \(\varphi\) and integrality of \(2^{m-n}k\) give \(\varphi(2^m x) = \varphi(2^{m-n}u)\), whence \(\sum_{m \ge n} 2^{-m}\varphi(2^m x) = 2^{-n}T(2^n x - k)\). Thus for some \(c_{n,k} \in \mathbb{R}\) and integer \(s_{n,k}\) (the sum of the \(n\) slopes \(\pm 1\)),

\begin{equation*} T(x) = c_{n,k} + s_{n,k}\Bigl( x - \frac{k}{2^n} \Bigr) + \frac{1}{2^n} T(2^n x - k) \quad \text{on } I_{n,k} , \tag{3} \end{equation*}

Halving steers \(s\) to zero: the \(m = n\) term contributes slope \(+1\) on the left half and \(-1\) on the right, so

\begin{equation*} s_{n+1,2k} = s_{n,k} + 1, \qquad s_{n+1,2k+1} = s_{n,k} - 1 , \end{equation*}

and passing \(|s_{n,k}|\) times to the half that decreases \(|s|\) reaches some \(J = I_{N,K} \subseteq I_{n,k}\) with \(s_{N,K} = 0\).

Now let \((\alpha,\beta)\) be a nonempty open interval; as monotonicity passes to subintervals we may take \(\alpha, \beta\) finite. With \(2^{-n} < (\beta - \alpha)/3\) and \(k\) least with \(k/2^n > \alpha\), minimality gives \(k/2^n \le \alpha + 2^{-n}\), so \(I_{n,k} \subseteq (\alpha,\beta)\); take \(J = I_{N,K} \subseteq I_{n,k}\) with \(s_{N,K} = 0\), so that (3) reads \(T(x) = c + 2^{-N}T(2^N x - K)\) on \(J\). Since \(\varphi\) vanishes at integers,

\begin{equation*} T(0) = 0, \qquad T(1/2) = \tfrac{1}{2}, \qquad T(1) = 0 , \end{equation*}

so the endpoints \(p < r\) and midpoint \(q\) of \(J\) satisfy \(T(p) = T( r) = c\) and \(T(q) = c + 2^{-N-1}\). Thus \(T\) is neither increasing nor decreasing on \((\alpha,\beta)\).

For \(\gamma \ge 1/2\) the triangle inequality suffices. Let \(L = \ell(I)\) with midpoint \(c\), pick \(y \in I \cap J\) and let \(x \in J\). Then \(|x - y| \le \ell(J) < L/\gamma \le 2L\) and \(|y - c| \le L/2\), so

\begin{equation*} |x - c| \le |x - y| + |y - c| < 2L + \frac{L}{2} = \frac{5L}{2} , \end{equation*}

that is, \(x \in 5 * I\). Hence \(J \subseteq 5 * I\).

No for \(0 < \gamma < 1/2\). Then \(1/\gamma > 2\); pick \(t\) with \(2 < t < 1/\gamma\) and take \(I = [0,1]\), \(J = [-t, 0]\). The hypotheses hold, since \(\gamma \ell(J) = \gamma t < 1 = \ell(I)\) and \(I \cap J = \{0\}\), but \(5 * I = [-2,3]\) while \(-t < -2\), so \(J \not\subseteq 5 * I\).

(\(\Leftarrow\)) Every \(x \in E\) lies in some \(I_k\) with \(k \ge n\), so \(E \subseteq \bigcup_{k \ge n} I_k\) for every \(n\); monotonicity and countable subadditivity of outer measure, with \(m^*(I) = \ell(I)\) (Proposition 1 of Chapter 2), give

\begin{equation*} m^*(E) \le \sum_{k=n}^{\infty} \ell(I_k) \longrightarrow 0 , \end{equation*}

the tails of a convergent series. Hence \(m^*(E) = 0\), and a set of outer measure zero is measurable of measure zero (Proposition 4 of Chapter 2).

(\(\Rightarrow\)) For each \(n\), the definition of outer measure supplies a countable cover \(E \subseteq \bigcup_k J_{n,k}\) by open bounded intervals with \(\sum_k \ell(J_{n,k}) < 2^{-n}\); adjoin \((0, 2^{-n})\) to make the \(n\)-th collection nonempty, at the cost of the bound \(2^{-n+1}\). The pairs \((n,k)\) form a countably infinite index set; enumerate the family as \(\{I_j\}_{j=1}^{\infty}\). All terms being nonnegative, the sum is order-independent and

\begin{equation*} \sum_{j=1}^{\infty} \ell(I_j) = \sum_{n=1}^{\infty} \sum_k \ell(J_{n,k}) \le \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = 2 < \infty . \end{equation*}

Each \(x \in E\) lies in one interval of every block \(n\), and distinct blocks give distinct \(I_j\), so \(x\) lies in infinitely many \(I_j\).

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 \(f = \sum_{k \ge 1} \varphi_k\), each \(\varphi_k\) increasing with

\begin{equation*} 0 \le \varphi_k(x) \le d_k - c_k \quad \text{on } (a,b) . \tag{\(\ast\)} \end{equation*}

By \((\ast)\) the series is dominated by the convergent \(\sum_k (d_k - c_k)\), so \(f\) is finite, and termwise subtraction of the two convergent series gives, for \(a < x < y < b\),

\begin{equation*} f(y) - f(x) = \sum_{k=1}^{\infty}\bigl[\varphi_k(y) - \varphi_k(x)\bigr] \ge 0 , \tag{\(\ast\ast\)} \end{equation*}

so \(f\) is increasing.

Now fix \(x_0 \in E\) and \(n \in \mathbb{N}\); we show \(\overline{D}f(x_0) = \infty\), which rules out differentiability. Choose \(n\) distinct indices \(k_1, \dots, k_n\) with \(x_0 \in (c_{k_i}, d_{k_i})\), and put \(\delta = \min\{b - x_0, \min_i (d_{k_i} - x_0)\} > 0\). For \(0 < t < \delta\) and \(y = x_0 + t\) each \(k_i\) has \(c_{k_i} < x_0 < y < d_{k_i}\), so the middle branch gives \(\varphi_{k_i}(y) - \varphi_{k_i}(x_0) = t\); discarding the remaining nonnegative terms of \((\ast\ast)\),

\begin{equation*} \frac{f(x_0 + t) - f(x_0)}{t} \ge n \quad \text{for all } 0 < t < \delta , \end{equation*}

whence \(\overline{D}f(x_0) \ge n\) for every \(n\).

Everything is the translation invariance of Lebesgue measure: \(A\) is measurable if and only if \(A + \gamma\) is, with \(m(A + \gamma) = m(A)\) (Proposition 10 of Chapter 2). Write \(I = [\alpha,\beta]\), \(I_\gamma = I + \gamma\) and \((Tg)(t) = g(t+\gamma)\) for \(t \in I\). Since \(\{t \in I : (Tg)(t) > c\} = \{s \in I_\gamma : g(s) > c\} - \gamma\), the map \(T\) is a measurability-preserving linear order-isomorphism from functions on \(I_\gamma\) onto functions on \(I\) (with inverse translation by \(-\gamma\)), carrying simple to simple, bounded to bounded, finite support to finite support, and \(g^{\pm}\) to \((Tg)^{\pm}\).

(i) Simple: \(T\chi_A = \chi_{A - \gamma}\), so for \(\psi = \sum_{i=1}^n a_i\chi_{A_i}\) in canonical form, linearity gives

\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*}

(ii) Bounded measurable: both \(I\) and \(I_\gamma\) have finite measure, so by Theorem 4 of Chapter 4 each integral is the supremum of \(\int \psi\) over simple \(\psi \le\) the function. As \(T\) is an order isomorphism carrying one such family onto the other and preserving integrals by (i), the suprema agree.

(iii) Nonnegative measurable: each integral is the supremum of \(\int h\) over bounded measurable \(h\) of finite support with \(0 \le h \le\) the function; \(T\) carries one family onto the other and preserves integrals by (ii), so the suprema agree (finite or infinite together). In particular \(g\) is integrable over \(I_\gamma\) exactly when \(Tg\) is over \(I\).

(iv) General integrable: \(g^{\pm}\) have finite integrals (Proposition 14 of Chapter 4) and \((Tg)^{\pm} = T(g^{\pm})\), so by (iii),

\begin{equation*} \int_{\alpha}^{\beta} g(t+\gamma)\,dt = \int_{\alpha}^{\beta}(Tg)^{+} - \int_{\alpha}^{\beta}(Tg)^{-} = \int_{\alpha+\gamma}^{\beta+\gamma} g . \end{equation*}

The sign of \(\gamma\) was never used.

For (14), extend \(f\) by the value \(f(b)\) on \((b, b+1]\), so it is integrable over \([a,b+1]\); fix \(0 < h \le 1\) and \(a \le u \le v \le b\). By (iv) with \(\gamma = h\), \(\mathrm{Diff}_h f\) is integrable over \([u,v]\) and by linearity (Theorem 17 of Chapter 4),

\begin{equation*} \int_u^v \mathrm{Diff}_h f = \frac{1}{h}\Bigl[ \int_u^v f(t + h)\, dt - \int_u^v f \Bigr] = \frac{1}{h}\Bigl[ \int_{u + h}^{v + h} f - \int_u^v f \Bigr], \end{equation*}

the last equality by the change of variables formula. Splitting \([u, v+h]\) at \(v\) and at \(u+h\) (Corollary 18 of Chapter 4, legitimate since \(f\) 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*}

so \(\int_{u+h}^{v+h} f - \int_u^v f = \int_v^{v+h} f - \int_u^{u+h} f\). 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).

For \(f = \chi_{\mathbb{Q}}\),

\begin{equation*} \overline{D}f(x) = +\infty , \qquad \underline{D}f(x) = -\infty \qquad \text{for every } x \in \mathbb{R} . \end{equation*}

The bracketed sup and inf in the definitions are monotone in \(h\), so it suffices to exhibit, for each \(h > 0\) and \(M > 0\), difference quotients above \(M\) and below \(-M\) with \(0 < |t| \le h\). Fix \(x, h, M\) and put \(\delta = \min\{h, 1/M\} > 0\).

(i) \(x\) 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*}

Here \(x + t\) is irrational exactly when \(t\) is, and the irrationals are dense: an irrational \(t \in (-\delta, 0)\) gives quotient \(1/|t| > 1/\delta \ge M\), and an irrational \(t \in (0,\delta)\) gives \(-1/t < -M\).

(ii) \(x\) 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 density of the rationals, a rational \(q \in (x, x+\delta)\) gives \(t = q - x \in (0,\delta)\) with quotient \(1/t > 1/\delta \ge M\), and a rational \(q \in (x - \delta, x)\) gives \(t = q - x \in (-\delta, 0)\) with quotient \(1/t < -M\).

Iterate the Vitali Covering Lemma on the leftover set, keeping the new intervals away from the closed union of the old ones.

The key observation: if \(F\) is closed and \(A \subseteq E \sim F\), then \(\mathcal{F}_F = \{I \in \mathcal{F} : I \cap F = \emptyset\}\) is a Vitali covering of \(A\). Indeed, for \(x \in A\) and \(\epsilon > 0\), openness of \(F^c\) gives \(\delta > 0\) with \((x - \delta, x+\delta) \cap F = \emptyset\), and an \(I \in \mathcal{F}\) with \(x \in I\), \(\ell(I) < \min\{\epsilon,\delta\}\) then lies inside \((x-\delta,x+\delta)\), so \(I \in \mathcal{F}_F\).

Construct inductively finite disjoint \(\{I_k\}_{k=1}^{n_j} \subseteq \mathcal{F}\), each stage extending the last, with

\begin{equation*} m^*\Bigl[E \sim \bigcup_{k=1}^{n_j} I_k\Bigr] < \frac{1}{j} . \tag{\(\dagger\)} \end{equation*}

Stage \(1\) is the Vitali Covering Lemma applied to \(E\) (of finite outer measure, Vitali-covered by \(\mathcal{F}\)) with \(\epsilon = 1\). Given stage \(j\), put \(F_j = \bigcup_{k \le n_j} I_k\), closed as a finite union of closed intervals, and \(E_j = E \sim F_j\), with \(m^*(E_j) < 1/j < \infty\) by \((\dagger)\). If \(E_j = \emptyset\) the finite collection already works. Otherwise \(\mathcal{F}_{F_j}\) Vitali-covers \(E_j\) by the observation, and the Vitali Covering Lemma with \(\epsilon = 1/(j+1)\) yields a finite disjoint \(I_{n_j+1},\dots,I_{n_{j+1}}\) in \(\mathcal{F}_{F_j}\) with \(m^*[E_j \sim \bigcup_{k > n_j} I_k] < 1/(j+1)\); adjoining one more interval of \(\mathcal{F}_{F_j}\) meeting \(E_j\) if need be makes \(n_{j+1} > n_j\). These new intervals miss \(F_j\), hence all earlier ones, and since \(E_j = E \sim F_j\),

\begin{equation*} E \sim \bigcup_{k=1}^{n_{j+1}} I_k = E_j \sim \bigcup_{k=n_j+1}^{n_{j+1}} I_k , \end{equation*}

which is \((\dagger)\) for \(j+1\).

The union \(\{I_k\}_{k=1}^{\infty}\) of all stages is a countable subcollection of \(\mathcal{F}\), disjoint because any two members lie in a common finite stage. For each \(j\),

\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^*\Bigl[E \sim \bigcup_{k=1}^{\infty} I_k\Bigr] \leq m^*\Bigl[E \sim \bigcup_{k=1}^{n_j} I_k\Bigr] < \frac{1}{j} . \end{equation*}

Since this holds for every \(j\), \(m^*[E \sim \bigcup_{k=1}^{\infty} I_k] = 0\), as required.

Write \(U = \bigcup_{I \in \mathcal{C}} I\). Since the measurable sets form a \(\sigma\)-algebra and \(U = \bigcup_n A_n\) with \(A_n = U \cap (-n,n)\), it is enough to show each \(A_n\) (of outer measure \(\le 2n < \infty\)) is measurable.

Let \(\mathcal{F}_n\) consist of the closed bounded nondegenerate \(J \subseteq [-n,n]\) lying inside some member of \(\mathcal{C}\). It Vitali-covers \(A_n\): for \(x \in A_n\) and \(\epsilon > 0\), pick \(I = [c,d] \in \mathcal{C}\) with \(c \le x \le d\), \(c < d\), and \(\delta > 0\) with \([x-\delta,x+\delta] \subseteq (-n,n)\); then (i) if \(x < d\), take \(J = [x, x+\eta]\) with \(\eta = \min\{\epsilon/2, \delta, d-x\}\); (ii) if \(x = d\), then \(c < x\) and \(J = [x - \eta, x]\) with \(\eta = \min\{\epsilon/2, \delta, x-c\}\) serves. (Nondegeneracy of \(I\) is exactly what makes \(\eta > 0\).)

For each \(j\) the Vitali Covering Lemma gives a finite disjoint \(\{J^{\,j}_k\}_{k=1}^{m_j} \subseteq \mathcal{F}_n\) with \(m^*[A_n \sim \bigcup_k J^{\,j}_k] < 1/j\). Put \(S = \bigcup_{j,k} J^{\,j}_k\), a countable union of closed intervals, hence measurable, with \(S \subseteq U \cap [-n,n]\). Then \(A_n \cap S = S \cap (-n,n)\) is measurable, while \(A_n \sim S \subseteq A_n \sim \bigcup_k J^{\,j}_k\) for every \(j\) forces \(m^*(A_n \sim S) = 0\), so \(A_n \sim S\) is measurable (Proposition 4 of Chapter 2). Hence

\begin{equation*} A_n = \bigl[S \cap (-n,n)\bigr] \cup (A_n \sim S) \end{equation*}

is measurable, and so is \(U\).

\(\overline{D}f(0) = 1\) and \(\underline{D}f(0) = -1\). Since \(f(0) = 0\), the difference quotient at \(0\) is \(\sin(1/t)\), so

\begin{equation*} \overline{D}f(0) = \lim_{h \to 0}\Bigl[\sup_{0 < |t| \le h}\sin(1/t)\Bigr], \qquad \underline{D}f(0) = \lim_{h \to 0}\Bigl[\inf_{0 < |t| \le h}\sin(1/t)\Bigr]. \end{equation*}

For any \(h > 0\), taking \(k\) large makes \(t_k = (\pi/2 + 2k\pi)^{-1} \le h\) with \(\sin(1/t_k) = 1\), and \(s_k = (-\pi/2 + 2k\pi)^{-1} \le h\) with \(\sin(1/s_k) = -1\); as \(|\sin| \le 1\), the sup is \(1\) and the inf is \(-1\) for every \(h\).

Split \(f\) as a difference of increasing functions and apply Lebesgue’s Theorem. Since \(0 \le g^{\pm} \le |g|\), both \(g^{+}\) and \(g^{-}\) are integrable over \([a,b]\), so

\begin{equation*} f_1(x) = \int_a^x g^{+} , \qquad f_2(x) = \int_a^x g^{-} \end{equation*}

are defined and \(f = f_1 - f_2\) by linearity. Each is increasing: for \(x < y\), additivity over domains gives \(f_i(y) - f_i(x) = \int_x^y g^{\pm} \ge 0\).

By Lebesgue’s Theorem each \(f_i\) is differentiable off a null set \(E_i \subseteq (a,b)\), and \(E = E_1 \cup E_2\) is null. For \(x \notin E\) both difference quotients converge finitely, and

\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*}

converges to \(f_1^{\prime}(x) - f_2^{\prime}(x)\). Hence \(f\) is differentiable a.e. on \((a,b)\).

Extend \(f\) to \([a,b]\) by its endpoint limits and cite Corollary 4. Boundedness makes

\begin{equation*} A = \inf_{(a,b)} f = \lim_{x \to a^{+}} f(x) , \qquad B = \sup_{(a,b)} f = \lim_{x \to b^{-}} f(x) \end{equation*}

finite, the two identifications holding because \(f\) is increasing: given \(\epsilon > 0\), some \(x_0\) has \(f(x_0) < A + \epsilon\), and then \(A \le f(x) \le f(x_0) < A + \epsilon\) on \((a, x_0)\); symmetrically at \(b\).

Set \(\tilde f(a) = A\), \(\tilde f = f\) on \((a,b)\), \(\tilde f(b) = B\); this is increasing on \([a,b]\) by the definitions of \(A\) and \(B\). Corollary 4 gives \(\tilde f\,’\) integrable over \([a,b]\) with \(\int_a^b \tilde f\,’ \le B - A\). Since \(\tilde f = f\) on \((a,b)\) and differentiability is a local property there, \(\tilde f\,’ = f^{\prime}\) a.e. on \((a,b)\) (Lebesgue’s Theorem), and \(\{a,b\}\) is null, so

\begin{equation*} \int_a^b f^{\prime} \le B - A = \lim_{x \to b^{-}} f(x) - \lim_{x \to a^{+}} f(x) , \end{equation*}

which is (15).

The two one-sided test points \(t = \pm h\) do it. Choose \(\delta > 0\) with \((c-\delta, c+\delta) \subseteq (a,b)\) and \(f \ge f( c)\) there, and for \(0 < h \le \delta/2\) write

\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 \(\overline{D}f( c) = \inf_h S(h)\) and \(\underline{D}f( c) = \sup_h I(h)\), \(S\) being nondecreasing and \(I\) nonincreasing in \(h\).

(i) \(t = h > 0\): the numerator \(f(c+h) - f( c) \ge 0\), so \(S(h) \ge 0\) for every \(h\), whence \(\overline{D}f( c) \ge 0\).

(ii) \(t = -h < 0\): again \(f(c-h) - f( c) \ge 0\), so the quotient is \(\le 0\) and \(I(h) \le 0\) for every \(h\), whence \(\underline{D}f( c) \le 0\).

The printed hypothesis must be read one-sidedly, since the Weierstrass function \(W(x) = \sum_{n \ge 0} 2^{-n}\cos(13^{n}\pi x)\) of the Remark in Section 6.2 is continuous with \(\overline{D}W \equiv +\infty\) yet has \(W(0) = 2 > -2 = W(1)\); so we prove the assertion for the right derivate \(D^{+}f( c) = \limsup_{t \to 0^{+}}[f(c+t)-f( c)]/t\), the case of the left derivate following on applying it to \(\tilde{f}(x) = -f(-x)\).

Following the hint, suppose first that \(g\) is continuous on \([a,b]\) with \(D^{+}g \ge \epsilon > 0\) on \((a,b)\), and that \(g(u) > g(v)\) for some \(a \le u < v \le b\). Fix \(\lambda\) with \(g(v) < \lambda < g(u)\) and put \(c = \sup\{x \in [u,v] : g(x) \ge \lambda\}\); that set is nonempty and closed by the continuity of \(g\), so \(g( c) \ge \lambda\), while \(g(v) < \lambda\) forces \(c < v\) and \(g < \lambda\) on \((c,v]\), whence \(g( c) \le \lambda\) by continuity again. Thus \(g( c) = \lambda\), and \(g(u) > \lambda\) gives \(c > u\), so \(c \in (a,b)\). Then 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 \(D^{+}g( c) \le 0\), contradicting \(D^{+}g( c) \ge \epsilon\). Hence every such \(g\) is increasing on \([a,b]\).

Now let \(f\) be as given and let \(\epsilon > 0\). The function \(g(x) = f(x) + \epsilon x\) is continuous with \(D^{+}g = D^{+}f + \epsilon \ge \epsilon > 0\) on \((a,b)\), so the previous paragraph makes \(g\) increasing: for \(a \le u < v \le b\),

\begin{equation*} f(v) - f(u) \ge \epsilon(u-v) \ge -\epsilon\,(b-a) . \end{equation*}

Letting \(\epsilon \to 0^{+}\) gives \(f(v) \ge f(u)\), so \(f\) is increasing.

Everything follows from the identity \(Q_{f+g}(t) = Q_f(t) + Q_g(t)\), where \(Q_{\varphi}(t) = [\varphi(c+t)-\varphi( c)]/t\). Fix \(c \in (a,b)\), let \(h_0 > 0\) satisfy \((c-h_0, c+h_0) \subseteq (a,b)\), and for \(0 < h \le h_0\) put

\begin{equation*} 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 \(\overline{D}\varphi( c) = \lim_{h \to 0^{+}} S_{\varphi}(h)\) and \(\underline{D}\varphi( c) = \lim_{h \to 0^{+}} I_{\varphi}(h)\), both limits existing in \([-\infty,+\infty]\) by monotonicity in \(h\).

For \(0 < |t| \le h\) we have \(Q_{f+g}(t) = Q_f(t) + Q_g(t) \le S_f(h) + S_g(h)\), so \(S_{f+g}(h) \le S_f(h) + S_g(h)\); letting \(h \to 0^{+}\), and noting that the limit of the right side is \(\overline{D}f( c) + \overline{D}g( c)\) whenever that sum is defined,

\begin{equation*} \overline{D}(f+g)( c) \le \overline{D}f( c) + \overline{D}g( c) . \end{equation*}

Replacing suprema by infima reverses each inequality and gives \(I_{f+g}(h) \ge I_f(h) + I_g(h)\), hence

\begin{equation*} \underline{D}(f+g)( c) \ge \underline{D}f( c) + \underline{D}g( c) . \end{equation*}

Finally \(I_{f+g}(h) \le S_{f+g}(h)\) for each \(h\), so in the limit \(\underline{D}(f+g)( c) \le \overline{D}(f+g)( c)\). As \(c \in (a,b)\) was arbitrary, the whole chain holds on \((a,b)\).

Write \(\sigma(t) = g(\gamma+t) - c\) and \(Q(\tau) = [f(c+\tau)-f( c)]/\tau\), so that whenever \(\sigma(t) \neq 0\),

\begin{equation*} \frac{(f \circ g)(\gamma+t) - (f\circ g)(\gamma)}{t} = Q(\sigma(t)) \cdot \frac{\sigma(t)}{t} . \tag{\(\dagger\)} \end{equation*}

Since \(c\) is interior to \([a,b]\) and \(g\) is continuous at \(\gamma\), the composition \(f \circ g\) is defined on a neighbourhood of \(\gamma\); and \(\sigma(t)/t \to g^{\prime}(\gamma)\) as \(t \to 0\). We use twice the elementary rule that if \(v(t) \to m > 0\) then \(\limsup_{t\to0} u(t)v(t) = m \cdot \limsup_{t\to0}u(t)\) in \([-\infty,+\infty]\) (Check!). Put \(S( r) = \sup_{0 < |\tau| \le r} Q(\tau)\), nondecreasing in \(r\) with \(S( r) \to \overline{D}f( c)\) as \(r \to 0^{+}\).

(i) Let \(m = g^{\prime}(\gamma) > 0\) and choose \(\delta > 0\) with \(m/2 \le \sigma(t)/t \le 2m\) for \(0 < |t| \le \delta\); then \(\sigma(t) \neq 0\) there, it has the sign of \(t\), and \(|\sigma(t)| \le 2m|t|\). So \((\dagger)\) applies and the limsup rule gives \(\overline{D}(f\circ g)(\gamma) = m \cdot \limsup_{t \to 0} Q(\sigma(t))\). For \(0 < h \le \delta\) every \(Q(\sigma(t))\) with \(0 < |t| \le h\) belongs to the family \(\{Q(\tau) : 0 < |\tau| \le 2mh\}\), so

\begin{equation*} \sup_{0 < |t| \le h} Q(\sigma(t)) \le S(2mh) ; \end{equation*}

conversely \(\sigma\) is continuous with \(\sigma(-h) < 0 < \sigma(h)\), so by the Intermediate Value Theorem it attains on \([-h,h]\) every value \(\tau\) with \(0 < |\tau| \le \rho(h) := \min\{\sigma(h),\ -\sigma(-h)\} > 0\), necessarily at some \(t \neq 0\) since \(\sigma(0) = 0\); hence

\begin{equation*} \sup_{0 < |t| \le h} Q(\sigma(t)) \ge S(\rho(h)) . \end{equation*}

Both \(2mh\) and \(\rho(h)\) tend to \(0^{+}\) with \(h\), so \(\limsup_{t\to0}Q(\sigma(t)) = \overline{D}f( c)\) and \(\overline{D}(f\circ g)(\gamma) = \overline{D}f( c)\cdot g^{\prime}(\gamma)\), an equality in the extended reals.

(ii) Finiteness of \(\overline{D}f( c)\) and \(\underline{D}f( c)\), together with the monotonicity of \(r \mapsto S( r)\) and of \(r \mapsto \inf_{0<|\tau| \le r}Q\), supplies \(\delta_0 > 0\) and \(M < \infty\) with \(|Q(\tau)| \le M\) for \(0 < |\tau| \le \delta_0\). Given \(\eta > 0\), continuity of \(g\) at \(\gamma\) and \(\sigma(t)/t \to 0\) supply \(\delta^{\prime} > 0\) with \(|\sigma(t)| \le \delta_0\) and \(|\sigma(t)|/|t| < \eta/M\) for \(0 < |t| \le \delta^{\prime}\); then by \((\dagger)\) the difference quotient of \(f \circ g\) at \(\gamma\) has modulus at most \(M \cdot \eta/M = \eta\), and it is \(0\) when \(\sigma(t) = 0\). As \(\eta\) was arbitrary, \(f \circ g\) is differentiable at \(\gamma\) with derivative \(0\), so \(\overline{D}(f \circ g)(\gamma) = 0\).

Both assertions are verified through the criterion of Proposition 1 of Chapter 3: \(\{x \in I : f(x) > c\}\) must be measurable for every \(c \in \mathbb{R}\).

Let \(f\) be strictly increasing on the interval \(I\), fix \(c\), and put \(E_c = \{x \in I : f(x) > c\}\). If \(E_c \neq \emptyset\), let \(\alpha = \inf E_c \in [-\infty,\infty)\). Then \(E_c \subseteq [\alpha,\infty) \cap I\) by definition of the infimum; and if \(x \in I\) with \(x > \alpha\), there is \(y \in E_c\) with \(\alpha \le y < x\), so \(f(x) > f(y) > c\) and \(x \in E_c\). Hence

\begin{equation*} (\alpha,\infty) \cap I \subseteq E_c \subseteq [\alpha,\infty) \cap I , \end{equation*}

so \(E_c\) is one of these two intervals, therefore Borel and measurable; and \(f\) is measurable on \(I\).

If \(f\) is merely increasing on \(I\), then \(g(x) = f(x) + x\) is strictly increasing, hence measurable by the above, and \(f = g - x\) is measurable by Theorem 6 of Chapter 3. If \(f\) is decreasing, apply this to \(-f\).

Let \(M \ge 0\) bound \(|\overline{D}f|\) and \(|\underline{D}f|\) on \((a,b)\); we show \(f\) is Lipschitz with constant \(M\). Fix \(\epsilon > 0\) and put \(K = M + \epsilon\). For \(x \in (a,b)\), \(\limsup_{t\to0}[f(x+t)-f(x)]/t = \overline{D}f(x) \le M < K\) and \(\liminf_{t\to0}[f(x+t)-f(x)]/t = \underline{D}f(x) \ge -M > -K\), so there is \(\delta_x > 0\) with \(|[f(x+t)-f(x)]/t| < K\) for \(0 < |t| < \delta_x\), \(x + t \in [a,b]\); multiplying by \(|t|\),

\begin{equation*} |f(y) - f(x)| \le K\,|y-x| \quad \text{for } y \in [a,b],\ |y-x| < \delta_x . \tag{\(\ast\)} \end{equation*}

Let \(a < u < w < b\) and put \(S = \{v \in [u,w] : |f(v)-f(u)| \le K(v-u)\}\), which contains \(u\) and is closed by the continuity of \(f\); let \(s = \sup S \in S\). If \(s < w\) then \(s \in (a,b)\), and for \(s < v < \min\{w, s+\delta_s\}\), \((\ast)\) at \(s\) gives

\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 \(v \in S\) with \(v > s\), a contradiction. Hence \(s = w\), i.e. \(|f(w)-f(u)| \le K(w-u)\) whenever \(a < u < w < b\).

For arbitrary \(u < v\) in \([a,b]\), apply this to \(u_n = \max\{u, a + 1/n\}\) and \(v_n = \min\{v, b - 1/n\}\) and let \(n \to \infty\), using the continuity of \(f\) on \([a,b]\): this gives \(|f(v)-f(u)| \le K(v-u) = (M+\epsilon)(v-u)\). Letting \(\epsilon \to 0^{+}\), \(|f(v)-f(u)| \le M|v-u|\) on \([a,b]\).

\(\int_{0}^{1}|f^{\prime}| = \infty\), so \(f^{\prime}\) is not integrable. For \(0 < x \le 1\),

\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*}

continuous there, while \(|[f(x)-f(0)]/x| = |x\sin(1/x^{2})| \le x\) gives \(f^{\prime}(0) = 0\); so \(f^{\prime}\) is measurable on \([0,1]\). Put

\begin{equation*} a_n = \left[\left(2n+\tfrac12\right)\pi\right]^{-1/2}, \qquad b_n = \left[\left(2n-\tfrac12\right)\pi\right]^{-1/2} , \end{equation*}

so that \(0 < a_n < b_n\) and \(b_{n+1} < a_n \le b_n \le b_1 < 1\); hence the \([a_n,b_n]\) are pairwise disjoint subintervals of \((0,1)\). Since \(\sin(1/a_n^{2}) = 1\) and \(\sin(1/b_n^{2}) = -1\), we get \(f(a_n) = a_n^{2}\) and \(f(b_n) = -b_n^{2}\). On \([a_n,b_n] \subset (0,1]\) the derivative \(f^{\prime}\) is continuous, so its Riemann and Lebesgue integrals agree (Theorem 3 of Chapter 4) and the Fundamental Theorem of Calculus gives

\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{2}{(2n+1)\pi} . \end{equation*}

By Additivity Over Domains of Integration (Theorem 11 of Chapter 4) applied to the disjoint sets \([a_n,b_n] \subseteq [0,1]\), together with the nonnegativity of \(|f^{\prime}|\) off their union,

\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} \qquad \text{for every } N, \end{equation*}

and \(\sum_n 1/(2n+1)\) diverges, since \(1/(2n+1) \ge 1/(3n)\).

No. Let \(w\) be continuous and nowhere differentiable on an open interval \((\alpha,\beta)\) – Weierstrass’s function, as recorded in the Remark following Lebesgue’s Theorem in Section 6.2 – fix \(\alpha < \alpha^{\prime} < \beta^{\prime} < \beta\), and put

\begin{equation*} f(x) = w\bigl(\alpha^{\prime} + (\beta^{\prime}-\alpha^{\prime})x\bigr), \qquad x \in [0,1] . \end{equation*}

Then \(f\) is continuous, being a composition of \(w\) with an affine map, and it is differentiable at no \(x_0 \in (0,1)\): with \(c = \beta^{\prime}-\alpha^{\prime} > 0\) and \(p = \alpha^{\prime} + cx_0 \in (\alpha,\beta)\),

\begin{equation*} \frac{w(p + cs) - w(p)}{cs} = \frac{1}{c}\cdot\frac{f(x_0+s) - f(x_0)}{s} , \end{equation*}

and \(t = cs\) runs over punctured neighbourhoods of \(0\) as \(s\) does, so a limit on the right would be \(w^{\prime}(p)\).

If \(f\) were of bounded variation on some nondegenerate \([a,b] \subseteq [0,1]\), then by Corollary 6 of this section it would be differentiable almost everywhere on \((a,b)\), and \(m((a,b)) = b-a > 0\) yields at least one such point \(x_0\); since the difference quotients of \(f\) and of its restriction agree for all small \(t\), \(f\) would be differentiable at \(x_0 \in (0,1)\) – a contradiction.

No: \(TV(f) = \infty\). Fix \(n\), set \(x_{2k} = k/n\) for \(0 \le k \le n\), and, the irrationals being dense, choose an irrational \(x_{2k+1} \in (k/n, (k+1)/n)\) for \(0 \le k \le n-1\). Then \(P_n = \{0 = x_0 < x_1 < \cdots < x_{2n} = 1\}\) is a partition of \([0,1]\) with \(f(x_i) = 1\) for \(i\) even and \(f(x_i) = 0\) for \(i\) odd, so every consecutive difference has absolute value \(1\):

\begin{equation*} V(f,P_n) = \sum_{i=1}^{2n}|f(x_i) - f(x_{i-1})| = 2n . \end{equation*}

Hence \(TV(f) \ge 2n\) for every \(n\), i.e. \(TV(f) = \infty\), and \(f\) is not of bounded variation on \([0,1]\).

Take

\begin{equation*} h(x) = x + \sin x, \qquad g(x) = x \qquad (0 \le x \le 2\pi) . \end{equation*}

Then \(h - g = \sin = f\); \(g\) is increasing, and so is \(h\), since \(h^{\prime}(x) = 1 + \cos x \ge 0\) on \([0,2\pi]\) and the Mean Value Theorem converts this into \(h(v) - h(u) \ge 0\) for \(u < v\).

Method (2): the canonical Jordan decomposition of Lemma 5, \(f = [f+V] - V\) with \(V(x) = TV(f_{[0,x]})\). As \(\sin\) is increasing on \([0,\pi/2]\), decreasing on \([\pi/2, 3\pi/2]\) and increasing on \([3\pi/2, 2\pi]\), the first Example of this section (the total variation of a monotone function is the absolute increment) together with the additivity (20) gives

\begin{equation*} g(x) = V(x) = \begin{cases} \sin x & 0 \le x \le \pi/2, \\[2pt] 2 - \sin x & \pi/2 \le x \le 3\pi/2, \\[2pt] 4 + \sin x & 3\pi/2 \le x \le 2\pi, \end{cases} \qquad h(x) = \begin{cases} 2\sin x & 0 \le x \le \pi/2, \\[2pt] 2 & \pi/2 \le x \le 3\pi/2, \\[2pt] 4 + 2\sin x & 3\pi/2 \le x \le 2\pi . \end{cases} \end{equation*}

Here \(V\) climbs \(0 \to 1 \to 3 \to 4\) and \(h\) climbs \(0 \to 2 \to 2 \to 4\) across the three subintervals, so both are increasing and agree at the junctions \(\pi/2\) and \(3\pi/2\); and \(h - g = \sin\) on each piece. (Check!)

With \(a = x_0 < x_1 < \cdots < x_n = b\) a partition on whose open intervals \(f\) is constant, write \(c_i\) for the value of \(f\) on \((x_{i-1},x_i)\) and \(d_i = f(x_i)\). Then

\begin{equation*} TV(f) = \sum_{i=1}^{n}\Bigl(|c_i - d_{i-1}| + |d_i - c_i|\Bigr) , \end{equation*}

the sum of the absolute jumps along \(d_0, c_1, d_1, \ldots, c_n, d_n\); in particular every step function is of bounded variation.

Lower bound: choose \(y_i \in (x_{i-1},x_i)\), so that \(P^{*} = \{x_0 < y_1 < x_1 < \cdots < y_n < x_n\}\) is a partition of \([a,b]\) with \(V(f,P^{*}) = \sum_{i}(|c_i - d_{i-1}| + |d_i - c_i|)\).

Upper bound: by Problem 6.31 refinement does not decrease the variation, so it suffices to bound \(V(f,P^{\prime})\) for partitions \(P^{\prime}\) containing every \(x_i\). Fix \(i\) and let \(x_{i-1} = s_0 < s_1 < \cdots < s_k = x_i\) be the points of \(P^{\prime}\) in \([x_{i-1},x_i]\), so that \(f(s_j) = c_i\) for \(0 < j < k\).

(i) If \(k \ge 2\) the inner terms vanish and \(\sum_{j=1}^{k}|f(s_j)-f(s_{j-1})| = |c_i - d_{i-1}| + |d_i - c_i|\).

(ii) If \(k = 1\) the triangle inequality gives \(|d_i - d_{i-1}| \le |c_i - d_{i-1}| + |d_i - c_i|\).

Each consecutive pair of \(P^{\prime}\) lies in exactly one \([x_{i-1},x_i]\), so summing over \(i\) gives \(V(f,P^{\prime}) \le \sum_{i}(|c_i-d_{i-1}| + |d_i-c_i|)\), and the two bounds agree.

(i) No: \(TV(f) = \infty\). Put \(x_n = (n\pi)^{-1/2}\), a sequence strictly decreasing to \(0\) with \(x_1 < 1\), so that \(f(x_n) = (-1)^{n}/(n\pi)\). For \(N \ge 2\), \(P_N = \{-1, 0, x_N, x_{N-1}, \ldots, x_1, 1\}\) is a partition of \([-1,1]\); discarding all terms of \(V(f,P_N)\) but those from the consecutive pairs \(x_{n+1} < x_n\), whose values have opposite signs,

\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} \ \longrightarrow\ \infty . \end{equation*}

(Note \(f\) is continuous on \([-1,1]\), since \(|f(x)| \le x^{2}\).)

(ii) Yes: \(g\) is Lipschitz, hence of bounded variation with \(TV(g) \le 6\). Indeed \(g^{\prime}(x) = 2x\cos(1/x) + \sin(1/x)\) for \(x \neq 0\), while \(|[g(h)-g(0)]/h| = |h\cos(1/h)| \le |h|\) gives \(g^{\prime}(0) = 0\); so \(|g^{\prime}| \le 2+1 = 3\) on \([-1,1]\) and the Mean Value Theorem yields \(|g(v)-g(u)| \le 3|v-u|\). Hence, for every partition \(P = \{x_0,\ldots,x_k\}\),

\begin{equation*} V(g,P) \le 3\sum_{i=1}^{k}(x_i - x_{i-1}) = 3 \cdot 2 = 6 , \end{equation*}

so \(TV(g) \le 6 < \infty\).

Both are of bounded variation. Let \(f, g\) be of bounded variation on \([a,b]\) and let \(P = \{x_0,\ldots,x_k\}\) be any partition. By the triangle inequality applied termwise,

\begin{equation*} \begin{aligned} V(\alpha f + \beta g, P) &\le |\alpha|\,V(f,P) + |\beta|\,V(g,P) \\ &\le |\alpha|\,TV(f) + |\beta|\,TV(g) , \end{aligned} \end{equation*}

a finite bound independent of \(P\), so \(TV(\alpha f + \beta g) \le |\alpha|TV(f) + |\beta|TV(g) < \infty\).

For the product, note first that bounded variation forces boundedness: for \(x \in [a,b]\), \(|f(x)-f(a)| \le V(f,\{a,x,b\}) \le TV(f)\), so \(|f| \le M := |f(a)| + TV(f)\); let \(N\) be the corresponding bound for \(g\). Adding and subtracting \(f(x_i)g(x_{i-1})\),

\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 \(i\) gives \(V(fg,P) \le M\,TV(g) + N\,TV(f)\), again independent of \(P\), so \(TV(fg) \le M\,TV(g) + N\,TV(f) < \infty\).

It suffices to treat \(P = P^{\prime} \cup \{c\}\) with \(c \notin P^{\prime}\), since \(P\) arises from \(P^{\prime}\) by adjoining its finitely many extra points one at a time. Write \(P^{\prime} = \{x_0 < x_1 < \cdots < x_k\}\) and let \(j\) be the unique index with \(x_{j-1} < c < x_j\). The sums \(V(f,P^{\prime})\) and \(V(f,P)\) share all their terms except that \(|f(x_j)-f(x_{j-1})|\) is replaced by the pair \(|f( c)-f(x_{j-1})|\), \(|f(x_j)-f( c)|\), and by the triangle inequality

\begin{equation*} |f(x_j) - f(x_{j-1})| \le |f(x_j)-f( c)| + |f( c)-f(x_{j-1})| . \end{equation*}

Adding the common terms \(\sum_{i \neq j}|f(x_i)-f(x_{i-1})|\) to both sides gives \(V(f,P^{\prime}) \le V(f,P)\). No hypothesis on \(f\) is used.

Take \(P_n = Q_1 \cup Q_2 \cup \cdots \cup Q_n\), where \(Q_n\) is a partition of \([a,b]\) with

\begin{equation*} TV(f) - \frac{1}{n} < V(f,Q_n) \le TV(f) , \end{equation*}

available because \(TV(f) < \infty\) is the supremum of the \(V(f,P)\). Each \(P_n\) is a finite subset of \([a,b]\) containing \(a\) and \(b\), hence a partition; and \(P_n \subseteq P_{n+1}\), \(Q_n \subseteq P_n\), so \(P_{n+1}\) refines \(P_n\) and \(P_n\) refines \(Q_n\). By Problem 6.31,

\begin{equation*} V(f,P_n) \le V(f,P_{n+1}), \qquad TV(f) - \frac{1}{n} < V(f,Q_n) \le V(f,P_n) \le TV(f) . \end{equation*}

Thus \(\{V(f,P_n)\}\) is increasing and \(|V(f,P_n) - TV(f)| < 1/n \to 0\), so it converges to \(TV(f)\).

Fix a partition \(P = \{x_0, x_1, \ldots, x_k\}\) of \([a,b]\). Since \(V(f,P)\) involves only the finitely many values \(f(x_i)\), and \(f_n(x_i) \to f(x_i)\) for each \(i\), the algebra of limits and the continuity of the absolute value give

\begin{equation*} \lim_{n \to \infty} V(f_n,P) = \sum_{i=1}^{k}\bigl|f(x_i) - f(x_{i-1})\bigr| = V(f,P) . \end{equation*}

As \(V(f_n,P) \le TV(f_n)\) for every \(n\), and a convergent sequence coincides with its limit inferior, which is monotone under termwise inequality,

\begin{equation*} V(f,P) = \liminf_{n \to \infty} V(f_n,P) \le \liminf_{n \to \infty} TV(f_n) . \end{equation*}

Taking the supremum over all partitions \(P\) of \([a,b]\) gives \(TV(f) \le \liminf_n TV(f_n)\), an inequality in \([0,\infty]\).

Let \(P = \{x_0, x_1, \ldots, x_k\}\) be an arbitrary partition of \([a,b]\). The triangle inequality applied termwise gives

\begin{equation*} V(f+g,P) \le V(f,P) + V(g,P) \le TV(f) + TV(g) , \end{equation*}

a bound independent of \(P\), so \(TV(f+g) \le TV(f) + TV(g)\). Next, for \(\alpha \in \mathbb{R}\),

\begin{equation*} V(\alpha f, P) = \sum_{i=1}^{k}|\alpha|\,\bigl|f(x_i)-f(x_{i-1})\bigr| = |\alpha|\,V(f,P) , \end{equation*}

so the set of variations of \(\alpha f\) is exactly \(|\alpha| \cdot S\), where \(S = \{V(f,P) : P \text{ a partition of } [a,b]\}\).

(i) If \(\alpha = 0\), both sides of the claimed identity are \(0\).

(ii) If \(\alpha \neq 0\), then \(\sup(cS) = c\,\sup S\) for the positive constant \(c = |\alpha|\) (apply \(s \le \sup S \Rightarrow cs \le c\sup S\), and the same with \(cS\) and \(1/c\)), whence \(TV(\alpha f) = |\alpha|\,TV(f)\).

The case \(\alpha > \beta\). For \(0 < x \le 1\) the product and chain rules give

\begin{equation*} f^{\prime}(x) = \alpha x^{\alpha-1}\sin(1/x^{\beta}) - \beta x^{\alpha-\beta-1}\cos(1/x^{\beta}) , \end{equation*}

continuous on \((0,1]\), hence measurable, and \(|\sin|, |\cos| \le 1\) yield \(|f^{\prime}(x)| \le \alpha x^{\alpha-1} + \beta x^{\alpha-\beta-1}\) there. Both exponents exceed \(-1\) (here \(\alpha - \beta > 0\) is used), so for every \(n\)

\begin{equation*} \int_{1/n}^{1}|f^{\prime}| \le \int_{0}^{1}\alpha x^{\alpha-1}\,dx + \int_{0}^{1}\beta x^{\alpha-\beta-1}\,dx = 1 + \frac{\beta}{\alpha-\beta} , \end{equation*}

the integrands being continuous on \([1/n,1]\), where Riemann and Lebesgue integrals agree. The functions \(|f^{\prime}|\chi_{[1/n,1]}\) are nonnegative, measurable, and increase pointwise on \((0,1]\) to \(|f^{\prime}|\), so the Monotone Convergence Theorem of Chapter 4 gives \(\int_{0}^{1}|f^{\prime}| \le 1 + \beta/(\alpha-\beta) < \infty\): \(f^{\prime}\) is integrable over \([0,1]\).

Bounded variation follows. Let \(0 = t_0 < t_1 < \cdots < t_k = 1\) be a partition. For \(i \ge 2\), \(f\) is continuously differentiable on \([t_{i-1},t_i] \subset (0,1]\), so \(|f(t_i)-f(t_{i-1})| \le \int_{t_{i-1}}^{t_i}|f^{\prime}|\) by the Fundamental Theorem of Calculus; for \(i = 1\), the same estimate on \([s,t_1]\) with \(s \to 0^{+}\), together with the continuity of \(f\) at \(0\) (from \(|f(x)| \le x^{\alpha}\)) and \(f(0) = 0\), gives \(|f(t_1)-f(t_0)| \le \int_{0}^{t_1}|f^{\prime}|\). Summing and using additivity of the integral,

\begin{equation*} V(f,P) \le \int_{0}^{1}|f^{\prime}| \le 1 + \frac{\beta}{\alpha-\beta} < \infty , \end{equation*}

a bound independent of \(P\), so \(TV(f) < \infty\).

The case \(\alpha \le \beta\). Put \(x_n = (\pi/2 + n\pi)^{-1/\beta}\), strictly decreasing to \(0\) with \(x_0 < 1\), so that \(f(x_n) = (-1)^{n}x_n^{\alpha}\), and let \(P_N = \{0, x_N, x_{N-1}, \ldots, x_0, 1\}\). Consecutive \(f(x_n)\) have opposite signs, so, writing \(p = \alpha/\beta \in (0,1]\) and using \(t^{-p} \ge t^{-1}\) for \(t \ge 1\) together with \(1 \le \pi/2 + n\pi \le \pi(n+1)\),

\begin{equation*} V(f,P_N) \ \ge\ \sum_{n=0}^{N-1}\bigl(x_n^{\alpha} + x_{n+1}^{\alpha}\bigr) \ \ge\ \sum_{n=0}^{N-1}\Bigl(\frac{\pi}{2}+n\pi\Bigr)^{-p} \ \ge\ \frac{1}{\pi}\sum_{m=1}^{N}\frac{1}{m} . \end{equation*}

The harmonic series diverges, so \(TV(f) = \infty\) and \(f\) is not of bounded variation on \([0,1]\).

Bisect repeatedly. 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*}

holds in \([0,\infty]\) with no finiteness hypothesis, so \(f\) fails to be of bounded variation on \([a,b]\) precisely when it fails on at least one of the two halves. Starting from \([a_0,b_0] = [0,1]\), let \([a_{n+1},b_{n+1}]\) be a closed half of \([a_n,b_n]\) on which \(f\) fails to be of bounded variation. These intervals are nested with \(b_n - a_n = 2^{-n}\), so by the Nested Set Theorem \(\bigcap_{n \ge 0}[a_n,b_n]\) is a single point \(x_0 \in [0,1]\).

Given \(\delta > 0\), choose \(n\) with \(2^{-n} < \delta\); since \(a_n \le x_0 \le b_n\), every point of \([a_n,b_n]\) is within \(\delta\) of \(x_0\), so \([a_n,b_n] \subseteq J_{\delta} := [x_0-\delta, x_0+\delta] \cap [0,1]\). Were \(f\) of bounded variation on \(J_{\delta}\), two applications of (20) (splitting at \(a_n\), then at \(b_n\)) would make it so on \([a_n,b_n]\). Hence \(f\) fails to be of bounded variation on \(J_{\delta}\) for every \(\delta > 0\), and therefore, by (20) again, on every nondegenerate closed \([c,d] \subseteq [0,1]\) containing \(x_0\) in its interior relative to \([0,1]\).

That relative interior cannot be dropped from the printed statement: with \(h\) the continuous function of the Example of Section 6.3, which is not of bounded variation on \([0,1/2]\), the function \(g(x) = h(1/2-x)\) on \([0,1/2)\), \(g \equiv 0\) on \([1/2,1]\), is continuous and not of bounded variation on \([0,1]\), yet every \(x_0 \in [0,1]\) lies in some nondegenerate closed subinterval on which \(g\) is of bounded variation – namely \([1/2,1]\) if \(x_0 \ge 1/2\), and \([0,(x_0+1/2)/2]\) if \(x_0 < 1/2\), on which \(g\) is Lipschitz.

(i) Take \(f(x) = x\cos(\pi/2x)\) for \(0 < x \le 1\) and \(f(0) = 0\). By the Example of Section 6.3 this \(f\) is continuous on \([0,1]\) with \(V(f,P_n) = 1 + 1/2 + \cdots + 1/n\) for \(P_n = \{0, 1/2n, 1/[2n-1], \ldots, 1/3, 1/2, 1\}\), so \(TV(f) = \infty\) and, by Theorem 8, \(f\) is not absolutely continuous on \([0,1]\). Yet on each \([\epsilon,1]\) it is continuously differentiable, with

\begin{equation*} f^{\prime}(x) = \cos(\pi/2x) + \frac{\pi}{2x}\sin(\pi/2x) , \end{equation*}

bounded on that compact interval; so \(f\) is Lipschitz there by the Mean Value Theorem and absolutely continuous there by Proposition 7.

(ii) Let \(\epsilon > 0\). Continuity at \(0\) supplies \(c \in (0,1)\) with \(f( c) - f(0) < \epsilon/2\); let \(\delta > 0\) answer the \(\epsilon/2\) challenge for the absolute continuity of \(f\) on \([c,1]\). Let \(\{(a_k,b_k)\}_{k=1}^{n}\) be disjoint open subintervals of \((0,1)\) with \(\sum_{k}[b_k-a_k] < \delta\), and cut each at \(c\): with \(a_k^{\prime} = \max\{a_k,c\}\), \(b_k^{\prime} = \max\{b_k,c\}\), \(a_k^{\prime\prime} = \min\{a_k,c\}\), \(b_k^{\prime\prime} = \min\{b_k,c\}\), inspection of the three positions of \((a_k,b_k)\) relative to \(c\) shows, \(f\) being increasing, that

\begin{equation*} |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 nondegenerate \((a_k^{\prime},b_k^{\prime})\) are disjoint in \((c,1)\) of total length \(< \delta\), so the first increments sum to \(< \epsilon/2\); the nondegenerate \((a_k^{\prime\prime},b_k^{\prime\prime})\) are disjoint in \((0,c)\), and the increments of an increasing function over disjoint subintervals of \([u,v]\) sum to at most \(f(v)-f(u)\) (the complementary increments are nonnegative and the total telescopes), so they sum to at most \(f( c)-f(0) < \epsilon/2\). Adding, \(\sum_{k}|f(b_k)-f(a_k)| < \epsilon\).

(iii) \(f(x) = \sqrt{x}\) is continuous and increasing on \([0,1]\), and on \([\epsilon,1]\) it has \(f^{\prime}(x) = 1/(2\sqrt{x}) \le 1/(2\sqrt{\epsilon})\), hence is Lipschitz there and, by Proposition 7, absolutely continuous there; so part (ii) makes \(f\) absolutely continuous on \([0,1]\). It is not Lipschitz: a constant \(c\) with \(\sqrt{x} = |f(x)-f(0)| \le cx\) for all \(x \in (0,1]\) would force \(c \ge 1/\sqrt{x} \to \infty\).

Both implications turn on padding a finite disjoint collection out to a countable one.

Assume the countable criterion, let \(\epsilon > 0\), take the corresponding \(\delta\), and shrink it so that \(\delta \le b-a\). Let \(\{(a_k,b_k)\}_{k=1}^{n}\) be disjoint open intervals in \((a,b)\) with \(\sigma = \sum_{k=1}^{n}[b_k-a_k] < \delta\). The set \(K = \bigcup_{k=1}^{n}[a_k,b_k]\) is closed with \(m(K) = \sigma\), so by excision the open set \((a,b) \sim K\) has measure at least \((b-a) - \sigma > 0\) and hence contains an open interval \((\alpha,\beta)\). Choose disjoint open intervals \(\{(a_k,b_k)\}_{k > n}\) inside \((\alpha,\beta)\) with \(\sum_{k>n}[b_k-a_k] < \delta - \sigma\) (for instance lengths \(2^{-j}\eta\) marching to the right). The full collection is then a countable disjoint family in \((a,b)\) of total length \(< \delta\), so

\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*}

and \(f\) is absolutely continuous on \([a,b]\).

Conversely, let \(f\) be absolutely continuous and let \(\delta > 0\) answer the \(\epsilon/2\) challenge in the definition. If \(\{(a_k,b_k)\}_{k=1}^{\infty}\) is a countable disjoint family in \((a,b)\) with \(\sum_{k=1}^{\infty}[b_k-a_k] < \delta\), then each initial segment is a finite disjoint family of total length \(< \delta\), so \(\sum_{k=1}^{n}|f(b_k)-f(a_k)| < \epsilon/2\) for every \(n\). The partial sums of a series of nonnegative terms being bounded by \(\epsilon/2\), the series converges and \(\sum_{k=1}^{\infty}|f(b_k)-f(a_k)| \le \epsilon/2 < \epsilon\).

Assume \(f\) increasing and absolutely continuous, and let \(\epsilon > 0\). By Problem 6.38 choose \(\delta > 0\) so that every countable disjoint collection \(\{(a_k,b_k)\}\) of open intervals in \((a,b)\) of total length \(< \delta\) satisfies \(\sum_{k}[f(b_k)-f(a_k)] < \epsilon/2\) (a finite collection being padded out as in that problem). Let \(E \subseteq [a,b]\) be measurable with \(m(E) < \delta\). By assertion (i) of Theorem 11 of Chapter 2 applied with \(\delta - m(E) > 0\), there is an open \(\mathcal{O} \supseteq E\) with \(m^{*}(\mathcal{O} \sim E) < \delta - m(E)\), so \(m(\mathcal{O}) < \delta\) by countable additivity; put \(U = \mathcal{O} \cap (a,b)\). By Proposition 9 of Chapter 1, \(U\) is a countable disjoint union of open intervals \((a_k,b_k)\) inside \((a,b)\), with \(\sum_{k}[b_k-a_k] = m(U) < \delta\). Since \(f\) is increasing, \(f((a_k,b_k)) \subseteq [f(a_k),f(b_k)]\), so monotonicity and countable subadditivity of outer measure give

\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*}

As \(f(E) \subseteq f(U) \cup \{f(a),f(b)\}\) and a two-point set is null, \(m^{*}(f(E)) \le \epsilon/2 < \epsilon\).

Conversely, let \(f\) be increasing and continuous with the stated property (the printed statement omits continuity, which is needed here and is no restriction since absolutely continuous functions are continuous), and let \(\delta\) answer the \(\epsilon\) challenge. Let \(\{(a_k,b_k)\}_{k=1}^{n}\) be disjoint open intervals in \((a,b)\) with \(\sum_{k}[b_k-a_k] < \delta\) and put \(E = \bigcup_{k=1}^{n}[a_k,b_k]\), so that \(m(E) = \sum_{k}[b_k-a_k] < \delta\). The Intermediate Value Theorem gives \(f([a_k,b_k]) = [f(a_k),f(b_k)]\), and these images are non-overlapping because \(f\) is increasing, so

\begin{equation*} \sum_{k=1}^{n}|f(b_k)-f(a_k)| = m\Bigl(\bigcup_{k=1}^{n}[f(a_k),f(b_k)]\Bigr) = m^{*}(f(E)) < \epsilon , \end{equation*}

using that finitely many non-overlapping closed intervals have union of measure the sum of their lengths. Hence \(f\) is absolutely continuous on \([a,b]\).

Let \(f\) be increasing and absolutely continuous on \([a,b]\) and let \(E \subseteq [a,b]\) have measure zero. For each \(\epsilon > 0\) the forward implication of Problem 6.39 – which uses only that \(f\) is increasing and absolutely continuous – supplies \(\delta > 0\) with \(m^{*}(f(A)) < \epsilon\) for every measurable \(A \subseteq [a,b]\) with \(m(A) < \delta\); taking \(A = E\) gives \(m^{*}(f(E)) < \epsilon\). Hence \(m^{*}(f(E)) = 0\), and \(f(E)\) is measurable of measure zero by Proposition 4 of Chapter 2.

Suppose now that the Cantor-Lebesgue function \(\varphi\) were absolutely continuous on \([0,1]\). The identity is Lipschitz, hence absolutely continuous by Proposition 7, and sums of absolutely continuous functions are absolutely continuous (Problem 6.42), so \(\psi(x) = x + \varphi(x)\) would be absolutely continuous; and \(\psi\) is increasing, being the sum of an increasing and a strictly increasing function. By the preceding paragraph \(\psi\) would carry the Cantor set \(C\), which is of measure zero by Proposition 19 of Chapter 2, to a set of measure zero. But Proposition 21 of Chapter 2 (page 52) gives \(m(\psi( C)) = 1\). Hence \(\varphi\) is not absolutely continuous on \([0,1]\).

(i) \(f\) is continuous, being absolutely continuous, so a closed \(F \subseteq [a,b]\) is compact by the Heine-Borel Theorem, its continuous image \(f(F)\) is compact and hence closed; and since \(f(\bigcup_n F_n) = \bigcup_n f(F_n)\), \(f\) carries \(F_{\sigma}\) subsets of \([a,b]\) to \(F_{\sigma}\) sets, which are Borel and hence measurable.

(ii) By Problem 6.40, \(f\) carries sets of measure zero to sets of measure zero.

Let \(E \subseteq [a,b]\) be measurable. By assertion (iv) of Theorem 11 of Chapter 2 there is an \(F_{\sigma}\) set \(F \subseteq E\) with \(m^{*}(E \sim F) = 0\); its closed pieces lie in \([a,b]\), so \(f(F)\) is measurable by (i). The set \(E \sim F\) is measurable of measure zero (Proposition 4 of Chapter 2), so \(f(E \sim F)\) is measurable of measure zero by (ii). Hence

\begin{equation*} f(E) = f(F) \cup f(E \sim F) \end{equation*}

is measurable.

Let \(f\) and \(g\) be absolutely continuous on \([a,b]\) and let \(\epsilon > 0\); in each case \(\{(a_k,b_k)\}_{k=1}^{n}\) denotes a finite disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^{n}[b_k-a_k] < \delta\).

The sum: take \(\delta = \min\{\delta_1,\delta_2\}\), where \(\delta_1, \delta_2\) answer the \(\epsilon/2\) challenge for \(f\) and for \(g\) respectively. By the triangle inequality,

\begin{equation*} \begin{aligned} \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{aligned} \end{equation*}

The product: \(f\) and \(g\) are continuous, hence bounded on the compact interval \([a,b]\) by the Extreme Value Theorem, say \(|f| \le M\) and \(|g| \le M\) with \(M \ge 1\); take \(\delta = \min\{\delta_1,\delta_2\}\), where \(\delta_1,\delta_2\) answer the \(\epsilon/(2M)\) challenge for \(f\) and for \(g\). 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*}

so summing over \(k\) gives \(\sum_{k=1}^{n}|(fg)(b_k)-(fg)(a_k)| < M\cdot\epsilon/(2M) + M\cdot\epsilon/(2M) = \epsilon\).

(i) \(g\) is Lipschitz, hence absolutely continuous by Proposition 7: for \(x \neq 0\),

\begin{equation*} g^{\prime}(x) = 2x\cos\!\left(\frac{\pi}{2x}\right) + \frac{\pi}{2}\sin\!\left(\frac{\pi}{2x}\right) , \end{equation*}

while \(|g(x)-g(0)| \le x^{2}\) gives \(g^{\prime}(0) = 0\); so \(|g^{\prime}| \le 2 + \pi/2\) on \([-1,1]\) and the Mean Value Theorem yields \(|g(u)-g(v)| \le (2+\pi/2)|u-v|\).

The function \(f(x) = x^{1/3}\) is continuous and increasing but not Lipschitz at \(0\), so we argue from the definition. Given \(\epsilon > 0\), choose \(\eta \in (0,1)\) with \(f(\eta) - f(-\eta) = 2\eta^{1/3} < \epsilon/2\); on \([-1,-\eta]\) and on \([\eta,1]\) we have \(f^{\prime}(x) = \tfrac13 |x|^{-2/3} \le M := \tfrac13 \eta^{-2/3}\), so by the Mean Value Theorem \(f\) is Lipschitz with constant \(M\) on each. Put \(\delta = \epsilon/(2M)\) and let \(\{(a_k,b_k)\}_{k=1}^{n}\) be disjoint open intervals in \((-1,1)\) with \(\sum_{k}[b_k-a_k] < \delta\). Cut each at whichever of \(\pm\eta\) it contains; since \(f\) is increasing, \(|f(b_k)-f(a_k)|\) is the sum of the increments of \(f\) over the resulting pieces, so \(\sum_{k}|f(b_k)-f(a_k)| = \Sigma_{\mathrm{mid}} + \Sigma_{\mathrm{out}}\), where

(a) the pieces inside \([-\eta,\eta]\) are non-overlapping, so their increments telescope: \(\Sigma_{\mathrm{mid}} \le f(\eta)-f(-\eta) < \epsilon/2\);

(b) the pieces inside \([-1,-\eta] \cup [\eta,1]\) have total length \(< \delta\), so \(\Sigma_{\mathrm{out}} \le M\delta = \epsilon/2\).

Hence \(\sum_{k}|f(b_k)-f(a_k)| < \epsilon\) and \(f\) is absolutely continuous.

(ii) \(V(f\circ g, P_n) = 2^{1/3}\sum_{m=1}^{n} m^{-2/3}\). Writing \(h = f \circ g\), we have \(h(1/k) = [k^{-2}\cos(k\pi/2)]^{1/3}\), which is \(0\) for \(k\) odd and \((-1)^{m}(2m)^{-2/3}\) for \(k = 2m\); also \(h(-1) = h(0) = 0\). So along \(P_n\) each of the \(n\) nonzero values is flanked by partition points where \(h\) vanishes, contributing its absolute value twice, and the increment from \(-1\) to \(0\) is \(0\):

\begin{equation*} V(f \circ g, P_n) = 2\sum_{m=1}^{n}\frac{1}{(2m)^{2/3}} = 2^{1/3}\sum_{m=1}^{n}\frac{1}{m^{2/3}} . \end{equation*}

(iii) Since \(t \mapsto t^{-2/3}\) decreases on \((0,\infty)\),

\begin{equation*} V(f\circ g, P_n) \ \ge\ 2^{1/3}\int_{1}^{n+1}\frac{dx}{x^{2/3}} = 3 \cdot 2^{1/3}\left[(n+1)^{1/3} - 1\right] \longrightarrow \infty , \end{equation*}

so \(TV(f\circ g) = \infty\) and \(f \circ g\) is not of bounded variation on \([-1,1]\); by Theorem 8 it is therefore not absolutely continuous there either.

The same \(\delta\) works, scaled by the Lipschitz constant. Let \(c > 0\) satisfy \(|f(s) - f(t)| \le c\,|s-t|\) for all \(s,t \in \mathbb{R}\), and given \(\epsilon > 0\) let \(\delta > 0\) respond to the \(\epsilon/c\) challenge for the absolute continuity of \(g\) on \([a,b]\). If \(\{(a_k,b_k)\}_{k=1}^n\) is a finite disjoint collection of open intervals in \((a,b)\) with \(\sum_{k=1}^n [b_k - a_k] < \delta\), then applying the Lipschitz estimate to the pair \(g(a_k), g(b_k)\) for each \(k\),

\begin{equation*} \begin{aligned} \sum_{k=1}^n |(f \circ g)(b_k) - (f \circ g)(a_k)| &\le c \sum_{k=1}^n |g(b_k) - g(a_k)| \\ &< c \cdot \frac{\epsilon}{c} = \epsilon . \end{aligned} \end{equation*}

Hence \(f \circ g\) is absolutely continuous on \([a,b]\).

Compose the two responses: strict monotonicity of \(g\) is exactly what keeps the image intervals disjoint, so the absolute continuity of \(f\) may be applied to them.

Assume \(g\) strictly increasing (the decreasing case is identical with \(c_k = g(b_k)\), \(d_k = g(a_k)\)). Being continuous and strictly increasing, \(g\) maps \([a,b]\) onto \([c,d] := [g(a), g(b)]\), on which \(f\) is absolutely continuous by hypothesis. Given \(\epsilon > 0\), let \(\delta_1 > 0\) respond to the \(\epsilon\) challenge for \(f\) on \([c,d]\), and let \(\delta_2 > 0\) respond to the \(\delta_1\) challenge for \(g\) on \([a,b]\).

Let \(\{(a_k,b_k)\}_{k=1}^n\) be disjoint open intervals in \((a,b)\) with \(\sum_{k=1}^n [b_k - a_k] < \delta_2\), and set \(c_k = g(a_k)\), \(d_k = g(b_k)\). Strict monotonicity gives \(c \le c_k < d_k \le d\), and, for \(j \ne k\) with \(b_j \le a_k\), \(d_j \le c_k\); so \(\{(c_k,d_k)\}_{k=1}^n\) is a finite disjoint collection of open intervals in \((c,d)\). Since

\begin{equation*} \sum_{k=1}^n [d_k - c_k] = \sum_{k=1}^n |g(b_k) - g(a_k)| < \delta_1 , \end{equation*}

the choice of \(\delta_1\) yields

\begin{equation*} \sum_{k=1}^n |(f \circ g)(b_k) - (f \circ g)(a_k)| = \sum_{k=1}^n |f(d_k) - f(c_k)| < \epsilon . \end{equation*}

Hence \(f \circ g\) is absolutely continuous on \([a,b]\).

Write \(\mathcal{F}_{Lip}\), \(\mathcal{F}_{AC}\), \(\mathcal{F}_{BV}\) for the three families on \([a,b]\).

(i) The inclusions (27). Proposition 7 gives \(\mathcal{F}_{Lip} \subseteq \mathcal{F}_{AC}\), and Theorem 8 gives \(\mathcal{F}_{AC} \subseteq \mathcal{F}_{BV}\). Both are strict:

• \(f(x) = \sqrt{x-a}\) lies in \(\mathcal{F}_{AC} \setminus \mathcal{F}_{Lip}\). It is absolutely continuous by Problem 37(iii) transplanted to \([a,b]\), while \(|f(a+t) - f(a)|/t = 1/\sqrt{t} \to \infty\) as \(t \to 0^+\), so no Lipschitz constant exists.
• \(\psi(x) = \varphi\bigl((x-a)/(b-a)\bigr)\), with \(\varphi\) the Cantor-Lebesgue function, lies in \(\mathcal{F}_{BV} \setminus \mathcal{F}_{AC}\). It is increasing, hence of bounded variation with \(TV(\psi) = \psi(b) - \psi(a)\). Were it absolutely continuous, then since the affine \(L(t) = a + (b-a)t\) carries disjoint open subintervals of \((0,1)\) to disjoint open subintervals of \((a,b)\) and scales lengths by \(b-a\), a \(\delta\) responding to the \(\epsilon\) challenge for \(\psi\) would make \(\delta/(b-a)\) respond to it for \(\varphi = \psi \circ L\), contradicting the Example on page 120.

(ii) Closure under linear combinations. Fix \(\alpha, \beta \in \mathbb{R}\); the triangle inequality \(|(\alpha f + \beta g)(u) - (\alpha f + \beta g)(v)| \le |\alpha||f(u)-f(v)| + |\beta||g(u)-g(v)|\) gives all three cases at once: for Lipschitz \(f,g\) with constants \(c_f, c_g\) it yields the constant \(|\alpha|c_f + |\beta|c_g\); summed over a partition \(P\) it yields \(V(\alpha f + \beta g, P) \le |\alpha| TV(f) + |\beta| TV(g)\), so \(\mathcal{F}_{BV}\) is closed; and summed over a disjoint collection of total length less than \(\delta = \min\{\delta_f, \delta_g\}\), where \(\delta_f\) responds to the \(\epsilon/[2(|\alpha|+1)]\) challenge for \(f\) and \(\delta_g\) to the \(\epsilon/[2(|\beta|+1)]\) challenge for \(g\), it gives

\begin{equation*} \sum_{k=1}^n |(\alpha f + \beta g)(b_k) - (\alpha f + \beta g)(a_k)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

(iii) The total variation function \(T_f(x) = TV(f_{[a,x]})\) stays in the collection. By (21), \(T_f(v) - T_f(u) = TV(f_{[u,v]}) \ge 0\) for \(a \le u < v \le b\), so \(T_f\) is real-valued and increasing.

• \(f \in \mathcal{F}_{BV}\): an increasing real-valued function is of bounded variation, so \(T_f \in \mathcal{F}_{BV}\).
• \(f \in \mathcal{F}_{AC}\): then \(f \in \mathcal{F}_{BV}\) by Theorem 8, and display (24) in the last paragraph of that theorem’s proof is verbatim the criterion for absolute continuity of \(T_f\); so \(T_f \in \mathcal{F}_{AC}\).
• \(f \in \mathcal{F}_{Lip}\) with constant \(c\): the second Example of Section 6.3, applied to \(f\) restricted to \([u,v]\) (Lipschitz there with the same \(c\)), gives \(TV(f_{[u,v]}) \le c[v-u]\); with (21) this reads \(0 \le T_f(v) - T_f(u) \le c\,[v-u]\), so \(T_f \in \mathcal{F}_{Lip}\).

(iv) Decomposition. Let \(\mathcal{F}\) be any one of the three and \(f \in \mathcal{F}\). By (i), \(f\) is of bounded variation, so Lemma 5 applies and (23) gives

\begin{equation*} f(x) = \bigl[f(x) + T_f(x)\bigr] - T_f(x), \qquad x \in [a,b]. \end{equation*}

By (iii) \(T_f \in \mathcal{F}\), and by (ii) \(f + T_f \in \mathcal{F}\); \(T_f\) is increasing as noted, and \(f + T_f\) is increasing by (22). So \(f\) is a difference of two increasing members of \(\mathcal{F}\).

The two criteria differ only by the freedom to pass to a subcollection, which preserves both disjointness and smallness of total length.

(i) Necessity. If \(f\) is absolutely continuous and \(\delta\) responds to the \(\epsilon\) challenge, the triangle inequality gives \(\bigl|\sum_{k=1}^n [f(b_k)-f(a_k)]\bigr| \le \sum_{k=1}^n |f(b_k)-f(a_k)| < \epsilon\), so the same \(\delta\) serves.

(ii) Sufficiency. Let \(\delta > 0\) respond to the \(\epsilon/2\) challenge in the displayed condition, and let \(\{(a_k,b_k)\}_{k=1}^n\) be disjoint open intervals in \((a,b)\) with \(\sum_{k=1}^n [b_k - a_k] < \delta\). Split the indices by the sign of the increment,

\begin{equation*} P = \{k : f(b_k) - f(a_k) \ge 0\}, \qquad N = \{k : f(b_k) - f(a_k) < 0\} . \end{equation*}

Each of \(\{(a_k,b_k)\}_{k \in P}\) and \(\{(a_k,b_k)\}_{k \in N}\) is again a finite disjoint collection of open intervals in \((a,b)\) whose lengths sum to less than \(\delta\) (the terms being nonnegative; an empty index set contributes \(0\)). Since the terms of each subsum have a constant sign, the absolute value may be moved outside, and the choice of \(\delta\) gives

\begin{equation*} \sum_{k \in P} |f(b_k) - f(a_k)| = \left| \sum_{k \in P} [f(b_k) - f(a_k)] \right| < \frac{\epsilon}{2}, \end{equation*}

and likewise for \(N\). Adding, \(\sum_{k=1}^n |f(b_k) - f(a_k)| < \epsilon\), so \(f\) is absolutely continuous on \([a,b]\).

\(\varphi^{\prime} = 0\) almost everywhere, so Theorem 10 would force \(\varphi(1) - \varphi(0) = 0\).

Indeed \(\varphi\) is constant on each of the countably many deleted middle-third intervals \(I_j\), hence differentiable there with \(\varphi^{\prime} = 0\); and \([0,1] \sim \bigcup_j I_j = C\) has measure zero, so \(\varphi^{\prime} = 0\) a.e. on \((0,1)\). If \(\varphi\) were absolutely continuous on \([0,1]\), Theorem 10 (whose hypothesis is exactly that absolute continuity) would give

\begin{equation*} 1 = \varphi(1) - \varphi(0) = \int_0^1 \varphi^{\prime} = \int_0^1 0 = 0 , \end{equation*}

a contradiction. So \(\varphi\) is not absolutely continuous.

Comparison with Problem 40. There one uses Problems 38 and 39 (an increasing absolutely continuous function carries null sets to null sets) applied to \(\psi(x) = x + \varphi(x)\), which is increasing and would be absolutely continuous, together with \(m(\psi( C)) = 1\) (page 52), to reach the same contradiction. Problem 40 is purely measure-theoretic and never differentiates, at the cost of the auxiliary \(\psi\); the present argument is shorter but invokes the whole of Theorem 10, which rests on Lebesgue’s differentiation theorem, Theorem 9, and the Vitali Convergence Theorem.

The right-hand limit always equals \(f(b) - f(a)\) and the left-hand integrand always equals \(f^{\prime}\) a.e., so the two assertions are the same statement.

With \(f\) extended to \([a,b+1]\) by \(f \equiv f(b)\) on \((b,b+1]\), as on page 113, the extension is continuous and each \(\mathrm{Diff}_{1/n} f\) is continuous, hence integrable, on \([a,b]\).

(i) The change of variables formula (14) with \(u = a\), \(v = b\) gives \(\int_a^b \mathrm{Diff}_h f = \mathrm{Av}_h f(b) - \mathrm{Av}_h f(a)\) for \(0 < h \le 1\). Here \(\mathrm{Av}_h f(b) = f(b)\), since \(f \equiv f(b)\) on \((b, b+h]\), while continuity of \(f\) at \(a\) gives

\begin{equation*} \bigl| \mathrm{Av}_h f(a) - f(a) \bigr| = \left| \frac{1}{h}\int_a^{a+h} [f(t) - f(a)]\,dt \right| \le \sup_{a \le t \le a+h} |f(t)-f(a)| \to 0 \end{equation*}

as \(h \to 0^+\). Taking \(h = 1/n\),

\begin{equation*} \lim_{n \to \infty} \left[ \int_a^b \mathrm{Diff}_{1/n} f \right] = f(b) - f(a), \end{equation*}

which is (29) in the proof of Theorem 10; only continuity was used.

(ii) At each \(x \in (a,b)\) at which \(f\) is differentiable – almost every \(x\), by hypothesis – we have \(x + 1/n \in (a,b)\) for large \(n\) and \(\mathrm{Diff}_{1/n} f(x) \to f^{\prime}(x)\). Hence \(\lim_n \mathrm{Diff}_{1/n} f = f^{\prime}\) a.e. on \([a,b]\), and two measurable functions equal a.e. have the same integral, one being integrable exactly when the other is:

\begin{equation*} \int_a^b \left[ \lim_{n \to \infty} \mathrm{Diff}_{1/n} f \right] = \int_a^b f^{\prime} . \end{equation*}

Substituting (i) and (ii) into the second displayed assertion turns it term by term into \(\int_a^b f^{\prime} = f(b) - f(a)\), and conversely. Both presuppose the same thing, that \(f^{\prime}\) be integrable over \([a,b]\).

The Vitali Convergence Theorem identifies the two sides.

As established in Problem 49 (using only the continuity of \(f\), via the change of variables formula (14) of Section 6.2 and \(\mathrm{Av}_h f(x) \to f(x)\) at \(x = a, b\)),

\begin{equation*} \lim_{n \to \infty} \int_a^b \operatorname{Diff}_{1/n} f = f(b) - f(a), \end{equation*}

and \(\{\operatorname{Diff}_{1/n} f\}\) converges pointwise to \(f^{\prime}\) at each point of differentiability of \(f\), hence a.e. on \([a,b]\). Each \(\operatorname{Diff}_{1/n} f\) is continuous, hence measurable, on \([a,b]\); by hypothesis the sequence is uniformly integrable; and \(m([a,b]) < \infty\). So the Vitali Convergence Theorem of Chapter 4 applies: \(f^{\prime}\) is integrable over \([a,b]\) and

\begin{equation*} \lim_{n \to \infty} \int_a^b \operatorname{Diff}_{1/n} f = \int_a^b f^{\prime} . \end{equation*}

Comparing the two limits of the same sequence of numbers, \(\int_a^b f^{\prime} = f(b) - f(a)\).

Domination by an integrable \(g\) makes \(\{\operatorname{Diff}_{1/n} f\}\) uniformly integrable, so the preceding problem applies.

Precisely: a sequence of measurable functions on the finite-measure set \([a,b]\) dominated by a single integrable function is uniformly integrable, by the absolute continuity of integration (Proposition 23 of Chapter 4). Each \(\operatorname{Diff}_{1/n} f\) is continuous, hence measurable, on \([a,b]\), and \(|\operatorname{Diff}_{1/n} f| \le g\) a.e. for every \(n\); so Problem 50 gives \(\int_a^b f^{\prime} = f(b) - f(a)\).

Method (2): directly. As in Problem 49, \(\lim_n \int_a^b \operatorname{Diff}_{1/n} f = f(b) - f(a)\) and \(\operatorname{Diff}_{1/n} f \to f^{\prime}\) a.e. on \([a,b]\). Discarding the countable union of the exceptional sets, \(|\operatorname{Diff}_{1/n} f| \le g\) off one set of measure zero for all \(n\) at once; letting \(n \to \infty\) gives \(|f^{\prime}| \le g\) a.e., so \(f^{\prime}\) is integrable by the integral comparison test. The Lebesgue Dominated Convergence Theorem of Chapter 4 then yields

\begin{equation*} f(b) - f(a) = \lim_{n \to \infty} \int_a^b \operatorname{Diff}_{1/n} f = \int_a^b f^{\prime} . \end{equation*}

Apply Theorem 10 to the product \(f \cdot g\), which is itself absolutely continuous.

Indeed \(f\) and \(g\) are continuous on \([a,b]\), hence bounded there, say \(|f|, |g| \le M\); given \(\epsilon > 0\), let \(\delta > 0\) respond to the \(\epsilon/(2M)\) challenge for both \(f\) and \(g\). For a disjoint collection \(\{(a_k,b_k)\}_{k=1}^n\) of open intervals in \((a,b)\) of total length less than \(\delta\), the identity \(fg(b_k) - fg(a_k) = f(b_k)[g(b_k)-g(a_k)] + g(a_k)[f(b_k)-f(a_k)]\) gives

\begin{equation*} \sum_{k=1}^n \bigl| f(b_k)g(b_k) - f(a_k)g(a_k) \bigr| \le M\cdot\frac{\epsilon}{2M} + M\cdot\frac{\epsilon}{2M} = \epsilon . \end{equation*}

By Theorem 10, each of \(f, g, f \cdot g\) is differentiable a.e. on \((a,b)\) with integrable derivative. At each of the almost all points where both \(f\) and \(g\) are differentiable, the elementary product rule holds (its proof needs only the continuity of \(f\) there), so

\begin{equation*} (f \cdot g)^{\prime} = f \cdot g^{\prime} + f^{\prime} \cdot g \quad \text{a.e. on } (a,b), \end{equation*}

and \(|f g^{\prime}| \le M|g^{\prime}|\), \(|f^{\prime} g| \le M|f^{\prime}|\) a.e. make both summands integrable, licensing the split by linearity. Theorem 10 applied to \(f \cdot g\) therefore gives

\begin{equation*} f(b)g(b) - f(a)g(a) = \int_a^b (f\cdot g)^{\prime} = \int_a^b f \cdot g^{\prime} + \int_a^b f^{\prime} \cdot g , \end{equation*}

which rearranges to the asserted formula.

Both directions run off the indefinite integral representation. By Theorem 11 (applicable since \(f\) is absolutely continuous, hence so is each restriction),

\begin{equation*} f(v) - f(u) = \int_u^v f^{\prime} \quad \text{for } a \le u \le v \le b. \tag{\(\dagger\)} \end{equation*}

(i) If \(f\) is Lipschitz with constant \(c\), then at each point \(x\) of differentiability, \(|[f(x+h)-f(x)]/h| \le c\) for small \(h \ne 0\); letting \(h \to 0\) gives \(|f^{\prime}(x)| \le c\), hence \(|f^{\prime}| \le c\) a.e. on \([a,b]\).

(ii) If \(|f^{\prime}| \le c\) a.e., then for \(a \le u < v \le b\), \((\dagger)\) and the monotonicity of integration give

\begin{equation*} |f(v) - f(u)| = \left| \int_u^v f^{\prime} \right| \le \int_u^v |f^{\prime}| \le c\,(v-u), \end{equation*}

the middle step legitimate because \(f^{\prime}\) is integrable over \([u,v]\) and the last because altering an integrand on a null set does not change its integral. The estimate is symmetric in \(u, v\) and trivial for \(u = v\), so \(f\) is Lipschitz with constant \(c\).

(i) Take the gaps complementary to a Vitali selection of intervals across which \(f\) barely grows.

Put \(\eta = \epsilon/(b-a)\) and let \(E\) be the set of \(x \in (a,b)\) with \(f^{\prime}(x) = 0\), so \(m(E) = b-a\) since \(f\) is singular. Let \(\mathcal{F}\) be the family of nondegenerate closed intervals \([c,d] \subseteq (a,b)\) with \(f(d) - f( c) < \eta\,(d-c)\). Then \(\mathcal{F}\) covers \(E\) in the sense of Vitali: for \(x \in E\) and \(\ell > 0\) pick \(h_0 > 0\) with \([x-h_0, x+h_0] \subseteq (a,b)\), \(2h_0 < \ell\) and \(|f(x+t) - f(x)| \le (\eta/2)|t|\) for \(|t| \le h_0\); then for \(0 < h \le h_0\),

\begin{equation*} f(x+h) - f(x-h) \le \tfrac{\eta}{2}h + \tfrac{\eta}{2}h = \eta h < \eta\,(2h), \end{equation*}

so \([x-h,x+h] \in \mathcal{F}\) contains \(x\) and has length \(< \ell\). Since \(m^*(E) < \infty\), the Vitali Covering Lemma yields a finite disjoint \(\{[c_k,d_k]\}_{k=1}^n \subseteq \mathcal{F}\) with \(m^*\bigl(E \sim \bigcup_k [c_k,d_k]\bigr) < \delta\). Relabel so \(a < c_1 < d_1 < \cdots < c_n < d_n < b\), set \(d_0 = a\), \(c_{n+1} = b\), and let \((a_j, b_j) = (d_{j-1}, c_j)\) for \(j = 1, \dots, n+1\) be the complementary gaps: a finite disjoint collection of open intervals in \((a,b)\). Since \((a,b) \sim \bigcup_k [c_k,d_k]\) differs from \(E \sim \bigcup_k [c_k,d_k]\) by a subset of the null set \((a,b) \sim E\),

\begin{equation*} \sum_{j=1}^{n+1} [b_j - a_j] = m\Bigl( (a,b) \sim \bigcup_{k=1}^n [c_k,d_k] \Bigr) = m^*\Bigl( E \sim \bigcup_{k=1}^n [c_k,d_k] \Bigr) < \delta . \end{equation*}

The gaps and the \([c_k,d_k]\) abut consecutively and exhaust \([a,b]\), so the increments telescope, and \(\sum_k [f(d_k) - f(c_k)] < \eta \sum_k [d_k - c_k] \le \eta\,(b-a) = \epsilon\) by membership in \(\mathcal{F}\); hence

\begin{equation*} \begin{aligned} \sum_{j=1}^{n+1} [f(b_j) - f(a_j)] &= [f(b) - f(a)] - \sum_{k=1}^n [f(d_k) - f(c_k)] \\ &> f(b) - f(a) - \epsilon . \end{aligned} \end{equation*}

(ii) We show \(\int_a^b f^{\prime} = 0\); since \(f\) is increasing, \(f^{\prime} \ge 0\) a.e. and is integrable by Corollary 4, so Proposition 9 of Chapter 4 then gives \(f^{\prime} = 0\) a.e., i.e. \(f\) singular. Let \(\epsilon > 0\), and let \(\delta > 0\) be as in the absolute continuity of integration (Proposition 23 of Chapter 4) for the integrable \(f^{\prime}\): \(\int_A f^{\prime} < \epsilon\) whenever \(m(A) < \delta\). Take \(\{(a_k,b_k)\}_{k=1}^n\) from the assumed property for this \(\epsilon, \delta\), and put \(U = \bigcup_k (a_k,b_k)\), so \(\int_U f^{\prime} < \epsilon\). The complement \([a,b] \sim U\) is a finite disjoint union of closed intervals \([\alpha_j, \beta_j]\) abutting the \((a_k,b_k)\) (degenerate ones may be discarded), so telescoping gives

\begin{equation*} \sum_{j=1}^p [f(\beta_j) - f(\alpha_j)] = [f(b)-f(a)] - \sum_{k=1}^n [f(b_k)-f(a_k)] < \epsilon , \end{equation*}

and Corollary 4 applied to \(f\) on each \([\alpha_j, \beta_j]\) gives \(\int_{[a,b] \sim U} f^{\prime} \le \sum_j [f(\beta_j)-f(\alpha_j)] < \epsilon\). Hence \(0 \le \int_a^b f^{\prime} < 2\epsilon\) for every \(\epsilon > 0\).

(iii) Each partial sum \(s_N = \sum_{n=1}^N f_n\) is increasing and singular (a finite sum of derivatives), and \(f = s_N + r_N\) with \(r_N = \sum_{n>N} f_n\) increasing; in particular \(f\) is real-valued and increasing. Given \(\epsilon, \delta > 0\), fix \(N\) with \(r_N(b) - r_N(a) = [f(b)-f(a)] - [s_N(b)-s_N(a)] < \epsilon/2\), and apply (i) to \(s_N\) with challenges \(\epsilon/2, \delta\) to get disjoint \(\{(a_k,b_k)\}_{k=1}^n\) in \((a,b)\) of total length \(< \delta\) with \(\sum_k [s_N(b_k)-s_N(a_k)] > s_N(b)-s_N(a) - \epsilon/2\). Since \(r_N\) is increasing, \(f(b_k)-f(a_k) \ge s_N(b_k)-s_N(a_k)\), so

\begin{equation*} \sum_{k=1}^n [f(b_k) - f(a_k)] > [f(b)-f(a)] - [r_N(b)-r_N(a)] - \frac{\epsilon}{2} > f(b) - f(a) - \epsilon . \end{equation*}

Thus \(f\) has the property of (i), and by (ii) it is singular.

(i) The total variation function dominates \(f\) increment by increment. By (21) of Section 6.3, \(v(w) - v(u) = TV(f_{[u,w]}) \ge 0\) for \(a \le u < w \le b\), so \(v\) is increasing with \(v(a) = 0\), \(v(b) = TV(f)\), and the crudest partition \(\{u,w\}\) gives

\begin{equation*} |f(w) - f(u)| \le TV(f_{[u,w]}) = v(w) - v(u) \quad \text{for } a \le u < w \le b. \tag{\(\ast\)} \end{equation*}

By Lebesgue’s Theorem \(v\) is differentiable a.e., and by Corollary 6 so is \(f\), with \(f^{\prime}\) integrable. At a point \(x\) where both derivatives exist, dividing \((\ast)\) (with \(u = x\), \(w = x+h\)) by \(h > 0\) and letting \(h \to 0^+\) gives \(|f^{\prime}(x)| \le v^{\prime}(x)\). Corollary 4 applied to the increasing \(v\) gives \(\int_a^b v^{\prime} \le v(b) - v(a) = TV(f)\), so by monotonicity of integration

\begin{equation*} \int_a^b |f^{\prime}| \le \int_a^b v^{\prime} \le TV(f). \tag{\(\ast\ast\)} \end{equation*}

(ii) Equality holds exactly when \(f\) is absolutely continuous.

If \(f\) is absolutely continuous, Theorem 11 makes \(f\) the indefinite integral of \(f^{\prime}\), so for any partition \(P = \{x_0, \dots, x_k\}\),

\begin{equation*} V(f,P) = \sum_{i=1}^k \left| \int_{x_{i-1}}^{x_i} f^{\prime} \right| \le \sum_{i=1}^k \int_{x_{i-1}}^{x_i} |f^{\prime}| = \int_a^b |f^{\prime}| ; \end{equation*}

taking the supremum, \(TV(f) \le \int_a^b |f^{\prime}|\), which with \((\ast\ast)\) gives equality.

Conversely, suppose \(\int_a^b |f^{\prime}| = TV(f)\) and fix \(x \in (a,b)\). Applying \((\ast\ast)\) to the restrictions of \(f\) to \([a,x]\) and to \([x,b]\), whose total variations are \(v(x)\) and \(v(b)-v(x)\) by (20), and adding,

\begin{equation*} v(b) = \int_a^x |f^{\prime}| + \int_x^b |f^{\prime}| \le v(x) + [v(b) - v(x)] = v(b), \end{equation*}

so both are equalities; in particular \(v(x) = \int_a^x |f^{\prime}|\) for all \(x \in [a,b]\) (the endpoints being immediate). Thus \(v\) is the indefinite integral of the integrable \(|f^{\prime}|\), hence absolutely continuous by Theorem 11, and \((\ast)\) transfers this to \(f\): if \(\delta\) responds to the \(\epsilon\) challenge for \(v\), then for a disjoint collection \(\{(a_k,b_k)\}_{k=1}^n\) of total length less than \(\delta\),

\begin{equation*} \sum_{k=1}^n |f(b_k) - f(a_k)| \le \sum_{k=1}^n |v(b_k) - v(a_k)| < \epsilon . \end{equation*}

(iii) For \(f\) increasing, \(TV(f) = f(b) - f(a)\) and \(|f^{\prime}| = f^{\prime}\) a.e., so (i) reads \(\int_a^b f^{\prime} \le f(b)-f(a)\), which is Corollary 4, and (ii) reads: \(f\) is absolutely continuous iff \(\int_a^b f^{\prime} = f(b)-f(a)\), which is Corollary 12. Parts (i) and (ii) are thus those corollaries extended from monotone to bounded variation, with \(f(b)-f(a)\) replaced by \(TV(f)\) and \(f^{\prime}\) by \(|f^{\prime}|\).

Throughout, \(c = g(a)\), \(d = g(b)\); \(g\) is a continuous strictly increasing bijection of \([a,b]\) onto \([c,d]\), so it carries \((u,w)\) onto \((g(u), g(w))\), preserves disjointness, and satisfies \(g(S \sim T) = g(S) \sim g(T)\) for \(T \subseteq S\). By Theorem 10, \(g^{\prime}\) exists a.e., is integrable, and \(g^{\prime} \ge 0\) a.e.

(i) Write \(\mathcal{O} = \bigcup_k (a_k,b_k)\) as the countable disjoint union of its components. Then \(g(\mathcal{O}) = \bigcup_k (g(a_k), g(b_k))\) is a countable disjoint union of open intervals, hence open, and by countable additivity of \(m\), Theorem 10 on each \([a_k,b_k]\), and the Monotone Convergence Theorem (the partial sums of \(\sum_k g^{\prime}\chi_{(a_k,b_k)}\) increase a.e. to \(g^{\prime}\chi_{\mathcal{O}}\)),

\begin{equation*} m(g(\mathcal{O})) = \sum_k [g(b_k) - g(a_k)] = \sum_k \int_{a_k}^{b_k} g^{\prime} = \int_{\mathcal{O}} g^{\prime} . \end{equation*}

(ii) Write \(E = \bigcap_n \mathcal{O}_n\) with \(\mathcal{O}_n\) open in \((a,b)\) and descending (replace \(\mathcal{U}_n\) by \((a,b) \cap \bigcap_{j \le n}\mathcal{U}_j\)). As \(g\) is one-to-one, \(g(E) = \bigcap_n g(\mathcal{O}_n)\) is a \(G_\delta\) set, and the \(g(\mathcal{O}_n)\) descend inside \([c,d]\), of finite measure; so continuity of measure and part (i) give \(m(g(E)) = \lim_n \int_{\mathcal{O}_n} g^{\prime}\). Since \(\chi_{\mathcal{O}_n} \to \chi_E\) pointwise with \(0 \le g^{\prime}\chi_{\mathcal{O}_n} \le g^{\prime}\) integrable, the Dominated Convergence Theorem gives \(\lim_n \int_{\mathcal{O}_n} g^{\prime} = \int_E g^{\prime}\).

(iii) Let \(m(E) = 0\) and \(\epsilon > 0\). Since \(g^{\prime}\) is integrable, the absolute continuity of integration (Proposition 23 of Chapter 4) supplies \(\delta > 0\) with \(\int_A g^{\prime} < \epsilon\) whenever \(m(A) < \delta\); by outer regularity choose open \(\mathcal{O}\) with \(E \cap (a,b) \subseteq \mathcal{O} \subseteq (a,b)\) and \(m(\mathcal{O}) < \delta\). By part (i),

\begin{equation*} m^*\bigl(g(E \cap (a,b))\bigr) \le m(g(\mathcal{O})) = \int_{\mathcal{O}} g^{\prime} < \epsilon . \end{equation*}

As \(\epsilon\) was arbitrary and \(g(E) \sim g(E \cap (a,b)) \subseteq \{c,d\}\) is finite, \(m(g(E)) = 0\); and \(\int_E g^{\prime} = 0\) since \(m(E) = 0\).

(iv) Let \(A\) be measurable and \(A_0 = A \cap (a,b)\); since \(A \sim A_0\) and \(g(A) \sim g(A_0)\) lie in two-point sets, it suffices to treat \(A_0\). By Theorem 11 of Chapter 2 there is a \(G_\delta\) set \(G\) with \(A_0 \subseteq G \subseteq (a,b)\) and \(m(G \sim A_0) = 0\) (intersect a \(G_\delta\) hull with \((a,b)\)). As \(g\) is one-to-one, \(g(A_0) = g(G) \sim g(G \sim A_0)\), where \(g(G)\) is measurable with \(m(g(G)) = \int_G g^{\prime}\) by (ii) and \(g(G \sim A_0)\) is null by (iii). By excision,

\begin{equation*} m(g(A_0)) = m(g(G)) = \int_G g^{\prime} = \int_{A_0} g^{\prime} + \int_{G \sim A_0} g^{\prime} = \int_{A_0} g^{\prime} . \end{equation*}

(v) Lemma: for measurable \(E \subseteq [c,d]\), the function \(x \mapsto \chi_E(g(x))g^{\prime}(x)\) agrees a.e. on \([a,b]\) with a measurable function and \(\int_a^b \chi_E(g(x))g^{\prime}(x)\,dx = m(E)\). (This is needed because \(g^{-1}\) need not carry measurable sets to measurable sets, so \(\varphi \circ g\) alone need not be measurable.) Choose Borel sets \(B \subseteq E \subseteq B \cup N\) with \(m(E \sim B) = 0\), \(m(N) = 0\), \(N \subseteq [c,d]\). As \(g\) is continuous, \(g^{-1}(B)\) and \(g^{-1}(N)\) are Borel, and \(g(g^{-1}(S)) = S\) for \(S \subseteq [c,d]\), so part (iv) gives \(\int_{g^{-1}(B)} g^{\prime} = m(B) = m(E)\) and \(\int_{g^{-1}(N)} g^{\prime} = 0\); since \(g^{\prime} \ge 0\) a.e., the latter forces \(g^{\prime} = 0\) a.e. on \(g^{-1}(N)\). From \(0 \le \chi_{g^{-1}(E)} - \chi_{g^{-1}(B)} \le \chi_{g^{-1}(N)}\), multiplying by \(g^{\prime} \ge 0\),

\begin{equation*} \chi_E(g(x))\,g^{\prime}(x) = \chi_{g^{-1}(B)}(x)\,g^{\prime}(x) \quad \text{for a.a. } x \in [a,b], \end{equation*}

whose integral is \(\int_{g^{-1}(B)} g^{\prime} = m(E)\). Now for \(\varphi = \sum_{i=1}^p \alpha_i \chi_{E_i}\) simple on \([c,d]\), linearity and the Lemma give

\begin{equation*} \int_a^b \varphi(g(x))g^{\prime}(x)\,dx = \sum_{i=1}^p \alpha_i\, m(E_i) = \int_c^d \varphi(y)\,dy , \end{equation*}

the integrand being integrable since \(|\varphi \circ g| \cdot g^{\prime} \le \bigl(\sum_i |\alpha_i|\bigr) g^{\prime}\) a.e.

(vi) By the Simple Approximation Theorem take nonnegative simple \(\varphi_n \uparrow f\) pointwise on \([c,d]\). By (v), each \((\varphi_n \circ g)g^{\prime}\) agrees a.e. with a measurable \(\psi_n\) having \(\int_a^b \psi_n = \int_c^d \varphi_n\). Off the single null set \(E_0\) obtained by discarding the countably many exceptional sets together with \(\{g^{\prime} \text{ undefined or } < 0\}\), the \(\psi_n\) are nonnegative, increasing in \(n\), and converge to \((f \circ g)g^{\prime}\). Two applications of the Monotone Convergence Theorem give

\begin{equation*} \int_a^b f(g(x))g^{\prime}(x)\,dx = \lim_{n\to\infty} \int_c^d \varphi_n = \int_c^d f(y)\,dy , \end{equation*}

finite since \(f\) is integrable; so \((f \circ g)g^{\prime}\) is integrable over \([a,b]\).

(vii) The function of (vi) whose composition with \(g\) is \(\chi_{\mathcal{O}}\) is \(f = \chi_{g(\mathcal{O})}\), which is nonnegative and integrable on \([c,d]\) because \(g(\mathcal{O})\) is open and bounded by (i). As \(g\) is one-to-one, \(g(x) \in g(\mathcal{O})\) iff \(x \in \mathcal{O}\), so \(f \circ g = \chi_{\mathcal{O}}\) and (vi) reads

\begin{equation*} m(g(\mathcal{O})) = \int_c^d \chi_{g(\mathcal{O})} = \int_a^b \chi_{\mathcal{O}}(x)g^{\prime}(x)\,dx = \int_{\mathcal{O}} g^{\prime} , \end{equation*}

which is part (i).

Yes; the formula holds verbatim for merely increasing absolutely continuous \(g\). Only the proof changes: the route of Problem 56 goes through \(m(g(\mathcal{O})) = \int_{\mathcal{O}} g^{\prime}\), whose proof needs \(g\) one-to-one, so we instead push Lebesgue measure back through \(g\).

Since \(g\) is increasing and continuous, \(g([a,b]) = [c,d]\); by Theorem 10, \(g^{\prime}\) exists a.e., is nonnegative a.e., is integrable, and \(\int_u^v g^{\prime} = g(v) - g(u)\) for \(a \le u \le v \le b\).

(i) A measure. For Borel \(E \subseteq [c,d]\) set \(\nu(E) = \int_a^b \chi_E(g(x))g^{\prime}(x)\,dx\); this is defined because \(g\) is continuous, hence Borel measurable, so \(\chi_E \circ g = \chi_{g^{-1}(E)}\) is measurable, and the integrand is nonnegative. Countable additivity of the integral of nonnegative measurable functions (Monotone Convergence Theorem) makes \(\nu\) a measure, and \(\nu([c,d]) = \int_a^b g^{\prime} = d - c < \infty\).

(ii) \(\nu = m\) on intervals. Given \(c \le \alpha \le \beta \le d\), put \(u = \inf\{x : g(x) \ge \alpha\}\) and \(v = \sup\{x : g(x) \le \beta\}\); both sets are nonempty since \(g(b) = d \ge \alpha\) and \(g(a) = c \le \beta\). Continuity and monotonicity give \(g(u) = \alpha\), \(g(v) = \beta\), \(u \le v\), and \(g^{-1}([\alpha,\beta]) = [u,v]\). Hence

\begin{equation*} \nu([\alpha,\beta]) = \int_u^v g^{\prime} = g(v) - g(u) = \beta - \alpha = m([\alpha,\beta]), \end{equation*}

and the degenerate case \(\beta = \alpha\) gives \(\nu(\{\alpha\}) = 0\), so \(\nu\) and \(m\) agree on all subintervals of \([c,d]\).

(iii) \(\nu = m\) on Borel sets. An open \(\mathcal{O} \subseteq (c,d)\) is a countable disjoint union of open intervals, so \(\nu(\mathcal{O}) = m(\mathcal{O})\) by countable additivity; a \(G_\delta\) set \(G \subseteq (c,d)\) is a decreasing intersection of such, and both measures are finite, so continuity from above gives \(\nu(G) = m(G)\). For Borel \(E\), discard \(\{c\},\{d\}\) (null for both by (ii)), take a \(G_\delta\) set \(G\) with \(E \subseteq G \subseteq (c,d)\), \(m(G \sim E) = 0\), and a \(G_\delta\) set \(H \subseteq (c,d)\) with \(G \sim E \subseteq H\), \(m(H) = 0\); then \(\nu(G \sim E) \le \nu(H) = m(H) = 0\), whence \(\nu(E) = \nu(G) = m(G) = m(E)\).

(iv) Null sets, and Lebesgue measurable \(E\). If \(m(N) = 0\), take Borel \(B \supseteq N\) with \(m(B) = 0\); by (iii) \(\int_a^b \chi_B(g(x))g^{\prime}(x)\,dx = 0\) with nonnegative integrand, so by Proposition 9 of Chapter 4,

\begin{equation*} g^{\prime} = 0 \ \text{ a.e. on } g^{-1}(B) \supseteq g^{-1}(N). \tag{\(\ast\)} \end{equation*}

So for Lebesgue measurable \(E \subseteq [c,d]\) and a Borel \(B_0\) with \(m(E\,\Delta\,B_0) = 0\), \((\ast)\) applied to \(E\,\Delta\,B_0\) gives \(\chi_E(g(x))g^{\prime}(x) = \chi_{B_0}(g(x))g^{\prime}(x)\) a.e.; by completeness of \(m\) the left side is measurable and

\begin{equation*} \int_a^b \chi_E(g(x))g^{\prime}(x)\,dx = \nu(B_0) = m(B_0) = m(E). \tag{\(\ast\ast\)} \end{equation*}

(In particular, by \((\ast)\) the right-hand side of the formula does not depend on the representative chosen for \(f\).)

(v) Simple functions, then \(f\). For nonnegative simple \(\varphi = \sum_{k=1}^n a_k \chi_{E_k}\) on \([c,d]\), linearity and \((\ast\ast)\) give \(\int_a^b \varphi(g(x))g^{\prime}(x)\,dx = \sum_k a_k\,m(E_k) = \int_c^d \varphi\). Taking nonnegative simple \(\varphi_n \uparrow f\) (Simple Approximation Theorem), for a.a. \(x\) the products \(\varphi_n(g(x))g^{\prime}(x)\) increase to \(f(g(x))g^{\prime}(x)\), so two applications of the Monotone Convergence Theorem give

\begin{equation*} \int_a^b f(g(x))g^{\prime}(x)\,dx = \lim_{n \to \infty}\int_c^d \varphi_n = \int_c^d f(y)\,dy . \end{equation*}

Take \(f(x) = \int_0^x \chi_E\), where \(E = [0,1] \setminus C\) and \(C\) is a generalized Cantor set with \(m( C) = 1/2\).

The set \(C\). Following Problem 39 of Chapter 2 with \(\alpha = 1/2\): set \(C_0 = [0,1]\) and obtain \(C_n\) from \(C_{n-1}\) by deleting from the interior of each of its \(2^{n-1}\) closed intervals the open middle interval of length \(\alpha 3^{-n}\). This is legitimate, since the common length of the intervals of \(C_n\) is

\begin{equation*} \ell_n = \frac{1}{2^n}\Bigl(1 - \alpha\bigl(1 - (2/3)^n\bigr)\Bigr) \ \ge\ \frac{1-\alpha}{2^n}, \end{equation*}

and \(\alpha 3^{-n} < (1-\alpha)2^{-(n-1)}\) for \(\alpha = 1/2\), \(n \ge 1\) (Check!). Put \(C = \bigcap_n C_n\), a closed set. The deleted intervals are pairwise disjoint of total length \(\sum_{n \ge 1} 2^{n-1}\alpha 3^{-n} = \alpha\), so \(m( C) = 1 - \alpha = 1/2\); and \(C \subseteq C_n\) forces any interval inside \(C\) to have length \(\le \ell_n \le 2^{-n}\) for every \(n\), so \(C\) has empty interior.

\(f\) is absolutely continuous, being the indefinite integral of the bounded, hence integrable, function \(\chi_E\) (Theorem 11).

\(f\) is strictly increasing: for \(0 \le x < y \le 1\), \(f(y) - f(x) = m(E \cap (x,y))\), and \(E \cap (x,y) = (x,y) \setminus C\) is nonempty (as \(C\) has empty interior) and open (as \(C\) is closed), hence of positive measure.

\(f^{\prime} = 0\) on a set of measure \(1/2\): by Theorem 14, \(f^{\prime} = \chi_E\) a.e. on \((0,1)\), say off a null set \(Z\), so \(f^{\prime} = 0\) at every point of \(C \cap (0,1) \setminus Z\), a set of measure \(m( C) = 1/2\).

No, not as it stands: the displayed derivative statement is true, but neither establishing it nor deducing the formula from it is possible without the machinery of Problem 56, which already proves the formula outright.

Write \(F(y) = \int_{g(a)}^y f\), \(\Phi(x) = \int_a^x f(g(t))g^{\prime}(t)\,dt\), and \(H = F \circ g - \Phi\), the bracketed expression.

(i) A vanishing derivative does not force constancy. The Cantor-Lebesgue function is continuous and increasing with derivative \(0\) a.e., yet not constant; to pass from \(H^{\prime} = 0\) a.e. to \(H \equiv H(a) = 0\) one needs \(H\) absolutely continuous, so that Theorem 10 gives \(H(x) - H(a) = \int_a^x H^{\prime} = 0\). Now \(F \circ g\) is absolutely continuous by the argument of Problem 45: \(F\) is absolutely continuous on \([g(a),g(b)]\) as an indefinite integral (Theorem 11), and \(g\) strictly increasing carries a disjoint collection in \((a,b)\) of small total length to a disjoint collection in \((g(a),g(b))\) of small total length. But \(\Phi\) is a real-valued absolutely continuous function only if \((f \circ g)\cdot g^{\prime}\) is integrable over \([a,b]\) – a priori this nonnegative measurable product could have integral \(+\infty\), in which case \(H\) is not even defined. That integrability is half of the conclusion of the theorem, so the scheme presupposes much of what it should prove.

(ii) The chain rule is unavailable on a set that can have positive measure. Granting integrability, the natural computation

\begin{equation*} H^{\prime}(x) = F^{\prime}(g(x))\,g^{\prime}(x) - f(g(x))\,g^{\prime}(x) = 0 \end{equation*}

uses Theorem 14 in the form \(F^{\prime} = f\), which holds only off some null \(N \subseteq [g(a),g(b)]\); so it is valid only for \(x \notin g^{-1}(N)\), and \(g^{-1}(N)\) can have positive measure. Take \(g\) as in Problem 58, so \(g\) is strictly increasing and absolutely continuous on \([0,1]\) with \(g^{\prime} = \chi_E\), \(E = [0,1]\setminus C\), \(m( C) = 1/2\), and \(g([0,1]) = [0,1/2]\). Then \(N := g( C)\) has \(m(N) = \int_C g^{\prime} = 0\) by Problem 56(iv), while \(g^{-1}(N) = C\) has measure \(1/2\) since \(g\) is one-to-one. For \(f = 1 + \chi_N\) on \([0,1/2]\) we get \(F(y) = y\), so \(F^{\prime} \equiv 1\), whereas \(f = 2\) on \(N\): the step \(F^{\prime}(g(x)) = f(g(x))\) is false for every \(x \in C\).

(iii) What the repair costs. On \(A := g^{-1}(N)\) both terms must be handled separately. For the second: \(m(g(A)) \le m(N) = 0\), so \(\int_A g^{\prime} = 0\) by Problem 56(iv) and hence \(g^{\prime} = 0\) a.e. on \(A\) (Proposition 9 of Chapter 4, \(g^{\prime} \ge 0\) a.e.). For the first one needs the same identity \(\int_A \psi^{\prime} = m(\psi(A))\) for the increasing absolutely continuous \(\psi := F \circ g\), together with \(m(\psi(A)) \le m(F(N)) = 0\), the latter because an increasing absolutely continuous function carries null sets to null sets (Problems 38 and 39). Granting these, \(H^{\prime} = 0\) a.e. on \(A\), the chain rule applies off \(A\), and \(H\) absolutely continuous with \(H(a) = 0\) gives \(H \equiv 0\), the change of variables formula.

So the scheme goes through only after assuming the integrability of \((f \circ g)g^{\prime}\) and invoking \(m(h(A)) = \int_A h^{\prime}\), that is, parts (i)-(iv) of Problem 56 – which already deliver the formula through parts (v) and (vi).

An absolutely continuous singular \(f\) is constant because Theorem 10, applied to \(f\) on \([a,x]\) (where it is again absolutely continuous), gives

\begin{equation*} f(x) - f(a) = \int_a^x f^{\prime} = 0 \qquad \text{for every } x \in [a,b], \end{equation*}

the integral vanishing since \(f^{\prime} = 0\) a.e.

Uniqueness. Let \(f = g_1 + h_1 = g_2 + h_2\) with \(g_i\) absolutely continuous, \(h_i\) singular, \(h_1(a) = h_2(a) = 0\), and set \(u = g_1 - g_2 = h_2 - h_1\). As \(u = g_1 - g_2\), it is absolutely continuous (take \(\delta = \min\{\delta_1,\delta_2\}\), where \(\delta_i\) responds to the \(\epsilon/2\) challenge for \(g_i\)). As \(u = h_2 - h_1\), it is of bounded variation and, at each of the almost all points where both \(h_i^{\prime}\) exist and vanish, \(u^{\prime} = h_2^{\prime} - h_1^{\prime} = 0\); so \(u\) is singular. By the first part \(u\) is constant, and \(u(a) = h_2(a) - h_1(a) = 0\); hence \(u \equiv 0\), that is \(g_1 = g_2\) and \(h_1 = h_2\).

Sufficiency is the case \(n = 2\): with \(\lambda_1 = \lambda\), \(\lambda_2 = 1-\lambda\) the inequality is exactly (38).

Necessity, by induction on \(n\). Note first that \(\sum_k \lambda_k x_k\) lies in \([\min_k x_k, \max_k x_k] \subseteq (a,b)\), so the left-hand side is defined. The cases \(n = 1, 2\) are the identity and (38). Let \(n \ge 3\), assume the result for \(n-1\) points, and let \(x_k \in (a,b)\), \(\lambda_k \ge 0\), \(\sum_{k=1}^n \lambda_k = 1\). If \(\lambda_n = 1\) the inequality is an identity, so let \(\mu = \sum_{k=1}^{n-1}\lambda_k = 1 - \lambda_n > 0\) and \(y = \sum_{k=1}^{n-1}(\lambda_k/\mu)x_k \in (a,b)\). Since \(\sum_{k=1}^n \lambda_k x_k = \mu y + (1-\mu)x_n\), (38) applied to \(y, x_n\) with \(\lambda = \mu\), then the induction hypothesis multiplied by \(\mu > 0\), give

\begin{equation*} \begin{aligned} \varphi\Bigl(\sum_{k=1}^n \lambda_k x_k\Bigr) &\le \mu\,\varphi(y) + (1-\mu)\varphi(x_n) \\ &\le \mu \sum_{k=1}^{n-1}\frac{\lambda_k}{\mu}\varphi(x_k) + \lambda_n \varphi(x_n) = \sum_{k=1}^n \lambda_k \varphi(x_k). \end{aligned} \end{equation*}

Jensen’s Inequality for simple \(f\). Let \(\varphi\) be convex on \((-\infty,\infty)\) and \(f = \sum_{k=1}^n c_k \chi_{E_k}\) the canonical form of a simple function on \([0,1]\), the \(E_k = f^{-1}(c_k)\) being disjoint with union \([0,1]\). Put \(\lambda_k = m(E_k) \ge 0\), so \(\sum_k \lambda_k = m([0,1]) = 1\) by finite additivity. Since \(\varphi \circ f = \sum_k \varphi(c_k)\chi_{E_k}\), the definition of the integral of a simple function and the inequality just proved give

\begin{equation*} \varphi\left(\int_0^1 f\right) = \varphi\Bigl(\sum_{k=1}^n \lambda_k c_k\Bigr) \le \sum_{k=1}^n \lambda_k \varphi(c_k) = \int_0^1 (\varphi \circ f), \end{equation*}

which is (42) for \(f\) simple.

Necessity is (38) with \(\lambda = 1/2\); sufficiency runs the inequality along dyadic \(\lambda\) and closes by continuity.

Fix \(x_1 \ne x_2\) in \((a,b)\) and put \(z(\lambda) = \lambda x_1 + (1-\lambda)x_2\), a point of the closed interval with endpoints \(x_1, x_2\), which lies in \((a,b)\); \(z\) is affine in \(\lambda\).

(i) Dyadic \(\lambda\). Let \(D_n = \{k/2^n : 0 \le k \le 2^n\}\); we show \(\varphi(z(\lambda)) \le \lambda\varphi(x_1) + (1-\lambda)\varphi(x_2)\) for \(\lambda \in D_n\), by induction on \(n\). For \(n = 0\) this is an identity. Given it for \(n\), let \(\lambda = k/2^{n+1}\); if \(k\) is even then \(\lambda \in D_n\), and if \(k\) is odd then \(\lambda^{\prime} = (k-1)/2^{n+1}\) and \(\lambda^{\prime\prime} = (k+1)/2^{n+1}\) lie in \(D_n\) with \(\lambda = (\lambda^{\prime}+\lambda^{\prime\prime})/2\), so \(z(\lambda) = [z(\lambda^{\prime}) + z(\lambda^{\prime\prime})]/2\) and midpoint convexity followed by the inductive hypothesis gives

\begin{equation*} \begin{aligned} \varphi(z(\lambda)) &\le \frac{\varphi(z(\lambda^{\prime})) + \varphi(z(\lambda^{\prime\prime}))}{2} \\ &\le \frac{\lambda^{\prime}+\lambda^{\prime\prime}}{2}\varphi(x_1) + \Bigl(1 - \frac{\lambda^{\prime}+\lambda^{\prime\prime}}{2}\Bigr)\varphi(x_2) = \lambda\varphi(x_1) + (1-\lambda)\varphi(x_2). \end{aligned} \end{equation*}

(ii) General \(\lambda\). The dyadic rationals are dense in \([0,1]\), so take \(\lambda_j \to \lambda\) with \(\lambda_j\) dyadic. Then \(z(\lambda_j) \to z(\lambda)\) inside the closed interval with endpoints \(x_1, x_2\), on which \(\varphi\) is continuous, so passing to the limit in (i) gives (38). Hence \(\varphi\) is convex.

No. Take \(\varphi(x) = -\sqrt{x}\) on \([0,1]\).

It is convex in the stated sense. It is continuous on \([0,1]\); on \((0,1)\) it has \(\varphi^{\prime\prime}(x) = \tfrac14 x^{-3/2} > 0\), so \(\varphi^{\prime}\) is increasing there and Proposition 15 gives (38) for all \(x_1, x_2 \in (0,1)\); and (38) extends to pairs involving the endpoints by approximating \(x_1, x_2 \in [0,1]\) by points of \((0,1)\) and passing to the limit, both sides being continuous in the points.

It is not Lipschitz: for \(0 < x \le 1\),

\begin{equation*} \left| \frac{\varphi(x) - \varphi(0)}{x - 0} \right| = \frac{1}{\sqrt{x}}, \end{equation*}

which exceeds any prescribed \(c > 0\) once \(x < \min\{1, c^{-2}\}\).

Corollary 17 is not contradicted: it gives Lipschitzness only on closed subintervals \([c,d] \subseteq (a,b)\), its proof using the finiteness of \(\varphi^{\prime}(c^+)\) and \(\varphi^{\prime}(d^-)\) guaranteed by Lemma 16 at interior points. Here \(\varphi^{\prime}(0^+) = -\infty\).

Both directions pass through the monotonicity of \(\varphi^{\prime}\), which exists on all of \((a,b)\) since \(\varphi^{\prime\prime}\) does.

(i) If \(\varphi^{\prime\prime} \ge 0\) on \((a,b)\), then for \(u < v\) in \((a,b)\) the Mean Value Theorem applied to \(\varphi^{\prime}\) gives \(\varphi^{\prime}(v) - \varphi^{\prime}(u) = \varphi^{\prime\prime}( c)(v-u) \ge 0\) for some \(c \in (u,v)\). So \(\varphi\) is differentiable with increasing derivative, and Proposition 15 makes \(\varphi\) convex.

(ii) If \(\varphi\) is convex, its one-sided derivatives coincide with \(\varphi^{\prime}\) at every point, so (40) of Lemma 16 reads, for \(u < v\) in \((a,b)\),

\begin{equation*} \varphi^{\prime}(u) \le \frac{\varphi(v)-\varphi(u)}{v-u} \le \varphi^{\prime}(v), \end{equation*}

whence \(\varphi^{\prime}\) is increasing. Fix \(x \in (a,b)\). For small \(h \ne 0\) the quotient \([\varphi^{\prime}(x+h) - \varphi^{\prime}(x)]/h\) has nonnegative numerator and denominator of the same sign, hence is nonnegative; letting \(h \to 0\), \(\varphi^{\prime\prime}(x) \ge 0\).

On \((0,\infty)\) we have \(a + bt > 0\), so \(\varphi\) is twice differentiable there with

\begin{equation*} \varphi^{\prime}(t) = p\,b\,(a+bt)^{p-1}, \qquad \varphi^{\prime\prime}(t) = p(p-1)\,b^2\,(a+bt)^{p-2} \ge 0 \end{equation*}

(the possibly negative exponents are harmless since the base is positive, and \(p(p-1) \ge 0\) as \(p \ge 1\)). By Proposition 15, \(\varphi\) is convex on \((0,\infty)\).

Convexity on \([0,\infty)\) in the sense of Problem 63 then follows by continuity: \(\varphi\) is continuous on \([0,\infty)\), being \(t \mapsto a+bt \ge 0\) followed by \(s \mapsto s^p\), which is continuous on \([0,\infty)\) for \(p \ge 1\). Only the case \(x_1 = 0 < x_2\), \(0 < \lambda < 1\), is not already covered (the other cases being identities or interior); applying the interior inequality to \(\varepsilon\) and \(x_2\) for \(0 < \varepsilon < x_2\) and letting \(\varepsilon \to 0^+\),

\begin{equation*} \varphi\bigl(\lambda \cdot 0 + (1-\lambda)x_2\bigr) \le \lambda\varphi(0) + (1-\lambda)\varphi(x_2). \end{equation*}

Hence (38) holds throughout \([0,\infty)\).

Precisely the affine functions \(\varphi(t) = mt + c\).

Sufficiency. Such a \(\varphi\) is convex, satisfying (38) with equality, and for \(f\) integrable over \([0,1]\), \(\varphi \circ f = mf + c\) is integrable and, by linearity and \(\int_0^1 1 = 1\),

\begin{equation*} \int_0^1 (\varphi \circ f) = m\int_0^1 f + c = \varphi\left(\int_0^1 f\right). \end{equation*}

Necessity. Two-valued simple functions already force it. For real \(t_1, t_2\) and \(\lambda \in [0,1]\) put \(f = t_1\chi_{[0,\lambda)} + t_2\chi_{[\lambda,1]}\), bounded and measurable, hence integrable, with \(\varphi \circ f = \varphi(t_1)\chi_{[0,\lambda)} + \varphi(t_2)\chi_{[\lambda,1]}\) likewise integrable. Since \(m([0,\lambda)) = \lambda\), equality in Jensen’s Inequality for this \(f\) reads

\begin{equation*} \varphi(\lambda t_1 + (1-\lambda)t_2) = \lambda\varphi(t_1) + (1-\lambda)\varphi(t_2). \end{equation*}

Set \(m = \varphi(1) - \varphi(0)\), \(c = \varphi(0)\), and let \(t \in \mathbb{R}\).

• \(0 \le t \le 1\): apply the equality to \(0, 1\) with \(\lambda = 1-t\), giving \(\varphi(t) = (1-t)\varphi(0) + t\varphi(1) = mt + c\).
• \(t > 1\): apply it to \(0, t\) with \(1 - \lambda = 1/t\), giving \(\varphi(1) = (1 - 1/t)\varphi(0) + \varphi(t)/t\), so \(\varphi(t) = t\varphi(1) - (t-1)\varphi(0) = mt+c\).
• \(t < 0\): apply it to \(t, 1\) with \(\lambda = 1/(1-t)\), which sends the combination to \(0\), giving \(\varphi(0) = \varphi(t)/(1-t) - t\varphi(1)/(1-t)\), so \(\varphi(t) = (1-t)\varphi(0) + t\varphi(1) = mt+c\).

Replace integrals by averages: \([0,1]\) enters (42) only through \(m([0,1]) = 1\).

Jensen’s Inequality on \([a,b]\). Let \(a < b\), let \(\varphi\) be convex on \((-\infty,\infty)\), let \(f\) be integrable over \([a,b]\), and suppose \(\varphi \circ f\) is integrable over \([a,b]\). Then

\begin{equation*} \varphi\left(\frac{1}{b-a}\int_a^b f(x)\,dx\right) \le \frac{1}{b-a}\int_a^b (\varphi \circ f)(x)\,dx . \end{equation*}

Proof. Put \(\alpha = \frac{1}{b-a}\int_a^b f\), a real number since \(f\) is integrable. By Lemma 16 the convex \(\varphi\) has one-sided derivatives at \(\alpha\) with \(\varphi^{\prime}(\alpha^-) \le \varphi^{\prime}(\alpha^+)\); choose \(m\) between them, so that the line through \((\alpha, \varphi(\alpha))\) of slope \(m\) supports the graph:

\begin{equation*} \varphi(t) \ge m(t-\alpha) + \varphi(\alpha) \qquad \text{for all } t \in \mathbb{R} \end{equation*}

(for \(t > \alpha\), Lemma 16 gives \([\varphi(t)-\varphi(\alpha)]/(t-\alpha) \ge \varphi^{\prime}(\alpha^+) \ge m\); for \(t < \alpha\), \([\varphi(\alpha)-\varphi(t)]/(\alpha-t) \le \varphi^{\prime}(\alpha^-) \le m\)). Since \(f\) is finite a.e., substituting \(t = f(x)\) gives \(\varphi(f(x)) \ge m[f(x)-\alpha] + \varphi(\alpha)\) a.e. on \([a,b]\), with both sides integrable (the left by hypothesis, the right because \(f\) and constants are). Monotonicity and linearity of the integral, with \(\int_a^b f = \alpha(b-a)\), then give

\begin{equation*} \int_a^b (\varphi \circ f) \ge m\bigl[\alpha(b-a) - \alpha(b-a)\bigr] + \varphi(\alpha)(b-a) = \varphi(\alpha)(b-a) , \end{equation*}

and dividing by \(b - a > 0\) yields the inequality. Taking \([a,b] = [0,1]\) recovers (42).

Apply the supporting-line inequality for the convex \(\varphi(t) = e^t\) at \(\alpha = \int_0^1 f\), a real number since \(f\) is integrable.

Convexity holds because \(\varphi^{\prime\prime}(t) = e^t > 0\) (Proposition 15), and \(\varphi\) being differentiable, its supporting line at \(\alpha\) is the tangent, of slope \(e^{\alpha}\):

\begin{equation*} e^{t} \ge e^{\alpha}(t-\alpha) + e^{\alpha} \qquad \text{for all } t \in \mathbb{R} . \end{equation*}

Since \(f\) is integrable it is finite a.e., so substituting \(t = f(x)\),

\begin{equation*} e^{f(x)} \ge e^{\alpha}\bigl(f(x) - \alpha\bigr) + e^{\alpha} \qquad \text{for a.a. } x \in [0,1]. \end{equation*}

The right-hand side is integrable, being \(e^{\alpha}f\) plus a constant; \(e^{f}\) is nonnegative and measurable, since \(\varphi\) is continuous and the composition of a continuous with a measurable function is measurable (Proposition 7 of Chapter 3). Monotonicity of the integral of nonnegative measurable functions, valid even when the value \(+\infty\) occurs, together with \(\int_0^1 f = \alpha\) and \(\int_0^1 1 = 1\), gives

\begin{equation*} \int_0^1 \exp(f(x))\,dx \ge e^{\alpha}\cdot 0 + e^{\alpha} = \exp\left[\int_0^1 f(x)\,dx\right] . \end{equation*}

Realize \(\{\zeta_n\}\) as a step function on \([0,1]\) with \(m(E_n)=\alpha_n\) and apply Problem 68. Since \(\sum_n\alpha_n=1\), some \(\alpha_{n_0}>0\), so \(S:=\sum_{n=1}^{\infty}\alpha_n\zeta_n\in(0,\infty]\); if \(S=\infty\) there is nothing to prove, so assume \(S<\infty\).

From \(e^{s}\ge 1+s\) we get \(\log\zeta\le\zeta-1\) for \(\zeta>0\), so that \((\log\zeta_n)^{+}\le\zeta_n\) and

\begin{equation*} \sum_{n=1}^{\infty}\alpha_n(\log\zeta_n)^{+} \le\sum_{n=1}^{\infty}\alpha_n\zeta_n=S<\infty . \end{equation*}

Hence \(\sum_n\alpha_n\log\zeta_n\) is defined in \([-\infty,\infty)\) and its partial sums converge to it in the extended sense; since \(\prod_{n=1}^{N}\zeta_n^{\alpha_n}=\exp\bigl(\sum_{n=1}^{N}\alpha_n\log\zeta_n\bigr)\) and \(\exp(s)\to 0\) as \(s\to-\infty\),

\begin{equation*} \prod_{n=1}^{\infty}\zeta_n^{\alpha_n} =\exp\left(\sum_{n=1}^{\infty}\alpha_n\log\zeta_n\right), \qquad \exp(-\infty):=0 . \end{equation*}

If that sum is \(-\infty\) the left-hand side is \(0<S\) and we are done; so assume it finite, which with the last display forces \(\sum_n\alpha_n|\log\zeta_n|<\infty\).

Put \(A_0=0\), \(A_N=\sum_{n=1}^{N}\alpha_n\), and \(E_n=[A_{n-1},A_n)\): these are disjoint with \(m(E_n)=\alpha_n\) and union \([0,1)\), which differs from \([0,1]\) by a null set. Let \(g=\zeta_n\) on \(E_n\), \(g(1)=1\), and \(f=\log g\). Countable additivity of integration over the \(E_n\) (Monotone Convergence Theorem of Chapter 4, applied to \(f^{+}\), \(f^{-}\) and to \(g\)) gives

\begin{equation*} \int_0^1|f|=\sum_{n=1}^{\infty}\alpha_n|\log\zeta_n|<\infty, \qquad \int_0^1 e^{f}=\int_0^1 g=S, \end{equation*}

and, \(f\) being thereby integrable, \(\int_0^1 f=\sum_{n}\alpha_n\log\zeta_n\). So Problem 68 applies to \(f\) and yields

\begin{equation*} \prod_{n=1}^{\infty}\zeta_n^{\alpha_n}=\exp\left[\int_0^1 f\right]\le\int_0^1 e^{f}=\sum_{n=1}^{\infty}\alpha_n\zeta_n . \end{equation*}

Method (2): apply \(\log\zeta\le\zeta-1\) to \(\zeta=\zeta_n/S\), multiply by \(\alpha_n\ge 0\) and sum over \(n\le N\):

\begin{equation*} \sum_{n=1}^{N}\alpha_n\log\zeta_n\le A_N\log S+\frac{1}{S}\sum_{n=1}^{N}\alpha_n\zeta_n-A_N . \end{equation*}

Letting \(N\to\infty\), the right-hand side tends to \(\log S+1-1=\log S\), so \(\sum_n\alpha_n\log\zeta_n\le\log S\), the same conclusion.

Apply Problem 68 to \(f=\log g\). First, \(\int_0^1 g>0\): the sets \(E_k=\{g>1/k\}\) increase to \([0,1]\), so \(m(E_{k_0})>0\) for some \(k_0\) by continuity of measure, whence \(\int_0^1 g\ge m(E_{k_0})/k_0>0\). Thus the left-hand side is defined in \((-\infty,\infty]\).

(i) \(\int_0^1 g=+\infty\). The left-hand side is \(+\infty\) and dominates whatever value the right-hand side takes.

(ii) \(A:=\int_0^1 g<\infty\), so \(g\) is integrable and finite a.e., and \(e^{f}=g\) a.e. By \(\log\zeta\le\zeta-1\) (Problem 69),

\begin{equation*} \int_0^1 f^{+}\le\int_0^1 (g-1)^{+}\le\int_0^1 g=A<\infty , \end{equation*}

so \(\int_0^1 f\) is defined in \([-\infty,\infty)\). If \(\int_0^1 f=-\infty\) the assertion reads \(\log A\ge-\infty\), true since \(A>0\). Otherwise \(f\) is integrable over \([0,1]\) and Problem 68 gives

\begin{equation*} \exp\left[\int_0^1 \log g\right]=\exp\left[\int_0^1 f\right]\le\int_0^1 e^{f}=\int_0^1 g=A . \end{equation*}

Since \(A>0\) and \(\log\) is increasing on \((0,\infty)\), taking logarithms yields

\begin{equation*} \int_0^1 \log(g(x))\,dx\le\log A=\log\left(\int_0^1 g(x)\,dx\right). \end{equation*}

Each half is a one-line domination, in opposite directions.

Sufficiency. Assume \(|\varphi(x)|\le c_1+c_2|x|\) on \(\mathbb{R}\) and let \(f\) be integrable over \([0,1]\). Then \(f\) is finite a.e. (Proposition 15 of Chapter 4), so redefining \(f\) to be \(0\) on the null set where \(|f|=\infty\) changes neither \(\int_0^1|f|\) nor \(\varphi\circ f\) a.e.; assume \(f\) real-valued. Since \(\varphi\) is continuous on \(\mathbb{R}\) and \(f\) is measurable and real-valued, \(\varphi\circ f\) is measurable (Proposition 7 of Chapter 3), and pointwise on \([0,1]\)

\begin{equation*} |(\varphi\circ f)(x)|\le c_1+c_2|f(x)|=:g(x) . \end{equation*}

Here \(g\ge 0\) is integrable over \([0,1]\), since \(|f|\) is and \(m([0,1])<\infty\) (Theorem 17 of Chapter 4), with \(\int_0^1 g=c_1+c_2\int_0^1|f|<\infty\). By the Integral Comparison Test (Proposition 16 of Chapter 4), \(\varphi\circ f\) is integrable over \([0,1]\).

Necessity. Suppose no such constants exist. Then for each \(n\) the choice \(c_1=c_2=n\) fails, giving \(x_n\in\mathbb{R}\) with \(|\varphi(x_n)|>n(1+|x_n|)\). Put

\begin{equation*} \alpha_n=\frac{1}{2n^2(1+|x_n|)}\in\left(0,\tfrac{1}{2n^2}\right], \qquad \sum_{n=1}^{\infty}\alpha_n\le\frac{\pi^2}{12}<1 , \end{equation*}

let \(E_n=[s_{n-1},s_n)\subseteq[0,1]\) with \(s_N=\sum_{n\le N}\alpha_n\), so \(m(E_n)=\alpha_n\), and set \(f=\sum_{n=1}^{\infty}x_n\chi_{E_n}\) (that is, \(f=x_n\) on \(E_n\) and \(f=0\) off \(\bigcup_n E_n\)). As a pointwise limit of simple functions \(f\) is measurable, and by the Monotone Convergence Theorem applied to \(\sum_{n\le N}|x_n|\chi_{E_n}\),

\begin{equation*} \int_0^1|f|=\sum_{n=1}^{\infty}\alpha_n|x_n| =\sum_{n=1}^{\infty}\frac{|x_n|}{2n^2(1+|x_n|)} \le\frac12\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty , \end{equation*}

so \(f\) is integrable over \([0,1]\). But \(|\varphi\circ f|=|\varphi(x_n)|\) on \(E_n\), so by additivity over the disjoint \(E_1,\dots,E_N\) (Theorem 11 of Chapter 4) and the choice of \(x_n\),

\begin{equation*} \int_0^1|\varphi\circ f|\ge\sum_{n=1}^{N}\alpha_n|\varphi(x_n)| >\sum_{n=1}^{N}\frac{1}{2n} \qquad\text{for every } N , \end{equation*}

and the harmonic series diverges, so \(\varphi\circ f\) is not integrable. This contradicts the hypothesis; hence constants \(c_1,c_2\) with \(|\varphi(x)|\le c_1+c_2|x|\) on \(\mathbb{R}\) do exist.

It is a norm; no \(c\) works in the first inequality, and \(c=b-a\) works in the second. Here \(a<b\), and \(\|f\|_1\) is finite because a continuous \(f\) is bounded and measurable on a set of finite measure.

Homogeneity and the Triangle Inequality are the monotonicity and linearity of integration applied to \(|\alpha f|=|\alpha||f|\) and \(|f+g|\le|f|+|g|\). (Check!) For definiteness, suppose \(\|f\|_1=0\) but \(|f(x_0)|=\epsilon>0\); by continuity \(|f|>\epsilon/2\) on some \([c,d]\subseteq[a,b]\) with \(d-c>0\), so \[ \|f\|_1\ge\int_c^d|f|\ge\frac{\epsilon}{2}(d-c)>0, \] a contradiction. Thus \(\|f\|_1=0\) exactly when \(f\) vanishes at every point, and \(\|\cdot\|_1\) is a norm.

No \(c\ge 0\) has \(\|f\|_{\max}\le c\|f\|_1\). For \(n>1/(b-a)\) let \(f_n\) be the triangular spike

\begin{equation*} f_n(x)=\begin{cases} n\bigl(1-n(x-a)\bigr) & a\le x\le a+1/n, \\ 0 & a+1/n<x\le b, \end{cases} \end{equation*}

which is continuous with \(\|f_n\|_{\max}=f_n(a)=n\), while the triangle of base \(1/n\) and height \(n\) has area \[ \|f_n\|_1=\int_a^{a+1/n} n\bigl(1-n(x-a)\bigr)\,dx=\frac12 . \] So \(\|f_n\|_{\max}=2n\|f_n\|_1\), and \(\|f_n\|_{\max}\le c\|f_n\|_1\) would force \(n\le c/2\) for all such \(n\).

Take \(c=b-a\). Since \(|f|\le\|f\|_{\max}\) on \([a,b]\), monotonicity of integration gives \[ \|f\|_1=\int_a^b|f|\le (b-a)\,\|f\|_{\max}. \]

Yes, it is a norm. \(X\) is a linear subspace of the space of real-valued functions on \(\mathbb{R}\): it contains \(0\), and padding the shorter coefficient list with zeros, \[ (\alpha p+\beta q)(x)=\sum_{k=0}^{n}(\alpha a_k+\beta b_k)x^k \] is again a polynomial with real coefficients.

The coefficients are determined by the function, so \(\|p\|=\sum_{k\ge 0}|a_k|\) is unambiguous: if two coefficient lists give the same function, their difference is a polynomial vanishing at every real number, hence not a nonzero polynomial of degree \(d\) (which would have at most \(d\) roots), so all its coefficients vanish. Each \(p\) therefore has a unique coefficient sequence \((a_k)\) with finitely many nonzero terms, and \(\|p\|<\infty\).

The three norm properties now read off the coefficient sequences \((a_k+b_k)\) of \(p+q\) and \((\alpha a_k)\) of \(\alpha p\), all sums being finite: \[ \|p+q\|=\sum_{k\ge0}|a_k+b_k|\le\sum_{k\ge0}|a_k|+\sum_{k\ge0}|b_k|=\|p\|+\|q\| , \] \[ \|\alpha p\|=\sum_{k\ge0}|\alpha a_k|=|\alpha|\,\|p\| , \] and \(\|p\|\ge 0\) with \(\|p\|=0\) if and only if every \(a_k=0\), that is, by the uniqueness just established, if and only if \(p\) is the zero function.

It is a norm, the only point of substance being that the weight \(x^2\) vanishes only on a null set. With \(M=\max\{a^2,b^2\}\), monotonicity of integration gives \[ 0\le\|f\|=\int_a^b x^2|f(x)|\,dx\le M\int_a^b|f|=M\|f\|_1<\infty , \] and \(f=g\) a.e. forces \(x^2|f|=x^2|g|\) a.e., so \(\|\cdot\|\) is a finite functional on \(L^1[a,b]\).

Multiplying \(|f+g|\le|f|+|g|\) and \(|\alpha f|=|\alpha||f|\) (valid a.e., where \(f\) and \(g\) are finite) by \(x^2\ge 0\) and integrating gives, by linearity and monotonicity of integration, \[ \|f+g\|\le\|f\|+\|g\| , \qquad \|\alpha f\|=|\alpha|\,\|f\| . \]

Definiteness. If \(\|f\|=0\) then the nonnegative integrable function \(h(x)=x^2|f(x)|\) has zero integral, so \(h=0\) a.e. on \([a,b]\) by Proposition 9 of Chapter 4; off the null set \(\{h\ne 0\}\cup\{0\}\) we have \(x^2>0\), whence \(f(x)=0\). Thus \(f=0\) a.e., that is, \(f\) is the zero element of \(L^1[a,b]\); and plainly \(\|0\|=0\).

The infimum defining \(\|f\|_\infty\) is attained, and continuity upgrades “essentially” to “everywhere”. Put \[ S=\Bigl\{ M\in\mathbb{R} \;\Big|\; m\bigl\{x\in[a,b] : |f(x)|>M\bigr\}=0 \Bigr\}. \] Saying \(m\{|f|>M\}=0\) is saying \(|f|\le M\) a.e. on \([a,b]\), so \(S\) is exactly the set of essential upper bounds for \(f\) and \(\|f\|_\infty=\inf S\); every member of \(S\) is \(\ge 0\) because \(m([a,b])=b-a>0\), and \(S\ne\emptyset\) since \(f\in L^\infty[a,b]\).

The infimum lies in \(S\). For each \(n\) pick \(M_n\in S\) with \(M_n<\|f\|_\infty+1/n\) and set \(E_n=\{|f|>M_n\}\), a null set. Then \(E_\infty=\bigcup_n E_n\) is null by countable subadditivity, and off \(E_\infty\) we have \(|f|\le\|f\|_\infty+1/n\) for every \(n\), hence \(|f|\le\|f\|_\infty\). So \[ \bigl\{x\in[a,b] : |f(x)|>\|f\|_\infty\bigr\}\subseteq E_\infty \] is null, \(\|f\|_\infty\in S\), and being a lower bound for \(S\) it is \(\min S\).

Now let \(f\) be continuous, so \(\|f\|_{\max}=\max_{[a,b]}|f|\) exists. Since \(|f|\le\|f\|_{\max}\) everywhere, \(\|f\|_{\max}\in S\) and \(\|f\|_\infty\le\|f\|_{\max}\). If the inequality were strict, put \(\epsilon=\|f\|_{\max}-\|f\|_\infty>0\) and choose \(x_0\) with \(|f(x_0)|=\|f\|_{\max}\); by continuity of \(|f|\) at \(x_0\),

\begin{equation*} |f|>\|f\|_{\max}-\frac{\epsilon}{2}=\|f\|_\infty+\frac{\epsilon}{2} \quad\text{on } (x_0-\delta,x_0+\delta)\cap[a,b] \end{equation*}

for some \(\delta>0\). That set contains an interval of positive length, yet it lies in \(\{|f|>\|f\|_\infty\}\), which is null. Hence \(\|f\|_\infty=\|f\|_{\max}\).

Both are subspaces of the space of all real sequences, with the coordinatewise operations, and both functionals inherit the three norm properties from the termwise triangle inequality \(|a_k+b_k|\le|a_k|+|b_k|\).

The space \(\ell^\infty\), with \(\|a\|_\infty=\sup_k|a_k|\). For all \(k\),

\begin{equation*} |a_k+b_k|\le|a_k|+|b_k|\le\|a\|_\infty+\|b\|_\infty, \qquad |\alpha a_k|=|\alpha||a_k|\le|\alpha|\|a\|_\infty , \end{equation*}

so \(a+b\) and \(\alpha a\) are bounded and, the supremum being the least upper bound, \(\|a+b\|_\infty\le\|a\|_\infty+\|b\|_\infty\). For \(\alpha\ne 0\), multiplication by \(|\alpha|\) is an increasing bijection of \([0,\infty)\), so \(\|\alpha a\|_\infty=|\alpha|\sup_k|a_k|=|\alpha|\|a\|_\infty\); for \(\alpha=0\) both sides vanish. Finally \(\|a\|_\infty\ge 0\), and \(\|a\|_\infty=0\) forces every \(a_k=0\).

The space \(\ell^1\), with \(\|a\|_1=\sum_{k=1}^{\infty}|a_k|\). For each \(n\),

\begin{equation*} \sum_{k=1}^{n}|a_k+b_k| \le\sum_{k=1}^{n}|a_k|+\sum_{k=1}^{n}|b_k|\le\|a\|_1+\|b\|_1 , \end{equation*}

so the increasing partial sums of \(\sum_k|a_k+b_k|\) are bounded above and converge to a limit at most \(\|a\|_1+\|b\|_1\): thus \(a+b\in\ell^1\) with \(\|a+b\|_1\le\|a\|_1+\|b\|_1\). Likewise \(\sum_{k=1}^n|\alpha a_k|=|\alpha|\sum_{k=1}^n|a_k|\le|\alpha|\|a\|_1\), so \(\alpha a\in\ell^1\) and, letting \(n\to\infty\), \(\|\alpha a\|_1=|\alpha|\|a\|_1\). Finally \(\|a\|_1\ge 0\) as a limit of nonnegative partial sums, and \(\|a\|_1=0\) gives \(|a_k|\le\|a\|_1=0\) for each \(k\).

Divide by the norms. Assume that \(\int_E|u\cdot v|\le 1\) whenever \(u\in L^p(E)\), \(v\in L^q(E)\) are normalized, and let \(f\in L^p(E)\), \(g\in L^q(E)\) be arbitrary; here \(1\le p<\infty\) and \(q\) is conjugate to \(p\), so both norms are finite.

(i) \(\|f\|_p=0\) or \(\|g\|_q=0\). Then that factor vanishes a.e. on \(E\) (Proposition 9 of Chapter 4 applied to \(|f|^p\) for a finite index; for the index \(\infty\), from \(|h|\le\|h\|_\infty\) a.e.), so \(f\cdot g=0\) a.e. and both sides of Holder’s Inequality are \(0\).

(ii) \(\|f\|_p>0\) and \(\|g\|_q>0\). Put \(f^\sharp=f/\|f\|_p\) and \(g^\sharp=g/\|g\|_q\). Positive homogeneity of the norms, \(\|\alpha h\|_p=|\alpha|\|h\|_p\) by the linearity of integration applied to \(|\alpha h|^p=|\alpha|^p|h|^p\) (and as verified in Section 7.1 for the index \(\infty\)), makes these normalized, so the hypothesis applies to them. Since \(|f\cdot g|=\|f\|_p\|g\|_q\,|f^\sharp\cdot g^\sharp|\) pointwise on \(E\), the product \(f\cdot g\) is a constant multiple of an integrable function, hence integrable, and

\begin{equation*} \int_E|f\cdot g|=\|f\|_p\,\|g\|_q\int_E|f^\sharp\cdot g^\sharp|\le\|f\|_p\cdot\|g\|_q . \end{equation*}

Everything follows from the power-integral lemma

\begin{equation*} \int_{(0,1]}x^\beta\,dx<\infty\iff\beta>-1, \qquad \int_{(1,\infty)}x^\beta\,dx<\infty\iff\beta<-1, \end{equation*}

the finite values being \(1/(\beta+1)\) and \(-1/(\beta+1)\). (Truncate to \([1/n,1]\) or \((1,n]\), where \(x^\beta\) is continuous, evaluate the Riemann integral \(\int_{1/n}^1 x^\beta=(1-n^{-(\beta+1)})/(\beta+1)\) and \(\int_1^n x^\beta=(n^{\beta+1}-1)/(\beta+1)\), or \(\ln n\) when \(\beta=-1\), and let \(n\to\infty\) by the Monotone Convergence Theorem, the two integrals agreeing by Theorem 3 of Chapter 4.)

Example (i). \(E=(0,1]\), \(1\le p_1<p_2\le\infty\), and \(-1/p_1<\alpha<-1/p_2\) (read \(-1/p_2=0\) when \(p_2=\infty\)); such \(\alpha<0\) exists since \(p_1<p_2\). Put \(f(x)=x^\alpha\), continuous hence measurable.

Multiplying by \(p_1>0\) turns \(\alpha>-1/p_1\) into \(\alpha p_1>-1\), so by the lemma

\begin{equation*} \int_{(0,1]}|f|^{p_1}=\int_{(0,1]}x^{\alpha p_1}\,dx=\frac{1}{\alpha p_1+1}<\infty . \end{equation*}

If \(p_2<\infty\), then \(\alpha p_2<-1\) and the lemma gives \(\int_{(0,1]}|f|^{p_2}=\infty\). If \(p_2=\infty\), then for every \(M>0\), since \(\alpha<0\), \[ \{x\in(0,1] : |f(x)|>M\}\supseteq\bigl(0,\min\{1,M^{1/\alpha}\}\bigr), \] a set of positive measure, so \(f\) has no essential upper bound. Either way \(f\in L^{p_1}(E)\sim L^{p_2}(E)\), and the inclusion \(L^{p_2}(E)\subseteq L^{p_1}(E)\) of Corollary 3 is proper.

Example (ii). \(E=(1,\infty)\), \(f(x)=x^{-1/2}/(1+\ln x)\), positive and continuous there since \(\ln x>0\). For \(p=2\), \(|f|^2=1/\bigl(x(1+\ln x)^2\bigr)\) has antiderivative \(-1/(1+\ln x)\), so truncating to \((1,n]\) as in the lemma,

\begin{equation*} \int_{(1,\infty)}|f|^2=\lim_{n\to\infty}\left[1-\frac{1}{1+\ln n}\right]=1<\infty . \end{equation*}

For \(1\le p<2\) put \(\delta=1-p/2>0\); substituting \(t=\ln x\), \(x^\delta/(1+\ln x)^p=e^{\delta t}/(1+t)^p\to\infty\), so \(x^\delta\ge(1+\ln x)^p\) for \(x\ge X\) and there \[ |f(x)|^p=\frac{x^{-p/2}}{(1+\ln x)^p}\ge x^{-p/2-\delta}=\frac{1}{x} , \] whence \(\int_{(1,\infty)}|f|^p\ge\int_{(X,\infty)}x^{-1}dx=\infty\) by the lemma.

The printed claim that \(f\in L^p(E)\) only for \(p=2\) is a misprint: for \(2<p<\infty\), \((1+\ln x)^p>1\) gives \(|f|^p\le x^{-p/2}\) with \(-p/2<-1\), so \(\int_{(1,\infty)}|f|^p\le 2/(p-2)<\infty\), and \(0<f<1\) on \(E\) puts \(f\) in \(L^\infty(E)\); the correct statement is \(f\in L^p\bigl((1,\infty)\bigr)\iff 2\le p\le\infty\). Membership exactly at \(p=2\) is achieved on \(E=(0,\infty)\) by \[ h(x)=\frac{x^{-1/2}}{1+|\ln x|} , \] which agrees with \(f\) on \((1,\infty)\), while on \((0,1)\) the antiderivative \(1/(1-\ln x)\) of \(|h|^2\) gives \(\int_{(0,1)}|h|^2=1\); for \(p>2\), with \(\delta=p/2-1>0\), \(x^{-\delta}\ge(1-\ln x)^p\) near \(0\) forces \(|h|^p\ge 1/x\) there, and \(h(x)=e^{t/2}/(1+t)\) with \(t=-\ln x\) is unbounded on sets of positive measure, so \(h\notin L^\infty\). Thus \(h\in L^p\bigl((0,\infty)\bigr)\iff p=2\).

The displayed inequality says the quadratic \(p(\lambda)=A\lambda^2+2B\lambda+C\) is nonnegative, and its discriminant condition is Cauchy-Schwarz. First, \(f\cdot g\) is integrable over \(E\): it is measurable (Theorem 6 of Chapter 3), \(f\) and \(g\) are finite a.e. (Proposition 13 of Chapter 4, \(f^2\) and \(g^2\) being integrable), and \(0\le(|f|-|g|)^2\) gives a.e. on \(E\)

\begin{equation*} |f\cdot g|\le\tfrac12\bigl(f^2+g^2\bigr), \end{equation*}

whose right-hand side is integrable; apply the integral comparison test. So

\begin{equation*} A=\int_E f^2\ge 0, \qquad B=\int_E f\cdot g, \qquad C=\int_E g^2\ge 0 \end{equation*}

are all finite. Expanding \((\lambda f+g)^2=\lambda^2f^2+2\lambda f\cdot g+g^2\) a.e. on \(E\) and using the linearity of integration on these three integrable functions, then the monotonicity of integration on \((\lambda f+g)^2\ge 0\),

\begin{equation*} p(\lambda)=\lambda^2\int_E f^2+2\lambda\int_E f\cdot g+\int_E g^2=\int_E(\lambda f+g)^2\ge 0 \end{equation*}

for every real \(\lambda\).

(i) \(A>0\). Were \(B^2-AC>0\), the quadratic formula would give two distinct real roots

\begin{equation*} \lambda_\pm=\frac{-2B\pm\sqrt{4B^2-4AC}}{2A} , \end{equation*}

and \(p<0\) strictly between them, contradicting \(p\ge 0\). Hence \(B^2\le AC\).

(ii) \(A=0\). Then \(\int_E f^2=0\) with \(f^2\ge 0\) forces \(f=0\) a.e. on \(E\) (Proposition 9 of Chapter 4), so \(B=0\) and \(B^2\le AC\) holds trivially; symmetrically for \(C=0\).

Thus \(\bigl|\int_E f\cdot g\bigr|\le\sqrt{\int_E f^2}\,\sqrt{\int_E g^2}\) for all \(f,g\in L^2(E)\). Applying this to the pair \(|f|,|g|\), which lies in \(L^2(E)\) since \(|f|^2=f^2\) and \(|g|^2=g^2\), and noting that the integrand \(|f|\cdot|g|\) is nonnegative,

\begin{equation*} \int_E|f\cdot g|\le\sqrt{\int_E f^2}\cdot\sqrt{\int_E g^2}, \end{equation*}

which with the integrability established at the outset is the Cauchy-Schwarz Inequality.

Equality holds if and only if \(a^p=b^q\); the printed condition \(a=b=1\) is a misprint, since for \(p=q=2\) Young’s Inequality \(ab\le(a^2+b^2)/2\) is an equality for every \(a=b>0\).

Sharpen the book’s proof. For \(x>0\) put \(g(x)=\frac1p x^p+\frac1q-x\), so \(g^{\prime}(x)=x^{p-1}-1\) is negative on \((0,1)\) and positive on \((1,\infty)\), because \(p-1>0\) makes \(x\mapsto x^{p-1}\) strictly increasing with value \(1\) at \(x=1\). Hence \(g\) attains its minimum on \((0,\infty)\) at the single point \(x=1\), where \(g(1)=\frac1p+\frac1q-1=0\):

\begin{equation*} x\le\frac{1}{p}x^p+\frac1q \quad (x>0), \qquad\text{with equality if and only if } x=1 . \tag{6} \end{equation*}

Transfer. Since \(p(q-1)=q\), dividing Young’s Inequality by \(b^q>0\) and setting \(x_0=a/b^{q-1}>0\) turns it into

\begin{equation*} \frac{ab}{b^q}=x_0\;\le\;\frac{a^p}{p\,b^{p(q-1)}}+\frac{b^q}{q\,b^q} =\frac{x_0^{\,p}}{p}+\frac1q , \end{equation*}

which is \((6)\) at \(x_0\). Division by a positive number preserves equality as well as inequality, so

\begin{equation*} ab=\frac{a^p}{p}+\frac{b^q}{q}\iff x_0=1\iff a=b^{q-1}\iff a^p=b^{p(q-1)}=b^q . \end{equation*}

In particular \(a=b=1\) does give equality, which is the correct half of the printed claim.

Both halves run off Young’s Inequality in the form settled by Problem 9, extended to \(a,b\ge 0\):

\begin{equation*} ab\le\frac{a^p}{p}+\frac{b^q}{q}, \quad\text{with equality if and only if } a^p=b^q . \tag{7} \end{equation*}

(For \(a,b>0\) this is Problem 9; if exactly one of \(a,b\) is \(0\) the inequality is strict and \(a^p\ne b^q\); if both are \(0\) both sides vanish.) Here \(1<p<\infty\), which the exponents in the asserted relation presuppose.

Sufficiency. Let \(\alpha|f|^p=\beta|g|^q\) a.e. on \(E\), with \(\alpha,\beta\) not both zero.

(i) One of \(\alpha,\beta\) is zero, or they are nonzero of opposite sign. If \(\alpha=0\) then \(\beta|g|^q=0\) a.e. forces \(g=0\) a.e., so \(\int_E|f\cdot g|=0=\|f\|_p\|g\|_q\); symmetrically if \(\beta=0\). If \(\alpha\beta<0\), then \(\alpha|f|^p=\beta|g|^q\) is simultaneously \(\ge 0\) and \(\le 0\), so \(f=0\) and \(g=0\) a.e. and both sides again vanish.

(ii) \(\alpha,\beta\) nonzero of the same sign. Replacing \((\alpha,\beta)\) by \((-\alpha,-\beta)\) if need be, take \(\alpha,\beta>0\) and put \(c=\alpha/\beta>0\), so \(|g|^q=c|f|^p\) a.e. If \(f=0\) a.e. then \(g=0\) a.e. and both sides are \(0\); otherwise \(\|f\|_p>0\) and, since \(1/p+1/q=1\) gives \(1+p/q=p\), a.e. on \(E\)

\begin{equation*} |f\cdot g|=|f|\bigl(c|f|^p\bigr)^{1/q}=c^{1/q}|f|^{\,1+p/q}=c^{1/q}|f|^{\,p} . \end{equation*}

Integrating, \(\int_E|f\cdot g|=c^{1/q}\|f\|_p^p\); and from \(\|g\|_q^q=c\|f\|_p^p\),

\begin{equation*} \|f\|_p\|g\|_q=\|f\|_p\bigl(c\|f\|_p^p\bigr)^{1/q} =c^{1/q}\|f\|_p^{\,1+p/q}=c^{1/q}\|f\|_p^{\,p} . \end{equation*}

The two agree, so there is equality.

Necessity. Suppose \(\int_E|f\cdot g|=\|f\|_p\|g\|_q\). If \(f=0\) a.e. take \((\alpha,\beta)=(1,0)\); if \(g=0\) a.e. take \((0,1)\). Otherwise \(A=\|f\|_p>0\) and \(B=\|g\|_q>0\) (Proposition 9 of Chapter 4), and the normalizations \(F=|f|/A\), \(G=|g|/B\) satisfy \(\int_EF^p=\int_EG^q=1\) and

\begin{equation*} \int_E F\cdot G=\frac{1}{AB}\int_E|f\cdot g|=1 . \end{equation*}

Applying \((7)\) pointwise to \(F(x),G(x)\ge 0\) (finite a.e., since \(F^p,G^q\) are integrable), the integrable function \(h=\frac{F^p}{p}+\frac{G^q}{q}-F\cdot G\) is \(\ge 0\) a.e. with

\begin{equation*} \int_E h=\frac1p+\frac1q-1=0 , \end{equation*}

so \(h=0\) a.e. on \(E\) by Proposition 9 of Chapter 4. The equality case of \((7)\) then gives \(F^p=G^q\) a.e., that is,

\begin{equation*} \alpha|f|^p=\beta|g|^q \quad\text{a.e. on } E, \qquad \alpha=B^q>0,\ \beta=A^p>0 . \end{equation*}

Everything transfers along the linear injection \(x\mapsto T_x\), because \(I=[1,n+1)\) is the disjoint union of the \(I_k=[k,k+1)\) with \(m(I_k)=1\). The simple function \(T_x=\sum_{k=1}^n x_k\chi_{I_k}\) is bounded and measurable on a set of finite measure, so \(T_x\in L^p(I)\) for \(1\le p\le\infty\), and additivity of integration over the \(I_k\) gives the explicit formulas

\begin{equation*} \|x\|_p=\Bigl[\sum_{k=1}^n|x_k|^p\Bigr]^{1/p}\ \ (p<\infty), \qquad \|x\|_\infty=\max_{1\le k\le n}|x_k| , \end{equation*}

the second because \(|T_x|\le\max_k|x_k|\) on \(I\) while \(|T_x|=|x_k|\) on the set \(I_k\) of measure \(1>0\). Evaluating on each \(I_k\) shows \(T_{x+y}=T_x+T_y\) and \(T_{\lambda x}=\lambda T_x\) on \(I\); and \(T_x=0\) a.e. forces each \(x_k=0\), again because \(m(I_k)>0\), so \(T\) is one-to-one modulo null sets.

Hence \(\|\cdot\|_p\) is a norm on \(\mathbb{R}^n\): it is finite and nonnegative by the formulas; \(\|x\|_p=0\) gives \(\|T_x\|_p=0\), so \(T_x=0\) a.e. (Proposition 9 of Chapter 4 applied to \(|T_x|^p\) when \(p<\infty\); by definition of the essential supremum when \(p=\infty\)) and \(x=0\) by injectivity; and

\begin{equation*} \|\lambda x\|_p=\|\lambda T_x\|_p=|\lambda|\,\|x\|_p, \qquad \|x+y\|_p=\|T_x+T_y\|_p\le\|x\|_p+\|y\|_p \end{equation*}

by the positive homogeneity of the \(L^p\) norm and Minkowski’s Inequality in \(L^p(I)\).

Holder’s Inequality in \(\mathbb{R}^n\). For \(1\le p\le\infty\) and \(q\) conjugate to \(p\),

\begin{equation*} \sum_{k=1}^n|x_ky_k| \le\Bigl[\sum_{k=1}^n|x_k|^p\Bigr]^{1/p}\Bigl[\sum_{k=1}^n|y_k|^q\Bigr]^{1/q} =\|x\|_p\|y\|_q , \end{equation*}

read as \(\bigl(\sum_k|x_k|\bigr)\max_k|y_k|\) when \(p=1\), \(q=\infty\). Indeed \(T_x\cdot T_y=T_z\) with \(z=(x_1y_1,\dots,x_ny_n)\), so the formula with \(p=1\) gives \(\int_I|T_x\cdot T_y|=\sum_k|x_ky_k|\), and Theorem 1 applied to \(T_x\in L^p(I)\), \(T_y\in L^q(I)\) bounds this by \(\|T_x\|_p\|T_y\|_q=\|x\|_p\|y\|_q\). Theorem 1 is stated for \(1\le p<\infty\); the case \(p=\infty\), \(q=1\) follows by exchanging the roles of \(T_x\) and \(T_y\).

Minkowski’s Inequality in \(\mathbb{R}^n\). For \(1\le p<\infty\),

\begin{equation*} \Bigl[\sum_{k=1}^n|x_k+y_k|^p\Bigr]^{1/p} \le\Bigl[\sum_{k=1}^n|x_k|^p\Bigr]^{1/p}+\Bigl[\sum_{k=1}^n|y_k|^p\Bigr]^{1/p}, \end{equation*}

and \(\max_k|x_k+y_k|\le\max_k|x_k|+\max_k|y_k|\) for \(p=\infty\): this is the triangle inequality just proved, with the explicit formulas inserted.

The map \(a\mapsto T_a\) is a linear isometry of \(\ell^p\) into \(L^p(I)\), \(I=[1,\infty)\) being the disjoint union of the \(I_k=[k,k+1)\) with \(m(I_k)=1\). It is measurable, since for each \(c\)

\begin{equation*} \{x\in I : T_a(x)>c\}=\bigcup\{\,I_k : a_k>c\,\} \end{equation*}

is a countable union of intervals; and applying the Monotone Convergence Theorem to the simple functions \(\varphi_n=\sum_{k=1}^n|a_k|^p\chi_{I_k}\), which increase pointwise to \(|T_a|^p\),

\begin{equation*} \int_I|T_a|^p=\lim_{n\to\infty}\sum_{k=1}^n|a_k|^p\,m(I_k) =\sum_{k=1}^\infty|a_k|^p<\infty , \qquad\text{so}\qquad \|T_a\|_p=\|a\|_p . \tag{8} \end{equation*}

Thus \(T_a\in L^p(I)\). The computation is valid for an arbitrary nonnegative sequence, with \(+\infty\) allowed on both sides, and evaluating on each \(I_k\) gives \(T_{a+b}=T_a+T_b\) and \(T_{\lambda a}=\lambda T_a\).

Holder’s Inequality in \(\ell^p\). For \(1<p<\infty\) and \(q\) conjugate to \(p\), if \(a\in\ell^p\) and \(b\in\ell^q\) then \((a_kb_k)\in\ell^1\) and

\begin{equation*} \sum_{k=1}^\infty|a_kb_k| \le\Bigl[\sum_{k=1}^\infty|a_k|^p\Bigr]^{1/p}\Bigl[\sum_{k=1}^\infty|b_k|^q\Bigr]^{1/q} =\|a\|_p\|b\|_q . \end{equation*}

Indeed \(T_a\cdot T_b=T_c\) with \(c=(a_kb_k)\), so by \((8)\) with \(p=1\) applied to \((|a_kb_k|)\) and by Theorem 1 applied to \(T_a\in L^p(I)\), \(T_b\in L^q(I)\),

\begin{equation*} \sum_{k=1}^\infty|a_kb_k|=\int_I|T_a\cdot T_b|\le\|T_a\|_p\|T_b\|_q=\|a\|_p\|b\|_q<\infty . \end{equation*}

The case \(p=1\), \(q=\infty\) is the same once one notes \(\|T_b\|_\infty=\sup_k|b_k|\): that number bounds \(|T_b|\) on \(I\), and any essential upper bound \(M\) satisfies \(|b_k|\le M\) for each \(k\), since \(|T_b|=|b_k|\) on \(I_k\) of measure \(1>0\).

Minkowski’s Inequality in \(\ell^p\). For \(1\le p<\infty\) and \(a,b\in\ell^p\), also \(a+b\in\ell^p\) and

\begin{equation*} \Bigl[\sum_{k=1}^\infty|a_k+b_k|^p\Bigr]^{1/p} \le\Bigl[\sum_{k=1}^\infty|a_k|^p\Bigr]^{1/p}+ \Bigl[\sum_{k=1}^\infty|b_k|^p\Bigr]^{1/p} . \end{equation*}

For by Minkowski’s Inequality in \(L^p(I)\), \(T_{a+b}=T_a+T_b\in L^p(I)\) with \(\|T_{a+b}\|_p\le\|T_a\|_p+\|T_b\|_p\); the extended form of \((8)\) identifies \(\int_I|T_{a+b}|^p\) with \(\sum_k|a_k+b_k|^p\), which is therefore finite, so \(a+b\in\ell^p\) and \(\|a+b\|_p=\|T_{a+b}\|_p\).

Split off the excess power: with \(|f|\le M\) on \(E\) and \(p_2>p_1\),

\begin{equation*} |f|^{p_2}=|f|^{p_1}\cdot|f|^{\,p_2-p_1}\le M^{\,p_2-p_1}|f|^{p_1} \qquad\text{on } E , \end{equation*}

since \(t\mapsto t^{\,p_2-p_1}\) is increasing on \([0,\infty)\). The right-hand side is integrable over \(E\), being a constant multiple of \(|f|^{p_1}\), so by the integral comparison test the nonnegative measurable function \(|f|^{p_2}\) is integrable, with

\begin{equation*} \int_E|f|^{p_2}\le M^{\,p_2-p_1}\int_E|f|^{p_1}<\infty, \qquad\text{that is}\qquad \|f\|_{p_2}\le M^{\,1-p_1/p_2}\|f\|_{p_1}^{\,p_1/p_2} . \end{equation*}

Hence \(f\in L^{p_2}(E)\). If \(p_2=\infty\) instead, \(M\) is an essential upper bound for the measurable function \(f\), so \(f\in L^\infty(E)\) with \(\|f\|_\infty\le M\). (Only \(|f|\le M\) a.e. is used, so \(M=\|f\|_\infty\) serves when \(f\in L^\infty(E)\); and no hypothesis on \(m(E)\) is needed.)

Dominate \(|f|^p\) by \((2p)^p x^{-1/2}\), which is integrable over \((0,1]\). The function \(f(x)=\ln(1/x)\) is continuous and nonnegative on \((0,1]\), hence measurable.

Not in \(L^\infty(0,1]\): for every \(M>0\),

\begin{equation*} \{x\in(0,1] : |f(x)|>M\}=\{x : \ln(1/x)>M\}=(0,e^{-M}), \end{equation*}

of measure \(e^{-M}>0\), so \(f\) has no essential upper bound.

In \(L^p(0,1]\) for every \(p<\infty\): from \(\ln v\le v-1\le v\) for \(v>0\) (the function \(v-1-\ln v\) has derivative \(1-1/v\), so its minimum is \(0\) at \(v=1\)), taking \(v=u^{\alpha}\) with \(\alpha=1/(2p)\) gives \(\alpha\ln u\le u^{\alpha}\), that is \(\ln u\le 2p\,u^{1/(2p)}\) for \(u>0\). With \(u=1/x\),

\begin{equation*} 0\le f(x)\le 2p\,x^{-1/(2p)} , \qquad\text{so}\qquad |f(x)|^p\le(2p)^p\,x^{-1/2} \quad\text{on } (0,1] . \tag{9} \end{equation*}

The majorant is integrable: \(x^{-1/2}\chi_{[1/n,1]}\) increases pointwise to \(x^{-1/2}\), and on \([1/n,1]\) the Lebesgue and Riemann integrals agree (Theorem 3 of Chapter 4), so by the Monotone Convergence Theorem

\begin{equation*} \int_{(0,1]}x^{-1/2}\,dx=\lim_{n\to\infty}\Bigl[2\sqrt{x}\Bigr]_{1/n}^{1} =\lim_{n\to\infty}\Bigl(2-\frac{2}{\sqrt n}\Bigr)=2 . \end{equation*}

By \((9)\) and the integral comparison test, \(|f|^p\) is integrable over \((0,1]\) with

\begin{equation*} \int_{(0,1]}|f|^{p}\le 2\,(2p)^{p}<\infty , \end{equation*}

so \(f\in L^p(0,1]\), with \(\|f\|_p\le 2^{1/p}\cdot 2p\), for every \(p\) with \(1\le p<\infty\).

If \(p,q,r\in[1,\infty]\) satisfy \(\frac1p+\frac1q+\frac1r=1\) (with \(1/\infty=0\)) and \(f\in L^p(E)\), \(g\in L^q(E)\), \(h\in L^r(E)\), then \(f\cdot g\cdot h\) is integrable over \(E\) and

\begin{equation*} \int_E|f\cdot g\cdot h|\le\|f\|_p\,\|g\|_q\,\|h\|_r . \end{equation*}

(i) Some exponent is \(\infty\), say \(r=\infty\). Then \(p,q\) are conjugate, and since \(|h|\le\|h\|_\infty\) a.e. on \(E\) (observation (2) of Section 7.1), monotonicity of integration and Theorem 1 of Chapter 7 applied to \(f,g\) give

\begin{equation*} \int_E|f\cdot g\cdot h|\le\|h\|_\infty\int_E|f\cdot g|\le\|f\|_p\|g\|_q\|h\|_\infty , \end{equation*}

integrability included. Relabelling covers \(p=\infty\) and \(q=\infty\); two infinite exponents force the third to be \(1\), and one exponent equal to \(1\) forces the other two to be \(\infty\), so this case is already covered.

(ii) \(1<p,q,r<\infty\). Define \(s\) by \(\frac1s=\frac1q+\frac1r=1-\frac1p\), so \(s\) is conjugate to \(p\) and \(1<s<q\), \(s<r\). Then \(\alpha=q/s\) and \(\beta=r/s\) exceed \(1\) and are conjugate, since

\begin{equation*} \frac1\alpha+\frac1\beta=s\left(\frac1q+\frac1r\right)=1 ; \end{equation*}

moreover \(|g|^s\in L^\alpha(E)\) because \(\int_E(|g|^s)^\alpha=\int_E|g|^q<\infty\), and likewise \(|h|^s\in L^\beta(E)\). Holder’s Inequality for this pair gives

\begin{equation*} \int_E|g\cdot h|^s=\int_E|g|^s|h|^s \le\Bigl[\int_E|g|^{q}\Bigr]^{s/q}\Bigl[\int_E|h|^{r}\Bigr]^{s/r}=\|g\|_q^s\,\|h\|_r^s , \end{equation*}

so \(g\cdot h\in L^s(E)\) with \(\|g\cdot h\|_s\le\|g\|_q\|h\|_r\). Applying Holder’s Inequality once more, to \(f\in L^p(E)\) and \(g\cdot h\in L^s(E)\) with \(p,s\) conjugate,

\begin{equation*} \int_E|f\cdot g\cdot h|\le\|f\|_p\,\|g\cdot h\|_s\le\|f\|_p\,\|g\|_q\,\|h\|_r . \end{equation*}

Method (2) for \(1<p,q,r<\infty\): normalize so that \(\|f\|_p=\|g\|_q=\|h\|_r=1\) (if one norm vanishes that function is \(0\) a.e. and the inequality is trivial; otherwise divide, both sides being positively homogeneous). At a point where \(a=|f|\), \(b=|g|\), \(c=|h|\) are positive and finite (which is a.e., the powers being integrable), concavity of \(\ln\) with the weights \(1/p,1/q,1/r\) summing to \(1\) gives

\begin{equation*} \ln\left(\frac{a^p}{p}+\frac{b^q}{q}+\frac{c^r}{r}\right)\ge\frac1p\ln a^p+\frac1q\ln b^q+\frac1r\ln c^r=\ln(abc) , \end{equation*}

so \(|f\cdot g\cdot h|\le\frac{|f|^p}{p}+\frac{|g|^q}{q}+\frac{|h|^r}{r}\) a.e. on \(E\) (trivially if some value is \(0\)). Integrating,

\begin{equation*} \int_E|f\cdot g\cdot h|\le\frac1p+\frac1q+\frac1r=1=\|f\|_p\|g\|_q\|h\|_r . \end{equation*}

No. Take \(f_n=n\cdot\chi_{[0,1/n]}\), a nonnegative simple function with

\begin{equation*} \|f_n\|_1=n\cdot m\bigl([0,1/n]\bigr)=1 \qquad\text{for every } n , \end{equation*}

so \(\{f_n\}\) is bounded in \(L^1[0,1]\). Given the challenge \(\epsilon=1/2\) and any \(\delta>0\), choose \(n\) with \(1/n<\delta\) and set \(A=[0,1/n]\); then \(m(A)<\delta\) while

\begin{equation*} \int_A|f_n|=n\cdot\frac1n=1>\frac12 . \end{equation*}

So no \(\delta\) responds, and \(\{f_n\}\) is not uniformly integrable over \([0,1]\). The restriction \(1<p<\infty\) in Corollary 2 of Chapter 7 is therefore essential: its estimate \(\int_A|f|\le\|f\|_p[m(A)]^{1/q}\) degenerates at \(p=1\), where \(q=\infty\) makes the factor \([m(A)]^{0}=1\).

No. Fix \(1<p<\infty\) and take \(f_n=f\) for all \(n\), where \(f(x)=1/(1+|x|)\) on \(\mathbb{R}\): this \(f\) is continuous, and

\begin{equation*} \int_{\mathbb{R}}|f|^p=2\int_0^\infty\frac{dx}{(1+x)^p}=\frac{2}{p-1}<\infty , \qquad \int_{\mathbb{R}}|f|=2\int_0^\infty\frac{dx}{1+x}=\infty , \end{equation*}

so \(\{f_n\}\) is bounded in \(L^p(\mathbb{R})\) while \(f\) is not integrable over \(\mathbb{R}\). Let \(E_0\) be measurable with \(m(E_0)<\infty\). Since \(0\le f\le 1\), monotonicity gives \(\int_{E_0}f\le m(E_0)<\infty\); and \(f\ge 0\) allows additivity over domains in the extended sense,

\begin{equation*} \int_{\mathbb{R}}f=\int_{E_0}f+\int_{\mathbb{R}\sim E_0}f \qquad\text{in } [0,\infty] , \end{equation*}

so \(\int_{\mathbb{R}\sim E_0}f\) must be \(\infty\), else \(f\) would be integrable. Hence no set of finite measure meets the challenge \(\epsilon=1\), and \(\{f_n\}\) is not tight.

Method (2), with integrable members of constant norm: \(f_n=\chi_{[n,n+1]}\) has \(\|f_n\|_r=1\) for every \(r\in[1,\infty)\). Given \(E_0\) of finite measure, the disjoint sets \(E_0\cap[n,n+1)\) satisfy \(\sum_n m\bigl(E_0\cap[n,n+1)\bigr)\le m(E_0)<\infty\), so their measures tend to \(0\); choosing \(N\) with \(m\bigl(E_0\cap[N,N+1]\bigr)<1/2\) (the endpoint being null),

\begin{equation*} \int_{\mathbb{R}\sim E_0}|f_N|=1-m\bigl(E_0\cap[N,N+1]\bigr)>\frac12 . \end{equation*}

Squeeze \(\|f\|_p\) between \((M-\epsilon)[m(E_\epsilon)]^{1/p}\) and \(M[m(E)]^{1/p}\), where \(M=\|f\|_\infty\). By observation (2) of Section 7.1 the infimum defining \(M\) is attained, so

\begin{equation*} |f|\le M \quad\text{a.e. on } E . \tag{$*$} \end{equation*}

If \(m(E)=0\) then \(M=0=\|f\|_p\) for every \(p\) and there is nothing to prove; assume \(0<m(E)<\infty\).

Upper bound. By \((*)\) and monotonicity of integration, \(\int_E|f|^p\le M^p\,m(E)<\infty\) (so each \(\|f\|_p\) is finite), and taking \(p\)-th roots,

\begin{equation*} \|f\|_p\le M\cdot[m(E)]^{1/p} . \end{equation*}

Since \([m(E)]^{1/p}=\exp\bigl(\tfrac1p\ln m(E)\bigr)\to 1\), we get \(\limsup_{p\to\infty}\|f\|_p\le M\).

Lower bound. If \(M=0\) this is trivial; otherwise fix \(0<\epsilon<M\) and put \(E_\epsilon=\{x\in E : |f(x)|>M-\epsilon\}\). Then \(m(E_\epsilon)>0\), for \(m(E_\epsilon)=0\) would make \(M-\epsilon<M\) an essential upper bound, contradicting \(M=\inf\); and \(m(E_\epsilon)\le m(E)<\infty\). Since \(t\mapsto t^p\) is increasing on \([0,\infty)\), monotonicity and additivity over domains give

\begin{equation*} \int_E|f|^p\ge\int_{E_\epsilon}|f|^p\ge(M-\epsilon)^p\,m(E_\epsilon), \qquad\text{so}\qquad \|f\|_p\ge(M-\epsilon)[m(E_\epsilon)]^{1/p} . \end{equation*}

Again \([m(E_\epsilon)]^{1/p}\to 1\), so \(\liminf_{p\to\infty}\|f\|_p\ge M-\epsilon\); as \(\epsilon\in(0,M)\) was arbitrary, \(\liminf_{p\to\infty}\|f\|_p\ge M\). Hence \(\lim_{p\to\infty}\|f\|_p=M=\|f\|_\infty\).

The maximum is attained at the conjugate function \(g=f^*\) of Section 7.2. Each competing integral is defined, since Holder’s Inequality (Theorem 1 of Chapter 7) makes \(f\cdot g\) integrable for \(f\in L^p(E)\), \(g\in L^q(E)\), and it supplies the upper bound: if \(\|g\|_q\le 1\) then

\begin{equation*} \int_E f\cdot g\le\int_E|f\cdot g|\le\|f\|_p\|g\|_q\le\|f\|_p . \end{equation*}

(i) \(f=0\) in \(L^p(E)\). Then \(\|f\|_p=0\) and \(g=0\) attains it.

(ii) \(f\ne 0\), so \(\|f\|_p>0\) and \(m(E)>0\). Take

\begin{equation*} f^*=\|f\|_p^{\,1-p}\cdot\operatorname{sgn}(f)\cdot|f|^{p-1} , \end{equation*}

with \(\operatorname{sgn}(f)=1\) where \(f\ge 0\) and \(-1\) where \(f<0\). Since \(|f|^p\) is integrable, \(f\) is finite a.e., so \(\operatorname{sgn}(f)\cdot f=|f|\) a.e. and

\begin{equation*} \int_E f\cdot f^*=\|f\|_p^{\,1-p}\int_E|f|^{p} =\|f\|_p^{\,1-p}\cdot\|f\|_p^{\,p}=\|f\|_p . \end{equation*}

Moreover \(\|f^*\|_q=1\): for \(p>1\) the exponent \(q\) is finite with \(q(p-1)=p\), so

\begin{equation*} \int_E|f^*|^q=\|f\|_p^{\,q(1-p)}\int_E|f|^{q(p-1)}=\|f\|_p^{\,-p}\int_E|f|^{p}=1 ; \end{equation*}

for \(p=1\), \(f^*=\operatorname{sgn}(f)\) has \(|f^*|=1\) on \(E\), so \(\|f^*\|_\infty=1\), no \(c<1\) being an essential upper bound because \(m(E)>0\).

So \(f^*\) is admissible and realizes the bound \(\|f\|_p\), which is therefore a maximum.

Both directions are immediate once the conjugate function is admitted as a test function; \(f=0\) means \(f=0\) a.e. on \(E\), and each \(\int_E f\cdot g\) is a real number by Holder’s Inequality (Theorem 1 of Chapter 7).

If \(f=0\) a.e. on \(E\), then \(f\cdot g=0\) a.e. for every \(g\in L^q(E)\), so \(\int_E f\cdot g=0\).

Conversely, suppose \(\int_E f\cdot g=0\) for every \(g\in L^q(E)\). Restricting to those \(g\) with \(\|g\|_q\le 1\), Problem 19 gives

\begin{equation*} \|f\|_p=\max_{g\in L^q(E),\ \|g\|_q\le 1}\int_E f\cdot g=0 , \end{equation*}

and \(\int_E|f|^p=0\) with \(|f|^p\ge 0\) forces \(|f|^p=0\) a.e. on \(E\) by Proposition 9 of Chapter 4. Hence \(f=0\) a.e. on \(E\).

Method (2): argue by contradiction without Problem 19. If \(\|f\|_p>0\), the conjugate function \(f^*=\|f\|_p^{\,1-p}\operatorname{sgn}(f)|f|^{p-1}\) belongs to \(L^q(E)\) (for \(p>1\), \(q(p-1)=p\) gives \(\int_E|f^*|^q=\|f\|_p^{-p}\int_E|f|^p=1\); for \(p=1\), \(f^*=\operatorname{sgn}(f)\) is measurable with \(|f^*|\le 1\)), and since \(f\) is finite a.e., \(\operatorname{sgn}(f)\cdot f=|f|\) a.e., so

\begin{equation*} \int_E f\cdot f^*=\|f\|_p^{\,1-p}\int_E|f|^{p}=\|f\|_p>0 , \end{equation*}

contradicting the hypothesis at \(g=f^*\).

With \(q\) the conjugate of \(p\), the property holds for all \(f\in L^p[0,1]\) precisely when

\begin{equation*} \lambda\le\frac1q=1-\frac1p \ \ (1\le p<\infty), \qquad \lambda<1 \ \ (p=\infty) ; \end{equation*}

so \(\lambda\le 0\) for \(p=1\), and at \(p=\infty\) the critical value \(\lambda=1\) is excluded. Take \(0<\epsilon\le 1\); every \(f\in L^p[0,1]\) is integrable over \([0,1]\) (Corollary 3 of Chapter 7 with \(p_1=1\le p_2=p\) when \(p<\infty\); essential boundedness on a set of finite measure when \(p=\infty\)), so \(\int_0^\epsilon f\) is a real number.

Sufficiency for \(1\le p<\infty\), \(\lambda\le 1/q\). Apply Holder’s Inequality to \(u=f\cdot\chi_{[0,\epsilon]}\in L^p[0,1]\) and \(v=\chi_{[0,\epsilon]}\in L^q[0,1]\), noting \(\|v\|_q=\epsilon^{1/q}\) (read as \(\|v\|_\infty=1=\epsilon^0\) when \(q=\infty\)):

\begin{equation*} \left|\frac{1}{\epsilon^\lambda}\int_0^\epsilon f\right|\le\frac{1}{\epsilon^\lambda}\int_0^1|u\cdot v| \le\epsilon^{\,1/q-\lambda}\,\bigl\|f\cdot\chi_{[0,\epsilon]}\bigr\|_p . \end{equation*}

Here \(\epsilon^{\,1/q-\lambda}\le 1\) since \(\lambda\le 1/q\) and \(\epsilon\le 1\), while

\begin{equation*} \bigl\|f\cdot\chi_{[0,\epsilon]}\bigr\|_p=\Bigl[\int_0^\epsilon|f|^p\Bigr]^{1/p} \longrightarrow 0 \quad\text{as } \epsilon\to 0^+ \end{equation*}

by Proposition 23 of Chapter 4 applied to the integrable function \(|f|^p\) (given \(\eta>0\) there is \(\delta>0\) with \(\int_A|f|^p<\eta^p\) whenever \(m(A)<\delta\); take \(A=[0,\epsilon]\)). Hence the limit is \(0\).

Sufficiency for \(p=\infty\), \(\lambda<1\). Since \(|f|\le\|f\|_\infty\) a.e.,

\begin{equation*} \left|\frac{1}{\epsilon^\lambda}\int_0^\epsilon f\right| \le\frac{1}{\epsilon^\lambda}\cdot\|f\|_\infty\cdot\epsilon =\|f\|_\infty\,\epsilon^{\,1-\lambda}\longrightarrow 0 , \end{equation*}

because \(1-\lambda>0\).

Necessity. For the remaining \(\lambda\) – namely \(\lambda>1-1/p\) when \(p<\infty\), and \(\lambda\ge 1\) when \(p=\infty\), all of which are positive – the single function \(f(x)=x^{\lambda-1}\) on \((0,1]\), \(f(0)=0\), defeats the limit. It lies in \(L^p[0,1]\): for \(p<\infty\), \(\lambda>1-1/p\) gives \((\lambda-1)p+1>0\), so by the Monotone Convergence Theorem on \([1/n,1]\), where the Riemann and Lebesgue integrals agree (Theorem 3 of Chapter 4),

\begin{equation*} \int_0^1|f|^p=\lim_{n\to\infty}\frac{1-(1/n)^{(\lambda-1)p+1}}{(\lambda-1)p+1} =\frac{1}{(\lambda-1)p+1}<\infty ; \end{equation*}

for \(p=\infty\), \(\lambda\ge 1\) makes \(0\le f\le 1\) on \((0,1]\). The same truncation, with \(\lambda>0\), computes

\begin{equation*} \int_0^\epsilon f=\lim_{n\to\infty}\frac{\epsilon^{\lambda}-(1/n)^{\lambda}}{\lambda} =\frac{\epsilon^{\lambda}}{\lambda}, \qquad\text{so}\qquad \frac{1}{\epsilon^\lambda}\int_0^\epsilon f=\frac1\lambda\ne 0 \end{equation*}

for every \(\epsilon\in(0,1]\). Hence no such \(\lambda\) has the required property.

Take \(M=\|f\|_p^p+1\), where \(f\in L^p[a,b]\) is the integrand of \(F\), so that \(F(x)=F(a)+\int_a^x f\) on \([a,b]\); this is finite and positive. Let \(q=p/(p-1)\), so \(p/q=p-1\).

Fix a partition \(a=x_0<\cdots<x_n=b\) and an index \(k\). By additivity over domains and monotonicity of integration, \(|F(x_k)-F(x_{k-1})|\le\int_{x_{k-1}}^{x_k}|f|\), and Holder’s Inequality (Theorem 1 of this chapter) applied on \([x_{k-1},x_k]\) to \(|f|\in L^p\) and the constant \(1\), which lies in \(L^q\) there because the interval has finite measure, gives

\begin{equation*} \int_{x_{k-1}}^{x_k}|f| \le\left[\int_{x_{k-1}}^{x_k}|f|^p\right]^{1/p}(x_k-x_{k-1})^{1/q} . \end{equation*}

Raising to the \(p\)-th power (legitimate, \(t\mapsto t^p\) being increasing on \([0,\infty)\)) and using \(p/q=p-1\),

\begin{equation*} |F(x_k)-F(x_{k-1})|^p\le\left[\int_{x_{k-1}}^{x_k}|f|^p\right](x_k-x_{k-1})^{p-1} , \end{equation*}

so dividing by \(|x_k-x_{k-1}|^{p-1}>0\) and summing over \(k\), the subintervals having pairwise disjoint interiors and union \([a,b]\),

\begin{equation*} \sum_{k=1}^{n}\frac{|F(x_k)-F(x_{k-1})|^p}{|x_k-x_{k-1}|^{p-1}} \le\sum_{k=1}^{n}\int_{x_{k-1}}^{x_k}|f|^p=\int_a^b|f|^p<M . \end{equation*}

Since \(M\) depends only on \(f\), it serves for every partition.

Take \(a_n = 1/n\) in \(X = \mathbb{R}\) normed by absolute value. It converges to \(0\), hence is Cauchy by Proposition 4.

It is not rapidly Cauchy. Suppose \(\sum_{k=1}^{\infty} \epsilon_k\) were a convergent series of positive numbers with \(|a_{k+1} - a_k| \le \epsilon_k^2\) for all \(k\). Since

\begin{equation*} |a_{k+1} - a_k| = \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)} \ge \frac{1}{(k+1)^2} , \end{equation*}

positivity of \(\epsilon_k\) forces \(\epsilon_k \ge 1/(k+1)\) for every \(k\), whence

\begin{equation*} \sum_{k=1}^{n} \epsilon_k \ge \sum_{k=1}^{n} \frac{1}{k+1} \to \infty \end{equation*}

by divergence of the harmonic series, contradicting convergence of \(\sum_k \epsilon_k\).

Regrouping and applying the triangle inequality and positive homogeneity of the norm,

\begin{equation*} \| (\alpha f_n + \beta g_n) - (\alpha f + \beta g) \| = \| \alpha [f_n - f] + \beta [g_n - g] \| \end{equation*}

\begin{equation*} \le |\alpha| \, \| f_n - f \| + |\beta| \, \| g_n - g \| . \end{equation*}

By hypothesis \(\|f_n - f\| \to 0\) and \(\|g_n - g\| \to 0\), so by the algebra of limits the right-hand side tends to \(|\alpha| \cdot 0 + |\beta| \cdot 0 = 0\); the squeeze principle then gives \(\| (\alpha f_n + \beta g_n) - (\alpha f + \beta g) \| \to 0\), that is, \(\{\alpha f_n + \beta g_n\} \to \alpha f + \beta g\) in \(X\).

Apply the norm comparison of Corollary 3 to \(h = f_n - f\), legitimate since \(m(E) < \infty\) and \(1 \le p_1 < p_2 \le \infty\) and since \(f_n - f\) lies in the linear space \(L^{p_2}(E)\). That corollary gives \(L^{p_2}(E) \subseteq L^{p_1}(E)\), so \(f\) and every \(f_n\) lie in \(L^{p_1}(E)\), and

\begin{equation*} \| f_n - f \|_{p_1} \le c \, \| f_n - f \|_{p_2} \quad \text{for all } n , \end{equation*}

where \(c = [m(E)]^{1/p_1 - 1/p_2}\) if \(p_2 < \infty\) and \(c = [m(E)]^{1/p_1}\) if \(p_2 = \infty\). Since \(\|f_n - f\|_{p_2} \to 0\) and \(c\) is a fixed finite constant, the squeeze principle gives \(\|f_n - f\|_{p_1} \to 0\), that is, \(\{f_n\} \to f\) in \(L^{p_1}(E)\).

Apply the Lebesgue Dominated Convergence Theorem of Chapter 4 to \(h_n = |f_n - f|^p\). Excise from \(E\) the measure-zero union of the exceptional sets for the pointwise convergence, for the countably many inequalities \(|f_n| \le g\), and for \(\{g = \infty\}\), so that all hypotheses hold at every point of \(E\) and no integral below changes.

Each \(f_n\) lies in \(L^p(E)\) since \(\int_E |f_n|^p \le \int_E g^p < \infty\); and \(f\) is measurable as a pointwise limit of measurable functions (Proposition 9 of Chapter 3), with \(|f| \le g\) on \(E\) on letting \(n \to \infty\) in \(|f_n| \le g\), so \(f \in L^p(E)\) too. The two hypotheses of the dominated convergence theorem now hold for \(\{h_n\}\):

(i) \(\{h_n\} \to 0\) pointwise on \(E\), since \(|f_n(x) - f(x)| \to 0\) and \(t \mapsto t^p\) is continuous on \([0, \infty)\);

(ii) \(0 \le h_n \le 2^p g^p\) on \(E\) for all \(n\), because \(|f_n - f| \le |f_n| + |f| \le 2g\), and \(2^p g^p\) is integrable over \(E\) since \(g \in L^p(E)\).

Therefore

\begin{equation*} \lim_{n \to \infty} \| f_n - f \|_p^p = \lim_{n \to \infty} \int_E h_n = \int_E 0 = 0 , \end{equation*}

and taking \(p\)-th roots gives \(\{f_n\} \to f\) in \(L^p(E)\).

Take \(g = |f| + \sum_{k=1}^{\infty} |f_{n_k} - f|\), where \(\{n_k\}\) is chosen strictly increasing with

\begin{equation*} \| f_{n_k} - f \|_p \le \frac{1}{2^k} \quad \text{for all } k ; \end{equation*}

such a choice is possible because \(\|f_n - f\|_p \to 0\), taking \(n_k = \max\{N, n_{k-1}+1\}\) for an \(N\) beyond which \(\|f_n - f\|_p \le 2^{-k}\). Excising a set of measure zero, \(f\) and each \(f_n\) are real-valued on \(E\).

The partial sums \(g_N = |f| + \sum_{k=1}^{N} |f_{n_k} - f|\) increase pointwise to \(g\), and Minkowski’s Inequality applied \(N\) times gives

\begin{equation*} \| g_N \|_p \le \| f \|_p + \sum_{k=1}^{N} \frac{1}{2^k} \le \| f \|_p + 1 =: C , \end{equation*}

a bound independent of \(N\). Since \(\{g_N^p\}\) increases pointwise to \(g^p\), the Monotone Convergence Theorem of Chapter 4 yields

\begin{equation*} \int_E g^p = \lim_{N \to \infty} \| g_N \|_p^p \le C^p < \infty . \end{equation*}

Hence \(g\) is finite a.e.; redefining it to be \(0\) on the measure-zero set where \(g = \infty\) leaves \(g \in L^p(E)\).

Finally, \(|f_{n_k}(x) - f(x)|\) is one of the nonnegative terms of the series defining \(g(x)\), so

\begin{equation*} |f_{n_k}| \le |f| + |f_{n_k} - f| \le g \quad \text{a.e. on } E, \ \text{for all } k . \end{equation*}

Apply the Vitali Convergence Theorem (Theorem 8 of this chapter) to \(\{f_n\}\). Write \(r = p + \theta\), so \(1 \le p < r < \infty\), and let \(M \ge 0\) satisfy \(\|f_n\|_r \le M\) for all \(n\); since \(m(E) < \infty\), Corollary 3 puts \(L^r(E) \subseteq L^p(E)\), so \(\{f_n\}\) is a sequence in \(L^p(E)\). Its three hypotheses:

(i) \(f \in L^p(E)\). Indeed \(f\) is measurable as a pointwise a.e. limit of measurable functions (Proposition 9 of Chapter 3), and \(\{|f_n|^r\} \to |f|^r\) pointwise a.e., so Fatou’s Lemma of Chapter 4 gives

\begin{equation*} \int_E |f|^{r} \le \liminf_{n \to \infty} \| f_n \|_{r}^{r} \le M^{r} < \infty , \end{equation*}

whence \(f \in L^r(E) \subseteq L^p(E)\), again by Corollary 3.

(ii) \(\{|f_n|^p\}\) is uniformly integrable over \(E\). With \(s = r/p > 1\),

\begin{equation*} \big\| \, |f_n|^p \, \big\|_{s} = \left[\int_E |f_n|^{r}\right]^{p/r} = \| f_n \|_{r}^{\,p} \le M^{p} \quad \text{for all } n , \end{equation*}

so the family is bounded in \(L^s(E)\) with \(1 < s < \infty\), and Corollary 2 of this chapter applies.

(iii) \(\{|f_n|^p\}\) is tight over \(E\): take \(E_0 = E\), of finite measure by hypothesis, so that \(E \sim E_0 = \emptyset\).

Theorem 8 therefore gives \(\{f_n\} \to f\) in \(L^p(E)\).

No: the space \(\mathcal{P}\) of polynomials on \([a,b]\), normed by \(\|p\|_{\max} = \max_{x \in [a,b]} |p(x)|\), is not complete. The partial sums of an exponential series witness this. For \(n \in \mathbb{N}\) put

\begin{equation*} p_n(x) = \sum_{k=0}^{n} \frac{1}{k!}\left(\frac{x-a}{b-a}\right)^{k} \in \mathcal{P} . \end{equation*}

With \(t = (x-a)/(b-a) \in [0,1]\), for \(m > n\),

\begin{equation*} \|p_m - p_n\|_{\max} = \max_{x} \left| \sum_{k=n+1}^{m} \frac{t^k}{k!} \right| \le \sum_{k=n+1}^{\infty} \frac{1}{k!} , \end{equation*}

and the tails of the convergent series \(\sum_k 1/k!\) tend to \(0\), so \(\{p_n\}\) is Cauchy in \(\mathcal{P}\).

Suppose \(\{p_n\} \to q\) in \(\mathcal{P}\). Maximum-norm convergence is uniform, hence pointwise, convergence on \([a,b]\), while \(p_n(x)\) is the \(n\)-th partial sum of the exponential series at \(t\); uniqueness of pointwise limits gives

\begin{equation*} q(x) = e^{(x-a)/(b-a)} \quad \text{for all } x \in [a,b]. \end{equation*}

The right side is everywhere positive, so \(q\) is not the zero polynomial; let \(d = \deg q\). Differentiating \(d+1\) times on \((a,b)\) makes the left side vanish identically while the right side equals \((b-a)^{-(d+1)} e^{(x-a)/(b-a)} > 0\), a contradiction. So \(\{p_n\}\) has no limit in \(\mathcal{P}\), and \(\mathcal{P}\) is not a Banach space.

Method (2): by Problem 31, \(C[a,b]\) with the maximum norm is complete, and by the Weierstrass Approximation Theorem of Section 7.4, \(\mathcal{P}\) is dense in \(C[a,b]\); but \(\mathcal{P} \ne C[a,b]\) since \(x \mapsto e^x\) is continuous and not a polynomial (by the derivative argument just given). A complete subspace is closed, and a dense proper subset is not, so \(\mathcal{P}\) is not complete.

The printed display has mismatched subscripts (with \(n > k\), telescoping from \(f_k\) only gives the tail \(\sum_{j=k}^{\infty} a_j\)); we prove inequality (6) of this section, namely the same statement with second index \(n\),

\begin{equation*} |f_{n+k}(x) - f_n(x)| \le \|f_{n+k} - f_n\|_{\max} \le \sum_{j=n}^{\infty} a_j = R_n , \end{equation*}

for all \(n, k\) and all \(x \in [a,b]\); the printed form follows whenever \(k \ge n\).

The first inequality is the definition of \(\|\cdot\|_{\max}\) applied to \(f_{n+k} - f_n \in C[a,b]\). For the second, telescope and use the triangle inequality together with the hypothesis \(\|f_{j+1} - f_j\|_{\max} \le a_j\):

\begin{equation*} \|f_{n+k} - f_n\|_{\max} = \Big\| \sum_{j=n}^{n+k-1} [f_{j+1} - f_j] \Big\|_{\max} \le \sum_{j=n}^{n+k-1} a_j \le R_n , \end{equation*}

the last step because the \(a_j\) are positive.

Since \(\sum_k a_k\) converges, \(R_n\) is finite and \(R_n \to 0\). Hence for each fixed \(x\) the real sequence \(\{f_n(x)\}\) is Cauchy, so by completeness of \(\mathbb{R}\) it converges; set \(f(x) = \lim_n f_n(x)\). Letting \(k \to \infty\) in \(|f_{n+k}(x) - f_n(x)| \le R_n\) gives

\begin{equation*} |f(x) - f_n(x)| \le R_n \quad \text{for all } x \in [a,b] \text{ and all } n , \end{equation*}

a bound independent of \(x\) tending to \(0\), so \(\{f_n\} \to f\) uniformly on \([a,b]\). A uniform limit of continuous functions is continuous (the \(\epsilon/3\) argument), so \(f \in C[a,b]\).

Let \(\{f_n\}\) be Cauchy in \(C[a,b]\). Choose inductively \(n_1 < n_2 < \cdots\) with

\begin{equation*} \|f_{n_{k+1}} - f_{n_k}\|_{\max} \le \left(\tfrac{1}{2}\right)^{k} \quad \text{for all } k : \end{equation*}

the Cauchy property with \(\epsilon = (1/2)^k\) supplies \(M_k\) with \(\|f_n - f_m\|_{\max} \le (1/2)^k\) for \(n, m \ge M_k\), and setting \(n_1 = M_1\), \(n_{k+1} = \max\{M_{k+1}, n_k + 1\}\) gives \(n_{k+1} > n_k \ge M_k\), so the estimate at index \(k\) applies.

With \(g_k = f_{n_k}\) and \(a_k = (1/2)^k\), the series \(\sum_k a_k\) of positive numbers converges and \(\|g_{k+1} - g_k\|_{\max} \le a_k\), so Problem 30 provides \(f \in C[a,b]\) with \(\{g_k\} \to f\) uniformly on \([a,b]\), that is, \(\{f_{n_k}\} \to f\) in \(C[a,b]\), since maximum-norm convergence is uniform convergence.

A Cauchy sequence in a normed linear space with a convergent subsequence converges to that subsequence’s limit (Proposition 4), so \(\{f_n\} \to f\) in \(C[a,b]\). Hence \(C[a,b]\) is complete: a Banach space.

As in Problem 30 the printed subscripts are misaligned; we prove the \(L^{\infty}\) analogue of (6) of this section, with second index \(n\),

\begin{equation*} |f_{n+k}(x) - f_n(x)| \le \|f_{n+k} - f_n\|_{\infty} \le \sum_{j=n}^{\infty} a_j = R_n \end{equation*}

for all \(n, k\) and all \(x \in E \sim E_0\), the printed form following for \(k \ge n\).

Take \(E_0 = \bigcup_n A_n \cup \bigcup_{n,k} B_{n,k}\), where

\begin{equation*} A_n = \{ |f_n| > \|f_n\|_{\infty} \}, \qquad B_{n,k} = \{ |f_{n+k} - f_n| > \|f_{n+k} - f_n\|_{\infty} \} . \end{equation*}

Each has measure zero because the essential supremum of a function in \(L^{\infty}(E)\) is itself an essential upper bound, being the union over \(M_i \downarrow \|g\|_{\infty}\) of the null sets \(\{|g| > M_i\}\); the differences \(f_{n+k} - f_n\) belong to \(L^{\infty}(E)\) since that space is linear. So \(m(E_0) = 0\) by countable subadditivity, and on \(E \sim E_0\) every \(f_n\) is real-valued, since there \(|f_n| \le \|f_n\|_{\infty} < \infty\).

Off \(E_0\) the first inequality holds by definition of \(B_{n,k}\); for the second, telescope and apply the triangle inequality with \(\|f_{j+1} - f_j\|_{\infty} \le a_j\):

\begin{equation*} \|f_{n+k} - f_n\|_{\infty} \le \sum_{j=n}^{n+k-1} \|f_{j+1} - f_j\|_{\infty} \le \sum_{j=n}^{n+k-1} a_j \le R_n , \end{equation*}

the last step because the \(a_j\) are positive.

Since \(\sum_k a_k\) converges, \(R_n\) is finite and \(R_n \to 0\); so for each \(x \in E \sim E_0\) the real sequence \(\{f_n(x)\}\) is Cauchy, hence convergent. Set \(f = \lim_n f_n\) on \(E \sim E_0\) and \(f = 0\) on \(E_0\); then \(f\) is the pointwise limit on \(E\) of the measurable functions \(f_n \cdot \chi_{E \sim E_0}\), hence measurable (Proposition 9 of Chapter 3). Letting \(k \to \infty\) above,

\begin{equation*} |f(x) - f_n(x)| \le R_n \quad \text{for all } x \in E \sim E_0 \text{ and all } n , \end{equation*}

so at \(n = 1\), \(|f| \le \|f_1\|_{\infty} + R_1 < \infty\) off \(E_0\) and \(f \in L^{\infty}(E)\). The bound \(R_n\) is independent of \(x\) and tends to \(0\), so \(\{f_n\} \to f\) uniformly on \(E \sim E_0\); as \(m(E_0) = 0\), \(R_n\) is an essential upper bound for \(|f - f_n|\), giving \(\|f - f_n\|_{\infty} \le R_n \to 0\).

Let \(\{f_n\}\) be Cauchy in \(L^{\infty}(E)\). Exactly as in Problem 31, the Cauchy property with \(\epsilon = (1/2)^k\) yields \(M_k\), and \(n_1 = M_1\), \(n_{k+1} = \max\{M_{k+1}, n_k+1\}\) give a strictly increasing \(\{n_k\}\) with \(n_{k+1} > n_k \ge M_k\), hence

\begin{equation*} \|f_{n_{k+1}} - f_{n_k}\|_{\infty} \le \left(\tfrac{1}{2}\right)^{k} \quad \text{for all } k . \end{equation*}

With \(g_k = f_{n_k}\) and \(a_k = (1/2)^k\), the series \(\sum_k a_k\) of positive numbers converges and \(\|g_{k+1} - g_k\|_{\infty} \le a_k\), so Problem 32 provides a null set \(E_0\) and \(f \in L^{\infty}(E)\) with \(\{g_k\} \to f\) uniformly on \(E \sim E_0\); its estimate gives

\begin{equation*} \|f - g_k\|_{\infty} \le \sum_{j=k}^{\infty} \left(\tfrac{1}{2}\right)^{j} = \left(\tfrac{1}{2}\right)^{k-1} \to 0 , \end{equation*}

so \(\{f_{n_k}\} \to f\) in \(L^{\infty}(E)\). A Cauchy sequence with a convergent subsequence converges to that limit (Proposition 4), so \(\{f_n\} \to f\) in \(L^{\infty}(E)\), and \(L^{\infty}(E)\) is a Banach space.

Coordinatewise limits supply the limit vector in both cases. Write \(a^{(n)} = (a^{(n)}_1, a^{(n)}_2, \ldots)\); for every \(p\) with \(1 \le p \le \infty\),

\begin{equation*} |a_k| \le \|a\|_{p} \quad \text{for all } a \in \ell^{p} \text{ and all } k , \end{equation*}

by definition of the supremum when \(p = \infty\), and because \(|a_k|^p \le \sum_i |a_i|^p\) with \(t \mapsto t^{1/p}\) increasing when \(p < \infty\). That \(\|\cdot\|_p\) is a norm on the linear space \(\ell^p\) is Problem 12 for \(p < \infty\), and immediate from \(\sup_k |a_k + b_k| \le \sup_k|a_k| + \sup_k|b_k|\) for \(p = \infty\).

(i) \(1 \le p < \infty\). Let \(\{a^{(n)}\}\) be Cauchy in \(\ell^p\). Applying the display to \(a^{(n)} - a^{(m)}\) makes each coordinate sequence \(\{a^{(n)}_k\}_n\) Cauchy in \(\mathbb{R}\), hence convergent; set \(a_k = \lim_n a^{(n)}_k\) and \(a = (a_1, a_2, \ldots)\). Given \(\epsilon > 0\), choose \(N\) with \(\sum_{k=1}^{\infty} |a^{(n)}_k - a^{(m)}_k|^p \le \epsilon^p\) for \(n, m \ge N\). Fix \(n \ge N\) and \(K\); discarding all but the first \(K\) nonnegative terms and letting \(m \to \infty\) in that finite sum (each summand converging, \(t \mapsto |t|^p\) being continuous),

\begin{equation*} \sum_{k=1}^{K} |a^{(n)}_k - a_k|^{p} \le \epsilon^{p} \quad \text{for every } K , \end{equation*}

so letting \(K \to \infty\) gives \(\|a^{(n)} - a\|_p \le \epsilon\) for all \(n \ge N\). In particular \(a^{(N)} - a \in \ell^p\), so \(a = a^{(N)} - (a^{(N)} - a) \in \ell^p\), and \(\{a^{(n)}\} \to a\) in \(\ell^p\).

(ii) \(p = \infty\). The same coordinatewise limit \(a\) exists, and choosing \(N\) with \(\|a^{(n)} - a^{(m)}\|_{\infty} \le \epsilon\) for \(n, m \ge N\) gives \(|a^{(n)}_k - a^{(m)}_k| \le \epsilon\) for every \(k\); letting \(m \to \infty\) and taking the supremum over \(k\),

\begin{equation*} \sup_{k} |a^{(n)}_k - a_k| \le \epsilon \quad \text{for all } n \ge N . \end{equation*}

So \(a^{(N)} - a \in \ell^{\infty}\), whence \(a \in \ell^{\infty}\), and \(\{a^{(n)}\} \to a\) in \(\ell^{\infty}\).

Thus \(\ell^p\) is complete for \(1 \le p \le \infty\), hence a Banach space.

Both are closed linear subspaces of the Banach space \(\ell^{\infty}\) (Problem 34), and a closed linear subspace \(Y\) of a Banach space \(X\) is itself complete: a sequence Cauchy in \(Y\) is Cauchy in \(X\), so converges in \(X\), and its limit lies in \(Y\) by closedness, the norms being the same. That \(c_0 \subseteq c \subseteq \ell^{\infty}\) are linear subspaces follows from boundedness of convergent sequences and the algebra of limits.

(i) \(c\) is closed. Let \(\{a^{(n)}\} \to a\) in \(\ell^{\infty}\) with each \(a^{(n)} \in c\), and put \(L_n = \lim_{k} a^{(n)}_k\). From \(|a^{(n)}_k - a^{(m)}_k| \le \|a^{(n)} - a^{(m)}\|_{\infty}\), letting \(k \to \infty\),

\begin{equation*} |L_n - L_m| \le \|a^{(n)} - a^{(m)}\|_{\infty} , \end{equation*}

and \(\{a^{(n)}\}\), being convergent, is Cauchy in \(\ell^{\infty}\) (Proposition 4); so \(\{L_n\}\) is Cauchy in \(\mathbb{R}\), say \(L_n \to L\). Given \(\epsilon > 0\), fix \(n\) with \(\|a - a^{(n)}\|_{\infty} < \epsilon/3\) and \(|L_n - L| < \epsilon/3\), then \(K\) with \(|a^{(n)}_k - L_n| < \epsilon/3\) for \(k \ge K\); for such \(k\),

\begin{equation*} |a_k - L| \le |a_k - a^{(n)}_k| + |a^{(n)}_k - L_n| + |L_n - L| < \epsilon . \end{equation*}

Hence \(a_k \to L\) and \(a \in c\).

(ii) \(c_0\) is closed. With \(\{a^{(n)}\} \to a\) in \(\ell^{\infty}\), each \(a^{(n)} \in c_0\), and \(\epsilon > 0\), choose \(n\) with \(\|a - a^{(n)}\|_{\infty} < \epsilon/2\) and then \(K\) with \(|a^{(n)}_k| < \epsilon/2\) for \(k \ge K\); for such \(k\),

\begin{equation*} |a_k| \le \|a - a^{(n)}\|_{\infty} + |a^{(n)}_k| < \epsilon , \end{equation*}

so \(a_k \to 0\) and \(a \in c_0\).

Therefore \(c\) and \(c_0\), normed by \(\|\cdot\|_{\infty}\), are Banach spaces.

Suppose \(\mathcal{S}\) is dense in \(X\) and let \(g \in X\). Denseness with \(\epsilon = 1/n\) supplies \(f_n \in \mathcal{S}\) with

\begin{equation*} \|f_n - g\| < \frac{1}{n} \quad \text{for each } n , \end{equation*}

and the squeeze principle gives \(\|f_n - g\| \to 0\), that is, \(\{f_n\} \to g\) in \(X\).

Conversely, given \(g \in X\) and \(\epsilon > 0\), take a sequence \(\{f_n\}\) in \(\mathcal{S}\) with \(\|f_n - g\| \to 0\) and an index \(N\) with \(\|f_N - g\| < \epsilon\); then \(f_N \in \mathcal{S}\) is the required approximant, so \(\mathcal{S}\) is dense in \(X\).

Let \(h \in \mathcal{H}\) and \(\epsilon > 0\). Denseness of \(\mathcal{G}\) in \(\mathcal{H}\) gives \(g \in \mathcal{G}\) with \(\|g - h\| < \epsilon/2\), and denseness of \(\mathcal{F}\) in \(\mathcal{G}\), applied to that \(g\), gives \(f \in \mathcal{F}\) with \(\|f - g\| < \epsilon/2\); by the triangle inequality,

\begin{equation*} \|f - h\| \le \|f - g\| + \|g - h\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{equation*}

Since also \(\mathcal{F} \subseteq \mathcal{G} \subseteq \mathcal{H}\), this is exactly the assertion that \(\mathcal{F}\) is dense in \(\mathcal{H}\).

The collection \(\mathcal{P}\) of polynomials with rational coefficients is the union over \(n \ge 0\) of

\begin{equation*} \mathcal{P}_n = \{\, c_0 + c_1 x + \cdots + c_n x^n \;:\; c_0, \dots, c_n \in \mathbb{Q} \,\} , \end{equation*}

since every polynomial has finite degree.

Each \(\mathcal{P}_n\) is countable: with \(\varphi\) a one-to-one correspondence of \(\mathbb{N}\) with \(\mathbb{Q}\) (Corollary 4 of Chapter 1), the map

\begin{equation*} \Phi(k_0, \dots, k_n) = \varphi(k_0) + \varphi(k_1) x + \cdots + \varphi(k_n) x^n \end{equation*}

carries \(\mathbb{N}^{n+1}\) onto the nonempty set \(\mathcal{P}_n\), and \(\mathbb{N}^{n+1}\) is countable by that same corollary, so Theorem 5 of Chapter 1 applies.

Hence \(\mathcal{P}\) is a countable union of countable sets, so is countable by Corollary 6 of Chapter 1.

Test \(g\) against its own conjugate function. First, the hypothesis extends from \(\mathcal{S}\) to all of \(L^q(E)\): given \(h \in L^q(E)\), Problem 36 supplies \(\{f_n\}\) in \(\mathcal{S}\) with \(\|f_n - h\|_q \to 0\), and Holder’s Inequality (Theorem 1), applicable since \(f_n - h \in L^q(E)\) and \(g \in L^p(E)\) with \(p, q\) conjugate, gives

\begin{equation*} \left| \int_E f_n \cdot g - \int_E h \cdot g \right| \le \int_E |(f_n - h) \cdot g| \le \|f_n - h\|_q \, \|g\|_p \to 0 , \end{equation*}

so \(\int_E h \cdot g = \lim_n \int_E f_n \cdot g = 0\).

Now suppose \(\|g\|_p > 0\). By Theorem 1 the conjugate function

\begin{equation*} g^{*} = \|g\|_p^{\,1-p} \cdot \operatorname{sgn}(g) \cdot |g|^{p-1} \end{equation*}

belongs to \(L^q(E)\) with \(\|g^*\|_q = 1\) and \(\int_E g \cdot g^{*} = \|g\|_p\) (the case \(p = 1\), \(q = \infty\), where \(g^* = \operatorname{sgn}(g)\), is included in that theorem). Taking \(h = g^{*}\) above forces \(\|g\|_p = 0\), a contradiction.

Hence \(\int_E |g|^p = 0\), so \(g = 0\) a.e. on \(E\) by Proposition 9 of Chapter 4, that is, \(g = 0\) in \(L^p(E)\).

Four details are left to the reader; \(1 \le p < \infty\) is fixed, \(\mathcal{S}[a,b]\) is the step functions on \([a,b]\), and \(\mathcal{S}’[a,b]\) those taking rational values and constant on the open intervals of a partition with rational interior points.

(i) \(\mathcal{S}’[a,b]\) is dense in \(\mathcal{S}[a,b]\) in the \(L^p\) norm. Let \(\psi = c_k\) on \((y_{k-1}, y_k)\), \(1 \le k \le n\), for a partition \(a = y_0 < \cdots < y_n = b\); the values at partition points are irrelevant, being a null set. By density of \(\mathbb{Q}\) pick rationals \(d_k\) with \(|d_k - c_k| < \epsilon / (2(b-a)^{1/p} + 2)\) and let \(\varphi = d_k\) on \((y_{k-1}, y_k)\), so by monotonicity of integration

\begin{equation*} \|\varphi - \psi\|_p \le \frac{\epsilon \, (b-a)^{1/p}}{2 (b-a)^{1/p} + 2} < \frac{\epsilon}{2}. \end{equation*}

Next, with \(M = \max_k |d_k|\) and \(\delta = \tfrac12 \min_k (y_k - y_{k-1}) > 0\), fix

\begin{equation*} 0 < \eta < \min\left\{ \delta, \ \frac{1}{2n}\left[\frac{\epsilon}{2(2M+1)}\right]^{p} \right\} \end{equation*}

and rationals \(x_k\) with \(|x_k - y_k| < \eta\) for \(1 \le k \le n-1\), \(x_0 = a\), \(x_n = b\); since \(\eta < \delta\) these still increase, so \(\varphi’ = d_k\) on \((x_{k-1}, x_k)\) lies in \(\mathcal{S}’[a,b]\). Now \(\varphi\) and \(\varphi’\) differ only on a set \(D\) of measure at most \(n\eta\) (finite subadditivity), where \(|\varphi - \varphi’| \le 2M\), so

\begin{equation*} \|\varphi - \varphi’\|_p \le 2M \, [m(D)]^{1/p} \le (2M+1)(n\eta)^{1/p} < \frac{\epsilon}{2} , \end{equation*}

and Minkowski’s Inequality gives \(\|\varphi’ - \psi\|_p < \epsilon\). With Proposition 10 (\(\mathcal{S}[a,b]\) dense in \(L^p[a,b]\)) and (13), \(\mathcal{S}’[a,b]\) is dense in \(L^p[a,b]\).

(ii) \(\mathcal{S}’[a,b]\) is countable. A member admitting a partition into exactly \(n\) subintervals is determined by the rational data

\begin{equation*} (x_1, \dots, x_{n-1}, d_1, \dots, d_n, e_0, \dots, e_n) \in \mathbb{Q}^{3n} , \end{equation*}

the \(x_k\) the interior partition points, the \(d_k\) the values on the open intervals and the \(e_k\) the values at the partition points; this is a map of a subset of \(\mathbb{Q}^{3n}\) onto \(\mathcal{S}’_n[a,b]\) (not injective, which is harmless). Since \(\mathbb{Q}^{3n}\) is countable (Corollary 4 of Chapter 1) and subsets of countable sets are countable (Theorem 3 of Chapter 1), Theorem 5 of Chapter 1 makes \(\mathcal{S}’_n[a,b]\) countable, and \(\mathcal{S}’[a,b] = \bigcup_n \mathcal{S}’_n[a,b]\) is countable by Corollary 6 of Chapter 1.

(iii) \(\mathcal{F} = \bigcup_n \mathcal{F}_n\) is countable and dense in \(L^p(\mathbb{R})\), where \(\mathcal{F}_n\) consists of the functions vanishing off \([-n,n]\) whose restriction lies in \(\mathcal{S}’[-n,n]\). Each \(\mathcal{F}_n\) is in one-to-one correspondence with \(\mathcal{S}’[-n,n]\) under restriction, hence countable by (ii), so \(\mathcal{F}\) is countable by Corollary 6 of Chapter 1; and each member is bounded and vanishes off a set of finite measure, so lies in \(L^p(\mathbb{R})\). Given \(f \in L^p(\mathbb{R})\) and \(\epsilon > 0\), the Monotone Convergence Theorem applied to \(\{|f|^p \chi_{[-n,n]}\}\) gives \(\int_{[-n,n]} |f|^p \to \int_{\mathbb{R}} |f|^p < \infty\), so for some \(n\), subtracting these finite numbers,

\begin{equation*} \|f - g\|_{L^p(\mathbb{R})}^p = \int_{\mathbb{R} \sim [-n,n]} |f|^p < \left(\frac{\epsilon}{2}\right)^p , \qquad g = f \cdot \chi_{[-n,n]} . \end{equation*}

By (i) choose \(\psi \in \mathcal{S}’[-n,n]\) with \(\|g - \psi\|_{L^p[-n,n]} < \epsilon/2\); extended by \(0\) it lies in \(\mathcal{F}_n\) and \(\|g - \psi\|_{L^p(\mathbb{R})} < \epsilon/2\), so \(\|f - \psi\|_{L^p(\mathbb{R})} < \epsilon\) by Minkowski’s Inequality.

(iv) General \(E\). Put \(\mathcal{F}_E = \{ f|_E : f \in \mathcal{F}\}\), countable as the image of a countable set (Theorem 5 of Chapter 1) and contained in \(L^p(E)\) since \(\int_E |f|^p \le \int_{\mathbb{R}} |f|^p\). For \(h \in L^p(E)\), extend by \(0\) to \(\hat h \in L^p(\mathbb{R})\) and pick \(f \in \mathcal{F}\) with \(\|\hat h - f\|_{L^p(\mathbb{R})} < \epsilon\) by (iii); monotonicity of integration over \(E \subseteq \mathbb{R}\) gives

\begin{equation*} \|h - f|_E\|_{L^p(E)} \le \|\hat h - f\|_{L^p(\mathbb{R})} < \epsilon . \end{equation*}

So \(L^p(E)\) has a countable dense subset and is separable.

No, unless \(m(E) = 0\) (when \(L^{p_2}(E) = \{0\}\) is trivially complete); assume \(m(E) > 0\). The setting is legitimate: \(m(E) < \infty\) and \(p_1 < p_2\) put \(L^{p_2}(E) \subseteq L^{p_1}(E)\) by Corollary 3, so \(\|\cdot\|_{p_1}\) is a finite norm on the linear subspace \(L^{p_2}(E)\).

Split \(E\) into disjoint pieces of prescribed measure. The function \(\mu(t) = m(E \cap (-\infty, t])\) is nondecreasing and continuous, since \(|\mu(t) - \mu(s)| = m(E \cap (s,t]) \le t-s\), with limits \(0\) and \(m(E)\) at \(\mp\infty\) by continuity of measure; so by the Intermediate Value Theorem there are \(c_k\) with \(\mu(c_k) = m(E) 2^{-k}\). Put \(A_k = E \cap (-\infty, c_k]\); then \(A_{k+1} \subseteq A_k\) (as \(\mu\) is nondecreasing) and, by excision,

\begin{equation*} E_k = A_k \sim A_{k+1}, \qquad m_k = m(E_k) = m(E) \, 2^{-(k+1)} > 0 , \end{equation*}

are disjoint measurable subsets of \(E\).

Now take \(a_k = [k^2 m_k]^{-1/p_1}\) and \(f = \sum_{k=1}^{\infty} a_k \chi_{E_k}\), the pointwise limit of the increasing simple functions \(f_n = \sum_{k=1}^{n} a_k \chi_{E_k}\). The Monotone Convergence Theorem and disjointness give \(\int_E |f|^r = \sum_k a_k^r m_k\) for every \(r > 0\), so

\begin{equation*} \int_E |f|^{p_1} = \sum_{k=1}^{\infty} \frac{1}{k^2} < \infty , \end{equation*}

while with \(\alpha = p_2/p_1 > 1\) and \(C = [m(E)/2]^{1-\alpha} > 0\),

\begin{equation*} a_k^{\,p_2} m_k = k^{-2\alpha} m_k^{\,1-\alpha} = C \cdot \frac{2^{\,k(\alpha - 1)}}{k^{2\alpha}} \to \infty , \end{equation*}

so \(\int_E |f|^{p_2} = \infty\). Thus \(f \in L^{p_1}(E) \sim L^{p_2}(E)\).

Each \(f_n\) is simple on a set of finite measure, hence in \(L^{p_2}(E)\), and \(f - f_n = \sum_{k>n} a_k \chi_{E_k}\) has the same form as \(f\), so

\begin{equation*} \|f - f_n\|_{p_1}^{p_1} = \sum_{k > n} \frac{1}{k^2} \longrightarrow 0 , \end{equation*}

the tail of a convergent series. Hence \(\{f_n\} \to f\) in \(L^{p_1}(E)\), so by Minkowski’s Inequality \(\{f_n\}\) is \(\|\cdot\|_{p_1}\)-Cauchy. Were \(\{f_n\} \to g\) in \(\|\cdot\|_{p_1}\) with \(g \in L^{p_2}(E) \subseteq L^{p_1}(E)\), Minkowski would give \(\|f-g\|_{p_1} = 0\), hence \(f = g\) a.e. by Proposition 9 of Chapter 4 and \(\int_E |f|^{p_2} < \infty\), a contradiction. So the space is not Banach.

Method (2): the simple functions are dense in \(L^{p_1}(E)\) (Proposition 9 of this chapter) and, \(m(E)\) being finite, all lie in \(L^{p_2}(E)\); so \(L^{p_2}(E)\) is a dense subspace of the Banach space \(L^{p_1}(E)\) (Riesz-Fischer), proper by the \(f\) above. By Problem 43 a dense subspace complete in the inherited norm is the whole space, so \(L^{p_2}(E)\) is not \(\|\cdot\|_{p_1}\)-complete.

Take \(E = \{0\}\), or any set of measure zero: every measurable \(f\) on \(E\) has \(|f| \le 0\) a.e., so \(\|f\|_{\infty} = 0\) and \(L^{\infty}(E) = \{0\}\), which has the countable dense subset \(\{0\}\).

Now let \(E\) contain a nondegenerate interval, hence a closed bounded \([a,b]\) with \(a < b\), and suppose \(\{f_n\}\) were a countable dense subset of \(L^{\infty}(E)\). For \(x \in [a,b]\) put \(g_x = \chi_{[a,x]}\) on \(E\), which lies in \(L^{\infty}(E)\), and pick \(\eta(x) \in \mathbb{N}\) with

\begin{equation*} \|g_x - f_{\eta(x)}\|_{\infty} < \tfrac{1}{2} . \end{equation*}

For \(a \le x_1 < x_2 \le b\) the difference \(g_{x_2} - g_{x_1}\) is \(\chi_{(x_1, x_2]}\) with \((x_1, x_2] \subseteq E\) of positive measure, so no \(M < 1\) is an essential upper bound while \(1\) is:

\begin{equation*} \|g_{x_1} - g_{x_2}\|_{\infty} = 1 . \end{equation*}

Hence \(\eta(x_1) = \eta(x_2) = k\) would give

\begin{equation*} 1 \le \|g_{x_1} - f_{k}\|_{\infty} + \|f_{k} - g_{x_2}\|_{\infty} < \tfrac{1}{2} + \tfrac{1}{2} = 1 , \end{equation*}

so \(\eta\) is one-to-one from \([a,b]\) into \(\mathbb{N}\), making \([a,b]\) countable by Theorem 3 of Chapter 1 and contradicting Theorem 7 of Chapter 1. So \(L^{\infty}(E)\) is not separable.

Only \(X \subseteq X_0\) needs proof. Let \(x \in X\). By denseness (Problem 36) there is a sequence \(\{x_n\}\) in \(X_0\) with \(\|x_n - x\| \to 0\), and such a sequence is Cauchy in \(X\), since

\begin{equation*} \|x_n - x_m\| \le \|x_n - x\| + \|x - x_m\| . \end{equation*}

The norm of \(X_0\) is the restriction of that of \(X\), so \(\{x_n\}\) is Cauchy in \(X_0\); as \(X_0\) is complete, \(\{x_n\} \to x_0\) for some \(x_0 \in X_0\), and this convergence is convergence in \(X\). Then

\begin{equation*} \|x - x_0\| \le \|x - x_n\| + \|x_n - x_0\| \longrightarrow 0 , \end{equation*}

so \(\|x - x_0\| = 0\) and \(x = x_0 \in X_0\).

\(\ell^p\) is separable for \(1 \le p < \infty\): the set \(\mathcal{D} = \bigcup_n \mathcal{D}_n\) of finitely supported rational sequences works, where

\begin{equation*} \mathcal{D}_n = \{ (q_1, \ldots, q_n, 0, 0, \ldots) : q_1, \ldots, q_n \in \mathbb{Q} \} \subseteq \ell^p . \end{equation*}

Each \(\mathcal{D}_n\) is in one-to-one correspondence with the countable set \(\mathbb{Q}^n\), so \(\mathcal{D}\) is countable. For density, let \(a \in \ell^p\) and \(\epsilon > 0\); since \(\sum_k |a_k|^p\) converges, choose \(N\) with \(\sum_{k>N} |a_k|^p < (\epsilon/2)^p\), so that \(a^{(N)} = (a_1, \ldots, a_N, 0, \ldots)\) satisfies \(\|a - a^{(N)}\|_p < \epsilon/2\). Choose rationals \(q_k\) with \(|a_k - q_k|^p < (\epsilon/2)^p / N\) for \(1 \le k \le N\) and set \(q = (q_1, \ldots, q_N, 0, \ldots) \in \mathcal{D}_N\); then

\begin{equation*} \|a^{(N)} - q\|_p = \left( \sum_{k=1}^N |a_k - q_k|^p \right)^{1/p} < \frac{\epsilon}{2} , \end{equation*}

and Minkowski’s Inequality gives \(\|a - q\|_p < \epsilon\).

\(\mathcal{P}(\mathbb{N})\) is uncountable: it is infinite (the singletons), and were it enumerated as \(\{A_n\}\), the set \(A = \{ n : n \notin A_n \}\) would equal some \(A_m\), whence \(m \in A \iff m \notin A_m = A\).

Hence \(\ell^{\infty}\) is not separable. For \(A \subseteq \mathbb{N}\) the sequence \(\chi_A\) lies in \(\ell^{\infty}\), and for \(A \ne B\) some index lies in exactly one of them while every term of \(\chi_A - \chi_B\) has absolute value at most \(1\), so \(\|\chi_A - \chi_B\|_{\infty} = 1\). If \(\{x_n\}\) were a countable dense subset, pick \(\eta(A)\) with \(\|\chi_A - x_{\eta(A)}\|_{\infty} < 1/2\); then \(\eta(A) = \eta(B) = m\) with \(A \ne B\) would give

\begin{equation*} 1 = \|\chi_A - \chi_B\|_{\infty} \le \|\chi_A - x_m\|_{\infty} + \|x_m - \chi_B\|_{\infty} < 1 , \end{equation*}

so \(\eta\) is one-to-one from \(\mathcal{P}(\mathbb{N})\) into \(\mathbb{N}\), making \(\mathcal{P}(\mathbb{N})\) countable, a contradiction.

Let \(f \in L^p(E)\) and \(\epsilon > 0\); we chain three \(\epsilon/3\) approximations, working on \(\mathbb{R}\). Extending \(f\) by \(0\) off \(E\) gives \(\tilde f \in L^p(\mathbb{R})\), and any continuous \(g\) on \(\mathbb{R}\) vanishing outside a bounded set restricts to a member of \(C_c(E)\) with

\begin{equation*} \|f - g\|_{L^p(E)} \le \|\tilde f - g\|_{L^p(\mathbb{R})} , \end{equation*}

so it suffices to approximate \(\tilde f\), which we rename \(f\).

(i) Truncation. The functions \(|f|^p \chi_{[-n,n]}\) increase pointwise to \(|f|^p\), so by the Monotone Convergence Theorem \(\int_{[-n,n]} |f|^p \to \int_{\mathbb{R}} |f|^p < \infty\); subtracting these finite numbers, some \(n\) has

\begin{equation*} \|f - f_1\|_{L^p(\mathbb{R})}^p = \int_{\mathbb{R} \sim [a,b]} |f|^p < (\epsilon/3)^p , \qquad f_1 = f \cdot \chi_{[a,b]} , \end{equation*}

with \(a = -n\), \(b = n\).

(ii) Step function. By Proposition 10 the step functions are dense in \(L^p[a,b]\), so some step function \(\psi\), extended by \(0\) off \([a,b]\) as is \(f_1\), has \(\|f_1 - \psi\|_{L^p(\mathbb{R})} < \epsilon/3\). Its partition points being a null set,

\begin{equation*} \psi = \sum_{k=1}^m c_k \, \chi_{(x_{k-1}, x_k)} \quad \text{in } L^p(\mathbb{R}) . \end{equation*}

(iii) Continuous approximation of \(\chi_{(c,d)}\). For \(0 < 2\delta < d - c\) let \(h_{\delta}\) be the piecewise-linear trapezoid that is \(0\) off \((c,d)\), \(1\) on \([c+\delta, d-\delta]\), and linear on \((c, c+\delta)\) and \((d-\delta, d)\); it is continuous (the pieces agree at the junctions) and vanishes off \([c,d]\). Since \(|\chi_{(c,d)} - h_{\delta}| \le 1\) and vanishes off a set of measure \(2\delta\),

\begin{equation*} \| \chi_{(c,d)} - h_{\delta} \|_{L^p(\mathbb{R})} \le (2\delta)^{1/p} \to 0 \quad \text{as } \delta \to 0^+ , \end{equation*}

as \(p < \infty\). So choose continuous \(g_k\), each vanishing off a bounded set, with \(\| \chi_{(x_{k-1},x_k)} - g_k \|_{L^p(\mathbb{R})} < \epsilon / (3 m (1 + |c_k|))\) and put \(g = \sum_{k=1}^m c_k g_k\), again continuous and vanishing off a bounded set. Minkowski’s Inequality gives

\begin{equation*} \|\psi - g\|_{L^p(\mathbb{R})} \le \sum_{k=1}^m |c_k| \, \| \chi_{(x_{k-1},x_k)} - g_k \|_{L^p(\mathbb{R})} \le \frac{\epsilon}{3} , \end{equation*}

and one more application of it yields \(\|f - g\|_{L^p(\mathbb{R})} < \epsilon\), hence \(\|f - g\|_{L^p(E)} < \epsilon\) with \(g|_E \in C_c(E)\).

With \(s = 1/p \in (0,1)\) and \(\varphi(t) = \operatorname{sgn}(t) |t|^{s}\), the claim is equivalent, on taking \(p\)-th roots of both nonnegative sides, to

\begin{equation*} |\varphi(a) - \varphi(b)| \le 2 \, |a - b|^{s} . \end{equation*}

Call this inequality \((\ast)\). We use subadditivity of \(t \mapsto t^s\): for \(u, v \ge 0\) not both \(0\), with \(T = u+v\), each of \((u/T)^s \ge u/T\) and \((v/T)^s \ge v/T\) holds since \(x^s = x \cdot x^{s-1} \ge x\) for \(0 \le x \le 1\) and \(s - 1 < 0\); adding gives

\begin{equation*} (u + v)^s \le u^s + v^s . \end{equation*}

Both sides of \((\ast)\) are unchanged under interchanging \(a, b\) and under \((a,b) \mapsto (-a,-b)\), \(\varphi\) being odd, so two cases suffice.

(i) \(a \ge b \ge 0\). Subadditivity with \(u = a-b\), \(v = b\) gives \(a^s \le (a-b)^s + b^s\), so

\begin{equation*} |\varphi(a) - \varphi(b)| = a^s - b^s \le (a-b)^s = |a-b|^s \le 2 |a-b|^s . \end{equation*}

(ii) \(a > 0 > b\). Here \(\varphi(a) - \varphi(b) = a^s + |b|^s\) and \(|a - b| = a + |b|\); since \(t \mapsto t^s\) is increasing and \(a, |b| \le a + |b|\),

\begin{equation*} a^s + |b|^s \le 2 (a + |b|)^s = 2 |a - b|^s . \end{equation*}

Raising \((\ast)\) to the power \(p\) gives the stated inequality.

Apply the Mean Value Theorem to \(\psi(t) = \operatorname{sgn}(t) |t|^{p}\), which is differentiable on \(\mathbb{R}\) with

\begin{equation*} \psi’(t) = p \, |t|^{p-1} . \end{equation*}

Indeed \(\psi = t^p\) near any \(t > 0\) and \(\psi = -(-t)^p\) near any \(t < 0\), giving \(p|t|^{p-1}\) in both cases by the chain rule; and at \(t = 0\), since \(\operatorname{sgn}(h)/h = 1/|h|\),

\begin{equation*} \frac{\psi(h) - \psi(0)}{h} = |h|^{p-1} \to 0 = p \cdot |0|^{p-1} \quad \text{as } h \to 0 , \end{equation*}

because \(p - 1 > 0\).

For \(a \ne b\) (the case \(a = b\) being trivial), the Mean Value Theorem supplies \(c\) strictly between \(a\) and \(b\) with \(\psi(a) - \psi(b) = p |c|^{p-1} (a-b)\). Since \(|c| \le \max\{|a|,|b|\} \le |a| + |b|\) and \(t \mapsto t^{p-1}\) is increasing on \([0,\infty)\),

\begin{equation*} \left| \operatorname{sgn}(a) |a|^{p} - \operatorname{sgn}(b) |b|^{p} \right| = p |c|^{p-1} |a-b| \le p \, |a - b| \, (|a| + |b|)^{p-1} . \end{equation*}

Pointwise \(|\Phi(f)|^p = |f|\), so \(\int_E |\Phi(f)|^p = \|f\|_1 < \infty\) and \(\Phi(f)\) lies in \(L^p(E)\) with \(\|\Phi(f)\|_p = \|f\|_1^{1/p}\). Write \(\varphi(t) = \operatorname{sgn}(t)|t|^{1/p}\) and \(\psi(t) = \operatorname{sgn}(t)|t|^{p}\), so \(\Phi(f) = \varphi \circ f\); each is a strictly increasing continuous bijection of \(\mathbb{R}\) and \(\varphi \circ \psi = \psi \circ \varphi = \mathrm{id}\) (Check!). Let \(q = p/(p-1)\).

Problem 46, applied at each \(x\) with \(a = f(x)\), \(b = g(x)\), gives \(|\Phi(f) - \Phi(g)|^p \le 2^p |f-g|\) on \(E\); integrating,

\begin{equation*} \|\Phi(f) - \Phi(g)\|_p^p \le 2^p \|f - g\|_1 , \qquad \|\Phi(f) - \Phi(g)\|_p \le 2 \, \|f - g\|_1^{1/p} . \end{equation*}

The printed inequality, lacking the exponent \(p\) on the left norm, is a misprint for the first of these: with \(E = [0,1]\), \(p = 2\), \(g \equiv 0\), \(f \equiv 1/100\) it would assert \(1/10 \le 4/100\).

Continuity of \(\Phi\) is immediate: \(\|f_n - f\|_1 \to 0\) forces \(\|\Phi(f_n) - \Phi(f)\|_p \le 2\|f_n - f\|_1^{1/p} \to 0\).

Since \(\varphi\) is one-to-one on \(\mathbb{R}\), \(\Phi(f) = \Phi(g)\) a.e. forces \(f = g\) a.e., so \(\Phi\) is one-to-one. Given \(h\) in \(L^p(E)\), set \(f = \psi \circ h\); then \(|f| = |h|^p\) is integrable, so \(f\) lies in \(L^1(E)\), and \(\Phi(f) = \varphi \circ \psi \circ h = h\). Hence \(\Phi\) is onto \(L^p(E)\) with

\begin{equation*} \Phi^{-1}(h)(x) = \operatorname{sgn}(h(x)) \, |h(x)|^{p} . \end{equation*}

For the continuity of \(\Phi^{-1}\), apply Problem 47 pointwise and then Holder with the conjugate exponents \(p, q\):

\begin{equation*} \begin{aligned} \left\| \Phi^{-1}(h) - \Phi^{-1}(k) \right\|_1 &\le p \int_E |h - k| \cdot (|h| + |k|)^{p-1} \\ &\le p \, \|h - k\|_p \cdot \big\| (|h| + |k|)^{p-1} \big\|_q \\ &\le p \, \big( \|h\|_p + \|k\|_p \big)^{p-1} \, \|h - k\|_p , \end{aligned} \end{equation*}

the last step because \((p-1)q = p\) gives \(\|(|h|+|k|)^{p-1}\|_q = \| \, |h|+|k| \, \|_p^{\,p-1}\), finite and bounded by \((\|h\|_p + \|k\|_p)^{p-1}\) by Minkowski. If \(\{h_n\} \to h\) in \(L^p(E)\) then \(\{\|h_n\|_p\}\) is bounded, say by \(M\), and

\begin{equation*} \left\| \Phi^{-1}(h_n) - \Phi^{-1}(h) \right\|_1 \le p \, \big( M + \|h\|_p \big)^{p-1} \, \|h_n - h\|_p \to 0 . \end{equation*}

If \(\mathcal{D} = \{f_n\}_{n=1}^\infty\) is countable and dense in \(L^1(E)\), then its image \(\Phi(\mathcal{D})\) under the Mazur mapping of the preceding problem is countable and dense in \(L^p(E)\).

Indeed, let \(h\) belong to \(L^p(E)\) and let \(\epsilon > 0\). By the preceding problem \(\Phi\) maps \(L^1(E)\) onto \(L^p(E)\), so \(h = \Phi(f)\) for \(f = \Phi^{-1}(h)\) in \(L^1(E)\); choose \(n\) with \(\|f - f_n\|_1 < (\epsilon/2)^p\). By the estimate of the preceding problem,

\begin{equation*} \|h - \Phi(f_n)\|_p = \|\Phi(f) - \Phi(f_n)\|_p \le 2 \, \|f - f_n\|_1^{1/p} < \epsilon . \end{equation*}

Hence \(L^p(E)\) has a countable dense subset, so it is separable.

A continuous surjection carries a countable dense set to a countable dense set, and \(L^1[a,b]\) is separable while \(L^\infty[a,b]\) is not.

Suppose \(\Phi\) were such a mapping. By Theorem 11 with \(p = 1\), choose a countable dense \(D = \{f_1, f_2, \dots\}\) in \(L^1[a,b]\); the countable set \(\Phi(D)\) is then dense in \(L^\infty[a,b]\). Indeed, given \(g\) in \(L^\infty[a,b]\) pick \(f\) with \(\Phi(f) = g\), then by Problem 36 a sequence \(\{g_k\}\) in \(D\) with \(\{g_k\} \to f\) in \(L^1[a,b]\); continuity of \(\Phi\) gives \(\{\Phi(g_k)\} \to g\), and Problem 36 again gives the density.

So \(L^\infty[a,b]\) would be separable, contradicting the Example following Theorem 11 (which uses \(a < b\): the uncountable family \(\{\chi_{[a,x)}\}_{a \le x \le b}\) is \(1\)-separated in the \(L^\infty\) norm). Hence no such \(\Phi\) exists.

Let \(f\) belong to \(L^p(E)\) and \(\epsilon > 0\); we produce \(g\) in \(C_c(E)\) with \(\|f - g\|_p < \epsilon\).

Since \(|f|^p\) is integrable, \(f\) is finite a.e. (Proposition 15 of Chapter 4); redefine it to be \(0\) on that null set, which changes no \(L^p\) distance and does not shrink \(E\). The truncations

\begin{equation*} h_n = f \cdot \chi_{\{x \in E \,:\, |x| \le n, \ |f(x)| \le n\}} \end{equation*}

satisfy \(|h_n - f|^p \le 2^p |f|^p\) with \(|f|^p\) integrable, and \(\{h_n\} \to f\) pointwise on \(E\), so the Lebesgue Dominated Convergence Theorem gives \(\{h_n\} \to f\) in \(L^p(E)\). Fix \(n \ge 1\) with \(\|f - h_n\|_p < \epsilon/2\) and write \(h = h_n\), \(M = n \ge 1\); thus \(h\) is measurable with \(|h| \le M\) on \(E\) and \(h = 0\) on \(\{x \in E : |x| > n\}\).

Put \(\eta = \tfrac12 (\epsilon/4M)^p\). Lusin’s Theorem, which by Problem 29 of Chapter 3 is available for \(E\) of infinite measure as well, applied to the real-valued measurable \(h\), yields a continuous \(g_0\) on \(\mathbb{R}\) and a closed \(F \subseteq E\) with \(h = g_0\) on \(F\) and \(m(E \sim F) < \eta\). Two continuous surgeries make \(g_0\) admissible without altering it on \(F\): with

\begin{equation*} \tau(t) = \max\{-M, \min\{t, M\}\}, \qquad \varphi(x) = \min\{1, \max\{0, n + 1 - |x|\}\} , \end{equation*}

set \(g = (\tau \circ g_0) \cdot \varphi\). Then \(g\) is continuous on \(\mathbb{R}\), \(|g| \le M\), and \(g\) vanishes off \([-n-1, n+1]\), so \(g\) belongs to \(C_c(E)\); being bounded of bounded support it also has \(\int_E |g|^p \le M^p(2n+2) < \infty\), so \(g\) lies in \(L^p(E)\) and Minkowski applies below. Also \(g = h\) on \(F\): (i) if \(|x| \le n\) then \(\varphi(x) = 1\) and \(\tau(h(x)) = h(x)\) since \(|h| \le M\); (ii) if \(|x| > n\) then \(h(x) = 0\), so \(\tau(g_0(x)) = 0\) and \(g(x) = 0\).

Hence \(h - g\) vanishes on \(F\) and \(|h - g| \le 2M\) on \(E\), so

\begin{equation*} \int_E |h - g|^p \le (2M)^p \, m(E \sim F) < (2M)^p \eta = \tfrac12 (\epsilon/2)^p < (\epsilon/2)^p , \end{equation*}

and by Minkowski \(\|f - g\|_p \le \|f - h\|_p + \|h - g\|_p < \epsilon\).

Write \(B = \{f \in X \mid \|f\| \le 1\}\), let \(\mathcal{M}\) be the set of \(M \ge 0\) for which (5) holds, so \(\|T\|_* = \inf \mathcal{M}\), and set \(S = \sup\{T(f) \mid f \in B\}\). Since \(B\) is symmetric and \(T\) is linear, \(\{T(f) \mid f \in B\}\) is symmetric about \(0\), whence

\begin{equation*} S = \sup\{\, |T(f)| \mid f \in B \,\} \ge T(0) = 0 . \end{equation*}

If \(M\) belongs to \(\mathcal{M}\) then \(|T(f)| \le M \|f\| \le M\) for \(f\) in \(B\), so \(S \le M\); taking the infimum, \(S \le \|T\|_*\).

Conversely \(S\) belongs to \(\mathcal{M}\): for \(f = 0\) this is (5) trivially since \(S \ge 0\), while for \(f \ne 0\) the vector \(h = f/\|f\|\) lies in \(B\), so

\begin{equation*} |T(f)| = \|f\| \cdot |T(h)| \le S \cdot \|f\| . \end{equation*}

Hence \(\|T\|_* \le S\), and \(\|T\|_* = S\), which is (8). In particular the infimum is attained, so (5) holds with \(M = \|T\|_*\) and, applied to \(f - h\), gives (6).

Throughout, \(\|T\|_* = \sup\{|T(f)| : \|f\| \le 1\}\) and the infimum defining it is attained, both by Problem 1.

With the pointwise operations, a linear combination \(\alpha T + \beta S\) of linear functionals is linear, and for \(f\) in \(X\)

\begin{equation*} |(\alpha T + \beta S)(f)| \le \big( |\alpha| \, \|T\|_* + |\beta| \, \|S\|_* \big) \|f\| , \end{equation*}

so \(\alpha T + \beta S\) is bounded: \(X^*\) is a linear space.

Nonnegativity is clear, \(\|T\|_*\) being the infimum of a nonempty set of nonnegative numbers, and \(\|0\|_* = 0\). If \(\|T\|_* = 0\), then since the infimum is attained (5) holds with \(M = 0\), so \(T = 0\).

Homogeneity and the triangle inequality follow from the supremum formula:

\begin{equation*} \|\alpha T\|_* = \sup_{\|f\| \le 1} |\alpha| \, |T(f)| = |\alpha| \, \|T\|_* , \end{equation*}

and, since \(|(T+S)(f)| \le |T(f)| + |S(f)| \le \|T\|_* + \|S\|_*\) for \(\|f\| \le 1\), the number \(\|T\|_* + \|S\|_*\) is an upper bound for the defining set of \(\|T+S\|_*\), so \(\|T+S\|_* \le \|T\|_* + \|S\|_*\).

If \(T\) is bounded and \(\{f_n\} \to f\) in \(X\), then (6) gives \(|T(f_n) - T(f)| \le \|T\|_* \|f_n - f\| \to 0\), which is (7).

Conversely, if \(T\) is unbounded then for each \(n\) the value \(M = n\) fails in (5), so there is \(f_n\) with \(|T(f_n)| > n \|f_n\|\); necessarily \(f_n \ne 0\), since otherwise linearity would make this read \(0 > 0\). Put \(h_n = f_n / (n\|f_n\|)\). Then

\begin{equation*} \|h_n\| = \frac{1}{n} \to 0 , \qquad |T(h_n)| = \frac{|T(f_n)|}{n \|f_n\|} > 1 \quad \text{for all } n , \end{equation*}

so \(\{h_n\} \to 0\) while \(\{T(h_n)\}\) does not converge to \(T(0) = 0\): (7) fails. By contraposition, (7) implies boundedness.

The two admissible sets of constants coincide: \(\mathcal{C} = \mathcal{M}\), where \(\mathcal{C}\) is the set of \(c \ge 0\) satisfying the Lipschitz inequality and \(\mathcal{M}\) the set of \(M \ge 0\) satisfying (5). Both assertions follow at once, since then one set is empty exactly when the other is and the infima agree.

If \(M\) belongs to \(\mathcal{M}\), then by linearity, for all \(g, h\) in \(X\),

\begin{equation*} |T(g) - T(h)| = |T(g-h)| \le M \cdot \|g - h\| , \end{equation*}

so \(M\) belongs to \(\mathcal{C}\). If \(c\) belongs to \(\mathcal{C}\), take \(h = 0\) and use \(T(0) = 0\):

\begin{equation*} |T(f)| = |T(f) - T(0)| \le c \cdot \|f\| \quad \text{for all } f \in X , \end{equation*}

so \(c\) belongs to \(\mathcal{M}\). Hence \(T\) is bounded if and only if it is Lipschitz, with Lipschitz constant \(\inf \mathcal{C} = \inf \mathcal{M} = \|T\|_*\).

The truncations \(f_n = f \cdot \chi_{E \cap [-n,\,n]}\) do it: each belongs to \(L^p(E)\) and vanishes outside \([-n,n]\), while

\begin{equation*} |f - f_n|^p = |f|^p \cdot \chi_{E \setminus [-n,\,n]} \to 0 \quad \text{pointwise on } E , \end{equation*}

dominated by the integrable function \(|f|^p\) (here \(p < \infty\) enters twice over), so the Lebesgue Dominated Convergence Theorem gives \(\|f - f_n\|_p \to 0\).

For \(L^\infty(\mathbb{R})\) the witness is \(f \equiv 1\). Let \(g\) in \(L^\infty(\mathbb{R})\) vanish a.e. off a bounded set \(B\) and choose \(n\) with \(B \subseteq [-n,n]\). Then \(|f - g| = 1\) a.e. on \(\mathbb{R} \setminus [-n,n]\), which has infinite, hence positive, measure, so no \(c < 1\) is an essential upper bound for \(|f - g|\) and

\begin{equation*} \|f - g\|_\infty \ge 1 . \end{equation*}

Thus no such \(g\) comes within \(1/2\) of \(f\).

With \(\Phi(x) = T(\chi_{[a,x)})\) as in the proof of Theorem 5, linearity gives \(\Phi(d) - \Phi( c) = T(\chi_{[c,d)})\) for \(a \le c \le d \le b\), and since \(\|\chi_{[c,d)}\|_1 = d - c\), boundedness of \(T\) gives

\begin{equation*} |\Phi(d) - \Phi( c)| \le \|T\|_* \cdot (d - c) . \end{equation*}

So \(\Phi\) is Lipschitz, hence absolutely continuous on \([a,b]\): for \(\epsilon > 0\) the choice \(\delta = \epsilon/(\|T\|_* + 1)\) serves. (For \(p > 1\) the text has instead the exponent \(1/p\) on \(\sum_k (b_k - a_k)\).)

By Theorem 10 of Chapter 6, \(g = \Phi^{\prime}\) is integrable over \([a,b]\) and \(\Phi(x) - \Phi(a) = \int_a^x g\); as \(\Phi(a) = T(0) = 0\), for \(a \le c \le d \le b\)

\begin{equation*} T(\chi_{[c,d)}) = \Phi(d) - \Phi( c) = \int_c^d g = \int_a^b g \cdot \chi_{[c,d)} . \end{equation*}

Every step function is, off a set of measure zero, a finite linear combination of such characteristic functions, and \(T\) respects equality a.e., so \(T(f) = \int_a^b g \cdot f\) for all step functions \(f\).

Let \(f\) be simple. By Proposition 10 of the preceding chapter and its proof there are step functions \(\{\varphi_n\} \to f\) in \(L^1[a,b]\) with \(|\varphi_n| \le C\) for all \(n\), and by the Riesz-Fischer Theorem a subsequence \(\{\varphi_{n_k}\} \to f\) pointwise a.e. Then \(T(\varphi_{n_k}) \to T(f)\) by (7), while \(|g \varphi_{n_k}| \le C|g|\) with \(C|g|\) integrable, so the Lebesgue Dominated Convergence Theorem gives \(\int_a^b g\varphi_{n_k} \to \int_a^b g f\). Hence

\begin{equation*} T(f) = \int_a^b g \cdot f \quad \text{for all simple functions } f \text{ on } [a,b] . \end{equation*}

Consequently \(|\int_a^b g \cdot f| = |T(f)| \le \|T\|_* \|f\|_1\) for simple \(f\), which is hypothesis (9) of Lemma 4 with \(p = 1\), \(M = \|T\|_*\); that lemma gives \(g\) in \(L^\infty[a,b]\) with \(\|g\|_\infty \le \|T\|_*\). By Proposition 2, \(\mathcal{R}_g \colon f \mapsto \int_a^b g \cdot f\) is then a bounded linear functional on \(L^1[a,b]\), and it agrees with \(T\) on the simple functions, a dense subspace by Proposition 9 of the preceding chapter; Proposition 3 gives \(T = \mathcal{R}_g\) on \(L^1[a,b]\).

The passage to a general measurable set \(E\) is the text’s, with one adjustment. Restricting \(T\) to each \(L^1[-n,n]\) produces mutually consistent \(g_n\) in \(L^\infty[-n,n]\) with \(\|g_n\|_\infty \le \|T\|_*\), hence a measurable \(g\) on \(\mathbb{R}\) agreeing with each a.e.; in place of Fatou’s Lemma, \(\{|g| > \|T\|_*\}\) is a countable union of null sets, so \(g\) lies in \(L^\infty(\mathbb{R})\) with \(\|g\|_\infty \le \|T\|_*\). Now \(T\) and \(\mathcal{R}_g\) agree on the functions vanishing outside a bounded set, dense in \(L^1(\mathbb{R})\) by the preceding problem, so Proposition 3 gives \(T = \mathcal{R}_g\); extension by zero carries this to arbitrary \(E\).

Theorem. Let \(1 \le p < \infty\) with conjugate \(q\). For \(y\) in \(\ell^q\) the formula

\begin{equation*} \mathcal{R}_y(x) = \sum_{k=1}^\infty y_k \cdot x_k \quad \text{for } x \in \ell^p \end{equation*}

defines a bounded linear functional on \(\ell^p\) with \(\|\mathcal{R}_y\|_* = \|y\|_q\), and \(y \mapsto \mathcal{R}_y\) is a linear isometry of \(\ell^q\) onto \((\ell^p)^*\). Here \(e_k\) denotes the sequence with \(k\)th term \(1\) and all others \(0\).

Holder’s Inequality for series (the counting-measure form of Theorem 1 of the preceding chapter, with the identical proof from Young’s Inequality) gives \(\sum_k |y_k x_k| \le \|y\|_q \|x\|_p\), so the series converges absolutely, \(\mathcal{R}_y\) is linear, and \(\|\mathcal{R}_y\|_* \le \|y\|_q\). For the reverse inequality:

(i) \(1 < p < \infty\), \(y \ne 0\). Put \(x^*_k = \mathrm{sgn}(y_k) |y_k|^{q-1}\). Since \(p(q-1) = q\) we get \(|x^*_k|^p = |y_k|^q\), so \(\|x^*\|_p = \|y\|_q^{q/p}\) is positive and finite, and by (8),

\begin{equation*} \|\mathcal{R}_y\|_* \ge \frac{\mathcal{R}_y(x^*)}{\|x^*\|_p} = \frac{\|y\|_q^{q}}{\|y\|_q^{q/p}} = \|y\|_q^{\,q - q/p} = \|y\|_q . \end{equation*}

(ii) \(p = 1\), \(q = \infty\). Here \(\mathrm{sgn}(y_k) e_k\) has norm at most \(1\) and \(\mathcal{R}_y(\mathrm{sgn}(y_k) e_k) = |y_k|\), so by (8) \(\|\mathcal{R}_y\|_* \ge \sup_k |y_k| = \|y\|_\infty\).

Since \(y \mapsto \mathcal{R}_y\) is linear and norm preserving, it is one-to-one. The span \(X_0\) of \(\{e_k\}\) is dense in \(\ell^p\), because \(\|x - \sum_{k \le n} x_k e_k\|_p^p = \sum_{k > n} |x_k|^p \to 0\) as a tail of a convergent series (here \(p < \infty\) enters).

For surjectivity, let \(T\) be a bounded linear functional on \(\ell^p\) and put \(y_k = T(e_k)\). If \(1 < p < \infty\), set \(A_n = \sum_{k=1}^n |y_k|^q\) and \(x^{(n)} = \sum_{k=1}^n \mathrm{sgn}(y_k)|y_k|^{q-1} e_k\); then \(T(x^{(n)}) = A_n\) and, by \(p(q-1) = q\), \(\|x^{(n)}\|_p = A_n^{1/p}\), so \(A_n \le \|T\|_* A_n^{1/p}\) and hence

\begin{equation*} A_n^{1/q} = A_n^{\,1 - 1/p} \le \|T\|_* \quad \text{for every } n \end{equation*}

(trivially when \(A_n = 0\); otherwise divide by the finite positive \(A_n^{1/p}\)). If \(p = 1\), then \(|y_k| = |T(e_k)| \le \|T\|_*\). Either way \(y\) belongs to \(\ell^q\) with \(\|y\|_q \le \|T\|_*\). Now \(\mathcal{R}_y(e_k) = y_k = T(e_k)\), so the two bounded functionals agree on the dense subspace \(X_0\), and Proposition 3 gives \(T = \mathcal{R}_y\), with \(\|T\|_* = \|y\|_q\).

Each of \(c_0^{\,*}\) and \(c^{\,*}\) is isometrically isomorphic to \(\ell^1\), but by different pairings: every \(T\) in \(c_0^{\,*}\) is \(x \mapsto \sum_{k \ge 1} a_k x_k\) for a unique \(a\) in \(\ell^1\), and every \(T\) in \(c^{\,*}\) is

\begin{equation*} S_a(x) = a_0 \cdot \lim_{k \to \infty} x_k + \sum_{k=1}^\infty a_k x_k \end{equation*}

for a unique summable \(a = (a_0, a_1, \dots)\), with \(\|T\|_* = \|a\|_1\) in both cases. Here \(e_k\) is the \(k\)th standard vector, \(\mathbf{1} = (1,1,\dots)\), and \(\operatorname{sgn}(t) = 1\) for \(t \ge 0\), \(= -1\) for \(t < 0\); uniqueness follows in each case from the norm identity, the correspondence being linear.

The dual of \(c_0\). For \(a\) in \(\ell^1\), \(\sum_k |a_k x_k| \le \|a\|_1 \|x\|_\infty\), so \(T_a(x) = \sum_k a_k x_k\) is a bounded linear functional on \(c_0\) with \(\|T_a\|_* \le \|a\|_1\). Conversely, given \(T\) in \(c_0^{\,*}\), put \(a_k = T(e_k)\); for fixed \(n\), the sequence \(x\) with \(x_k = \operatorname{sgn}(a_k)\) for \(k \le n\) and \(x_k = 0\) otherwise lies in \(c_0\) with \(\|x\|_\infty \le 1\), so

\begin{equation*} \sum_{k=1}^n |a_k| = T(x) \le \|T\|_* , \end{equation*}

whence \(a\) lies in \(\ell^1\) with \(\|a\|_1 \le \|T\|_*\). Moreover \(T = T_a\): for \(x\) in \(c_0\) and \(s_n = \sum_{k \le n} x_k e_k\) we have \(\|x - s_n\|_\infty = \sup_{k>n} |x_k| \to 0\), so continuity of \(T\) gives \(T(x) = \lim_n T(s_n) = T_a(x)\). Hence \(\|T_a\|_* = \|a\|_1\).

The dual of \(c\). For summable \(a\), the bound \(|\lim_k x_k| \le \|x\|_\infty\) gives \(|S_a(x)| \le \|a\|_1 \|x\|_\infty\), so \(S_a\) lies in \(c^{\,*}\) with \(\|S_a\|_* \le \|a\|_1\). Conversely, let \(T\) belong to \(c^{\,*}\). Its restriction to \(c_0\) is bounded, so by the first part \(T(y) = \sum_{k \ge 1} a_k y_k\) on \(c_0\) with \(a_k = T(e_k)\) summable; set \(a_0 = T(\mathbf{1}) - \sum_{k \ge 1} a_k\). For \(x\) in \(c\) with \(\ell = \lim_k x_k\), the sequence \(y = x - \ell \mathbf{1}\) lies in \(c_0\), so by linearity and absolute convergence

\begin{equation*} T(x) = \sum_{k=1}^\infty a_k (x_k - \ell) + \ell \, T(\mathbf{1}) = \sum_{k=1}^\infty a_k x_k + \ell \, a_0 = S_a(x) . \end{equation*}

For the reverse norm inequality, let \(\epsilon > 0\), choose \(N\) with \(\sum_{k > N} |a_k| < \epsilon\), and let \(x_k = \operatorname{sgn}(a_k)\) for \(k \le N\), \(x_k = \operatorname{sgn}(a_0)\) for \(k > N\); this \(x\) is eventually constant, so it lies in \(c\) with \(\lim_k x_k = \operatorname{sgn}(a_0)\) and \(\|x\|_\infty \le 1\), and

\begin{equation*} S_a(x) \ge |a_0| + \sum_{k=1}^N |a_k| - \sum_{k > N} |a_k| > \|a\|_1 - 2\epsilon . \end{equation*}

As \(\epsilon\) is arbitrary, \(\|S_a\|_* = \|a\|_1\).

The representing function is the jump function at \(x_0\):

\begin{equation*} g = \chi_{[x_0,\,b]} \ \text{ if } x_0 > a , \qquad g = \chi_{(a,\,b]} \ \text{ if } x_0 = a . \end{equation*}

Boundedness first: \(|T(f)| = |f(x_0)| \le \|f\|_{\max}\), with equality for \(f \equiv 1\), so \(\|T\|_* = 1\). In either case \(g\) is nondecreasing with values \(0\) and \(1\), so the variation over any partition telescopes to \(g(b) - g(a) = 1\) and \(TV(g) = 1\).

Let \(f\) be in \(C[a,b]\) and let \(P \colon a = t_0 < \cdots < t_n = b\) carry tags \(\xi_i\) in \([t_{i-1}, t_i]\). Exactly one difference \(g(t_i) - g(t_{i-1})\) is nonzero and it equals \(1\), so the Riemann-Stieltjes sum is \(S(f,g,P,\xi) = f(\xi_i)\) for that index \(i\), and \(x_0\) lies in \([t_{i-1}, t_i]\) too: (i) if \(x_0 > a\), then \(g(t_{i-1}) = 0\) forces \(t_{i-1} < x_0\) and \(g(t_i) = 1\) forces \(t_i \ge x_0\); (ii) if \(x_0 = a\), then \(i = 1\) and \(x_0 = t_0\). Hence \(|\xi_i - x_0| \le \operatorname{mesh}(P)\), so by the uniform continuity of \(f\) on \([a,b]\), given \(\epsilon > 0\) there is \(\delta > 0\) with

\begin{equation*} \big| S(f, g, P, \xi) - f(x_0) \big| = |f(\xi_i) - f(x_0)| < \epsilon \quad \text{whenever } \operatorname{mesh}(P) < \delta , \end{equation*}

for every choice of tags. Therefore \(\int_a^b f \, dg\) exists and \(T(f) = f(x_0) = \int_a^b f \, dg\).

Take \(g = \sigma g_0\), where \(x_0\) in \([a,b]\) maximizes the continuous function \(|f|\) (Extreme Value Theorem), \(\sigma = \pm 1\) is the sign of \(f(x_0)\), and \(g_0\) is the jump function at \(x_0\) of the preceding problem. The printed requirement \(TV(f) = 1\) is a misprint for \(TV(g) = 1\), since a continuous \(f\) need not be of bounded variation at all.

By the preceding problem, \(TV(g_0) = 1\) and \(\int_a^b h \, dg_0 = h(x_0)\) for every \(h\) in \(C[a,b]\). Since every variation sum and every Riemann-Stieltjes sum for \((f, g)\) is \(\sigma\) times, respectively \(|\sigma| = 1\) times, the corresponding sum for \((f, g_0)\), the function \(g\) is of bounded variation with \(TV(g) = TV(g_0) = 1\), the integral \(\int_a^b f \, dg\) exists, and

\begin{equation*} \int_a^b f \, dg = \sigma \int_a^b f \, dg_0 = \sigma f(x_0) = |f(x_0)| = \|f\|_{\max} . \end{equation*}

\(\Phi\) is Lipschitz with constant at most \(\|T\|_*\), because \(\|g_x - g_y\|_{\max} \le |x - y|\).

Indeed the two linear pieces assemble into \(g_x(t) = \min\{t, x\} - a\), continuous on \([a,b]\). For \(y \le x\) and \(t\) in \([a,b]\),

\begin{equation*} g_x(t) - g_y(t) = \min\{t,x\} - \min\{t,y\} = \begin{cases} 0, & a \le t \le y, \\ t - y, & y \le t \le x, \\ x - y, & x \le t \le b , \end{cases} \end{equation*}

which in every case lies in \([0, x-y]\); hence \(\|g_x - g_y\|_{\max} \le |x-y|\). Since \(T\) is linear and bounded,

\begin{equation*} |\Phi(x) - \Phi(y)| = |T(g_x - g_y)| \le \|T\|_* \cdot \|g_x - g_y\|_{\max} \le \|T\|_* \cdot |x - y| . \end{equation*}

The Radamacher functions have \(\|f_n - f\|_p = 1\) for every \(n\), so they cannot converge strongly to \(f \equiv 0\).

Indeed \(f_n = (-1)^k\) on \([k/2^n, (k+1)/2^n)\) for \(0 \le k \le 2^n - 1\) (the printed restriction \(k < 2^n - 1\) is a misprint: only this reading has the intervals exhaust \([0,1)\)), so \(|f_n| = 1\) a.e. on \(I\) and

\begin{equation*} \|f_n - f\|_p^p = \int_0^1 |f_n|^p = 1 \quad \text{for all } n . \end{equation*}

The same computation applies to every subsequence.

The defining formula gives \(f_n = \beta\) for \(k\) even and \(f_n = \alpha\) for \(k\) odd, so with \(r_n\) the \(n\)th Radamacher function of the first example of this section and \(f \equiv (\alpha+\beta)/2\),

\begin{equation*} f_n = \frac{\alpha+\beta}{2} + \frac{\beta - \alpha}{2} \, r_n \quad \text{a.e. on } I . \end{equation*}

(As in the preceding problem, \(k\) runs over \(0 \le k \le 2^n - 1\); the printed restriction is a misprint.) Hence \(|f_n| \le M = \max\{|\alpha|, |\beta|\}\) a.e., so \(\|f_n\|_p \le M\) for all \(n\) and \(\{f_n\}\) is bounded in \(L^p(I)\).

The engine is the estimate \(|\int_0^x r_n| \le 2^{-n}\) for all \(x\) in \([0,1]\): with \(m = \lfloor 2^n x \rfloor\) and \(\delta = x - m/2^n \in [0, 2^{-n})\),

\begin{equation*} \int_0^x r_n = \frac{1}{2^n}\sum_{k=0}^{m-1} (-1)^k + (-1)^m \delta , \end{equation*}

whose first term is \(0\) for \(m\) even and \(2^{-n}\) for \(m\) odd, so the total is \(\delta\) or \(2^{-n} - \delta\). Consequently, for each \(x\) in \([0,1]\),

\begin{equation*} \Big| \int_0^x f_n - \int_0^x f \Big| = \frac{|\beta - \alpha|}{2} \Big| \int_0^x r_n \Big| \le \frac{|\beta - \alpha|}{2^{n+1}} \to 0 , \end{equation*}

and since \(\{f_n\}\) is bounded in \(L^p(I)\) with \(f\) in \(L^p(I)\) and \(1 < p < \infty\), Theorem 11 gives \(\{f_n\} \rightharpoonup f\).

Now let \(\alpha \ne \beta\). Exactly half of the \(2^n\) indices \(k\) are even, so \(f_n = \beta\) on a set of measure \(1/2\) and \(f_n = \alpha\) on its complement, and

\begin{equation*} \|f_n\|_p^p = \frac{|\alpha|^p + |\beta|^p}{2} \quad \text{for all } n , \qquad \|f\|_p^p = \Big|\frac{\alpha+\beta}{2}\Big|^p . \end{equation*}

These differ: (i) if \(|\alpha| \ne |\beta|\), then \(s \mapsto s^p\) is strictly convex and nondecreasing on \([0,\infty)\) for \(p > 1\), so

\begin{equation*} \Big|\frac{\alpha+\beta}{2}\Big|^p \le \Big(\frac{|\alpha| + |\beta|}{2}\Big)^p < \frac{|\alpha|^p + |\beta|^p}{2} ; \end{equation*}

(ii) if \(|\alpha| = |\beta|\) and \(\alpha \ne \beta\), then \(\alpha = -\beta \ne 0\) and the left side is \(0 < |\alpha|^p\). Hence \(\|f\|_p < \liminf_n \|f_n\|_p\), so by Corollary 13 no subsequence converges strongly to \(f\). Nor to anything else: a subsequence converging strongly to some \(g\) would converge weakly to \(g\), while every subsequence converges weakly to \(f\), and weak limits are unique.

By Proposition 6 it suffices to show \(\int_a^b g \cdot f_n \to 0\) for every \(g\) in \(L^q[a,b]\), \(q\) the conjugate of \(p\); we prove this for every \(g\) in \(L^1[a,b]\), which covers all \(1 \le p < \infty\) at once since \([a,b]\) has finite measure and so \(L^q[a,b] \subseteq L^1[a,b]\) by Corollary 3 of the preceding chapter.

Being continuous and \(T\)-periodic, \(h\) is bounded by \(K = \max_{[0,T]} |h|\) on \(\mathbb{R}\), so \(|f_n| \le K\) on \([a,b]\) and each \(f_n\) belongs to \(L^p[a,b]\). Its indefinite integral \(H(u) = \int_0^u h\) is continuous and also \(T\)-periodic: the substitution \(t = s - T\) with \(h(t+T) = h(t)\) gives \(\int_T^{u+T} h = \int_0^u h\), so

\begin{equation*} H(u+T) - H(u) = \int_u^{T} h + \int_T^{u+T} h = \int_0^T h = 0 . \end{equation*}

Hence \(|H| \le M = \max_{[0,T]} |H|\) on \(\mathbb{R}\), and for \(a \le c \le d \le b\) the change of variable \(u = nx\) gives

\begin{equation*} \int_c^d f_n = \frac{H(nd) - H(nc)}{n} , \qquad \Big| \int_c^d f_n \Big| \le \frac{2M}{n} . \end{equation*}

If \(\varphi\) is a step function taking the values \(c_1, \dots, c_N\) on the subintervals of a partition of \([a,b]\), then, the partition points being a null set,

\begin{equation*} \Big| \int_a^b \varphi \cdot f_n \Big| \le \frac{2M}{n}\sum_{i=1}^N |c_i| \to 0 . \end{equation*}

Finally let \(g\) belong to \(L^1[a,b]\) and \(\epsilon > 0\). The step functions are dense in \(L^1[a,b]\) by Proposition 10 of the preceding chapter, so choose \(\varphi\) with \(\|g - \varphi\|_1 < \epsilon/(2(K+1))\); then for every \(n\)

\begin{equation*} \Big| \int_a^b (g - \varphi)\cdot f_n \Big| \le K \|g - \varphi\|_1 < \frac{\epsilon}{2} , \end{equation*}

while \(|\int_a^b \varphi \cdot f_n| < \epsilon/2\) for all large \(n\). Hence \(\int_a^b g \cdot f_n \to 0 = \int_a^b g \cdot f\).

By translation invariance \(\|f_n\|_p = \|f_0\|_p\) for all \(n\), so \(\{f_n\}\) is bounded in \(L^p(\mathbb{R})\); as \(f \equiv 0\) lies in \(L^p(\mathbb{R})\) and \(p > 1\), Theorem 10 reduces the claim to showing \(\int_A f_n \to 0\) for every measurable \(A\) of finite measure. Let \(q < \infty\) be the conjugate of \(p\).

Fix such an \(A\) and let \(\epsilon > 0\). Translation invariance gives \(\int_A f_n = \int_{A-n} f_0\), and the continuity of measure provides \(M > 0\) with \(m(A \setminus [-M,M]) < \epsilon\); write \(A_M = A \cap [-M,M]\) and \(B_M = A \setminus [-M,M]\), so that

\begin{equation*} \left| \int_A f_n \right| \le \int_{A_M - n} |f_0| + \int_{B_M - n} |f_0| . \end{equation*}

Holder’s Inequality against a characteristic function bounds the second term by \(\|f_0\|_p [m(B_M)]^{1/q} \le \|f_0\|_p \epsilon^{1/q}\); this is the decisive use of \(q < \infty\). For the first, \(A_M - n \subseteq [-M-n, M-n]\), so Holder gives

\begin{equation*} \int_{A_M - n} |f_0| \le \left[ \int_{-M-n}^{M-n} |f_0|^p \right]^{1/p} \cdot (2M)^{1/q} \to 0 , \end{equation*}

since \(A \mapsto \int_A |f_0|^p\) is a finite measure and the sets \((-\infty, M-n]\) decrease to \(\emptyset\). Hence \(\limsup_n |\int_A f_n| \le \|f_0\|_p \epsilon^{1/q}\), and \(\epsilon\) was arbitrary.

For \(p = 1\) the assertion fails: take \(f_0 = \chi_{[0,1]}\), so \(f_n = \chi_{[n,n+1]}\), and test against \(g \equiv 1\) in \(L^\infty(\mathbb{R})\):

\begin{equation*} \int_{\mathbb{R}} g \cdot f_n = 1 \text{ for all } n , \qquad \text{while} \qquad \int_{\mathbb{R}} g \cdot f = 0 , \end{equation*}

so by Proposition 6 there is no weak convergence to \(f \equiv 0\) in \(L^1(\mathbb{R})\).

Expand the square. Each of \(f_n^2\), \(f^2\) and \(f_n \cdot f\) is integrable over \(E\), the last by Holder’s Inequality with \(p = q = 2\), so linearity of integration gives

\begin{equation*} \|f_n - f\|_2^2 = \int_E f_n^2 - 2 \int_E f_n \cdot f + \int_E f^2 . \end{equation*}

By hypothesis the first two integrals both converge to the finite number \(\int_E f^2\), so

\begin{equation*} \lim_{n \to \infty} \|f_n - f\|_2^2 = \int_E f^2 - 2\int_E f^2 + \int_E f^2 = 0 . \end{equation*}

Yes for five of the six pairs; no for the pair (i), (ii). (The printed (iii) reads “converges to \(\{\|f_n\|_p\}\)”, a misprint for \(\|f\|_p\).)

Three implications do the work, for \(\{f_n\}\) bounded in \(L^p(E)\), \(1 < p < \infty\), and \(f\) in \(L^p(E)\):

(A) (i) implies (ii), by Theorem 12.

(B) (ii) and (iii) together imply (iv), by the Radon-Riesz Theorem.

(C) (iv) implies (ii) and (iii), and forces some subsequence to have (i): strong convergence implies weak convergence since \(|T(f_n) - T(f)| \le \|T\|_* \|f_n - f\|\); the norms converge since \(|\,\|f_n\|_p - \|f\|_p| \le \|f_n - f\|_p\); and the Riesz-Fischer Theorem of the preceding chapter supplies a subsequence converging pointwise a.e.

The pairs containing (iv). By (C) the whole sequence has (ii), (iii), (iv), and some subsequence has (i) as well; that subsequence retains (ii), (iii), (iv), since it too converges strongly to \(f\). Yes.

The pair (i), (iii). By (A) property (ii) holds, then (B) gives (iv): the whole sequence has all four. Yes.

The pair (ii), (iii). By (B) the sequence has (iv), and by (C) a subsequence gains (i). Yes.

The pair (i), (ii). No: on \(E = [0,1]\) with \(f \equiv 0\), take

\begin{equation*} f_n = n^{1/p} \cdot \chi_{(0, 1/n)} , \qquad \|f_n\|_p = 1 \text{ for all } n . \end{equation*}

This sequence is bounded, converges to \(0\) pointwise on \((0,1]\), hence has (i), and has (ii) by Theorem 12. But every subsequence has \(\|f_{n_k}\|_p = 1 \ne 0 = \|f\|_p\), so no subsequence has (iii), nor (iv) by (C).

Two facts drive the selections: a subsequence of a weakly convergent sequence converges weakly to the same limit, each \(\{T(f_{n_k})\}\) being a subsequence of the convergent \(\{T(f_n)\}\); and since \(\{\|f_n\|\}\) is unbounded, \(\{n : \|f_n\| > M\}\) is infinite for every \(M\) (a finite such set would bound the whole sequence).

So choose inductively \(n_1 < n_2 < \cdots\) with \(\|f_{n_k}\| \ge k \cdot 3^k\) and relabel: we may suppose \(\|f_n\| \ge \alpha_n = n \cdot 3^n\) for all \(n\). In particular \(\|f_n\| \ge 3 > 0\), so no \(f_n\) is zero and the quotients below are defined.

For the second selection, note that relabeling \(f_{n_k}\) as the \(k\)th term attaches to it the ratio \(\|f_{n_k}\|/\alpha_k\), not \(\|f_{n_k}\|/\alpha_{n_k}\), and since \(n_k \ge k\) relabeling can only raise the ratio; so a finite limit extracted from \(\{\|f_n\|/\alpha_n\}\) by Bolzano-Weierstrass need not survive the reindexing, and finiteness of \(\alpha\) cannot be arranged in general (for \(\|f_n\| = 3^{n^2}\), every relabeled subsequence has \(\|f_{n_k}\|/\alpha_k \ge 3^{k^2}/(k 3^k) \to \infty\)). What is always available suffices: \(\|f_n\| \ge n \cdot 3^n \to \infty\), so one may choose inductively \(n_1 < n_2 < \cdots\) with \(\|f_{n_k}\| \ge k \cdot \alpha_k\). After relabeling, the sequence still converges weakly to \(f\), still satisfies \(\|f_n\| \ge \alpha_n\), and

\begin{equation*} \frac{\|f_n\|}{\alpha_n} \ge n \to \alpha = \infty \in [1, \infty] . \end{equation*}

Now put \(c_n = \alpha_n/\|f_n\|\) and \(g_n = c_n f_n\). By homogeneity of the norm \(\|g_n\| = \alpha_n = n \cdot 3^n\) for all \(n\), and

\begin{equation*} \lim_{n \to \infty} c_n = c , \quad \text{where } c = 1/\alpha \in (0,1] \text{ if } \alpha < \infty, \ \ c = 0 \text{ if } \alpha = \infty . \end{equation*}

Hence for each \(T\) in \(X^*\), by linearity and the product rule for real sequences,

\begin{equation*} \lim_{n \to \infty} T(g_n) = \lim_{n \to \infty} c_n \, T(f_n) = c \cdot T(f) = T(c \cdot f) , \end{equation*}

so \(\{g_n\} \rightharpoonup c \cdot f\) in \(X\).

By Problem 7, weak convergence in \(\ell^p\) is exactly the pairing criterion

\begin{equation*} \lim_{n \to \infty} \sum_{k=1}^\infty \eta^k \zeta_n^k = \sum_{k=1}^\infty \eta^k \zeta^k \quad \text{for every } \eta \in \ell^q , \end{equation*}

\(q\) the conjugate of \(p\).

Weak convergence gives componentwise convergence for every \(1 \le p < \infty\): the coordinate functional \(T_k(\xi) = \xi^k\) is linear and bounded, since \(|\xi^k| \le \|\xi\|_p\), so \(\zeta_n^k = T_k(\zeta_n) \to T_k(\zeta) = \zeta^k\).

Conversely, assume componentwise convergence and \(1 < p < \infty\), so \(q < \infty\). Boundedness of \(\{\zeta_n\}\) makes \(M = \sup_n \|\zeta_n - \zeta\|_p\) finite. Given \(\eta\) in \(\ell^q\) and \(\epsilon > 0\), choose \(N\) with \([\sum_{k > N} |\eta^k|^q]^{1/q} < \epsilon\); then

\begin{equation*} \left| \sum_{k=1}^\infty \eta^k (\zeta_n^k - \zeta^k) \right| \le \sum_{k=1}^{N} |\eta^k| \, |\zeta_n^k - \zeta^k| + \left[ \sum_{k=N+1}^\infty |\eta^k|^q \right]^{1/q} \|\zeta_n - \zeta\|_p , \end{equation*}

the tail bounded by Holder’s Inequality for series (the counting-measure form of Theorem 1 of the preceding chapter). The second term is at most \(\epsilon M\) and the first is a fixed finite sum of terms each tending to \(0\), so the limit superior is at most \(\epsilon M\), and \(\epsilon\) was arbitrary.

This half genuinely needs \(p > 1\), and the statement fails as printed at \(p = 1\): there \(\zeta_n = e_n\) is bounded in \(\ell^1\) and converges componentwise to \(0\), yet pairing against \(\eta = (1,1,\ldots)\) in \(\ell^\infty\) gives \(\sum_k \eta^k \zeta_n^k = 1\) for all \(n\), so weak convergence fails (Problem 21).

Weak convergence in \(L^{p_2}[0,1]\) implies weak convergence in \(L^{p_1}[0,1]\); the converse fails in general but holds when \(\{f_n\}\) is bounded in \(L^{p_2}[0,1]\).

Let \(q_1, q_2\) be the conjugates, so \(1 < q_2 < q_1 \le \infty\). Since \(m([0,1]) = 1\), Corollary 3 of the preceding chapter gives \(L^{p_2}[0,1] \subseteq L^{p_1}[0,1]\) and \(L^{q_1}[0,1] \subseteq L^{q_2}[0,1]\).

For the implication, let \(\{f_n\} \rightharpoonup f\) in \(L^{p_2}[0,1]\) and let \(g\) belong to \(L^{q_1}[0,1]\). Then \(g\) lies in \(L^{q_2}[0,1]\), so \(\int_0^1 g f_n \to \int_0^1 g f\) by Proposition 6; as \(g\) was arbitrary, Proposition 6 with exponent \(p_1\) gives \(\{f_n\} \rightharpoonup f\) in \(L^{p_1}[0,1]\).

For the failure of the converse take \(\theta = \tfrac12(1/p_1 + 1/p_2)\), so \(1/p_2 < \theta < 1/p_1\), and \(f_n = n^{\theta} \chi_{(0,1/n)}\), \(f \equiv 0\). Since \(\|f_n\|_r = n^{\theta - 1/r}\) for \(1 \le r < \infty\),

\begin{equation*} \|f_n - f\|_{p_1} = n^{\theta - 1/p_1} \to 0 , \qquad \|f_n\|_{p_2} = n^{\theta - 1/p_2} \to \infty , \end{equation*}

so \(\{f_n\}\) converges strongly, hence weakly, in \(L^{p_1}[0,1]\), while being unbounded in \(L^{p_2}[0,1]\) it converges weakly there to nothing at all, by Theorem 7.

Finally suppose \(\{f_n\} \rightharpoonup f\) in \(L^{p_1}[0,1]\) and \(\{f_n\}\) is bounded in \(L^{p_2}[0,1]\). For measurable \(A \subseteq [0,1]\) the function \(\chi_A\) belongs to \(L^{q_1}[0,1]\), the measure being finite, so Proposition 6 with exponent \(p_1\) gives

\begin{equation*} \lim_{n \to \infty} \int_A f_n = \int_A f , \end{equation*}

which for the bounded sequence \(\{f_n\}\) with \(f\) in \(L^{p_2}[0,1]\) and \(1 < p_2 < \infty\) is equivalent, by Theorem 10, to \(\{f_n\} \rightharpoonup f\) in \(L^{p_2}[0,1]\).

Each \(e_n\) is a unit vector, \(\|e_n\|_p = 1\) for every \(1 \le p < \infty\), so \(\{e_n\}\) is bounded in \(\ell^p\) and no subsequence can converge strongly to \(0\): \(\|e_{n_k} - 0\|_p = 1\) for all \(k\).

Let \(p > 1\), so the conjugate \(q\) is finite. By Problem 7 every \(T\) in \((\ell^p)^*\) is \(T(\xi) = \sum_k \eta^k \xi^k\) for some \(\eta\) in \(\ell^q\), whence

\begin{equation*} T(e_n) = \eta^n \to 0 = T(0) , \end{equation*}

the terms of the convergent series \(\sum_k |\eta^k|^q\) tending to \(0\). Hence \(\{e_n\} \rightharpoonup 0\) in \(\ell^p\).

In \(\ell^1\), suppose \(\{e_n\} \rightharpoonup \zeta\). The coordinate functional \(T_k(\xi) = \xi^k\) is bounded on \(\ell^1\), since \(|\xi^k| \le \|\xi\|_1\), and \(e_n^k = 0\) for \(n \ne k\), so \(\zeta^k = \lim_n e_n^k = 0\) for every \(k\), that is, \(\zeta = 0\). But \(T(\xi) = \sum_{k=1}^\infty \xi^k\) is also a bounded linear functional on \(\ell^1\), with \(|T(\xi)| \le \|\xi\|_1\), and

\begin{equation*} T(e_n) = 1 \text{ for all } n, \quad \text{while} \quad T(\zeta) = 0 , \end{equation*}

a contradiction. So \(\{e_n\}\) converges weakly in \(\ell^1\) to nothing.

Radon-Riesz Theorem in \(\ell^2\). If \(\{\zeta_n\} \rightharpoonup \zeta\) in \(\ell^2\), then \(\{\zeta_n\} \to \zeta\) strongly if and only if \(\lim_{n\to\infty}\|\zeta_n\|_2 = \|\zeta\|_2\). Here \(\langle \xi,\eta\rangle = \sum_{k} \xi^k\eta^k\), absolutely convergent by the Cauchy-Schwarz Inequality.

By Problem 7 of this chapter (the Riesz Representation Theorem for \((\ell^p)^*\)) with \(p = q = 2\), every \(T \in (\ell^2)^*\) is \(T(\xi) = \langle \xi,\eta\rangle\) for a unique \(\eta \in \ell^2\), so weak convergence says

\begin{equation*} \lim_{n \to \infty} \langle \zeta_n, \eta \rangle = \langle \zeta, \eta \rangle \quad \text{for all } \eta \in \ell^2 . \tag{\(\ast\)} \end{equation*}

(\(\Rightarrow\)) The triangle inequality gives \(\bigl|\|\zeta_n\|_2 - \|\zeta\|_2\bigr| \le \|\zeta_n - \zeta\|_2 \to 0\).

(\(\Leftarrow\)) Expanding the square (the three series converge absolutely, the cross term by Cauchy-Schwarz), then taking \(\eta = \zeta\) in \((\ast)\) and using \(\|\zeta_n\|_2^2 \to \|\zeta\|_2^2\),

\begin{equation*} \begin{aligned} \|\zeta_n - \zeta\|_2^2 &= \|\zeta_n\|_2^2 - 2 \langle \zeta_n, \zeta \rangle + \|\zeta\|_2^2 \\ &\longrightarrow \|\zeta\|_2^2 - 2\|\zeta\|_2^2 + \|\zeta\|_2^2 = 0 . \end{aligned} \end{equation*}

Thus \(\{\zeta_n\} \to \zeta\) strongly in \(\ell^2\).

Test the weak convergence against the evaluation functionals. Fix \(x_0 \in [a,b]\) and put \(T_{x_0}(g) = g(x_0)\) for \(g \in C[a,b]\). Then \(T_{x_0}\) is linear, and since \(x_0\) is one of the points over which the maximum norm is taken,

\begin{equation*} |T_{x_0}(g)| = |g(x_0)| \le \max_{x \in [a,b]}|g(x)| = \|g\|_{\max} , \end{equation*}

so \(T_{x_0} \in (C[a,b])^*\), with \(\|T_{x_0}\|_* = 1\) as witnessed by \(g \equiv 1\) (Problem 9 of this chapter). Hence, by the definition of weak convergence applied to this \(T_{x_0}\),

\begin{equation*} \lim_{n \to \infty} f_n(x_0) = \lim_{n \to \infty} T_{x_0}(f_n) = T_{x_0}(f) = f(x_0) . \end{equation*}

As \(x_0 \in [a,b]\) was arbitrary, \(\{f_n\} \to f\) pointwise on \([a,b]\).

Fix \(x \in [a,b]\) and test the weak convergence against the single functional \(T_g\), \(g = \chi_{[a,x]}\). Since \(g \in L^1[a,b]\), with \(\|g\|_1 = x - a\), Hoelder’s Inequality with \(p = \infty\), \(q = 1\) gives

\begin{equation*} |T_g(h)| = \left| \int_a^b g \cdot h \right| \le \|g\|_1 \cdot \|h\|_\infty \quad \text{for all } h \in L^\infty[a,b] , \end{equation*}

so \(T_g\), linear by the linearity of integration, belongs to \((L^\infty[a,b])^*\). (No representation theorem for \((L^\infty)^*\) is needed: one functional suffices.) Since \(T_g(h) = \int_a^b \chi_{[a,x]}\cdot h = \int_a^x h\), the definition of weak convergence applied to \(T_g\) yields

\begin{equation*} \lim_{n \to \infty} \int_a^x f_n = \lim_{n\to\infty} T_g(f_n) = T_g(f) = \int_a^x f , \end{equation*}

and \(x \in [a,b]\) was arbitrary.

Both parts follow by applying the hypothesis to a single well-chosen vector.

(i) Apply the hypothesis to \(h = f_1 - f_2\): there is \(T \in X^*\) with \(T(h) = \|h\|\) and \(\|T\|_* = 1\). Since \(\{T(f_n)\}\) converges to both \(T(f_1)\) and \(T(f_2)\), and a real sequence has at most one limit,

\begin{equation*} \|f_1 - f_2\| = T(h) = T(f_1) - T(f_2) = 0 , \end{equation*}

so \(f_1 = f_2\).

(ii) Apply the hypothesis to \(f\) itself: there is \(T \in X^*\) with \(T(f) = \|f\|\) and \(\|T\|_* = 1\), whence \(T(f_n) \le |T(f_n)| \le \|T\|_* \|f_n\| = \|f_n\|\) for every \(n\). Since \(\{T(f_n)\} \to T(f)\), its limit inferior equals its limit, and the limit inferior respects termwise inequalities:

\begin{equation*} \|f\| = T(f) = \liminf_{n\to\infty} T(f_n) \le \liminf_{n \to \infty} \|f_n\| . \end{equation*}

Run the gliding-hump construction from the proof of Theorem 7, which uses weak convergence only through the boundedness of \(\left\{\int_E g\cdot f_n\right\}\). Throughout, \(f^* = \|f\|_p^{1-p}\operatorname{sgn}(f)|f|^{p-1}\) is the conjugate function of Theorem 1 of Chapter 7, so that for \(f \ne 0\) in \(L^p(E)\),

\begin{equation*} f^* \in L^q(E), \qquad \int_E f^* \cdot f = \|f\|_p , \qquad \|f^*\|_q = 1 . \tag{1} \end{equation*}

Normalization. Suppose \(\{\|f_n\|_p\}\) is unbounded. Then infinitely many terms exceed any given bound (only finitely many indices precede any \(N\)), so choose \(n_1 < n_2 < \cdots\) with \(\|f_{n_k}\|_p \ge k\cdot 3^k\) and set \(c_k = k\cdot 3^k/\|f_{n_k}\|_p \in (0,1]\), \(h_k = c_k f_{n_k}\). Then

\begin{equation*} \|h_n\|_p = n \cdot 3^n \quad \text{for all } n , \tag{2} \end{equation*}

each \(h_n \ne 0\), and since \(0 < c_k \le 1\) the sequence \(\left\{\int_E g \cdot h_n\right\}\) is still bounded for every \(g \in L^q(E)\) (Problem 18 of this chapter).

Signs. Set \(A_n = \int_E \left[\sum_{k=1}^{n-1}\epsilon_k (h_k)^*\right]\cdot h_n\) (finite by Hoelder, with \(A_1 = 0\)) and choose inductively \(\epsilon_n = 1/3^n\) if \(A_n \ge 0\), \(\epsilon_n = -1/3^n\) if \(A_n < 0\). Then \(A_n\) and \(\epsilon_n \|h_n\|_p\) have the same sign in

\begin{equation*} \int_E \left[ \sum_{k=1}^{n} \epsilon_k (h_k)^* \right] \cdot h_n = A_n + \epsilon_n \int_E (h_n)^* \cdot h_n = A_n + \epsilon_n \|h_n\|_p , \end{equation*}

and hence, by (1) and (2),

\begin{equation*} \left| \int_E \left[ \sum_{k=1}^{n} \epsilon_k (h_k)^* \right] \cdot h_n \right| \ge \frac{\|h_n\|_p}{3^n} = n \quad \text{for all } n . \tag{3} \end{equation*}

The function \(g\). By (1), \(\|\epsilon_k (h_k)^*\|_q = 1/3^k\), whose sum is finite, so the partial sums of \(\sum_k \epsilon_k (h_k)^*\) are Cauchy in \(L^q(E)\); that space is complete by the Riesz-Fischer Theorem of Chapter 7 for \(q < \infty\) and by Problem 33 of Chapter 7 for \(q = \infty\). Hence \(g = \sum_{k=1}^\infty \epsilon_k (h_k)^*\) lies in \(L^q(E)\), with remainder \(r_n = g - \sum_{k=1}^n \epsilon_k (h_k)^*\) satisfying \(\|r_n\|_q \le \sum_{k>n} 3^{-k} = 1/(2\cdot 3^n)\).

Contradiction. By the triangle inequality, (3), Hoelder’s Inequality and (2),

\begin{equation*} \begin{aligned} \left| \int_E g \cdot h_n \right| &\ge \left| \int_E \left[\sum_{k=1}^n \epsilon_k (h_k)^*\right]\cdot h_n \right| - \left| \int_E r_n \cdot h_n \right| \\ &\ge n - \frac{1}{2 \cdot 3^n}\cdot n \cdot 3^n = \frac{n}{2} , \end{aligned} \end{equation*}

so \(\left\{\int_E g\cdot h_n\right\}\) is unbounded, contradicting the previous paragraph. Therefore \(\{f_n\}\) is bounded in \(L^p(E)\).

Take disjointly supported tents. With \(\ell = b - a > 0\), put \(\alpha_n = a + \ell/2^{n+1}\), \(\beta_n = a + \ell/2^n\), \(m_n = (\alpha_n + \beta_n)/2\), and

\begin{equation*} f_n(x) = \max\left\{ 0,\ 1 - \frac{2\,|x - m_n|}{\beta_n - \alpha_n} \right\} \quad \text{for } x \in [a,b] . \end{equation*}

Each \(f_n\) is continuous (a maximum of two continuous functions) with \(0 \le f_n \le 1\) and \(f_n(m_n) = 1\), so \(\|f_n\|_{\max} = 1\) and the sequence is bounded. Since \(\{a + \ell/2^n\}\) strictly decreases, the intervals \((\alpha_n,\beta_n) \subseteq (a,b]\) are pairwise disjoint, and \(f_n\) vanishes off \((\alpha_n,\beta_n)\) (there \(1 - 2|x-m_n|/(\beta_n-\alpha_n) \le 0\)). Hence for \(m \ne n\) the point \(m_n\) lies outside \((\alpha_m,\beta_m)\), so

\begin{equation*} \|f_n - f_m\|_{\max} \ge |f_n(m_n) - f_m(m_n)| = 1 . \end{equation*}

A strongly convergent subsequence would be Cauchy in the maximum norm, hence would have two distinct-index terms within distance \(1\), which the display forbids. So no subsequence of \(\{f_n\}\) converges strongly in \(C[a,b]\).

Take \(e_n\) to be the sequence with \(n\)th component \(1\) and all other components \(0\). Exactly one component is nonzero, so \(\|e_n\|_p = 1\) for every \(p\) with \(1 \le p \le \infty\), and \(\{e_n\}\) is bounded in \(\ell^p\). For \(m \ne n\) the difference \(e_n - e_m\) has components \(1\) and \(-1\) and no others, so

\begin{equation*} \|e_n - e_m\|_p = 2^{1/p} \ge 1 \quad (p < \infty), \qquad \|e_n - e_m\|_\infty = 1 . \end{equation*}

A strongly convergent subsequence would be Cauchy, hence would have two distinct-index terms at distance less than \(1\), which the display forbids. So \(\{e_n\}\) has no strongly convergent subsequence in \(\ell^p\).

Take the concentrating sequence of the Example of this section, transplanted to an interval \([a,b] \subseteq E\) with \(a < b\): with \(I_n = [a, a + (b-a)/n]\), set \(f_n = \frac{n}{b-a}\chi_{I_n}\) on \(E\), so that \(\|f_n\|_1 = 1\). For the second question, take any \(E\) with \(m(E) = 0\).

No weakly convergent subsequence. Suppose \(\{f_{n_k}\} \rightharpoonup f\) in \(L^1(E)\), so \(\int_E f_{n_k}\cdot g \to \int_E f \cdot g\) for every \(g \in L^\infty(E)\). For a measurable \(A \subseteq E\) the test function \(g = \chi_A \operatorname{sgn}(f)\) lies in \(L^\infty(E)\) and satisfies \(\int_E f\cdot g = \int_A |f|\); we use it for two choices of \(A\).

(i) \(A = E \setminus [a,b]\): each \(f_{n_k}\) vanishes there, so \(\int_A |f| = \lim_k \int_E f_{n_k}\cdot g = 0\), and \(f = 0\) a.e. on \(A\) by Proposition 9 of Chapter 4.

(ii) \(A = J_j = [a + (b-a)/j,\, b]\) for \(j \ge 2\): once \(n_k > j\) we have \(I_{n_k} \cap J_j = \varnothing\), so likewise \(\int_{J_j}|f| = 0\) and \(f = 0\) a.e. on \(J_j\). As \((a,b] = \bigcup_{j \ge 2} J_j\) is a countable union, \(f = 0\) a.e. on \([a,b]\).

Hence \(f = 0\) a.e. on \(E\), and testing against \(g = \chi_E \in L^\infty(E)\) gives

\begin{equation*} 0 = \int_E f = \lim_{k \to \infty} \int_E f_{n_k} = 1 , \end{equation*}

a contradiction.

The second set. If \(m(E) = 0\) then \(L^1(E) = \{0\}\) and \(\int_E f\cdot g = 0\) always, so every sequence in \(L^1(E)\) converges weakly to \(0\).

Both directions come from the identity

\begin{equation*} \|T_n - T\|_* = \sup_{f \in B} |T_n(f) - T(f)| , \qquad B = \{ f \in X \mid \|f\| \le 1 \} . \tag{\(\ast\)} \end{equation*}

Indeed, for \(S \in X^*\) with \(s = \sup_{f \in B}|S(f)|\): the inequality \(|S(f)| \le \|S\|_*\|f\| \le \|S\|_*\) on \(B\) gives \(s \le \|S\|_*\), while \(|S(f)| \le s\|f\|\) for all \(f \in X\) (clear at \(f = 0\); otherwise apply \(s\) to \(f/\|f\| \in B\) and use homogeneity), so \(s\) is admissible in the definition of \(\|S\|_*\) as an infimum, whence \(\|S\|_* \le s\). Apply this to \(S = T_n - T \in X^*\).

(\(\Rightarrow\)) Given \(\epsilon > 0\), pick \(N\) with \(\|T_n - T\|_* < \epsilon\) for \(n \ge N\); then \(|T_n(f) - T(f)| \le \|T_n - T\|_*\|f\| < \epsilon\) for all \(f \in B\), with \(N\) independent of \(f\).

(\(\Leftarrow\)) Given \(\epsilon > 0\), pick \(N\) with \(|T_n(f) - T(f)| < \epsilon/2\) for all \(n \ge N\) and all \(f \in B\); taking the supremum over \(B\) in \((\ast)\) gives \(\|T_n - T\|_* \le \epsilon/2 < \epsilon\).

No. The sequence of that Example is \(f_n = n\cdot\chi_{[0,1/n]}\) on \([0,1]\). Take the challenge \(\epsilon = 1/2\); given any \(\delta > 0\), choose \(n\) with \(1/n < \delta\) and test on \(A = [0,1/n]\), a measurable set of measure \(1/n < \delta\):

\begin{equation*} \int_A |f_n| = n \cdot \frac{1}{n} = 1 > \frac{1}{2} . \end{equation*}

So no \(\delta\) answers \(\epsilon = 1/2\), and \(\{f_n\}\) is not uniformly integrable over \([0,1]\).

Method (2): \(\{f_n\} \to 0\) pointwise on \((0,1]\), hence a.e. on \([0,1]\), while \(\int_0^1 f_n = 1\) for every \(n\). Since \(m[0,1] < \infty\), uniform integrability would force these limits to agree by the Vitali Convergence Theorem of Chapter 4.

At the appeal to the separability of \(X = L^q(E)\), and again at the appeal to the Riesz Representation Theorem for the dual of \(L^q(E)\): both require \(q < \infty\), whereas \(p = 1\) forces \(q = \infty\).

(i) Separability. The proof cites Theorem 11 of Chapter 7 to make \(X = L^q(E)\) separable, the hypothesis of Helley’s Theorem; that theorem covers only \(1 \le q < \infty\), and for \(q = \infty\) separability is false by the Example of Section 7.4, since on a nondegenerate \([a,b] \subseteq E\)

\begin{equation*} \| \chi_{[a,x_1)} - \chi_{[a,x_2)} \|_{\infty} = 1 \quad \text{whenever } a \le x_1 < x_2 \le b \end{equation*}

exhibits uncountably many functions pairwise at distance \(1\). Helley’s diagonal argument then controls the subsequence at only countably many elements of \(L^\infty(E)\).

(ii) Representation. Even granted such a subsequence with \(T_{n_k}(g) \to T(g)\) for all \(g \in L^\infty(E)\), the proof next produces \(f \in L^1(E)\) with \(T(g) = \int_E f\cdot g\); this is a representation theorem for \((L^\infty(E))^*\), which Section 8.1 establishes only for \(L^r(E)\) with \(r < \infty\) and which the Remark closing that section warns is false for \(r = \infty\).

Only the preliminary step survives, and it needs a repaired citation: the proof gets \(\|T_n\|_* = \|f_n\|_1\) for \(T_n(g) = \int_E f_n \cdot g\) from Proposition 2 with \(p\) and \(q\) interchanged, which assumes a finite exponent. It holds directly: Hoelder’s Inequality gives \(\|T_n\|_* \le \|f_n\|_1\), and \(g = \operatorname{sgn}(f_n)\) has \(\|g\|_\infty \le 1\) with \(T_n(g) = \|f_n\|_1\).

Neither failure is reparable, since the conclusion itself is false for \(p = 1\) by the Example of this section.

The printed range must be narrowed to \(1 < p < \infty\); the assertion fails for \(p = 1\), as the last paragraph shows. By Problem 7 of this chapter, \((\ell^p)^*\) consists exactly of the functionals \(T_g(x) = \sum_j x_j g_j\) with \(g \in \ell^q\), \(q\) conjugate to \(p\), so weak convergence in \(\ell^p\) is convergence of all such sums.

Let \(\|x^{(n)}\|_p \le M\) for all \(n\), and take \(1 < p < \infty\).

A coordinatewise limit. Since \(|x^{(n)}_j| \le \|x^{(n)}\|_p \le M\) for every \(j\), each coordinate sequence is bounded, so the Bolzano-Weierstrass Theorem and the diagonal argument of Helley’s Theorem yield a subsequence with \(\lim_{k} x^{(n_k)}_j = a_j\) for every \(j\). For each fixed \(N\) the finite sum passes to the limit termwise,

\begin{equation*} \sum_{j=1}^{N} |a_j|^p = \lim_{k \to \infty} \sum_{j=1}^{N} |x^{(n_k)}_j|^p \le M^p , \end{equation*}

so \(a \in \ell^p\) with \(\|a\|_p \le M\), and \(\|x^{(n_k)} - a\|_p \le 2M\) by Minkowski’s Inequality.

Weak convergence to \(a\). Let \(g \in \ell^q\) and \(\epsilon > 0\). As \(\sum_j |g_j|^q < \infty\), choose \(N\) with \(\left(\sum_{j>N}|g_j|^q\right)^{1/q} < \epsilon/(2(2M+1))\); Hoelder’s Inequality on the tail then gives, for every \(k\),

\begin{equation*} \left| \sum_{j > N} \left( x^{(n_k)}_j - a_j \right) g_j \right| \le \| x^{(n_k)} - a \|_p \cdot \frac{\epsilon}{2(2M+1)} < \frac{\epsilon}{2} , \end{equation*}

while the remaining sum \(\sum_{j \le N}(x^{(n_k)}_j - a_j)g_j\) has only \(N\) terms, each tending to \(0\), so is under \(\epsilon/2\) for \(k\) large. Hence \(\sum_j x^{(n_k)}_j g_j \to \sum_j a_j g_j\) for every \(g \in \ell^q\), that is, \(\{x^{(n_k)}\} \rightharpoonup a\).

Failure at \(p = 1\). The tail estimate is unavailable, since \(q = \infty\), and indeed \(\{e_n\}\) (the \(n\)th standard unit sequence) is bounded in \(\ell^1\) with no weakly convergent subsequence: given \(\{e_{n_k}\}\), the choice \(g_j = (-1)^k\) if \(j = n_k\) and \(g_j = 0\) otherwise lies in \(\ell^\infty\) and gives \(\sum_j (e_{n_k})_j g_j = (-1)^k\), which diverges.

Apply Helley’s Theorem to the Riemann-Stieltjes functionals on \(X = C[0,1]\) with the maximum norm. Fix \(M\) with \(TV(f_n) \le M\) for all \(n\), and set

\begin{equation*} T_n(g) = \int_0^1 g(x) \, df_n(x) \quad \text{for } g \in C[0,1] . \end{equation*}

By the second example of Section 8.1, each \(f_n\) being of bounded variation makes this integral well defined and \(T_n\) linear, and every Riemann-Stieltjes sum obeys

\begin{equation*} \left| \sum_{i=1}^{m} g(\xi_i) \left[ f_n(x_i) - f_n(x_{i-1}) \right] \right| \le \|g\|_{\max} \cdot TV(f_n) \le M \|g\|_{\max} , \end{equation*}

so \(T_n \in X^*\) and \(\{T_n\}\) satisfies the boundedness hypothesis (28) of Helley’s Theorem with the single constant \(M\). The other hypothesis, separability of \(X\), holds because the polynomials with rational coefficients form a countable dense subset of \(C[0,1]\) by the Weierstrass Approximation Theorem. Helley’s Theorem therefore supplies a subsequence \(\{T_{n_k}\}\) and a \(T \in X^*\) with \(\lim_k T_{n_k}(g) = T(g)\) for every \(g \in C[0,1]\). A convergent real sequence is Cauchy, so \(\left\{\int_0^1 g\, df_{n_k}\right\}\) is Cauchy for each continuous \(g\), the one subsequence serving all \(g\) at once.

Both parts rest on the uniform estimate \(|T_n(h)| \le \|T_n\|_* \|h\| \le M\|h\|\), valid for all \(h \in X\) and all \(n\).

(i) Given \(g \in X\) and \(\epsilon > 0\), use the density of \(\mathcal{S}\) to pick \(h \in \mathcal{S}\) with \(\|g - h\| < \epsilon/(3(M+1))\), and then, \(\{T_n(h)\}\) being Cauchy, an index \(N\) with \(|T_n(h) - T_m(h)| < \epsilon/3\) for \(n, m \ge N\). For such \(n, m\), by linearity and the estimate,

\begin{equation*} \begin{aligned} |T_n(g) - T_m(g)| &\le |T_n(g-h)| + |T_n(h) - T_m(h)| + |T_m(h-g)| \\ &\le 2M\|g-h\| + \frac{\epsilon}{3} < \frac{2\epsilon}{3} + \frac{\epsilon}{3} = \epsilon , \end{aligned} \end{equation*}

using \(M/(M+1) \le 1\). So \(\{T_n(g)\}\) is Cauchy for every \(g \in X\).

(ii) By (i) and the completeness of \(\mathbb{R}\), \(T(g) = \lim_n T_n(g)\) exists for every \(g \in X\). For \(g, h \in X\) and real \(\alpha, \beta\), each \(T_n\) is linear, and passing to the limit in \(T_n(\alpha g + \beta h) = \alpha T_n(g) + \beta T_n(h)\), the left side converging by (i) applied to \(\alpha g + \beta h \in X\), gives \(T(\alpha g + \beta h) = \alpha T(g) + \beta T(h)\). Finally each \(|T_n(g)| \le M\|g\|\), and a limit of reals bounded by \(M\|g\|\) obeys the same bound, so

\begin{equation*} |T(g)| \le M \cdot \|g\| \quad \text{for all } g \in X , \end{equation*}

whence \(T \in X^*\) with \(\|T\|_* \le M\).

Take the averaging functionals \(T_n(f) = n\int_0^{1/n} f\) on \(X = L^\infty[0,1]\). Each is linear, and \(|f| \le \|f\|_\infty\) a.e. gives \(|T_n(f)| \le \|f\|_\infty\), so \(\{T_n\}\) lies in \(X^*\) and meets hypothesis (28) of Helley’s Theorem with \(M = 1\). (Only separability fails, \(L^\infty[0,1]\) not being separable.) Yet no subsequence converges pointwise on \(X\).

Let \(\{T_{n_k}\}\) be a subsequence. Since \(n_k \to \infty\), thin it to \(\{m_j\}\) with \(m_{j+1} \ge j\, m_j\), so \(m_j/m_{j+1} \le 1/j\), and put

\begin{equation*} f = \sum_{j=1}^\infty (-1)^j \chi_{A_j} , \qquad A_j = \left( \frac{1}{m_{j+1}}, \frac{1}{m_j} \right] , \end{equation*}

with \(f = 0\) on \((1/m_1, 1]\) and at \(0\). The \(A_j\) are pairwise disjoint, so \(|f| \le 1\) and \(f \in L^\infty[0,1]\). Splitting \((0, 1/m_j]\) disjointly as \(A_j \cup (0, 1/m_{j+1}]\),

\begin{equation*} T_{m_j}(f) = (-1)^j\left( 1 - \frac{m_j}{m_{j+1}} \right) + m_j \int_0^{1/m_{j+1}} f , \end{equation*}

whose second term is at most \(m_j/m_{j+1} \le 1/j\) in absolute value, so \(\left|T_{m_j}(f) - (-1)^j\right| \le 2/j\). Thus \(\{T_{m_j}(f)\}\) has subsequential limits \(1\) and \(-1\) and diverges; since a convergent sequence has all subsequences converging to one limit, \(\{T_{n_k}(f)\}\) diverges too.

The answers, “yes” meaning the property holds for a suitable subsequence:

• (i) \(\Rightarrow\) (ii), (iii), (iv): yes for every \(p\), the whole sequence for (ii) and (iv).
• (ii) \(\Rightarrow\) (i), (iii), (iv): no for every \(p\).
• (iii) \(\Rightarrow\) (iv): yes (whole sequence); (iii) \(\Rightarrow\) (ii): yes (whole sequence) if \(p > 1\), no if \(p = 1\); (iii) \(\Rightarrow\) (i): no.
• (iv) \(\Rightarrow\) (iii): yes (subsequence); (iv) \(\Rightarrow\) (ii): yes (whole sequence) if \(p > 1\), no if \(p = 1\); (iv) \(\Rightarrow\) (i): no.

Here \(q\) is conjugate to \(p\) and \(\|f_n\|_p \le M\).

The positive implications. Assume (i). For \(h \in L^q(E)\), Hoelder’s Inequality gives \(\left|\int_E (f_n - f)h\right| \le \|f_n - f\|_p\|h\|_q \to 0\), so (ii) holds by Proposition 6; the Riesz-Fischer Theorem gives a subsequence with (iii); and Chebychev’s Inequality applied to \(|f_n - f|^p\) gives, for \(\eta > 0\),

\begin{equation*} m \{ x \in E : |f_n(x) - f(x)| \ge \eta \} \le \frac{1}{\eta^p} \int_E |f_n - f|^p \to 0 , \end{equation*}

which is (iv). (iii) \(\Rightarrow\) (iv) is Proposition 3 of Chapter 5 (using \(m(E) < \infty\)), and (iv) \(\Rightarrow\) (iii) along a subsequence is Theorem 4 of Chapter 5.

For (iii) \(\Rightarrow\) (ii) with \(p > 1\), so \(q < \infty\): given \(h \in L^q(E)\) and \(\varepsilon > 0\), the continuity of integration of \(|h|^q\) yields \(\delta > 0\) with \(\int_A |h|^q < \varepsilon^q\) whenever \(m(A) < \delta\), and Egoroff’s Theorem (applicable since \(m(E) < \infty\)) yields \(F \subseteq E\) with \(m(E\setminus F) < \delta\) and \(\{f_n\} \to f\) uniformly on \(F\). Then \(\int_F |h| \le \|h\|_q m(E)^{1/p} < \infty\), so

\begin{equation*} \begin{aligned} \left| \int_F (f_n - f) h \right| &\le \left( \sup_F |f_n - f| \right) \int_F |h| \to 0 , \\ \left| \int_{E \setminus F} (f_n - f) h \right| &\le \|f_n - f\|_p \left( \int_{E\setminus F}|h|^q \right)^{1/q} \le (M + \|f\|_p)\,\varepsilon \end{aligned} \end{equation*}

by Hoelder’s and Minkowski’s Inequalities, whence \(\int_E f_n h \to \int_E f h\) and (ii) follows from Proposition 6. For (iv) \(\Rightarrow\) (ii) with \(p > 1\): every subsequence converges in measure, so by Theorem 4 of Chapter 5 has a further subsequence converging a.e., hence by the last paragraph one along which \(\int_E f_n h \to \int_E f h\); a real sequence all of whose subsequences have a further subsequence with the same limit converges to it.

The counterexamples, on \(E = [0,1]\) with \(f \equiv 0\).

(a) The moving spike \(g_n = n^{1/p}\chi_{(0,1/n)}\) has \(\|g_n\|_p = 1\), tends to \(0\) pointwise on \((0,1]\) and hence in measure, so it satisfies (iii) and (iv) but no subsequence satisfies (i). For \(p = 1\) it also refutes (iii) \(\Rightarrow\) (ii) and (iv) \(\Rightarrow\) (ii): the constant \(1\) lies in \(L^\infty[0,1]\) and \(\int_0^1 g_n \cdot 1 = 1 \not\to 0\), so by Proposition 6 no subsequence converges weakly to \(0\).

(b) The Rademacher functions \(r_n = \pm 1\) on alternate dyadic intervals of length \(2^{-n}\) have \(\|r_n\|_p = 1\) and satisfy (ii) but none of (i), (iii), (iv) along any subsequence. Weak convergence: \(0 \le \int_0^x r_n \le 2^{-n}\) for all \(x\) (on each pair of adjacent dyadic intervals the primitive rises then falls back to \(0\)), so \(\left|\int_c^d r_n\right| \le 2^{-n}\) and \(\int_0^1 r_n \psi \to 0\) for every step function \(\psi\); given \(h \in L^q[0,1] \subseteq L^1[0,1]\) and \(\varepsilon > 0\), Proposition 10 of Chapter 7 supplies a step \(\psi\) with \(\|h - \psi\|_1 < \varepsilon\), and \(|r_n| \le 1\) gives

\begin{equation*} \left| \int_0^1 r_n h \right| \le \int_0^1 |h - \psi| + \left| \int_0^1 r_n \psi \right| < \varepsilon + \left| \int_0^1 r_n \psi \right| , \end{equation*}

so \(\limsup_n \left|\int_0^1 r_n h\right| \le \varepsilon\) for all \(\varepsilon\). Failure of the rest: \(\|r_{n_k}\|_p = 1\) rules out (i); a subsequence converging to \(0\) a.e. would, by the Dominated Convergence Theorem with dominator \(1\), force \(\|r_{n_k}\|_p \to 0\); and a subsequence converging in measure would have, by Theorem 4 of Chapter 5, a further subsequence converging a.e.

The full sequence \(\{e_n\}\) already has arithmetic means converging strongly to \(0\), so it serves as its own subsequence. Let \(q\) be conjugate to \(p\), so \(1 < q < \infty\) and \((q-1)p = q\).

Weak convergence. Let \(T \in (\ell^p)^*\) and \(y_n = T(e_n)\). For fixed \(N\) the vector \(x = \sum_{n \le N}\operatorname{sgn}(y_n)|y_n|^{q-1}e_n\) has finite support, so lies in \(\ell^p\), and \((q-1)p = q\) gives \(\|x\|_p = S_N^{1/p}\) with \(S_N = \sum_{n \le N}|y_n|^q\). By linearity \(T(x) = S_N\), so

\begin{equation*} S_N = T(x) \le \|T\|_* \cdot \|x\|_p = \|T\|_* \cdot S_N^{1/p} , \end{equation*}

whence \(S_N^{1/q} \le \|T\|_*\) (trivially if \(S_N = 0\)). Thus \(\sum_n |y_n|^q \le \|T\|_*^q < \infty\), and the terms of a convergent series tend to \(0\), so \(T(e_n) \to 0 = T(0)\).

No strong subsequence. \(\|e_{n_k} - 0\|_p = 1\) for every \(k\).

Arithmetic means. For any subsequence, the vectors \(e_{n_1},\dots,e_{n_k}\) have disjoint supports, so \(\sigma_k = (e_{n_1}+\cdots+e_{n_k})/k\) has exactly \(k\) components, each \(1/k\), and

\begin{equation*} \|\sigma_k\|_p = \left( k \cdot k^{-p} \right)^{1/p} = k^{-(1 - 1/p)} \longrightarrow 0 \end{equation*}

since \(p > 1\).

Replacing \(a_n\) by \(a_n - a\) replaces \(\sigma_n = (a_1 + \cdots + a_n)/n\) by \(\sigma_n - a\), so assume \(a = 0\); then \(\{a_n\}\), being convergent, is bounded by some \(M\). Given \(\varepsilon > 0\), choose \(N\) with \(|a_n| < \varepsilon/2\) for \(n > N\) and split the sum there: for \(n > N\),

\begin{equation*} |\sigma_n| \le \frac{1}{n}\sum_{k=1}^{N}|a_k| + \frac{1}{n}\sum_{k=N+1}^{n}|a_k| \le \frac{NM}{n} + \frac{\varepsilon}{2} . \end{equation*}

With \(N\) and \(M\) fixed, \(NM/n < \varepsilon/2\) for all \(n\) beyond \(2NM/\varepsilon\), so \(|\sigma_n| < \varepsilon\) there. Hence \(\sigma_n \to 0\), that is, \(\sigma_n \to a\) in general.

Banach-Saks Theorem in \(\ell^2\). If \(\{x_n\} \rightharpoonup x\) in \(\ell^2\), then some subsequence has arithmetic means converging strongly to \(x\):

\begin{equation*} \lim_{k \to \infty} \left\| \frac{x_{n_1} + \cdots + x_{n_k}}{k} - x \right\|_2 = 0 . \end{equation*}

Write \(\langle u, v\rangle = \sum_n u_n v_n\), absolutely convergent by Cauchy-Schwarz, with \(\langle u,u\rangle = \|u\|_2^2\). By Problem 7 of this chapter (\(p = q = 2\)), every \(T \in (\ell^2)^*\) is \(T(u) = \langle u, y\rangle\) for some \(y \in \ell^2\), so \(\{x_n\} \rightharpoonup x\) says exactly that \(\langle x_n, y\rangle \to \langle x, y\rangle\) for every \(y \in \ell^2\). Replacing \(x_n\) by \(x_n - x\) translates the means by \(-x\), so assume \(x = 0\).

Boundedness. Each \(\{\langle y, x_n\rangle\}\) converges, hence is bounded, for every \(y \in \ell^2\); the gliding-hump argument of Problem 26 of this chapter then gives \(\sup_n \|x_n\|_2 < \infty\), since it uses only the conjugate-function identities (here \(x^* = x/\|x\|_2\), so \(\langle x^*, x\rangle = \|x\|_2\) and \(\|x^*\|_2 = 1\)) and the completeness of \(\ell^2\). Fix \(M\) with \(\|x_n\|_2^2 \le M\) for all \(n\).

The subsequence. Choose \(n_1 = 1\) and inductively \(n_{k+1} > n_k\) with \(|\langle s_k, x_{n_{k+1}}\rangle| \le 1\), where \(s_k = x_{n_1} + \cdots + x_{n_k}\); this is possible because \(s_k \in \ell^2\) and so \(\langle s_k, x_n\rangle \to 0\) as \(n \to \infty\). Then \(\|s_1\|_2^2 \le M \le 2 + M\), and expanding \(s_{k+1} = s_k + x_{n_{k+1}}\),

\begin{equation*} \|s_{k+1}\|_2^2 = \|s_k\|_2^2 + 2\langle s_k, x_{n_{k+1}}\rangle + \|x_{n_{k+1}}\|_2^2 \le (2+M)(k+1) , \end{equation*}

so by induction \(\|s_k\|_2^2 \le (2+M)k\) for every \(k\). Hence

\begin{equation*} \left\| \frac{s_k}{k} \right\|_2^2 = \frac{\|s_k\|_2^2}{k^2} \le \frac{2+M}{k} \longrightarrow 0 , \end{equation*}

and in the general case the means converge strongly to \(x\).

Take \(f_0\) to be the conjugate function of the representing kernel: with \(q\) the conjugate of \(p\) and \(g \in L^q(E)\) the function furnished by the Riesz Representation Theorem for the Dual of \(L^p(E)\), so that \(T(f) = \int_E f\cdot g\) and \(\|T\|_* = \|g\|_q\),

\begin{equation*} f_0 = \frac{\operatorname{sgn}(g)\,|g|^{q-1}}{\|g\|_q^{\,q-1}} \qquad (\text{and } f_0 = 0 \text{ if } g = 0 \text{ a.e.}) . \end{equation*}

(The printed range \(1 \le p < \infty\) must be read as \(1 < p < \infty\); see the last paragraph.) Since \((q-1)p = q\), \(\int_E |f_0|^p = \|g\|_q^{-q}\int_E |g|^q = 1\), so \(f_0 \in K\); and \(f_0 \cdot g = |g|^q/\|g\|_q^{q-1}\) gives

\begin{equation*} T(f_0) = \frac{1}{\|g\|_q^{\,q-1}} \int_E |g|^q = \|g\|_q = \|T\|_* \ge \|T\|_* \|f\|_p \ge T(f) \end{equation*}

for every \(f \in K\), the last two steps by \(\|f\|_p \le 1\) and the definition of \(\|T\|_*\).

Method (2): \(K\) is closed (by \(\left|\|f_n\|_p - \|f\|_p\right| \le \|f_n - f\|_p\)), bounded, and convex (by Minkowski’s Inequality), and \(-T\) is continuous, since \(T\) is bounded, and convex, being linear; so Theorem 17, whose hypothesis \(1 < p < \infty\) holds, supplies \(f_0 \in K\) minimizing \(-T\) over \(K\).

Why \(p = 1\) is excluded. On \(E = [0,1]\) let \(T(f) = \int_0^1 x f(x)\,dx\), so \(\|T\|_* = 1\). For \(f \in K\) not zero a.e., \(\int_0^1 (1-x)|f| > 0\), so

\begin{equation*} T(f) \le \int_0^1 |f| - \int_0^1 (1-x)|f| < \int_0^1 |f| \le 1 , \end{equation*}

and \(T(0) = 0 < 1\); yet \(f_n = n\chi_{[1-1/n,1]}\) lies in \(K\) with \(T(f_n) = 1 - 1/(2n) \to 1\). So the supremum \(1\) is not attained.

Dominated convergence along a rapidly Cauchy subsequence does it. As in Corollary 18 of this section we take \(m(E) < \infty\) or else \(c_1 = 0\); without one of these the conclusion is false (on \(E = \mathbb{R}\), \(p_1 = p_2 = 1\), with \(\varphi(s) = \min\{|s|^{1/2}, 1\}\) and \(f_n = n^{-2}\chi_{[0,n]} \to 0\) in \(L^1\), one has \(\|\varphi\circ f_n\|_1 = 1\)). We may take \(c_1, c_2 \ge 0\), and use repeatedly that \((a+b)^r \le 2^r(a^r + b^r)\) for \(r \ge 1\), \(a, b \ge 0\).

The target space. For \(h \in L^{p_1}(E)\), the growth bound gives

\begin{equation*} |\varphi \circ h|^{p_2} \le \left( c_1 + c_2 |h|^{p_1/p_2} \right)^{p_2} \le 2^{p_2}\left( c_1^{p_2} + c_2^{p_2} |h|^{p_1} \right) \quad \text{a.e. on } E , \tag{1} \end{equation*}

whose right side is integrable (the constant by the standing assumption, \(|h|^{p_1}\) since \(h \in L^{p_1}(E)\)), so \(\varphi\circ h \in L^{p_2}(E)\) by the integral comparison test.

Reduction. If the conclusion fails there are \(\varepsilon > 0\) and a subsequence with \(\|\varphi\circ f_{n_j} - \varphi\circ f\|_{p_2} \ge \varepsilon\) for all \(j\); it suffices to find a further subsequence along which this norm tends to \(0\).

A dominated subsequence. Since \(\{f_{n_j}\} \to f\) in \(L^{p_1}(E)\), extract a rapidly Cauchy subsequence \(\{h_k\}\), so \(\|h_{k+1} - h_k\|_{p_1} \le 2^{-k}\), and set \(g = |h_1| + \sum_{k\ge 1}|h_{k+1} - h_k|\). Minkowski’s Inequality bounds every partial sum \(g_m\) by \(\|h_1\|_{p_1} + 1\) in \(L^{p_1}\), so the Monotone Convergence Theorem applied to \(\{g_m^{p_1}\}\) gives \(g \in L^{p_1}(E)\); in particular \(g < \infty\) a.e., so the telescoping series converges absolutely a.e. and \(\{h_k\}\) converges pointwise a.e., with \(|h_k| \le g\) a.e. for every \(k\). Its \(L^{p_1}\) limit is \(f\), and the Riesz-Fischer Theorem gives a further subsequence converging a.e. to \(f\), so the pointwise limit is \(f\) a.e. and \(|f| \le g\) a.e.

The limit. As \(\varphi\) is continuous, \(|\varphi\circ h_k - \varphi\circ f|^{p_2} \to 0\) a.e. on \(E\), and by the growth bound with \(|h_k| \le g\), \(|f| \le g\), together with the inequality above,

\begin{equation*} \left| \varphi \circ h_k - \varphi \circ f \right|^{p_2} \le \left( 2c_1 + 2c_2\, g^{\,p_1/p_2} \right)^{p_2} \le 2^{p_2}\left( (2c_1)^{p_2} + (2c_2)^{p_2} g^{\,p_1} \right) , \end{equation*}

which is integrable exactly as in (1). The Dominated Convergence Theorem now gives \(\|\varphi\circ h_k - \varphi\circ f\|_{p_2} \to 0\), the required contradiction.

Apply Theorem 17 to the distance functional \(T(g) = \|g - f_0\|_p\) on the nonempty closed bounded convex set \(C\). It is convex, since \(\lambda g + (1-\lambda)h - f_0 = \lambda(g - f_0) + (1-\lambda)(h-f_0)\) and Minkowski’s Inequality give

\begin{equation*} T(\lambda g + (1-\lambda)h) \le \lambda T(g) + (1-\lambda) T(h) , \end{equation*}

and continuous, since \(|T(g) - T(h)| \le \|g - h\|_p\). Theorem 17, whose remaining hypothesis \(1 < p < \infty\) we note below, therefore yields \(g_0 \in C\) minimizing \(T\).

The printed range must exclude \(p = 1\). Theorem 17 rests on Theorem 14, which fails in \(L^1\), and the assertion itself fails: on \(E = [0,1]\) with \(f_0 \equiv 0\) take

\begin{equation*} C = \Big\{ f \in L^1[0,1] \ \Big| \ \int_0^1 x f(x)\, dx = 1 \ \text{ and } \ \|f\|_1 \le 3 \Big\} , \end{equation*}

which is bounded, convex (an intersection of a ball with a level set of a linear functional), closed (since \(\left|\int_0^1 x(f_n - f)\right| \le \|f_n - f\|_1\) and norms converge under strong convergence), and nonempty, containing \((8/3)\chi_{[1/2,1]}\). Every \(f \in C\) has

\begin{equation*} 1 = \int_0^1 x f \le \int_0^1 x|f| \le \|f\|_1 , \end{equation*}

while \(f_n = a_n \chi_{[1-1/n,1]}\) with \(a_n = (1/n - 1/(2n^2))^{-1}\) lies in \(C\) with \(\|f_n\|_1 = (1 - 1/(2n))^{-1} \to 1\), so the distance from \(f_0\) to \(C\) is \(1\). It is not attained: \(\|f\|_1 = 1\) for \(f \in C\) would force

\begin{equation*} 0 = \|f\|_1 - \int_0^1 x f = \int_0^1 (|f| - xf) \ge \int_0^1 (1-x)|f| \ge 0 , \end{equation*}

so \(f = 0\) a.e. on \([0,1)\), contradicting \(\int_0^1 xf = 1\).

Write, for \(0 \le k \le 2^n - 1\),

\begin{equation*} I_k = \big[ k/2^n, (k+1)/2^n \big), \qquad J_k = \big[ k/2^n, \ k/2^n + 1/2^{2n+1} \big) \subseteq I_k , \end{equation*}

so that \(f_n = 1 - 2^{n+1}\) on \(J_k\) and \(f_n = 1\) on \(I_k \setminus J_k\); the point \(x = 1\) lies in no \(I_k\), so \(f_n(1) = 1 - 2^{n+1}\) too. Put \(B_n = \{1\} \cup \bigcup_k J_k\) and \(A_n = [0,1]\setminus B_n\), the sets where \(f_n\) takes the two values, with

\begin{equation*} m(B_n) = 2^n \cdot \frac{1}{2^{2n+1}} = \frac{1}{2^{n+1}} , \qquad m(A_n) = 1 - \frac{1}{2^{n+1}} . \end{equation*}

Everything rests on \(f_n\) having mean zero on each \(I_k\):

\begin{equation*} \int_{I_k} f_n = \frac{1}{2^n} - \frac{1}{2^{2n+1}} + \frac{1 - 2^{n+1}}{2^{2n+1}} = \frac{1}{2^n} - \frac{1}{2^n} = 0 . \tag{\(\ast\)} \end{equation*}

(i) For \(x = 1\), \((\ast)\) summed over \(k\) gives \(\int_0^1 f_n = 0\). Otherwise \(x = k/2^n + t \in I_k\) with \(0 \le t < 1/2^n\), and by \((\ast)\) the earlier blocks contribute nothing, so \(\int_0^x f_n = \varphi(t) := \int_{k/2^n}^{k/2^n + t} f_n\). Two cases:

(a) \(0 \le t \le 1/2^{2n+1}\), where \(f_n = 1 - 2^{n+1}\): \(\varphi(t) = (1-2^{n+1})t\) decreases from \(0\) to \(1/2^{2n+1} - 1/2^n\).

(b) \(1/2^{2n+1} \le t < 1/2^n\), where \(f_n = 1\) beyond \(J_k\): \(\varphi(t) = (1/2^{2n+1} - 1/2^n) + (t - 1/2^{2n+1})\) increases back up to \(0\) at \(t = 1/2^n\).

So \(\varphi\) ranges between \(1/2^{2n+1} - 1/2^n\) and \(0\), giving \(\left|\int_0^x f_n\right| \le 1/2^n\), and since \(f \equiv 0\) and \(1/2^n \to 0\),

\begin{equation*} \lim_{n \to \infty} \int_0^x f_n = 0 = \int_0^x f \ \text{ for all } x \in [0,1]. \end{equation*}

(ii) Since \(1 - 2^{n+1} \ne 1\), the set on which \(f_n = 1\) is exactly \(A_n\). Therefore

\begin{equation*} E = \bigcap_{n=1}^{\infty} A_n = [0,1] \setminus \bigcup_{n=1}^{\infty} B_n , \end{equation*}

which is measurable, and by countable subadditivity \(m\left(\bigcup_n B_n\right) \le \sum_n 2^{-(n+1)} = 1/2\), so

\begin{equation*} m(E) \ge \frac{1}{2} > 0 , \qquad \int_E f_n = m(E) \ \text{ for all } n , \end{equation*}

the last because \(f_n \equiv 1\) on \(E\).

(iii) Splitting over \(A_n\) and \(B_n\),

\begin{equation*} \|f_n\|_1 = \Big( 1 - \frac{1}{2^{n+1}} \Big) + (2^{n+1} - 1)\cdot \frac{1}{2^{n+1}} = 2 - \frac{1}{2^n} \le 2 , \end{equation*}

so \(\{f_n\}\) is bounded in \(L^1[0,1]\). Testing weak convergence against \(\chi_E \in L^\infty[0,1]\), the conjugate space, would force \(\int_E f_n \to 0\), whereas it equals \(m(E) \ge 1/2\) by (ii). So \(\{f_n\} \not\rightharpoonup f\) in \(L^1[0,1]\).

No contradiction with Theorem 11, which is stated only for \(1 < p < \infty\); parts (i) and (iii) show it genuinely fails at \(p = 1\), the step functions not being dense in \(L^\infty[0,1]\).

(iv) Each \(f_n\) is bounded and measurable on a set of finite measure, so lies in \(L^p[0,1]\), and \(\chi_E \in L^q[0,1]\); the argument of (iii) applies verbatim, so \(\{f_n\} \not\rightharpoonup f\) in \(L^p[0,1]\).

No contradiction, but now for a different reason: Theorem 11 requires \(\{f_n\}\) to be bounded in \(L^p[0,1]\), and it is not. Indeed, using \(2^{n+1} - 1 \ge 2^n\),

\begin{equation*} \|f_n\|_p^p \ge \int_{B_n} (2^{n+1}-1)^p = \frac{(2^{n+1}-1)^p}{2^{n+1}} \ge 2^{\,n(p-1)-1} \longrightarrow \infty \end{equation*}

as \(n \to \infty\), since \(p > 1\).

Take \(g_n = 2^{n/2}\chi_{J_n}\) on the disjoint intervals \(J_n = (2^{-n}, 2^{-n+1}]\), and then continuous \(h_n\) with \(\|h_n - g_n\|_2 < 1/n\).

Each \(g_n\) is bounded and measurable, with \(\|g_n\|_2^2 = 2^n \cdot 2^{-n} = 1\), and disjointness of supports kills the cross term:

\begin{equation*} \|g_n - g_m\|_2^2 = \|g_n\|_2^2 + \|g_m\|_2^2 = 2 \quad \text{for } n \ne m . \end{equation*}

A Cauchy subsequence would have two distinct-index terms within distance \(1\), which this forbids.

The \(h_n\) exist because the continuous functions are dense in \(L^2[0,1]\) by Theorem 12 of Chapter 7 (\([0,1]\) being bounded, \(C_c[0,1]\) is all of \(C[0,1]\)). If some \(\{h_{n_k}\}\) converged in \(L^2[0,1]\), it would be Cauchy, and then for all \(j, k\),

\begin{equation*} \|g_{n_j} - g_{n_k}\|_2 \le \frac{1}{n_j} + \|h_{n_j} - h_{n_k}\|_2 + \frac{1}{n_k} , \end{equation*}

whose right side is eventually below any \(\epsilon > 0\) since \(n_k \ge k\). That would make \(\{g_{n_k}\}\) Cauchy, contradicting the previous paragraph.

Take \(c = \max\{1/c_1,\, c_2\}\) one way and \(c_1 = 1/c\), \(c_2 = c\) the other, where by definition \(\rho\) and \(\tau\) are equivalent provided \(c_1 \tau \le \rho \le c_2 \tau\) on \(X \times X\) for some \(c_1, c_2 > 0\).

(\(\Rightarrow\)) With that \(c > 0\) we have \(c \ge c_2\) and \(1/c \le c_1\), so, \(\tau\) being nonnegative,

\begin{equation*} \frac{1}{c}\,\tau(u,v) \le c_1 \tau(u,v) \le \rho(u,v) \le c_2 \tau(u,v) \le c\,\tau(u,v) . \end{equation*}

(\(\Leftarrow\)) The hypothesis with \(c_1 = 1/c > 0\) and \(c_2 = c > 0\) is verbatim the definition.

They are equivalent: \(\rho^{+} \le \rho^{*} \le n\,\rho^{+}\) throughout \(\mathbb{R}^n \times \mathbb{R}^n\), so \(c_1 = 1\) and \(c_2 = n\) serve.

Both are metrics, being induced by norms through formula (1) of Section 9.1: for \(\|u\|_* = |u_1| + \dots + |u_n|\) and \(\|u\|_+ = \max_i |u_i|\), nonnegativity, vanishing only at \(u = 0\), and homogeneity are read off from the corresponding facts for \(|\cdot|\), while summing \(|u_i + v_i| \le |u_i| + |v_i|\) over \(i\) gives the triangle inequality for \(\|\cdot\|_*\) and maximizing the bound \(|u_i + v_i| \le \|u\|_+ + \|v\|_+\) over \(i\) gives it for \(\|\cdot\|_+\). (Check!)

For the comparison, fix \(x, y\), put \(t_i = |x_i - y_i| \ge 0\), and choose \(k\) with \(t_k = \max_i t_i\). Then

\begin{equation*} \rho^{+}(x,y) = t_k \le \sum_{i=1}^{n} t_i = \rho^{*}(x,y) \le n\,t_k = n\,\rho^{+}(x,y), \end{equation*}

the first inequality because each \(t_i \ge 0\), the last because each \(t_i \le t_k\).

Take the discrete metric of Section 9.1 on \(X = \mathbb{R}^n\):

\begin{equation*} \rho(x, y) = \begin{cases} 0, & x = y, \\ 1, & x \ne y. \end{cases} \end{equation*}

It is a metric: only the triangle inequality needs a word, and if \(x \ne y\) then \(z\) differs from at least one of \(x, y\), so \(\rho(x,z) + \rho(z,y) \ge 1 = \rho(x,y)\). (Check!)

It is not equivalent to \(\rho^{*}\), because equivalence would give a \(c_1 > 0\) with

\begin{equation*} c_1 \rho^{*}(x,y) \le \rho(x,y) \le 1 \quad \text{for all } x, y \in \mathbb{R}^n, \end{equation*}

forcing \(\rho^{*} \le 1/c_1\) throughout; but \(\rho^{*}(0, (t,0,\dots,0)) = t\) is unbounded in \(t > 0\). The same points give \(\rho^{+}(0, (t,0,\dots,0)) = t\), so \(\rho\) is not equivalent to \(\rho^{+}\) either.

The spikes

\begin{equation*} f_n(x) = \max\{\, 0,\ 1 - n(x-a) \,\}, \qquad n \ge N \text{ where } 1/N < b - a, \end{equation*}

defeat equivalence. Each \(f_n\) is continuous, hence lies in \(X\), and it vanishes on \([a + 1/n, b]\), a nondegenerate interval since \(a + 1/n \le a + 1/N < b\).

Measure them against \(g \equiv 0\). Since \(f_n(a) = 1\) and \(0 \le f_n \le 1\),

\begin{equation*} \rho_{\max}(f_n, g) = 1, \qquad \rho_1(f_n, g) = \int_a^{a + 1/n} \bigl(1 - n(x-a)\bigr)\,dx = \frac{1}{2n}, \end{equation*}

the integral being \(\bigl[(x-a) - \tfrac{n}{2}(x-a)^2\bigr]_a^{a+1/n} = 1/n - 1/(2n)\).

Were the two metrics equivalent, some \(c > 0\) would satisfy \(\rho_{\max} \le c\,\rho_1\) on \(X \times X\), whence \(1 \le c/(2n)\), i.e. \(n \le c/2\), for every \(n \ge N\), which the Archimedean property of \(\mathbb{R}\) forbids.

Everything rests on the set inclusion \[ A \Delta B \subseteq [A \Delta C] \cup [C \Delta B], \quad A, B, C \in X. \] Indeed, if \(x \in A \sim B\), then according as \(x \in C\) or \(x \notin C\) we get \(x \in C \sim B \subseteq C \Delta B\) or \(x \in A \sim C \subseteq A \Delta C\); the case \(x \in B \sim A\) is this argument with \(A\) and \(B\) interchanged, using \(A \Delta C = C \Delta A\) and \(C \Delta B = B \Delta C\).

\(\rho\) is a pseudometric. It is nonnegative and finite, since \(A \Delta B\) is measurable and \(A \Delta B \subseteq E\) gives \(\rho(A,B) \le m(E) < \infty\); \(\rho(A,A) = m(\emptyset) = 0\); \(\rho\) is symmetric because \(A \Delta B = B \Delta A\); and by the monotonicity and finite subadditivity of \(m\) applied to the inclusion,

\begin{equation*} \rho(A,B) \le m\bigl([A \Delta C] \cup [C \Delta B]\bigr) \le \rho(A,C) + \rho(C,B). \end{equation*}

It is not a metric: for \(E = [0,1]\), \(A = [0,1]\) and \(B = (0,1]\), \(\rho(A,B) = m(\{0\}) = 0\) with \(A \ne B\).

The induced metric. Write \(A \cong B\) for \(\rho(A,B) = 0\); this is an equivalence relation, transitivity coming from \(0 \le \rho(A,C) \le \rho(A,B) + \rho(B,C) = 0\). Setting \(\widetilde{\rho}([A],[B]) = \rho(A,B)\) is representative-independent, since \(A \cong A^{\prime}\) and \(B \cong B^{\prime}\) give

\begin{equation*} \rho(A,B) \le \rho(A,A^{\prime}) + \rho(A^{\prime},B^{\prime}) + \rho(B^{\prime},B) = \rho(A^{\prime},B^{\prime}), \end{equation*}

and symmetrically the reverse. Nonnegativity, symmetry and the triangle inequality descend from \(\rho\), while \(\widetilde{\rho}([A],[B]) = 0\) says \(A \cong B\), that is, \([A] = [B]\).

The integral formula. At each \(x \in E\) the difference \(\chi_A(x) - \chi_B(x)\) is nonzero exactly when \(x\) lies in exactly one of \(A, B\), and then has modulus \(1\), so \(|\chi_A - \chi_B| = \chi_{A \Delta B}\) on \(E\); this simple function is integrable over \(E\) (bounded, \(m(E) < \infty\)) with

\begin{equation*} \int_E |\chi_A - \chi_B| = \int_E \chi_{A \Delta B} = m(A \Delta B) = \rho(A,B). \end{equation*}

The function \(\varphi(t) = t/(1+t) = 1 - 1/(1+t)\) is increasing on \([0,\infty)\), since for \(0 \le s \le t\),

\begin{equation*} \varphi(t) - \varphi(s) = \frac{1}{1+s} - \frac{1}{1+t} = \frac{t-s}{(1+s)(1+t)} \ge 0 . \end{equation*}

Hence \(a \le b + c\) with \(a, b, c \ge 0\) gives

\begin{equation*} \frac{a}{1+a} = \varphi(a) \le \varphi(b+c) = \frac{b}{1+b+c} + \frac{c}{1+b+c} \le \frac{b}{1+b} + \frac{c}{1+c}, \end{equation*}

the last step because \(1 + b + c \ge 1 + b > 0\) and \(1 + b + c \ge 1 + c > 0\) while the numerators \(b, c\) are nonnegative.

Write \(h_{f,g} = \varphi \circ |f-g|\) with \(\varphi(t) = t/(1+t)\), so \(\rho(f,g) = \int_E h_{f,g}\); the pseudometric axioms are Problem 6 integrated.

\(\rho\) is finite: \(h_{f,g}\) is measurable (composition of the continuous \(t \mapsto t/(1+|t|)\) with \(|f-g|\), by Proposition 7 of Chapter 3), and \(0 \le h_{f,g} < 1\) on the finite-measure set \(E\), so it is integrable by Theorem 4 of Chapter 4 with \(0 \le \rho(f,g) \le m(E)\).

Symmetry holds since \(|f-g| = |g-f|\), and \(\rho(f,f) = 0\) since \(\varphi(0) = 0\). For the triangle inequality, apply Problem 6 at each \(x \in E\) to \(a = |f(x)-k(x)|\), \(b = |f(x)-g(x)|\), \(c = |g(x)-k(x)|\), which satisfy \(a \le b + c\); this gives \(h_{f,k} \le h_{f,g} + h_{g,k}\) pointwise, and integrating over \(E\),

\begin{equation*} \rho(f,k) \le \int_E h_{f,g} + \int_E h_{g,k} = \rho(f,g) + \rho(g,k). \end{equation*}

The null pairs are exactly the a.e.-equal pairs: \(h_{f,g} \ge 0\), so by Proposition 9 of Chapter 4, \(\rho(f,g) = 0\) if and only if \(h_{f,g} = 0\) a.e. on \(E\), and \(\varphi(t) = 0\) with \(t \ge 0\) forces \(t = 0\), so this says \(f = g\) a.e. on \(E\). In particular \(\rho\) is not a metric: on \(E = [0,1]\) take \(f \equiv 0\) and \(g = \chi_{\{0\}}\).

The induced metric. Equality a.e. is an equivalence relation (the set where \(f \ne k\) lies in the union of the two null sets where \(f \ne g\) and \(g \ne k\)), and by the previous paragraph it is precisely the relation \(\rho(f,g) = 0\) attached to the pseudometric in Section 9.1. So \(\widetilde{\rho}([f],[g]) = \rho(f,g)\) is representative-independent:

\begin{equation*} \rho(f,g) \le \rho(f,f^{\prime}) + \rho(f^{\prime},g^{\prime}) + \rho(g^{\prime},g) = \rho(f^{\prime},g^{\prime}) \end{equation*}

when \(f \cong f^{\prime}\) and \(g \cong g^{\prime}\), and symmetrically the reverse. Nonnegativity, symmetry and the triangle inequality descend from \(\rho\), and \(\widetilde{\rho}([f],[g]) = 0\) holds exactly when \(f = g\) a.e., that is, when \([f] = [g]\).

The whole point is that \(t \le t^p\) for \(0 \le t \le 1\): this is clear at \(t = 0\), and for \(0 < t \le 1\) we have \(t^{p}/t = t^{-(1-p)} \ge 1\) since \(1 - p > 0\).

For \(a = b = 0\) the assertion reads \(0 \le 0\). Otherwise \(a + b > 0\), and applying the above to \(t = a/(a+b)\) and \(1 - t = b/(a+b)\), both in \([0,1]\),

\begin{equation*} 1 = t + (1-t) \le t^p + (1-t)^p = \frac{a^p + b^p}{(a+b)^p} . \end{equation*}

Multiplying by \((a+b)^p > 0\) gives \((a+b)^p \le a^p + b^p\).

Let \(X\) be the collection in question. The subadditivity \((a+b)^p \le a^p + b^p\) of Problem 8, combined with the monotonicity of \(t \mapsto t^p\) on \([0,\infty)\), supplies both the finiteness of \(\rho_p\) and its triangle inequality.

Finiteness: \(|g-h|^p\) is measurable (composition of the continuous \(t \mapsto |t|^p\) with \(g - h\), by Proposition 7 of Chapter 3), and pointwise on \(E\),

\begin{equation*} |g-h|^p \le (|g| + |h|)^p \le |g|^p + |h|^p , \end{equation*}

whose right side is integrable, so \(0 \le \rho_p(h,g) \le \int_E |g|^p + \int_E |h|^p < \infty\).

Symmetry is \(|g-h| = |h-g|\), and \(\rho_p(h,h) = 0\). For the triangle inequality, \(|g-h| \le |g-f| + |f-h|\) pointwise gives, by the same two facts,

\begin{equation*} |g-h|^p \le |g-f|^p + |f-h|^p , \end{equation*}

and integrating over \(E\) yields \(\rho_p(h,g) \le \rho_p(f,g) + \rho_p(h,f)\). So \(\rho_p\) is a pseudometric.

The null pairs are the a.e.-equal pairs: \(|g-h|^p \ge 0\) is integrable, so by Proposition 9 of Chapter 4 its integral vanishes exactly when \(|g-h|^p = 0\) a.e. on \(E\), and \(t^p = 0\) only at \(t = 0\). In particular \(\rho_p\) is not a metric (the zero function and \(\chi_{\{x_0\}}\) for \(x_0 \in E\) are at distance \(0\)).

The induced metric. Equality a.e. is an equivalence relation, since \(h = k\) off the union of two null sets, and by the last paragraph it coincides with \(\rho_p(h,g) = 0\). Hence \(\widetilde{\rho_p}([h],[g]) = \rho_p(h,g)\) is representative-independent: if \(h^{\prime} = h\) and \(g^{\prime} = g\) off a null set \(Z\), then \(|g^{\prime}-h^{\prime}|^p = |g-h|^p\) on \(E \sim Z\), and both being integrable, Proposition 15 of Chapter 4 gives

\begin{equation*} \rho_p(h^{\prime},g^{\prime}) = \int_{E \sim Z} |g^{\prime}-h^{\prime}|^p = \int_{E \sim Z} |g-h|^p = \rho_p(h,g). \end{equation*}

Nonnegativity, symmetry and the triangle inequality descend from \(\rho_p\), while \(\widetilde{\rho_p}([h],[g]) = 0\) says \(h = g\) a.e., that is, \([h] = [g]\).

Write \(d_n(x,y) = \rho_n(x_n,y_n)/(1 + \rho_n(x_n,y_n))\), so that \(\rho_* = \sum_n 2^{-n} d_n\); Problem 6 makes each \(d_n\) obey the triangle inequality, and the geometric weights make the sum finite.

Finiteness: \(0 \le d_n < 1\), so the series is dominated termwise by \(\sum 2^{-n}\) and

\begin{equation*} 0 \le \rho_*(x,y) \le \sum_{n=1}^{\infty} 2^{-n} = 1 . \end{equation*}

Vanishing: a series of nonnegative terms vanishes only when every term does, so \(\rho_*(x,y) = 0\) iff \(d_n(x,y) = 0\) for all \(n\), iff \(\rho_n(x_n,y_n) = 0\) for all \(n\) (as \(t/(1+t)\) vanishes only at \(t = 0\)), iff \(x_n = y_n\) for all \(n\), i.e. \(x = y\).

Symmetry is inherited termwise from the symmetry of each \(\rho_n\).

Triangle inequality: fixing \(n\) and applying Problem 6 to \(a = \rho_n(x_n,z_n) \le b + c\) with \(b = \rho_n(x_n,y_n)\) and \(c = \rho_n(y_n,z_n)\) gives

\begin{equation*} d_n(x,z) \le d_n(x,y) + d_n(y,z) ; \end{equation*}

multiplying by \(2^{-n}\) and summing, all three series converging by the first step,

\begin{equation*} \rho_*(x,z) \le \rho_*(x,y) + \rho_*(y,z). \end{equation*}

The metric is the pullback \[ \sigma(a_1, a_2) = \rho(f(a_1), f(a_2)), \qquad a_1, a_2 \in A. \]

Uniqueness is immediate: the defining condition of an isometry in Section 9.1 is exactly this identity, so any metric making \(f\) an isometry equals \(\sigma\).

That \(\sigma\) is a metric: nonnegativity, real-valuedness and symmetry are those of \(\rho\), and for the triangle inequality apply that of \(\rho\) to \(f(a_1), f(a_2), f(a_3)\),

\begin{equation*} \sigma(a_1,a_3) \le \rho(f(a_1),f(a_2)) + \rho(f(a_2),f(a_3)) = \sigma(a_1,a_2) + \sigma(a_2,a_3). \end{equation*}

For separation, \(\sigma(a_1,a_2) = 0\) forces \(f(a_1) = f(a_2)\) since \(\rho\) is a metric, hence \(a_1 = a_2\) since \(f\) is one-to-one; this is the only use of injectivity, and without it \(\sigma\) would be a mere pseudometric. Since \(f\) is onto \(X\) by hypothesis, \(f\) is an isometry of \((A,\sigma)\) onto \((X,\rho)\).

Embed \(\mathbb{R}^n\) isometrically in \(L^2[0,1]\) by \[ \Phi(x) = \sqrt{n} \sum_{k=1}^{n} x_k \, \chi_{I_k}, \] where \(I_k = [(k-1)/n, k/n)\) for \(k < n\) and \(I_n = [(n-1)/n, 1]\) partition \([0,1]\) into sets of measure \(1/n\). Each \(\Phi(x)\) is simple and bounded on a set of finite measure, hence lies in \(L^2[0,1]\), and \(\Phi\) is linear because the coefficients add coordinatewise.

Disjointness of the \(I_k\) gives \(\Phi(x)^2 = n \sum_k x_k^2 \chi_{I_k}\) pointwise, so

\begin{equation*} \|\Phi(x)\|_2^2 = n \sum_{k=1}^{n} x_k^2 \, m(I_k) = \sum_{k=1}^{n} x_k^2 = \|x\|^2 . \end{equation*}

Hence the Minkowski Inequality with \(p = 2\) transfers to \(\mathbb{R}^n\): for \(x, y \in \mathbb{R}^n\),

\begin{equation*} \|x + y\| = \|\Phi(x) + \Phi(y)\|_2 \le \|\Phi(x)\|_2 + \|\Phi(y)\|_2 = \|x\| + \|y\| , \end{equation*}

and applying this to \(x - y\) and \(y - z\) gives \(\rho(x,z) \le \rho(x,y) + \rho(y,z)\).

Yes in a general metric space; no in \(\mathbb{R}^n\); no in any normed linear space.

(i) Possible in general. On any set with at least two points carry the discrete metric and take \(r = 2\): every distance is at most \(1 < 2\), so \(B(x,2) = X\) for every \(x\), and \(B(u,2) = X = B(v,2)\) for distinct \(u, v\).

(ii) Impossible in a normed linear space \(X\) (hence in \(\mathbb{R}^n\), whose metric is induced by the Euclidean norm). Let \(u \ne v\), put \(d = \|u - v\| > 0\) and \(e = (u-v)/d\), so \(\|e\| = 1\), and choose \(s\) with \(\max\{0, r-d\} < s < r\), possible since \(r - d < r\) and \(0 < r\). Then \(w = u + s\,e\) satisfies

\begin{equation*} \|w - u\| = s\|e\| = s < r, \qquad \|w - v\| = \|(d+s)e\| = d + s > r , \end{equation*}

the last since \(s > r - d\), using \(u - v = d\,e\). So \(w \in B(u,r) \sim B(v,r)\) and the balls differ.

The quadrilateral inequality \[ \big|\rho(a,b) - \rho(c,d)\big| \le \rho(a,c) + \rho(b,d), \quad a,b,c,d \in X, \] does it. Two applications of the triangle inequality and symmetry give

\begin{equation*} \rho(a,b) \le \rho(a,c) + \rho(c,d) + \rho(d,b) = \rho(a,c) + \rho(c,d) + \rho(b,d), \end{equation*}

i.e. \(\rho(a,b) - \rho(c,d) \le \rho(a,c) + \rho(b,d)\); interchanging the pairs \((a,b)\) and \((c,d)\) bounds the negative of the left side by the same quantity, which is the claim.

Taking \(a = u_n\), \(b = v_n\), \(c = u\), \(d = v\),

\begin{equation*} \big|\rho(u_n,v_n) - \rho(u,v)\big| \le \rho(u_n,u) + \rho(v_n,v) \to 0 , \end{equation*}

since \(\{u_n\} \to u\) and \(\{v_n\} \to v\) mean exactly that the two summands tend to \(0\). Hence \(\{\rho(u_n,v_n)\} \to \rho(u,v)\).

(i) The inclusion \(B(x,r) \subseteq \overline{B}(x,r)\) is the implication \(\rho < r \Rightarrow \rho \le r\). For closedness, by Proposition 4 it suffices that \(X \sim \overline{B}(x,r)\) be open: given \(y\) there, \(\delta = \rho(y,x) - r > 0\) and any \(z \in B(y,\delta)\) has, by the triangle inequality,

\begin{equation*} \rho(z,x) \ge \rho(y,x) - \rho(y,z) > \rho(y,x) - \delta = r , \end{equation*}

so \(B(y,\delta) \subseteq X \sim \overline{B}(x,r)\).

(ii) In a normed linear space the two agree. By (i) and Proposition 3 (the closure is the smallest closed superset), \(\overline{B(x,r)} \subseteq \overline{B}(x,r)\). Conversely, if \(\|y - x\| \le r\), the points \(y_n = x + (1 - 1/n)(y-x)\) of \(X\) satisfy

\begin{equation*} \|y_n - x\| = \Bigl(1 - \frac{1}{n}\Bigr)\|y-x\| < r, \qquad \|y_n - y\| = \frac{1}{n}\|y-x\| \le \frac{r}{n} , \end{equation*}

so \(\{y_n\}\) lies in \(B(x,r)\) and converges to \(y\); by Proposition 6, \(y \in \overline{B(x,r)}\).

This fails in general: on a set with at least two points carrying the discrete metric, \(B(x,1) = \{x\}\) while \(\overline{B}(x,1) = X\), and \(\{x\}\) is already closed because every subset is open there (each \(u\) has \(B(u,1/2) = \{u\}\)). Hence \(\overline{B(x,1)} = \{x\} \ne X\).

Everything follows from the ball identity \[ B_X(x,r) = X \cap B_Y(x,r) \qquad (x \in X,\ r > 0), \tag{\(\ast\)} \] immediate since the metric of the subspace \(X\) is the restriction of the metric \(\rho\) of \(Y\).

If \(E = X \cap \mathcal{O}\) with \(\mathcal{O}\) open in \(Y\): each \(x \in E\) lies in \(\mathcal{O}\), so \(B_Y(x,r) \subseteq \mathcal{O}\) for some \(r > 0\), and by \((\ast)\),

\begin{equation*} B_X(x,r) = X \cap B_Y(x,r) \subseteq X \cap \mathcal{O} = E , \end{equation*}

so \(E\) is open in \(X\).

Conversely, if \(E\) is open in \(X\), choose for each \(x \in E\) an \(r_x > 0\) with \(B_X(x,r_x) \subseteq E\), and set \(\mathcal{O} = \bigcup_{x \in E} B_Y(x,r_x)\), open in \(Y\) by Proposition 1 since open balls are open (for \(E = \emptyset\) take \(\mathcal{O} = \emptyset\)). Distributing the intersection over the union and using \((\ast)\),

\begin{equation*} X \cap \mathcal{O} = \bigcup_{x \in E} B_X(x, r_x) = E , \end{equation*}

the last equality because each \(B_X(x,r_x) \subseteq E\) while each \(x \in E\) lies in \(B_X(x,r_x)\).

Equivalence supplies \(c_1, c_2 > 0\) with \(c_1 \sigma \le \rho \le c_2 \sigma\) on \(X \times X\), and hence each kind of ball contains a concentric ball of the other kind:

\begin{equation*} B_\sigma(x, r/c_2) \subseteq B_\rho(x,r), \qquad B_\rho(x, c_1 r) \subseteq B_\sigma(x,r). \end{equation*}

Indeed \(\sigma(x^{\prime},x) < r/c_2\) gives \(\rho(x^{\prime},x) \le c_2 \sigma(x^{\prime},x) < r\), and \(\rho(x^{\prime},x) < c_1 r\) gives \(c_1 \sigma(x^{\prime},x) \le \rho(x^{\prime},x) < c_1 r\), whence \(\sigma(x^{\prime},x) < r\) upon dividing by \(c_1 > 0\).

Now let \(\mathcal{O}\) be open in \((X,\rho)\) and \(x \in \mathcal{O}\), say \(B_\rho(x,r) \subseteq \mathcal{O}\). Then

\begin{equation*} B_\sigma(x, r/c_2) \subseteq B_\rho(x,r) \subseteq \mathcal{O} , \end{equation*}

so \(\mathcal{O}\) is open in \((X,\sigma)\). Symmetrically, if \(B_\sigma(x,r) \subseteq \mathcal{O}\) then \(B_\rho(x, c_1 r) \subseteq \mathcal{O}\), so openness in \((X,\sigma)\) implies openness in \((X,\rho)\).

This is Proposition 2 read through complements, using \(X \cap (Y \sim S) = X \sim S\) for \(S \subseteq Y\) and \(X \sim (X \sim A) = A\) for \(A \subseteq X\).

If \(A\) is closed in \(X\), then \(X \sim A\) is open in \(X\) by Proposition 4, so by Proposition 2 there is an \(\mathcal{O}\) open in \(Y\) with \(X \sim A = X \cap \mathcal{O}\). Then \(F = Y \sim \mathcal{O}\) is closed in \(Y\) by Proposition 4, and

\begin{equation*} X \cap F = X \sim \mathcal{O} = X \sim (X \cap \mathcal{O}) = X \sim (X \sim A) = A , \end{equation*}

the second equality because a point of \(X\) lies in \(\mathcal{O}\) exactly when it lies in \(X \cap \mathcal{O}\).

Conversely, if \(A = X \cap F\) with \(F\) closed in \(Y\), then \(\mathcal{O} = Y \sim F\) is open in \(Y\) and, by the same two identities,

\begin{equation*} X \cap \mathcal{O} = X \sim F = X \sim (X \cap F) = X \sim A . \end{equation*}

So \(X \sim A\) is open in \(X\) by Proposition 2, and \(A\) is closed in \(X\) by Proposition 4.

No in general, yes under the extra hypothesis, in both parts.

(i) With \(Y = \mathbb{R}\) and \(X = [0,1]\), the set \(\mathcal{O} = X\) is open in \(X\) by Proposition 1 but not in \(\mathbb{R}\), since every \((-\epsilon,\epsilon)\) contains \(-\epsilon/2 \notin [0,1]\). If \(X\) is open in \(Y\), however, Proposition 2 writes \(\mathcal{O} = X \cap \mathcal{U}\) with \(\mathcal{U}\) open in \(Y\), and Proposition 1 makes this intersection of two open subsets of \(Y\) open in \(Y\).

(ii) With \(Y = \mathbb{R}\) and \(X = (0,1)\), the set \(F = X\) is closed in \(X\) by Proposition 5 but not in \(\mathbb{R}\), since \(\{1/n\}_{n \ge 2}\) lies in \(F\) and converges to \(0 \notin F\) (Proposition 6). If \(X\) is closed in \(Y\), however, Problem 18 writes \(F = X \cap G\) with \(G\) closed in \(Y\), and Proposition 5 makes this intersection of closed subsets of \(Y\) closed in \(Y\).

The point is that an open ball inside \(E\) consists of interior points of \(E\).

\(\operatorname{int} E\) is open: given \(x \in \operatorname{int} E\), pick \(r > 0\) with \(B(x,r) \subseteq E\), and let \(x^{\prime} \in B(x,r)\). Since an open ball is open, with \(r^{\prime} = r - \rho(x^{\prime},x) > 0\),

\begin{equation*} B(x^{\prime}, r^{\prime}) \subseteq B(x,r) \subseteq E , \end{equation*}

so \(x^{\prime} \in \operatorname{int} E\). Thus \(B(x,r) \subseteq \operatorname{int} E\), and \(\operatorname{int} E\) is open.

Also \(\operatorname{int} E \subseteq E\) always, since \(x \in B(x,r) \subseteq E\) whenever \(x\) is an interior point.

Hence if \(E\) is open then every \(x \in E\) is an interior point, giving \(E \subseteq \operatorname{int} E\) and so \(E = \operatorname{int} E\); and if \(E = \operatorname{int} E\) then \(E\) is open, \(\operatorname{int} E\) being open.

Both assertions are Problem 20 applied to \(X \sim E\), because the two definitions coincide: \[ \operatorname{ext} E = \operatorname{int}(X \sim E). \] Indeed, an open ball centered at \(x\) lies in \(X \sim E\) precisely when \(x\) is an interior point of \(X \sim E\).

Openness: by Problem 20 the interior of any subset is open, so \(\operatorname{ext} E = \operatorname{int}(X \sim E)\) is open.

Closedness: Problem 20 applied to \(X \sim E\) gives that \(X \sim E\) is open if and only if \(X \sim E = \operatorname{int}(X \sim E) = \operatorname{ext} E\), and by Proposition 4 (with \(X \sim (X \sim E) = E\)) the set \(X \sim E\) is open if and only if \(E\) is closed. Chaining,

\begin{equation*} E \text{ is closed} \iff X \sim E \text{ is open} \iff X \sim E = \operatorname{ext} E . \end{equation*}