Solutions to Axler’s Measure, Integration & Real Analysis

Write \(P\) as \(a = x_0 < x_1 < \cdots < x_n = b\). By 1.3,

\begin{equation*} U(f, P, [a,b]) - L(f, P, [a,b]) = \sum_{j=1}^{n} (x_j - x_{j-1}) \Big( \sup_{[x_{j-1}, x_j]} f - \inf_{[x_{j-1}, x_j]} f \Big), \end{equation*}

and the hypothesis says this equals \(0\). Every summand is nonnegative, so every summand vanishes; since \(x_j - x_{j-1} > 0\), this forces

\begin{equation*} \sup_{[x_{j-1}, x_j]} f = \inf_{[x_{j-1}, x_j]} f \quad \text{for } j = 1, \ldots, n . \end{equation*}

Hence \(f\) is constant on each \([x_{j-1}, x_j]\), say with value \(c_j\). Consecutive subintervals share the point \(x_j\), so \(c_j = f(x_j) = c_{j+1}\) for \(1 \le j \le n-1\), and all the \(c_j\) equal a common value \(c\). The subintervals cover \([a,b]\), so \(f\) is constant on \([a,b]\).

Squeeze with the partition \(P_\varepsilon\) of \([a,b]\) formed by the distinct elements of \(\{a,\, s,\, s+\varepsilon,\, t-\varepsilon,\, t,\, b\}\) in increasing order, where \(0 < \varepsilon < \frac{t-s}{2}\). Its subintervals are \([a,s]\) (present only if \(a < s\)), \([s, s+\varepsilon]\), \([s+\varepsilon, t-\varepsilon]\), \([t-\varepsilon, t]\), and \([t,b]\) (present only if \(t < b\)). Now \(f \equiv 1\) on \([s+\varepsilon, t-\varepsilon]\), while \(f \equiv 0\) on \([a,s]\) and on \([t,b]\), and \(0 \le f \le 1\) throughout; hence

\begin{equation*} \begin{aligned} L(f, P_\varepsilon, [a,b]) &\ge (t-\varepsilon) - (s+\varepsilon) = (t-s) - 2\varepsilon , \\ U(f, P_\varepsilon, [a,b]) &\le \varepsilon + \big( (t-\varepsilon) - (s+\varepsilon) \big) + \varepsilon = t - s . \end{aligned} \end{equation*}

Letting \(\varepsilon \to 0\) in the first estimate and using 1.7 and 1.8,

\begin{equation*} t - s \le L(f, [a,b]) \le U(f, [a,b]) \le t - s . \end{equation*}

By 1.9, \(f\) is Riemann integrable on \([a,b]\) with \(\int_a^b f = t - s\).

Both directions run off 1.7, which gives \(L(f,P,[a,b]) \le L(f,[a,b])\) and \(U(f,[a,b]) \le U(f,P,[a,b])\) for every partition \(P\) of \([a,b]\).

(i) The \(\varepsilon\)-condition implies integrability. Given \(\varepsilon > 0\), choose \(P\) as in the statement; then with 1.8,

\begin{equation*} 0 \le U(f,[a,b]) - L(f,[a,b]) \le U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon . \end{equation*}

As \(\varepsilon > 0\) was arbitrary, \(L(f,[a,b]) = U(f,[a,b])\), so \(f\) is Riemann integrable by 1.9.

(ii) Integrability implies the \(\varepsilon\)-condition. Let \(\varepsilon > 0\). Since \(L(f,[a,b])\) is a supremum and \(U(f,[a,b])\) an infimum over partitions (1.7), there are partitions \(P_1, P_2\) of \([a,b]\) with

\begin{equation*} L(f, P_1, [a,b]) > L(f,[a,b]) - \tfrac{\varepsilon}{2} , \qquad U(f, P_2, [a,b]) < U(f,[a,b]) + \tfrac{\varepsilon}{2} . \end{equation*}

Let \(P\) merge the lists defining \(P_1\) and \(P_2\); each of those lists is a sublist of \(P\), so 1.5 gives \(L(f,P,[a,b]) \ge L(f,P_1,[a,b])\) and \(U(f,P,[a,b]) \le U(f,P_2,[a,b])\). Hence, using \(L(f,[a,b]) = U(f,[a,b])\),

\begin{equation*} \begin{aligned} U(f, P, [a,b]) - L(f, P, [a,b]) &\le U(f, P_2, [a,b]) - L(f, P_1, [a,b]) \\ &< \Big( U(f,[a,b]) + \tfrac{\varepsilon}{2} \Big) - \Big( L(f,[a,b]) - \tfrac{\varepsilon}{2} \Big) = \varepsilon . \end{aligned} \end{equation*}

Write \(L(h,P)\) and \(U(h,P)\) for \(L(h,P,[a,b])\) and \(U(h,P,[a,b])\). For every nonempty \(A \subseteq [a,b]\),

\begin{equation*} \inf_A f + \inf_A g \le \inf_A (f+g) \le \sup_A (f+g) \le \sup_A f + \sup_A g , \end{equation*}

since \(\inf_A f + \inf_A g\) is a lower bound and \(\sup_A f + \sup_A g\) an upper bound for the values of \(f+g\) on \(A\). Applying this with \(A = [x_{j-1},x_j]\) for each subinterval of a partition \(P\), multiplying by \(x_j - x_{j-1} > 0\) and summing gives, by 1.3,

\begin{equation*} L(f,P) + L(g,P) \le L(f+g,P) \le U(f+g,P) \le U(f,P) + U(g,P) . \tag{$\ast$} \end{equation*}

Let \(\varepsilon > 0\). By Exercise 3 in this section there are partitions \(P_1, P_2\) of \([a,b]\) with \(U(f,P_1) - L(f,P_1) < \frac{\varepsilon}{2}\) and \(U(g,P_2) - L(g,P_2) < \frac{\varepsilon}{2}\); let \(P\) merge their lists, so by 1.5 both estimates persist at \(P\). Then \((\ast)\) gives

\begin{equation*} U(f+g,P) - L(f+g,P) \le \big( U(f,P) - L(f,P) \big) + \big( U(g,P) - L(g,P) \big) < \varepsilon , \end{equation*}

so \(f+g\) is Riemann integrable by Exercise 3 again. For its value, keep that \(P\) and use \((\ast)\) together with \(L(f,P) \le L(f,[a,b]) = \int_a^b f\) and \(U(f,P) \ge U(f,[a,b]) = \int_a^b f\) from 1.7 and 1.9 (and the same for \(g\)):

\begin{equation*} \begin{aligned} \int_a^b (f+g) \le U(f+g,P) &\le U(f,P) + U(g,P) \\ &< \Big( L(f,P) + \tfrac{\varepsilon}{2} \Big) + \Big( L(g,P) + \tfrac{\varepsilon}{2} \Big) \\ &\le \int_a^b f + \int_a^b g + \varepsilon , \end{aligned} \end{equation*}

and symmetrically \(\int_a^b (f+g) \ge L(f+g,P) \ge L(f,P) + L(g,P) > \int_a^b f + \int_a^b g - \varepsilon\). As \(\varepsilon > 0\) was arbitrary, \(\int_a^b (f+g) = \int_a^b f + \int_a^b g\).

Negation swaps infima and suprema: for every nonempty \(A \subseteq [a,b]\),

\begin{equation*} \inf_A (-f) = -\sup_A f , \qquad \sup_A (-f) = -\inf_A f , \end{equation*}

because \(\alpha\) is an upper bound for \(\{f(x) : x \in A\}\) exactly when \(-\alpha\) is a lower bound for \(\{-f(x) : x \in A\}\), and conversely. Applying this with \(A = [x_{j-1},x_j]\) on each subinterval of a partition \(P\) of \([a,b]\) gives, by 1.3,

\begin{equation*} L(-f, P, [a,b]) = -\,U(f, P, [a,b]) , \qquad U(-f, P, [a,b]) = -\,L(f, P, [a,b]) . \end{equation*}

The same swap, applied now to the set of upper Riemann sums of \(f\) and to the set of lower ones, turns 1.7 into

\begin{equation*} L(-f, [a,b]) = -\,U(f, [a,b]) , \qquad U(-f, [a,b]) = -\,L(f, [a,b]) . \end{equation*}

Since \(f\) is Riemann integrable, \(L(f,[a,b]) = U(f,[a,b]) = \int_a^b f\), so both left sides equal \(-\int_a^b f\). By 1.9, \(-f\) is Riemann integrable on \([a,b]\) with \(\int_a^b (-f) = -\int_a^b f\).

By Exercise 4 in this section it suffices to prove that \(h = g - f\), which vanishes off a finite set, is Riemann integrable with \(\int_a^b h = 0\); then \(g = f + h\) is Riemann integrable with \(\int_a^b g = \int_a^b f\).

Let \(c_1 < \cdots < c_m\) be the points where \(h \ne 0\) (if there are none, every Riemann sum of \(h\) is \(0\) and we are done), and put \(M = \max\{|h(c_1)|, \ldots, |h(c_m)|\} > 0\), so \(|h| \le M\) on \([a,b]\). For \(\delta > 0\) let \(P_\delta\) be the partition of \([a,b]\) whose points are the distinct elements of

\begin{equation*} \{a, b\} \cup \big( \{ c_i - \delta,\; c_i + \delta : i = 1, \ldots, m \} \cap (a,b) \big) \end{equation*}

in increasing order. If a subinterval \([u,w]\) of \(P_\delta\) contains some \(c_i\), then \([u,w] \subseteq [c_i - \delta, c_i + \delta]\): otherwise \(c_i - \delta\) or \(c_i + \delta\) would lie in \((u,w) \subseteq (a,b)\) and hence be a point of \(P_\delta\) strictly between the consecutive points \(u\) and \(w\).

On a subinterval containing no \(c_i\) we have \(h \equiv 0\), so it contributes \(0\) to both \(L(h,P_\delta,[a,b])\) and \(U(h,P_\delta,[a,b])\). For each \(i\) at most two subintervals contain \(c_i\), and they have disjoint interiors and lie in \([c_i - \delta, c_i + \delta]\), so their lengths sum to at most \(2\delta\); hence the subintervals meeting \(\{c_1, \ldots, c_m\}\) have total length at most \(2m\delta\), and there \(|\sup h| \le M\), \(|\inf h| \le M\). Therefore

\begin{equation*} |L(h, P_\delta, [a,b])| \le 2m\delta M \quad\text{and}\quad |U(h, P_\delta, [a,b])| \le 2m\delta M . \end{equation*}

Given \(\varepsilon > 0\), take \(\delta = \frac{\varepsilon}{2mM}\); then 1.7 gives \(U(h,[a,b]) \le \varepsilon\) and \(L(h,[a,b]) \ge -\varepsilon\). As \(\varepsilon > 0\) was arbitrary and \(L(h,[a,b]) \le U(h,[a,b])\) by 1.8, both are \(0\). By 1.9, \(h\) is Riemann integrable with \(\int_a^b h = 0\).

Put \(M = \sup_{[a,b]} |f| < \infty\), and let \(P_n\) have the points \(x_j^{(n)} = a + \frac{j(b-a)}{2^n}\) for \(j = 0, \ldots, 2^n\), so each of its subintervals has length \(\frac{b-a}{2^n}\).

Refinement estimate: if every subinterval of a partition \(Q\) has length at most \(\ell\) and \(Q^{\prime}\) is obtained by adjoining \(p\) points to the list defining \(Q\), then

\begin{equation*} L(f, Q^{\prime}, [a,b]) \le L(f, Q, [a,b]) + 2pM\ell . \end{equation*}

For \(p = 1\) the adjoined point \(v\) either lies in \(Q\) already, when the two sums agree, or lies interior to one subinterval \([u,w]\) of \(Q\), when all other terms agree and, using \(|\inf_A f| \le M\),

\begin{equation*} (v-u) \inf_{[u,v]} f + (w-v) \inf_{[v,w]} f - (w-u) \inf_{[u,w]} f \le 2M(w-u) \le 2M\ell . \end{equation*}

For general \(p\), adjoin the points one at a time; each intermediate partition refines \(Q\), so its subintervals still have length at most \(\ell\).

Since \(x_j^{(n)} = x_{2j}^{(n+1)}\), the list defining \(P_n\) is a sublist of that defining \(P_{n+1}\), so 1.5 makes \(\big( L(f,P_n,[a,b]) \big)_{n}\) increasing while 1.7 bounds it above by \(L(f,[a,b])\); hence it converges, with limit at most \(L(f,[a,b])\).

For the reverse inequality let \(\varepsilon > 0\) and use 1.7 to choose a partition \(P\), say \(a = y_0 < \cdots < y_k = b\), with \(L(f,P,[a,b]) > L(f,[a,b]) - \varepsilon\). Let \(Q_n\) merge the lists of \(P_n\) and \(P\); this adjoins at most \(k-1\) points to \(P_n\), whose subintervals have length \(\frac{b-a}{2^n}\), so the refinement estimate together with \(L(f,P,[a,b]) \le L(f,Q_n,[a,b])\) from 1.5 gives

\begin{equation*} L(f, P, [a,b]) \le L(f, P_n, [a,b]) + \frac{2(k-1)M(b-a)}{2^n} . \end{equation*}

Letting \(n \to \infty\) gives \(\lim_{n \to \infty} L(f,P_n,[a,b]) \ge L(f,P,[a,b]) > L(f,[a,b]) - \varepsilon\); as \(\varepsilon > 0\) was arbitrary, \(\lim_{n \to \infty} L(f,P_n,[a,b]) = L(f,[a,b])\).

For the upper integral apply this to the bounded function \(-f\), then use the identities \(L(-f,P,[a,b]) = -U(f,P,[a,b])\) and \(L(-f,[a,b]) = -U(f,[a,b])\) from Exercise 5 in this section, whose proofs use only boundedness:

\begin{equation*} U(f, [a,b]) = -L(-f, [a,b]) = -\lim_{n \to \infty} L(-f, P_n, [a,b]) = \lim_{n \to \infty} U(f, P_n, [a,b]) . \end{equation*}

Squeeze the sum between the Riemann sums of the equally spaced partition. Let \(P_n\) have points \(x_j = a + \frac{j(b-a)}{n}\) for \(j = 0, \ldots, n\), and write \(R_n = \frac{b-a}{n} \sum_{j=1}^{n} f(x_j)\). Since \(x_j \in [x_{j-1},x_j]\), multiplying \(\inf_{[x_{j-1},x_j]} f \le f(x_j) \le \sup_{[x_{j-1},x_j]} f\) by \(x_j - x_{j-1} = \frac{b-a}{n} > 0\) and summing gives

\begin{equation*} L(f, P_n, [a,b]) \le R_n \le U(f, P_n, [a,b]) . \end{equation*}

Both outer terms converge to \(\int_a^b f\). Put \(M = \sup_{[a,b]} |f| < \infty\) (finite since \(f\) is Riemann integrable, hence bounded), let \(\varepsilon > 0\), and use 1.7 to choose a partition \(P\), say \(a = y_0 < \cdots < y_k = b\), with \(L(f,P,[a,b]) > L(f,[a,b]) - \varepsilon\). Merging \(P\) into \(P_n\) adjoins at most \(k-1\) points to a partition of mesh \(\frac{b-a}{n}\), so the refinement estimate proved in the solution to Exercise 7 in this section, together with 1.5, gives

\begin{equation*} L(f,[a,b]) - \varepsilon < L(f, P, [a,b]) \le L(f, P_n, [a,b]) + \frac{2(k-1)M(b-a)}{n} , \end{equation*}

while \(L(f,P_n,[a,b]) \le L(f,[a,b])\) by 1.7. Hence \(L(f,P_n,[a,b])\) lies within \(2\varepsilon\) of \(L(f,[a,b])\) for all large \(n\), and \(\varepsilon > 0\) was arbitrary, so \(L(f,P_n,[a,b]) \to L(f,[a,b])\). Applying this to \(-f\) and using \(L(-f,P,[a,b]) = -U(f,P,[a,b])\) and \(L(-f,[a,b]) = -U(f,[a,b])\) from Exercise 5 in this section gives \(U(f,P_n,[a,b]) \to U(f,[a,b])\).

Since \(f\) is Riemann integrable, \(L(f,[a,b]) = U(f,[a,b]) = \int_a^b f\) by 1.9, so the squeeze yields \(R_n \to \int_a^b f\).

Cut a good partition of \([a,b]\) at \(c\) and \(d\) and keep the middle piece; Exercise 3 in this section does the rest.

Let \(\varepsilon > 0\). By Exercise 3 there is a partition \(P\) of \([a,b]\) with \(U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon\). Let \(P^{\prime}\), say \(a = x_0 < \cdots < x_n = b\), be obtained by adjoining \(c\) and \(d\) to the list defining \(P\); since that list is a sublist of the new one, 1.5 gives

\begin{equation*} U(f,P^{\prime},[a,b]) - L(f,P^{\prime},[a,b]) \le U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon . \end{equation*}

Choose \(p < q\) with \(x_p = c\) and \(x_q = d\), and let \(Q\) be the partition \(x_p, x_{p+1}, \ldots, x_q\) of \([c,d]\). Each subinterval \([x_{j-1},x_j]\) with \(p < j \le q\) lies inside \([c,d]\), so restricting \(f\) to \([c,d]\) changes none of the suprema and infima below, and every omitted term is nonnegative:

\begin{equation*} \begin{aligned} U(f,Q,[c,d]) - L(f,Q,[c,d]) &= \sum_{j=p+1}^{q} (x_j - x_{j-1}) \Big( \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \Big) \\ &\le \sum_{j=1}^{n} (x_j - x_{j-1}) \Big( \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \Big) \\ &= U(f,P^{\prime},[a,b]) - L(f,P^{\prime},[a,b]) < \varepsilon . \end{aligned} \end{equation*}

By Exercise 3 again, \(f\) is Riemann integrable on \([c,d]\).

Everything comes from the splitting identity. If \(P\), say \(x_0 < \cdots < x_n\), is a partition of \([a,b]\) with \(c = x_p\) for some \(0 < p < n\), let \(P_1 : x_0, \ldots, x_p\) and \(P_2 : x_p, \ldots, x_n\) be the induced partitions of \([a,c]\) and \([c,b]\). Each subinterval of \(P\) lies wholly in \([a,c]\) or wholly in \([c,b]\), so restricting \(f\) changes none of the infima and suprema, and splitting the sums of 1.3 at \(j = p\) gives

\begin{equation*} \begin{aligned} L(f,P,[a,b]) &= L(f,P_1,[a,c]) + L(f,P_2,[c,b]) , \\ U(f,P,[a,b]) &= U(f,P_1,[a,c]) + U(f,P_2,[c,b]) . \end{aligned} \tag{$\ast$} \end{equation*}

Conversely, any partition \(P_1\) of \([a,c]\) and \(P_2\) of \([c,b]\) concatenate (at the shared endpoint \(c\)) to such a \(P\), and \((\ast)\) again holds.

Lower integrals add. For \(\ge\): given \(P_1, P_2\) with concatenation \(P\), \((\ast)\) and 1.7 give \(L(f,P_1,[a,c]) + L(f,P_2,[c,b]) \le L(f,[a,b])\); take suprema over \(P_1\) and over \(P_2\), which vary independently. For \(\le\): given any partition \(P\) of \([a,b]\), adjoin \(c\) to get \(P^{\prime}\) and apply 1.5, then \((\ast)\), then 1.7,

\begin{equation*} L(f,P,[a,b]) \le L(f,P^{\prime},[a,b]) = L(f,P^{\prime}_1,[a,c]) + L(f,P^{\prime}_2,[c,b]) \le L(f,[a,c]) + L(f,[c,b]) , \end{equation*}

and take the supremum over \(P\). The mirror-image argument gives the same for upper integrals, so

\begin{equation*} L(f,[a,b]) = L(f,[a,c]) + L(f,[c,b]) , \qquad U(f,[a,b]) = U(f,[a,c]) + U(f,[c,b]) . \end{equation*}

Subtracting,

\begin{equation*} U(f,[a,b]) - L(f,[a,b]) = \Big( U(f,[a,c]) - L(f,[a,c]) \Big) +\Big( U(f,[c,b]) - L(f,[c,b]) \Big) , \end{equation*}

where by 1.8 all three differences are nonnegative; so the left side vanishes exactly when both terms on the right do, which by 1.9 is precisely the asserted equivalence. Granting it, 1.9 and the additivity of the lower integral give

\begin{equation*} \int_a^b f = L(f,[a,b]) = L(f,[a,c]) + L(f,[c,b]) = \int_a^c f + \int_c^b f . \end{equation*}

\(F\) is Lipschitz with constant \(M = \sup_{[a,b]} |f| < \infty\), hence uniformly continuous on \([a,b]\).

Let \(a \le s < t \le b\). By Exercise 9 in this section, \(f\) is Riemann integrable on \([a,t]\) and on \([s,t]\), so \(F\) is well defined and

\begin{equation*} F(t) - F(s) = \int_s^t f : \end{equation*}

for \(s = a\) this is \(F(t) = \int_a^t f\) with \(F(a) = 0\), and for \(s > a\) it is Exercise 10 in this section applied on \([a,t]\) at the interior point \(s\). Since \(-M \le \inf_{[s,t]} f\) and \(\sup_{[s,t]} f \le M\), applying 1.13 on \([s,t]\) gives

\begin{equation*} |F(t) - F(s)| = \left| \int_s^t f \right| \le M(t-s) . \end{equation*}

Thus \(|F(t) - F(s)| \le M|t-s|\) for all \(s,t \in [a,b]\). Given \(\varepsilon > 0\), any \(s,t\) with \(|s-t| < \frac{\varepsilon}{M+1}\) then satisfy \(|F(t) - F(s)| \le \frac{M}{M+1}\varepsilon < \varepsilon\), so \(F\) is continuous on \([a,b]\).

Taking absolute values cannot increase oscillation: for nonempty \(A \subseteq [a,b]\) and \(x, y \in A\),

\begin{equation*} |f(x)| - |f(y)| \le |f(x) - f(y)| = \max\{ f(x)-f(y),\, f(y)-f(x) \} \le \sup_A f - \inf_A f , \end{equation*}

and taking the supremum over \(x\) and over \(y\) gives \(\sup_A |f| - \inf_A |f| \le \sup_A f - \inf_A f\). Applying this with \(A = [x_{j-1},x_j]\) on each subinterval of a partition \(P\) of \([a,b]\), multiplying by \(x_j - x_{j-1} > 0\) and summing gives, by 1.3,

\begin{equation*} U(|f|,P,[a,b]) - L(|f|,P,[a,b]) \le U(f,P,[a,b]) - L(f,P,[a,b]) . \end{equation*}

Given \(\varepsilon > 0\), Exercise 3 in this section supplies \(P\) making the right side less than \(\varepsilon\), hence the left side too; by Exercise 3 again, \(|f|\) is Riemann integrable.

For the inequality, note first that Riemann integration is monotone: if \(g \le h\) are Riemann integrable on \([a,b]\), then \(\inf_{[x_{j-1},x_j]} g \le \inf_{[x_{j-1},x_j]} h\) on every subinterval, so \(L(g,P,[a,b]) \le L(h,P,[a,b])\) for every \(P\), and taking suprema, by 1.7 and 1.9,

\begin{equation*} \int_a^b g = L(g,[a,b]) \le L(h,[a,b]) = \int_a^b h . \end{equation*}

By Exercise 5 in this section \(-|f|\) is Riemann integrable with \(\int_a^b (-|f|) = -\int_a^b |f|\), so applying monotonicity to \(-|f| \le f\) and to \(f \le |f|\),

\begin{equation*} -\int_a^b |f| \le \int_a^b f \le \int_a^b |f| , \quad\text{that is,}\quad \left| \int_a^b f \right| \le \int_a^b |f| . \end{equation*}

On an equally spaced partition the oscillation sum telescopes. Monotonicity gives \(f(a) \le f \le f(b)\), so \(f\) is bounded. Fix \(n \in \mathbb{Z}^+\) and let \(P_n\) be the partition with points \(x_j = a + \frac{j(b-a)}{n}\), \(j = 0, \ldots, n\). On \([x_{j-1},x_j]\) monotonicity gives \(f(x_{j-1}) \le f \le f(x_j)\), and both bounds are attained, so

\begin{equation*} \inf_{[x_{j-1},x_j]} f = f(x_{j-1}) \qquad\text{and}\qquad \sup_{[x_{j-1},x_j]} f = f(x_j) . \end{equation*}

Hence, by 1.3 and telescoping,

\begin{equation*} U(f,P_n,[a,b]) - L(f,P_n,[a,b]) = \frac{b-a}{n} \sum_{j=1}^{n} \big( f(x_j) - f(x_{j-1}) \big) = \frac{(b-a)\big( f(b) - f(a) \big)}{n} . \end{equation*}

Given \(\varepsilon > 0\), the Archimedean property supplies \(n\) making the right side less than \(\varepsilon\), so \(f\) is Riemann integrable on \([a,b]\) by Exercise 3 in this section.

Put \(\varepsilon_n = \sup_{[a,b]} |f_n - f|\); uniform convergence says exactly that \(\varepsilon_n \to 0\). Choosing \(N\) with \(\varepsilon_N \le 1\) and using that \(f_N\) is bounded (being Riemann integrable) gives \(|f| \le 1 + \sup_{[a,b]}|f_N| < \infty\), so \(f\) is bounded and every \(\varepsilon_n\) is finite.

On a nonempty \(A \subseteq [a,b]\) we have \(f_n - \varepsilon_n \le f \le f_n + \varepsilon_n\), so \(\sup_A f \le \sup_A f_n + \varepsilon_n\) and \(\inf_A f \ge \inf_A f_n - \varepsilon_n\); subtracting and applying the result with \(A = [x_{j-1},x_j]\) on the subintervals of a partition \(P\) of \([a,b]\), whose lengths total \(b-a\), gives by 1.3

\begin{equation*} U(f,P,[a,b]) - L(f,P,[a,b]) \le \Big( U(f_n,P,[a,b]) - L(f_n,P,[a,b]) \Big) + 2(b-a)\varepsilon_n . \tag{$\star$} \end{equation*}

Let \(\eta > 0\). Choose \(n\) with \(\varepsilon_n < \frac{\eta}{4(b-a)}\), then use Exercise 3 in this section to choose a partition \(P\) with \(U(f_n,P,[a,b]) - L(f_n,P,[a,b]) < \frac{\eta}{2}\); now \((\star)\) gives \(U(f,P,[a,b]) - L(f,P,[a,b]) < \eta\), so \(f\) is Riemann integrable by Exercise 3 again.

Hence Exercises 4, 5 and 12 in this section make \(f - f_n\) and \(|f - f_n|\) Riemann integrable, and

\begin{equation*} \left| \int_a^b f - \int_a^b f_n \right| = \left| \int_a^b (f - f_n) \right| \le \int_a^b |f - f_n| \le (b-a)\varepsilon_n , \end{equation*}

the last step by 1.13 applied to \(|f - f_n|\). Since \(\varepsilon_n \to 0\), \(\int_a^b f = \lim_{n \to \infty} \int_a^b f_n\).

\(f\) is Riemann integrable with \(\int_0^1 f = 0\).

Every subinterval of positive length contains an irrational, where \(f\) vanishes, and \(f \ge 0\); so every lower Riemann sum of \(f\) is \(0\) and \(L(f,[0,1]) = 0\) by 1.7.

For the upper integral, note \(0 \le f \le 1\) and that for each \(N \in \mathbb{Z}^+\) the set \(A_N = \{ a \in [0,1] : f(a) \ge \frac1N \}\) is finite: such an \(a\) is rational, its least denominator \(n\) satisfies \(\frac1n = f(a) \ge \frac1N\), and \(a = \frac{m}{n} \in [0,1]\) forces \(0 \le m \le n\), so

\begin{equation*} A_N \subseteq \Big\{ \tfrac{m}{n} : n \in \{1,\ldots,N\},\ m \in \{0,1,\ldots,n\} \Big\} . \end{equation*}

Write \(K\) for the number of its elements (\(K \ge 1\), as \(0 \in A_N\)); off \(A_N\) we have \(f < \frac1N\).

Let \(\varepsilon > 0\). Choose \(N\) with \(\frac1N < \frac{\varepsilon}{2}\), then \(n > \frac{4K}{\varepsilon}\), and let \(P_n\) split \([0,1]\) into the \(n\) equal subintervals \(I_j\) of length \(\frac1n\). Each point of \(A_N\) lies in at most two of the \(I_j\), so at most \(2K\) indices \(j\) have \(I_j \cap A_N \ne \emptyset\); for those \(\sup_{I_j} f \le 1\), and for the others \(\sup_{I_j} f \le \frac1N\). Since the lengths of the others total at most \(1\),

\begin{equation*} U(f, P_n, [0,1]) \le \frac{2K}{n} + \frac{1}{N} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon . \end{equation*}

So \(U(f,[0,1]) < \varepsilon\) by 1.7 for every \(\varepsilon > 0\), giving \(U(f,[0,1]) \le 0\); with \(L(f,[0,1]) = 0\) and 1.8 this forces \(L(f,[0,1]) = U(f,[0,1]) = 0\), so 1.9 gives the claim.

Everything follows from the identity

\begin{equation*} U(f,[a,b]) = -L(-f,[a,b]) . \tag{$\ast$} \end{equation*}

For a nonempty \(E \subseteq [a,b]\) we have \(\sup_E f = -\inf_E (-f)\), since \(\alpha\) is an upper bound for the values of \(f\) on \(E\) exactly when \(-\alpha\) is a lower bound for the values of \(-f\) on \(E\). Applying this on each subinterval of a partition \(P\) of \([a,b]\) gives, by 1.3,

\begin{equation*} U(f, P, [a,b]) = -L(-f, P, [a,b]) , \end{equation*}

and the same swap, applied now to the (nonempty, bounded) set of these numbers, turns 1.7 into

\begin{equation*} U(f,[a,b]) = \inf_{P} U(f,P,[a,b]) = -\sup_{P} L(-f,P,[a,b]) = -L(-f,[a,b]) , \end{equation*}

which is \((\ast)\). By 1.9, \(f\) is Riemann integrable if and only if \(L(f,[a,b]) = U(f,[a,b])\), which by \((\ast)\) reads \(L(f,[a,b]) = -L(-f,[a,b])\), that is, \(L(-f,[a,b]) = -L(f,[a,b])\).

Both inequalities come from the single-partition versions plus a common refinement. For nonempty \(E \subseteq [a,b]\), \(\inf_E f + \inf_E g\) is a lower bound and \(\sup_E f + \sup_E g\) an upper bound for the values of \(f+g\) on \(E\), so

\begin{equation*} \inf_E f + \inf_E g \le \inf_E (f+g) , \qquad \sup_E (f+g) \le \sup_E f + \sup_E g . \end{equation*}

Applying these with \(E = [x_{j-1},x_j]\) on the subintervals of a partition \(P\) of \([a,b]\), multiplying by \(x_j - x_{j-1} > 0\) and summing gives, by 1.3,

\begin{equation*} L(f,P,[a,b]) + L(g,P,[a,b]) \le L(f+g,P,[a,b]) , \tag{1} \end{equation*}

\begin{equation*} U(f+g,P,[a,b]) \le U(f,P,[a,b]) + U(g,P,[a,b]) . \tag{2} \end{equation*}

Let \(\varepsilon > 0\). By 1.7 there are partitions \(P_1, P_2\) of \([a,b]\) with \(L(f,P_1,[a,b]) > L(f,[a,b]) - \frac{\varepsilon}{2}\) and \(L(g,P_2,[a,b]) > L(g,[a,b]) - \frac{\varepsilon}{2}\); let \(P\) merge their lists, so that 1.5 gives \(L(f,P,[a,b]) \ge L(f,P_1,[a,b])\) and \(L(g,P,[a,b]) \ge L(g,P_2,[a,b])\). (The refinement is what lets the two suprema be approached along one partition.) Then, using 1.7 and (1),

\begin{equation*} \begin{aligned} L(f+g,[a,b]) &\ge L(f+g,P,[a,b]) \ge L(f,P,[a,b]) + L(g,P,[a,b]) \\ &\ge L(f,P_1,[a,b]) + L(g,P_2,[a,b]) > L(f,[a,b]) + L(g,[a,b]) - \varepsilon , \end{aligned} \end{equation*}

and \(\varepsilon > 0\) was arbitrary. The mirror-image argument, with partitions \(Q_1, Q_2\) chosen by 1.7 so that \(U(f,Q_1,[a,b]) < U(f,[a,b]) + \frac{\varepsilon}{2}\) and \(U(g,Q_2,[a,b]) < U(g,[a,b]) + \frac{\varepsilon}{2}\), their merge \(Q\), 1.5 and (2), gives \(U(f+g,[a,b]) \le U(f,[a,b]) + U(g,[a,b])\).

Take \(f\) to be the function of 1.14 and \(g = 1 - f\):

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x \text{ is rational,} \\ 0 & \text{if } x \text{ is irrational,} \end{cases} \qquad g(x) = \begin{cases} 0 & \text{if } x \text{ is rational,} \\ 1 & \text{if } x \text{ is irrational.} \end{cases} \end{equation*}

Every \([c,d] \subseteq [0,1]\) with \(c < d\) contains a rational and an irrational, so \(\inf_{[c,d]} f = \inf_{[c,d]} g = 0\) and \(\sup_{[c,d]} f = \sup_{[c,d]} g = 1\). Since the subinterval lengths of a partition of \([0,1]\) total \(1\), every lower sum of \(f\) and of \(g\) is \(0\) and every upper sum is \(1\), so 1.7 gives

\begin{equation*} L(f,[0,1]) = L(g,[0,1]) = 0 , \qquad U(f,[0,1]) = U(g,[0,1]) = 1 . \end{equation*}

But \(f + g\) is the constant function \(1\), so all its lower and upper sums equal \(1\) and \(L(f+g,[0,1]) = U(f+g,[0,1]) = 1\). Hence

\begin{equation*} \begin{aligned} L(f,[0,1]) + L(g,[0,1]) &= 0 < 1 = L(f+g,[0,1]) , \\ U(f+g,[0,1]) = 1 &< 2 = U(f,[0,1]) + U(g,[0,1]) . \end{aligned} \end{equation*}

Take \(f = 0\) and let \(f_k\) be the triangular spike of height \(2k\) and area \(1\) on \(\bigl[0, \frac1k\bigr]\):

\begin{equation*} f_k(x) = \begin{cases} 4k^2 x & \text{if } 0 \le x \le \frac{1}{2k}, \\[2pt] 4k^2 \bigl( \tfrac{1}{k} - x \bigr) & \text{if } \frac{1}{2k} \le x \le \frac{1}{k}, \\[2pt] 0 & \text{if } \frac{1}{k} \le x \le 1 . \end{cases} \end{equation*}

The formulas agree where the cases overlap, both giving \(2k\) at \(x = \frac{1}{2k}\) and \(0\) at \(x = \frac1k\), so each \(f_k\) is continuous and hence Riemann integrable by 1.11, as is the constant \(f\).

Pointwise convergence: \(f_k(0) = 0\) for every \(k\), and for \(x > 0\) we have \(f_k(x) = 0\) as soon as \(\frac1k < x\), which happens for all large \(k\) by the Archimedean property. So \(f_k(x) \to 0 = f(x)\) for every \(x \in [0,1]\).

The integrals: \(\int_0^1 f = 0\) by 1.13. For \(f_k\), split \([0,1]\) at \(\frac1k\) and \(\bigl[0,\frac1k\bigr]\) at \(\frac{1}{2k}\) using Exercises 9 and 10 in Section 1A (for \(k = 1\) only the split at \(\frac12\) is needed), and integrate the two linear pieces by the fundamental theorem of calculus, with antiderivatives \(2k^2x^2\) and \(-2k^2\bigl(\frac1k - x\bigr)^2\):

\begin{equation*} \int_0^1 f_k = 2k^2 \Bigl( \frac{1}{2k} \Bigr)^{\!2} + \Bigl[ -2k^2 \Bigl( \tfrac1k - x \Bigr)^{\!2} \Bigr]_{x=1/(2k)}^{x=1/k} + 0 = \tfrac12 + \tfrac12 = 1 . \end{equation*}

Hence \(\int_0^1 f = 0 \ne 1 = \lim_{k \to \infty} \int_0^1 f_k\).

Two inequalities. Since \(A \subseteq A \cup B\), order preservation (2.5) gives \(|A| \le |A \cup B|\). In the other direction, countable subadditivity (2.8) applied to the sequence \(A, B, \emptyset, \emptyset, \ldots\), whose union is \(A \cup B\), gives

\begin{equation*} |A \cup B| \le |A| + |B| = |A| + 0 = |A| , \end{equation*}

using \(|\emptyset| = 0\) from 2.3. Hence \(|A \cup B| = |A|\).

(i) \(t = 0\). Then \(tA\) is \(\emptyset\) or \(\{0\}\), so \(|tA| = 0\) by 2.3, while \(|t|\,|A| = 0 \cdot |A| = 0\) by the stated convention.

(ii) \(t \ne 0\). Dilation carries open intervals to open intervals with

\begin{equation*} \ell(tI) = |t|\,\ell(I) , \end{equation*}

since \(t(\alpha,\beta)\) is \((t\alpha, t\beta)\) or \((t\beta, t\alpha)\) according to the sign of \(t\), of length \(|t|(\beta - \alpha)\), while \(\emptyset\) maps to \(\emptyset\) and an unbounded open interval maps to an unbounded one, both sides then being \(\infty = |t| \cdot \infty\). So if \(I_1, I_2, \ldots\) are open intervals with \(A \subseteq \bigcup_{k=1}^{\infty} I_k\), then \(tA \subseteq \bigcup_{k=1}^{\infty} tI_k\) (as \(x \mapsto tx\) is a bijection of \(\mathbb{R}\)), and 2.2 gives

\begin{equation*} |tA| \le \sum_{k=1}^{\infty} \ell(tI_k) = |t| \sum_{k=1}^{\infty} \ell(I_k) . \end{equation*}

Taking the infimum over all such covers yields \(|tA| \le |t|\,|A|\). Applying this inequality with \(\frac1t\) in place of \(t\) and \(tA\) in place of \(A\), and using \(\frac1t (tA) = A\), gives \(|A| \le \frac{1}{|t|} |tA|\), that is, \(|t|\,|A| \le |tA|\). Hence \(|tA| = |t|\,|A|\).

Since \(B \subseteq (B \setminus A) \cup A\), order preservation (2.5) and subadditivity (2.8, applied to \(B \setminus A,\, A,\, \emptyset,\, \emptyset, \ldots\)) give

\begin{equation*} |B| \le \bigl| (B \setminus A) \cup A \bigr| \le |B \setminus A| + |A| . \end{equation*}

If \(|B \setminus A| = \infty\) the desired inequality is trivial. Otherwise both terms on the right are finite, since \(|A| < \infty\) by hypothesis, so subtracting \(|A|\) is legitimate and gives \(|B| - |A| \le |B \setminus A|\).

Two open covers, chosen so that a finite subcover collapses to a single set.

(i) \(F\) is bounded. The intervals \((-k,k)\), \(k \in \mathbb{Z}^+\), are open and their union is \(\mathbb{R} \supseteq F\), so they form an open cover of \(F\) in the sense of 2.10. A finite subcover gives \(k_1, \ldots, k_n\) with \(F \subseteq (-k_1,k_1) \cup \cdots \cup (-k_n,k_n) = (-K,K)\), where \(K = \max\{k_1,\ldots,k_n\}\), the union collapsing because these intervals increase with \(k\). (An empty subcover would make \(F = \emptyset\), which is bounded.) So \(|x| < K\) for all \(x \in F\).

(ii) \(F\) is closed. Let \(b \in \mathbb{R} \setminus F\) and put

\begin{equation*} G_k = \Bigl( -\infty,\, b - \tfrac1k \Bigr) \cup \Bigl( b + \tfrac1k,\, \infty \Bigr) = \Bigl\{ x \in \mathbb{R} : |x - b| > \tfrac1k \Bigr\} , \end{equation*}

an open set, with \(G_1 \subseteq G_2 \subseteq \cdots\) and \(\bigcup_{k=1}^{\infty} G_k = \mathbb{R} \setminus \{b\} \supseteq F\). So \(\{G_k\}\) is an open cover of \(F\), and a finite subcover collapses, as in (i), to \(F \subseteq G_K\) for some \(K\) (an empty subcover again forcing \(F = \emptyset\), which is closed). Thus

\begin{equation*} \Bigl( b - \tfrac1K,\, b + \tfrac1K \Bigr) \subseteq \mathbb{R} \setminus F . \end{equation*}

Hence \(\mathbb{R} \setminus F\) is open, so \(F\) is closed.

Apply Heine–Borel to a bounded \(F_0 \in \mathcal{A}\), covered by the complements of the members of \(\mathcal{A}\).

If \(F_0 = \emptyset\), take \(n = 1\) and \(F_1 = F_0\). So assume \(F_0 \ne \emptyset\); being a member of \(\mathcal{A}\) it is closed, and it is bounded by choice. Put \(\mathcal{C} = \{ \mathbb{R} \setminus F : F \in \mathcal{A} \}\), a collection of open sets. It covers \(F_0\): each \(x \in F_0\) fails to lie in some \(F \in \mathcal{A}\), since \(\bigcap_{F \in \mathcal{A}} F = \emptyset\), and then \(x \in \mathbb{R} \setminus F \in \mathcal{C}\).

By the Heine–Borel Theorem (2.12) applied to the closed bounded set \(F_0\), there are \(F_1, \ldots, F_m \in \mathcal{A}\) with

\begin{equation*} F_0 \subseteq (\mathbb{R} \setminus F_1) \cup \cdots \cup (\mathbb{R} \setminus F_m) = \mathbb{R} \setminus (F_1 \cap \cdots \cap F_m) \end{equation*}

by De Morgan (the subcover is nonempty because \(F_0 \ne \emptyset\)). That says exactly \(F_0 \cap F_1 \cap \cdots \cap F_m = \emptyset\), so the \(m+1\) sets \(F_0, F_1, \ldots, F_m\) of \(\mathcal{A}\) are as required.

Each of the three sets becomes \([a,b]\) after adjoining a finite set, which has outer measure \(0\) by 2.3, so Exercise 1 in this section and \(|[a,b]| = b - a\) from 2.14 give

\begin{equation*} \begin{aligned} |(a,b)| &= \bigl| (a,b) \cup \{a,b\} \bigr| = |[a,b]| = b - a , \\ |[a,b)| &= \bigl| [a,b) \cup \{b\} \bigr| = |[a,b]| = b - a , \\ |(a,b]| &= \bigl| (a,b] \cup \{a\} \bigr| = |[a,b]| = b - a . \end{aligned} \end{equation*}

Method (2): for \(0 < \varepsilon < \frac{b-a}{2}\) we have \([a+\varepsilon, b-\varepsilon] \subseteq (a,b) \subseteq [a,b]\), so 2.5 and 2.14 give

\begin{equation*} b - a - 2\varepsilon = \bigl| [a+\varepsilon, b-\varepsilon] \bigr| \le |(a,b)| \le |[a,b]| = b - a , \end{equation*}

and letting \(\varepsilon \to 0\) gives \(|(a,b)| = b-a\); then 2.5 squeezes \(|[a,b)|\) and \(|(a,b]|\) between \(|(a,b)|\) and \(|[a,b]|\).

Write \(E = (a,b) \cup (c,d)\); the inequality \(|E| \le (b-a)+(d-c)\) always holds, by subadditivity (2.8) and \(|(a,b)| = b-a\), \(|(c,d)| = d-c\) from Exercise 6 of this section. So only the equality case is at issue. Since \(x\) lies in both intervals exactly when \(u < x < v\), where

\begin{equation*} u = \max\{a,c\}, \qquad v = \min\{b,d\}, \end{equation*}

we have \((a,b) \cap (c,d) = (u,v)\), nonempty precisely when \(u < v\), i.e. (given \(a<b\) and \(c<d\)) precisely when \(c < b\) and \(a < d\).

(i) Disjoint case. Then \(b \le c\) or \(d \le a\); interchanging the two intervals changes neither \(E\) nor \((b-a)+(d-c)\), so assume \(b \le c\). Splitting \((a,d)\) at \(b\) and at \(c\) gives

\begin{equation*} (a,d) \subseteq (a,b) \cup [b,c] \cup (c,d) = E \cup [b,c] \end{equation*}

(Check!), and \(|[b,c]| = c-b\) by 2.14 if \(b<c\), by 2.3 if \(b=c\). Since \(a<d\), Exercise 6 gives \(|(a,d)| = d-a\), so 2.5 and 2.8 yield

\begin{equation*} d - a = |(a,d)| \le |E| + (c-b) . \end{equation*}

All terms are finite, so \(|E| \ge (b-a)+(d-c)\); with the upper bound, equality.

(ii) Overlapping case. Put \(p = \min\{a,c\}\) and \(q = \max\{b,d\}\). Because the two intervals meet, their union is the single interval \(E = (p,q)\) (Check!), so \(|E| = q-p\) by Exercise 6. As \(\{p,u\} = \{a,c\}\) and \(\{v,q\} = \{b,d\}\) as multisets,

\begin{equation*} (b-a)+(d-c) = (v+q) - (p+u) = (q-p) + (v-u) = |E| + (v-u) , \end{equation*}

and \(v - u > 0\) with \(|E|\) finite, so \(|E| < (b-a)+(d-c)\).

With \(B = A \cap (-t,t)\) and \(C = A \setminus (-t,t)\), the inequality \(|A| \le |B| + |C|\) is subadditivity (2.8) applied to \(B, C, \emptyset, \dots\), so only \(|B| + |C| \le |A|\) needs proof; assume \(|A| < \infty\), fix \(\varepsilon > 0\), and by 2.2 choose open intervals \(I_1, I_2, \dots\) with

\begin{equation*} A \subseteq \bigcup_{k=1}^{\infty} I_k \qquad\text{and}\qquad \sum_{k=1}^{\infty} \ell(I_k) \le |A| + \varepsilon . \end{equation*}

Cut each \(I_k\) at \(-t\) and at \(t\), putting

\begin{equation*} J_k = I_k \cap (-t,t), \qquad K_k = I_k \cap (-\infty,-t), \qquad L_k = I_k \cap (t,\infty), \end{equation*}

again open intervals. Then

\begin{equation*} \ell(J_k) + \ell(K_k) + \ell(L_k) = \ell(I_k) \qquad \text{for each } k : \end{equation*}

cutting an open interval \(I\) at a single point \(c\) gives \(\ell\bigl(I \cap (-\infty,c)\bigr) + \ell\bigl(I \cap (c,\infty)\bigr) = \ell(I)\) (Check! the cases \(c \le a\), \(a < c < b\), \(c \ge b\), and \(I\) unbounded), and applying this to \(I_k\) at \(c = -t\) and then to \(I_k \cap (-t,\infty)\) at \(c = t\) and adding gives the display.

Now \(B \subseteq \bigcup_k J_k\), so \(|B| \le \sum_k \ell(J_k)\) by 2.2; and a point of \(A\) outside \((-t,t)\) lies in some \(I_k\), hence in \(K_k\) or \(L_k\) unless it is \(-t\) or \(t\), so

\begin{equation*} C \subseteq \Bigl(\bigcup_{k=1}^{\infty} K_k\Bigr) \cup \Bigl(\bigcup_{k=1}^{\infty} L_k\Bigr) \cup \{-t,t\}, \end{equation*}

whence \(|C| \le \sum_k \bigl(\ell(K_k) + \ell(L_k)\bigr)\) by 2.2, 2.3 and 2.8. Adding,

\begin{equation*} |B| + |C| \le \sum_{k=1}^{\infty} \ell(I_k) \le |A| + \varepsilon , \end{equation*}

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

By 2.5 the function \(t \mapsto |A \cap (-t,t)|\) is nondecreasing on \((0,\infty)\), so the limit exists in \([0,\infty]\) and

\begin{equation*} L := \lim_{t \to \infty} |A \cap (-t,t)| = \sup_{t>0} |A \cap (-t,t)| = \lim_{n \to \infty} |A \cap (-n,n)| . \end{equation*}

Since \(A \cap (-t,t) \subseteq A\), 2.5 gives \(|A \cap (-t,t)| \le |A|\) for every \(t\), hence \(L \le |A|\).

For the reverse, cut \(A\) into integer annuli: \(C_1 = A \cap (-1,1)\) and \(C_n = A \cap \{x : n-1 \le |x| < n\}\) for \(n \ge 2\). Applying Exercise 8 of this section to the set \(A \cap (-n,n)\) with \(t = n-1 > 0\), whose two pieces are \(A \cap (-(n-1),n-1)\) and \(C_n\) (Check!), gives

\begin{equation*} |A \cap (-n,n)| = |A \cap (-(n-1), n-1)| + |C_n| \qquad (n \ge 2) , \end{equation*}

so induction from \(|A \cap (-1,1)| = |C_1|\) yields \(|A \cap (-n,n)| = \sum_{k=1}^{n} |C_k|\) and therefore \(L = \sum_{k=1}^{\infty} |C_k|\). Each \(x \in A\) lies in \(C_m\) for the least \(m\) with \(|x| < m\), so \(A = \bigcup_{k=1}^{\infty} C_k\) and 2.8 gives

\begin{equation*} |A| \le \sum_{k=1}^{\infty} |C_k| = L . \end{equation*}

Write \(E = [0,1] \setminus \mathbb{Q}\). Since \(E \subseteq [0,1]\), the inequality \(|E| \le |[0,1]| = 1\) holds by 2.5 and 2.14. Conversely \([0,1] \cap \mathbb{Q}\) is countable, so \(|[0,1] \cap \mathbb{Q}| = 0\) by 2.4, and applying subadditivity (2.8) to \(E,\ [0,1] \cap \mathbb{Q},\ \emptyset, \dots\), whose union is \([0,1]\), gives

\begin{equation*} 1 = |[0,1]| \le |E| + 0 = |E| . \end{equation*}

Hence \(|[0,1] \setminus \mathbb{Q}| = 1\).

Write \(A = \bigcup_{k=1}^{\infty} I_k\). The bound \(|A| \le \sum_{k=1}^{\infty} \ell(I_k)\) is immediate from 2.2 (disjointness unused), so the content is the reverse bound, which comes from splitting outer measure at the endpoints of \(I_1, \dots, I_n\). Recorded once: \(|I| = \ell(I)\) for every open interval \(I\), by Exercise 6 of this section when \(I\) is bounded and nonempty, and because an unbounded \(I\) contains closed intervals of every length, so \(|I| = \infty\) by 2.5 and 2.14.

Splitting lemma. For every \(S \subseteq \mathbb{R}\) and \(c \in \mathbb{R}\),

\begin{equation*} |S| = \bigl|S \cap (-\infty,c]\bigr| + \bigl|S \cap (c,\infty)\bigr| . \end{equation*}

Here \(\le\) is 2.8, and \(\ge\) follows exactly as in Exercise 8 of this section: assuming \(|S| < \infty\), take open intervals \(J_k\) with \(S \subseteq \bigcup_k J_k\) and \(\sum_k \ell(J_k) \le |S| + \varepsilon\), cut each at \(c\) into \(J_k^{\prime} = J_k \cap (-\infty,c)\) and \(J_k^{\prime\prime} = J_k \cap (c,\infty)\) with \(\ell(J_k^{\prime}) + \ell(J_k^{\prime\prime}) = \ell(J_k)\) (Check!), and apply 2.2, 2.3 and 2.8 to

\begin{equation*} S \cap (-\infty,c] \subseteq \Bigl( \bigcup_{k=1}^{\infty} J_k^{\prime} \Bigr) \cup \{c\} \qquad\text{and}\qquad S \cap (c,\infty) \subseteq \bigcup_{k=1}^{\infty} J_k^{\prime\prime} . \end{equation*}

Iterating the lemma on \(S \cap (c_1,\infty)\) and inducting on \(m\) gives, for \(c_1 < \cdots < c_m\),

\begin{equation*} |S| = \bigl|S \cap (-\infty,c_1]\bigr| + \sum_{j=1}^{m-1} \bigl|S \cap (c_j, c_{j+1}]\bigr| + \bigl|S \cap (c_m,\infty)\bigr| . \end{equation*}

The bound \(|A| \ge \sum_{k=1}^{n} \ell(I_k)\). Fix \(n\). If some \(I_k\) with \(k \le n\) is unbounded then \(|A| \ge |I_k| = \infty\) by 2.5, so assume \(I_1, \dots, I_n\) bounded and discard the empty ones (they contribute \(0\)), leaving \((a_1,b_1), \dots, (a_p,b_p)\) with \(a_j < b_j\); the case \(p = 0\) is trivial. Disjoint nonempty open intervals \((a,b)\), \((a^{\prime},b^{\prime})\) satisfy \(b \le a^{\prime}\) or \(b^{\prime} \le a\), since otherwise \(\max\{a,a^{\prime}\} < \min\{b,b^{\prime}\}\) and they meet; so after relabelling

\begin{equation*} a_1 < b_1 \le a_2 < b_2 \le \cdots \le a_p < b_p . \end{equation*}

Let \(c_1 < \cdots < c_m\) list the distinct endpoints. No endpoint lies strictly between \(a_j\) and \(b_j\), so \(a_j, b_j\) are consecutive in that list and each \((a_j,b_j]\) is one of the intervals \((c_l,c_{l+1}]\) above. Applying the displayed identity with \(S = A\) and discarding the remaining terms (all nonnegative, and distinct from the kept ones),

\begin{equation*} |A| \ge \sum_{j=1}^{p} \bigl|A \cap (a_j,b_j]\bigr| \ge \sum_{j=1}^{p} (b_j - a_j) = \sum_{k=1}^{n} \ell(I_k) , \end{equation*}

the second inequality because \((a_j,b_j) \subseteq A \cap (a_j,b_j]\) has outer measure \(b_j - a_j\). Letting \(n \to \infty\) gives \(|A| \ge \sum_{k=1}^{\infty} \ell(I_k)\), hence equality.

Write \(G = \bigcup_{k=1}^{\infty} \bigl( r_k - 2^{-k},\ r_k + 2^{-k} \bigr)\), so that \(F = \mathbb{R} \setminus G\).

(a) \(G\) is a union of open sets, hence open, so its complement \(F\) is closed.

(b) If an interval \(I \subseteq F\) contained \(x < y\), then \([x,y] \subseteq I\), and by density of \(\mathbb{Q}\) there is a rational \(q\) with \(x < q < y\), so \(q \in F\). But \(q = r_j\) for some \(j\), and

\begin{equation*} q = r_j \in \bigl( r_j - 2^{-j},\ r_j + 2^{-j} \bigr) \subseteq G , \end{equation*}

contradicting \(q \notin G\).

(c) By 2.2 applied to the defining intervals,

\begin{equation*} |G| \le \sum_{k=1}^{\infty} \frac{2}{2^k} = 2 . \end{equation*}

Fix \(n \in \mathbb{Z}^{+}\). Since \([-n,n] \subseteq F \cup G\), 2.5 and 2.8 (applied to \(F, G, \emptyset, \dots\)) with \(|[-n,n]| = 2n\) from 2.14 give

\begin{equation*} 2n \le |F| + |G| \le |F| + 2 , \end{equation*}

and \(|G| \le 2 < \infty\) may be subtracted, so \(|F| \ge 2n - 2\) for every \(n\). Hence \(|F| = \infty\).

Take \(F = [0,1] \setminus G\), where \(q_1, q_2, \dots\) lists the countably infinite set \([0,1] \cap \mathbb{Q}\) and

\begin{equation*} G = \bigcup_{k=1}^{\infty} \Bigl( q_k - \frac{\varepsilon}{2^{k+2}}, \ q_k + \frac{\varepsilon}{2^{k+2}} \Bigr) . \end{equation*}

Then \(G\) is open, so \(F = [0,1] \cap (\mathbb{R} \setminus G)\) is an intersection of two closed sets, hence closed, and \(F \subseteq [0,1]\). Every element of \(F\) is irrational, since a rational \(x \in [0,1]\) equals some \(q_k\) and therefore lies in \(G\). Finally 2.2 gives

\begin{equation*} |G| \le \sum_{k=1}^{\infty} \frac{2\varepsilon}{2^{k+2}} = \frac{\varepsilon}{2} , \end{equation*}

and \([0,1] \subseteq F \cup G\), so 2.5, 2.8 and \(|[0,1]| = 1\) (2.14) give

\begin{equation*} 1 \le |F| + |G| \le |F| + \frac{\varepsilon}{2} . \end{equation*}

Since \(|G| \le \varepsilon/2 < \infty\) may be subtracted, \(|F| \ge 1 - \varepsilon/2 > 1 - \varepsilon\).

(a) The triangle \(T\) has its right angle at \((20,0)\), with legs \(20\) and \(9\), so \(\operatorname{area}(T) = \tfrac12 \cdot 20 \cdot 9 = 90\).

(b) Right angle at \((11,0)\), legs \(11\) and \(5\): area \(= \tfrac12 \cdot 11 \cdot 5 = 27.5\).

(c) Sides \(20 - 11 = 9\) and \(5 - 0 = 5\): area \(= 9 \cdot 5 = 45\).

(d) Right angle at \((20,5)\), legs \(20 - 11 = 9\) and \(9 - 5 = 4\): area \(= \tfrac12 \cdot 9 \cdot 4 = 18\).

(e) The three pieces meet only in the segments \(\{11\} \times [0,5]\) and \([11,20] \times \{5\}\) and the point \((11,5)\), which contribute no area, so the colored region has area \(27.5 + 45 + 18 = 90.5\).

(f) The colored region is not the triangle \(T\): its upper boundary bends upward at \((11,5)\). Indeed the line through \((0,0)\) and \((20,9)\) is \(y = \tfrac{9}{20}x\), of height \(\tfrac{99}{20} = 4.95 < 5\) at \(x = 11\); equivalently the three slopes \(\tfrac{5}{11}\), \(\tfrac{9}{20}\), \(\tfrac{4}{9}\) are distinct. So the union \(Q\) of the pieces is the quadrilateral \((0,0)\), \((20,0)\), \((20,9)\), \((11,5)\), which strictly contains \(T\), and \(Q \setminus T\) is the thin triangle with vertices \((0,0)\), \((11,5)\), \((20,9)\): base \(\sqrt{20^2+9^2} = \sqrt{481}\) and height the distance from \((11,5)\) to \(9x - 20y = 0\), namely

\begin{equation*} \frac{|9 \cdot 11 - 20 \cdot 5|}{\sqrt{481}} = \frac{1}{\sqrt{481}} , \end{equation*}

hence area \(\tfrac12\). Thus \(\operatorname{area}(Q) = 90 + 0.5 = 90.5\), agreeing with (e). Nonadditivity (2.18) is not at issue: the pieces do not overlap and their areas do add to the area of their union.

Everything rests on the fact that the intervals \((n,n+1]\), \(n \in \mathbb{Z}\), partition \(\mathbb{R}\): each \(x\) lies in \((n_x, n_x+1]\) where \(n_x = \lceil x \rceil - 1\) (Check! separately for \(x \in \mathbb{Z}\) and \(x \notin \mathbb{Z}\)), and in no other such interval, since \(m < n\) in \(\mathbb{Z}\) forces \(x \le m+1 \le n\) for \(x \in (m,m+1]\). Hence, writing \(E_K = \bigcup_{n \in K} (n,n+1]\) so that \(\mathcal{S} = \{E_K : K \subseteq \mathbb{Z}\}\),

\begin{equation*} x \in E_K \iff n_x \in K . \end{equation*}

The three conditions of 2.23 follow. First, \(\emptyset = E_\emptyset \in \mathcal{S}\). Second,

\begin{equation*} \mathbb{R} \setminus E_K = E_{\mathbb{Z} \setminus K} , \end{equation*}

because \(x \notin E_K\) iff \(n_x \notin K\) iff \(n_x \in \mathbb{Z} \setminus K\). Third, given \(K_1, K_2, \ldots \subseteq \mathbb{Z}\) and \(K = \bigcup_{j=1}^\infty K_j\),

\begin{equation*} x \in \bigcup_{j=1}^\infty E_{K_j} \iff n_x \in K_j \text{ for some } j \iff x \in E_K , \end{equation*}

so \(\bigcup_{j=1}^\infty E_{K_j} = E_K \in \mathcal{S}\). Thus \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\).

In each part we exhibit the asserted collection \(\mathcal{T}\) and check (i) \(\mathcal{T}\) is a \(\sigma\)-algebra, (ii) \(\mathcal{A} \subseteq \mathcal{T}\), (iii) every \(\sigma\)-algebra containing \(\mathcal{A}\) contains \(\mathcal{T}\); by 2.27 these three facts identify \(\mathcal{T}\) as the smallest \(\sigma\)-algebra containing \(\mathcal{A}\).

Part (a). Let

\begin{equation*} \mathcal{T} = \{E \subseteq X : E \text{ is countable} \text{ or } X \setminus E \text{ is countable}\}. \end{equation*}

(i) \(\emptyset\) is countable; the defining condition is symmetric in \(E\) and \(X \setminus E\), so \(\mathcal{T}\) is closed under complementation. If \(E_1, E_2, \ldots \in \mathcal{T}\) and \(E = \bigcup_{k=1}^\infty E_k\), then either every \(E_k\) is countable, whence \(E\) is countable; or some \(E_m\) is uncountable, whence \(X \setminus E_m\) is countable and \(X \setminus E \subseteq X \setminus E_m\) is countable. Either way \(E \in \mathcal{T}\).

(ii) Each \(\{x\}\) is finite, hence countable.

(iii) Let \(\mathcal{S} \supseteq \mathcal{A}\) be a \(\sigma\)-algebra and \(E \in \mathcal{T}\). If \(E\) is countable then \(E = \bigcup_k \{x_k\} \in \mathcal{S}\), a countable union of elements of \(\mathcal{A}\) (pad a finite list with copies of \(\emptyset\)). If \(X \setminus E\) is countable, that case gives \(X \setminus E \in \mathcal{S}\), so \(E \in \mathcal{S}\).

Part (b). Set

\begin{equation*} A_1 = (0,1), \qquad A_2 = [1,\infty), \qquad A_3 = (-\infty, 0]. \end{equation*}

These are nonempty, pairwise disjoint, and partition \(\mathbb{R}\). For \(K \subseteq \{1,2,3\}\) put \(E_K = \bigcup_{i \in K} A_i\); the eight subsets \(K\) give

\begin{equation*} \begin{aligned} E_\emptyset &= \emptyset, & E_{\{1\}} &= (0,1),\\ E_{\{2\}} &= [1,\infty), & E_{\{3\}} &= (-\infty,0],\\ E_{\{1,2\}} &= (0,\infty), & E_{\{1,3\}} &= (-\infty,1),\\ E_{\{2,3\}} &= (-\infty,0] \cup [1,\infty), & E_{\{1,2,3\}} &= \mathbb{R}, \end{aligned} \end{equation*}

so \(\mathcal{T} = \{E_K : K \subseteq \{1,2,3\}\}\) is exactly the eight-element collection displayed in the exercise (the eight sets are distinct, since \(i \in K \iff A_i \subseteq E_K\)).

(i) Each \(x\) lies in exactly one \(A_{i(x)}\), and \(x \in E_K \iff i(x) \in K\); exactly as in Exercise 1 of this section this gives \(\emptyset = E_\emptyset \in \mathcal{T}\) together with

\begin{equation*} \mathbb{R} \setminus E_K = E_{\{1,2,3\} \setminus K} \qquad\text{and}\qquad \bigcup_{j=1}^\infty E_{K_j} = E_{K_1 \cup K_2 \cup \cdots}. \end{equation*}

(ii) \((0,1) = E_{\{1\}}\) and \((0,\infty) = E_{\{1,2\}}\).

(iii) If \(\mathcal{S}\) is a \(\sigma\)-algebra containing \((0,1)\) and \((0,\infty)\), then by 2.25(a) and 2.25(b)

\begin{equation*} A_2 = (0,\infty) \setminus (0,1) \in \mathcal{S}, \qquad A_3 = \mathbb{R} \setminus (0,\infty) \in \mathcal{S}, \end{equation*}

and \(A_1 = (0,1) \in \mathcal{S}\); closure under finite unions (2.25(b)) puts every \(E_K\) in \(\mathcal{S}\).

Let \(\mathcal{B}\) be the collection of Borel subsets of \(\mathbb{R}\), the smallest \(\sigma\)-algebra containing the open sets (2.29).

\(\mathcal{S} \subseteq \mathcal{B}\). Every generator is Borel: \((r,s] = \emptyset\) if \(r \ge s\), while if \(r < s\) then

\begin{equation*} (r,s] = \bigcap_{k=1}^\infty \left(r,\ s + \tfrac{1}{k}\right) \end{equation*}

(since \(x < s + \frac1k\) for all \(k\) forces \(x \le s\)), a countable intersection of open sets, which lies in \(\mathcal{B}\) by 2.25(c). So \(\mathcal{B}\) is a \(\sigma\)-algebra containing the generating collection and hence contains the smallest one, \(\mathcal{S}\).

\(\mathcal{B} \subseteq \mathcal{S}\). By 2.29 it suffices that every open \(G \subseteq \mathbb{R}\) lie in \(\mathcal{S}\). With \(P = \{(r,s) \in \mathbb{Q} \times \mathbb{Q} : (r,s] \subseteq G\}\),

\begin{equation*} G = \bigcup_{(r,s) \in P} (r,s] : \end{equation*}

the inclusion \(\supseteq\) is the definition of \(P\), and for \(\subseteq\), given \(x \in G\) pick \(\varepsilon > 0\) with \((x-\varepsilon, x+\varepsilon) \subseteq G\) and then, by density of \(\mathbb{Q}\), rationals \(r \in (x - \varepsilon, x)\) and \(s \in (x, x+\varepsilon)\), so that \(x \in (r,s] \subseteq G\). Since \(P \subseteq \mathbb{Q} \times \mathbb{Q}\) is countable, this is a countable union of elements of \(\mathcal{S}\) (and \(G = \emptyset \in \mathcal{S}\) if \(P = \emptyset\)), so \(G \in \mathcal{S}\).

Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\).

\(\mathcal{S} \subseteq \mathcal{B}\). Every integer is rational, so each generator \((r,n]\) is among the intervals \((r,s]\) with \(r,s \in \mathbb{Q}\) shown to be Borel in Exercise 3 of this section. Hence \(\mathcal{B}\) is a \(\sigma\)-algebra containing the generating collection, so it contains \(\mathcal{S}\).

\(\mathcal{B} \subseteq \mathcal{S}\). By Exercise 3, \(\mathcal{B}\) is the smallest \(\sigma\)-algebra containing all \((r,s]\) with \(r,s \in \mathbb{Q}\), so it suffices to put those intervals in \(\mathcal{S}\). If \(r \ge s\) then \((r,s] = \emptyset \in \mathcal{S}\). If \(r < s\), choose \(n \in \mathbb{Z}\) with \(n \ge s\) (Archimedean property); then

\begin{equation*} (r,s] = (r,n] \setminus (s,n] , \end{equation*}

since for \(r < x \le n\) the failure of \(s < x\) is exactly \(x \le s\). Both sets on the right are generators of \(\mathcal{S}\), and \(\sigma\)-algebras are closed under differences by 2.25(b), so \((r,s] \in \mathcal{S}\).

Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\).

\(\mathcal{S} \subseteq \mathcal{B}\). Each generator \((r,r+1)\) is open, hence Borel, so \(\mathcal{B}\) is a \(\sigma\)-algebra containing the generating collection and therefore contains \(\mathcal{S}\).

Short rational intervals lie in \(\mathcal{S}\). If \(p,q \in \mathbb{Q}\) with \(p < q\) and \(q - p \le 1\), then \(q - 1 \le p\) and \(q \le p+1\), so

\begin{equation*} (p, p+1) \cap (q-1, q) = \bigl(\max\{p,\ q-1\},\ \min\{p+1,\ q\}\bigr) = (p, q) , \end{equation*}

and both factors are generators (\(p\) and \(q-1\) being rational), so \((p,q) \in \mathcal{S}\) by closure under finite intersections (2.25(b)).

\(\mathcal{B} \subseteq \mathcal{S}\). By 2.29 it suffices that every open \(G \subseteq \mathbb{R}\) lie in \(\mathcal{S}\). Put

\begin{equation*} P = \{(p,q) \in \mathbb{Q} \times \mathbb{Q} : p < q,\ q - p \le 1,\ (p,q) \subseteq G\} , \end{equation*}

so that \(G = \bigcup_{(p,q) \in P} (p,q)\): the inclusion \(\supseteq\) is the definition of \(P\), and for \(\subseteq\), given \(x \in G\) choose \(\varepsilon \in \bigl(0,\tfrac12\bigr)\) with \((x-\varepsilon, x+\varepsilon) \subseteq G\) and rationals \(p \in (x-\varepsilon, x)\), \(q \in (x, x+\varepsilon)\), whence \(q - p < 2\varepsilon < 1\) and \(x \in (p,q) \subseteq G\). As \(P \subseteq \mathbb{Q} \times \mathbb{Q}\) is countable, the previous paragraph and closure under countable unions give \(G \in \mathcal{S}\) (and \(G = \emptyset \in \mathcal{S}\) if \(P = \emptyset\)).

Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\).

\(\mathcal{S} \subseteq \mathcal{B}\). Each generator \([r,\infty)\) is closed, its complement \((-\infty,r)\) being open, hence Borel by the first bullet point of 2.30; so \(\mathcal{B}\) is a \(\sigma\)-algebra containing the generating collection and therefore contains \(\mathcal{S}\).

Bounded rational open intervals lie in \(\mathcal{S}\). For \(p \in \mathbb{Q}\),

\begin{equation*} (p, \infty) = \bigcup_{k=1}^\infty \left[p + \tfrac1k,\ \infty\right) \in \mathcal{S} , \end{equation*}

since \(x > p\) gives \(\frac1k \le x - p\) for some \(k\) by the Archimedean property, each \(p + \frac1k\) is rational, and \(\mathcal{S}\) is closed under countable unions. Also \((-\infty, q) = \mathbb{R} \setminus [q,\infty) \in \mathcal{S}\) for \(q \in \mathbb{Q}\), so

\begin{equation*} (p,q) = (p,\infty) \cap (-\infty, q) \in \mathcal{S} \end{equation*}

for all \(p,q \in \mathbb{Q}\), by closure under finite intersections (2.25(b)).

\(\mathcal{B} \subseteq \mathcal{S}\). Exactly as in Exercise 5 of this section, every open \(G \subseteq \mathbb{R}\) is the union of the countably many intervals \((p,q)\) with \(p,q \in \mathbb{Q}\), \(p<q\) and \((p,q) \subseteq G\); by the previous paragraph and closure under countable unions, \(G \in \mathcal{S}\). Now 2.29 gives \(\mathcal{B} \subseteq \mathcal{S}\).

Fix \(t \in \mathbb{R}\) and let \(f : \mathbb{R} \to \mathbb{R}\) be \(f(x) = x - t\), which is continuous and hence Borel measurable by 2.41. Since \(x - t \in B\) exactly when \(x = t + b\) for some \(b \in B\),

\begin{equation*} f^{-1}(B) = \{x \in \mathbb{R} : x - t \in B\} = t + B , \end{equation*}

so \(t + B\) is a Borel set for every Borel set \(B\), by the definition 2.40 of Borel measurability.

Method (2) (avoiding 2.41). Let \(\mathcal{B}\) be the collection of Borel subsets of \(\mathbb{R}\) and put \(\mathcal{T} = \{A \subseteq \mathbb{R} : t + A \in \mathcal{B}\}\). Because \(x \in t+A\) iff \(x - t \in A\),

\begin{equation*} t + (\mathbb{R} \setminus A) = \mathbb{R} \setminus (t + A), \qquad t + \bigcup_{k=1}^\infty A_k = \bigcup_{k=1}^\infty (t + A_k), \end{equation*}

and \(t + \emptyset = \emptyset\); since \(\mathcal{B}\) is a \(\sigma\)-algebra, so is \(\mathcal{T}\). If \(G\) is open then \(t + G\) is open, translation being an isometry (\(|y - x| < \varepsilon\) iff \(|(y-t)-(x-t)| < \varepsilon\)), so \(t + G \in \mathcal{B}\) and \(G \in \mathcal{T}\). Thus \(\mathcal{T}\) is a \(\sigma\)-algebra containing every open set, and 2.29 gives \(\mathcal{B} \subseteq \mathcal{T}\).

Suppose \(B \subseteq \mathbb{R}\) is a Borel set and \(t \in \mathbb{R}\).

(i) \(t = 0\). Then \(tB\) is \(\emptyset\) or \(\{0\}\), both Borel (\(\{0\}\) is closed, hence Borel by 2.30).

(ii) \(t \ne 0\). The function \(m(x) = x/t\) is continuous, hence Borel measurable by 2.41, so \(m^{-1}(B)\) is a Borel set by the definition 2.40. And \(m^{-1}(B) = tB\): if \(x/t \in B\) then \(x = t(x/t) \in tB\), while if \(x = tb\) with \(b \in B\) then \(x/t = b \in B\), using \(t \ne 0\).

Method (2) (avoiding 2.41). Fix \(t \ne 0\) and put \(\mathcal{S} = \{B \subseteq \mathbb{R} : tB \text{ is Borel}\}\). Multiplication by \(t\) is a bijection of \(\mathbb{R}\), so

\begin{equation*} t(\mathbb{R} \setminus B) = \mathbb{R} \setminus tB, \qquad t\Bigl(\bigcup_{k=1}^{\infty} B_k\Bigr) = \bigcup_{k=1}^{\infty} tB_k , \end{equation*}

and \(t\emptyset = \emptyset\); hence \(\mathcal{S}\) is a \(\sigma\)-algebra. It contains every open \(G\), since multiplication by \(t\) is a homeomorphism and so \(tG\) is open, hence Borel. By 2.29, \(\mathcal{S}\) contains every Borel set.

Take \(X = \mathbb{R}\), \(\mathcal{S} = \{\emptyset, \mathbb{R}\}\) (a \(\sigma\)-algebra; Check!), and

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x \geq 0,\\ -1 & \text{if } x < 0. \end{cases} \end{equation*}

Then \(|f|\) is constant with value \(1\), so \(|f|^{-1}(B)\) is \(\mathbb{R}\) or \(\emptyset\) according as \(1 \in B\) or not; either way it lies in \(\mathcal{S}\), so \(|f|\) is \(\mathcal{S}\)-measurable. But \((0,\infty)\) is open, hence Borel, and

\begin{equation*} f^{-1}\bigl((0,\infty)\bigr) = [0, \infty) \notin \mathcal{S} , \end{equation*}

so \(f\) fails the definition 2.35 of \(\mathcal{S}\)-measurability.

The set in question is \(A = A_+ \cup (-A_+)\), where

\begin{equation*} A_+ = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} D_k , \qquad D_k = \bigcup_{m=0}^{\infty} \left[ \frac{10m + 5}{10^{k}},\ \frac{10m + 6}{10^{k}} \right) , \end{equation*}

and this is a Borel set. Here a decimal expansion of \(x \ge 0\) means a representation \(x = N + \sum_{j=1}^{\infty} e_j 10^{-j}\) with \(N \in \mathbb{Z}\), \(N \ge 0\) and \(e_j \in \{0,1,\ldots,9\}\), written \(x = N.e_1e_2e_3\ldots\), while a negative number carries the expansion of its absolute value; the integer part is a finite string, so the digit 5 can recur infinitely often only among \(e_1, e_2, \ldots\).

Reduction to \([0,\infty)\). Since \(x\) and \(-x\) carry the same digit string, \(x \in A\) iff \(|x| \in A\), so \(A = A_+ \cup (-A_+)\) with \(A_+ = A \cap [0,\infty)\), and \(-A_+\) is Borel whenever \(A_+\) is, by Exercise 8 of this section with \(t = -1\).

Canonical digits. For \(x \ge 0\) and \(k \in \mathbb{Z}^+\) set \(d_k(x) = \lfloor 10^k x \rfloor - 10 \lfloor 10^{k-1} x \rfloor\). Writing \(q = \lfloor 10^{k-1}x\rfloor\) gives \(10q \le 10^k x < 10q + 10\), hence \(d_k(x) \in \{0,1,\ldots,9\}\), and the sum telescopes to

\begin{equation*} \lfloor x \rfloor + \sum_{k=1}^{n} d_k(x) 10^{-k} = \frac{\lfloor 10^n x \rfloor}{10^n} \longrightarrow x \end{equation*}

because \(x - 10^{-n} < \lfloor 10^n x \rfloor/10^n \le x\). So \(\lfloor x \rfloor . d_1(x) d_2(x) \ldots\) is a decimal expansion of \(x\), the canonical one.

An expansion with infinitely many \(5\)s is the canonical one. Let \(x = N.e_1e_2e_3\ldots\) with \(e_j = 5\) for infinitely many \(j\), and fix \(k \ge 0\). Some \(i > k\) has \(e_i = 5\), so the digits beyond position \(k\) are not all \(9\) and

\begin{equation*} 0 \le \sum_{j > k} e_j 10^{-j} \leq 10^{-k} - 4 \cdot 10^{-i} < 10^{-k} . \end{equation*}

With \(M_k = 10^k N + \sum_{j=1}^{k} e_j 10^{k-j} \in \mathbb{Z}\) this reads \(M_k \le 10^k x < M_k + 1\), so \(\lfloor 10^k x \rfloor = M_k\) for every \(k \ge 0\), whence \(\lfloor x \rfloor = M_0 = N\) and \(d_k(x) = M_k - 10M_{k-1} = e_k\). The converse direction is immediate, the canonical string being itself an expansion, so

\begin{equation*} A_+ = \{ x \geq 0 : d_k(x) = 5 \text{ for infinitely many } k \in \mathbb{Z}^+ \}. \end{equation*}

The sets \(D_k\). Fix \(k\). Then \(d_k(x) = 5\) iff \(\lfloor 10^k x \rfloor \in \{10m+5 : m \in \mathbb{Z}\}\): forwards because \(\lfloor 10^k x \rfloor = 10\lfloor 10^{k-1}x \rfloor + 5\); backwards because \(\lfloor 10^k x \rfloor = 10m+5\) gives \(m + \tfrac12 \le 10^{k-1}x < m + \tfrac35\), so \(\lfloor 10^{k-1}x \rfloor = m\) and \(d_k(x) = 5\). Since \(x \ge 0\) forces \(\lfloor 10^k x \rfloor \ge 0\), only \(m \ge 0\) occurs, so \(\{x \ge 0 : d_k(x) = 5\}\) is exactly the set \(D_k\) displayed at the start, a countable union of half-open intervals and hence Borel by 2.30.

Assembling. The digit occurs infinitely often exactly when for every \(n\) some \(k \ge n\) has \(d_k(x) = 5\), which is the displayed formula for \(A_+\); it is Borel by 2.23 (countable unions) and 2.25(c) (countable intersections), and hence so is \(A = A_+ \cup (-A_+)\).

(a) If \(E \in \mathcal{S}\) then \(E \in \mathcal{T}\) and \(E = E \cap X\), so \(E\) has the required form. Conversely, if \(F \in \mathcal{T}\) then \(F \cap X \in \mathcal{T}\) by 2.25(b) (using \(X \in \mathcal{T}\)) and \(F \cap X \subseteq X\), so \(F \cap X \in \mathcal{S}\).

(b) Each element of \(\mathcal{S}\) is a subset of \(X\), and complementation is taken in \(X\). The three conditions of 2.23 hold: \(\emptyset \in \mathcal{T}\) and \(\emptyset \subseteq X\); for \(E \in \mathcal{S}\), both \(E\) and \(X\) lie in \(\mathcal{T}\), so \(X \setminus E \in \mathcal{T}\) by 2.25(b) and \(X \setminus E \subseteq X\); and for \(E_1, E_2, \ldots \in \mathcal{S}\), the union \(\bigcup_{k=1}^{\infty} E_k\) lies in \(\mathcal{T}\) and is contained in \(X\). Hence \(\mathcal{S}\) is a \(\sigma\)-algebra on \(X\).

(a) Let \(a \in G_k\), with \(\delta > 0\) as in the definition. Then \(\bigl(a - \tfrac{\delta}{2}, a + \tfrac{\delta}{2}\bigr) \subseteq G_k\): for \(a^{\prime}\) in that interval put \(\delta^{\prime} = \tfrac{\delta}{2} - |a^{\prime} - a| > 0\), so that

\begin{equation*} |y - a| \leq |y - a^{\prime}| + |a^{\prime} - a| < \delta^{\prime} + |a^{\prime} - a| = \tfrac{\delta}{2} < \delta \end{equation*}

for \(y \in (a^{\prime} - \delta^{\prime}, a^{\prime} + \delta^{\prime})\); hence that interval sits inside \((a-\delta, a+\delta)\), where the oscillation bound \(\tfrac1k\) already holds, and \(a^{\prime} \in G_k\). So \(G_k\) contains an open interval about each of its points and is open.

(b) Write \(C\) for the set of continuity points. If \(a \in C\) and \(k \in \mathbb{Z}^+\), continuity at \(a\) with \(\varepsilon = \tfrac{1}{2k}\) gives \(\delta > 0\) with \(|f(y) - f(a)| < \tfrac{1}{2k}\) on \((a-\delta,a+\delta)\), so

\begin{equation*} |f(b) - f( c)| \leq |f(b) - f(a)| + |f(a) - f( c)| < \tfrac{1}{k} \end{equation*}

there, and \(a \in G_k\). Conversely, if \(a \in \bigcap_{k=1}^{\infty} G_k\) and \(\varepsilon > 0\), choose \(k\) with \(\tfrac1k < \varepsilon\) and take \(c = a\) in the defining property of \(G_k\): then \(|f(b) - f(a)| < \tfrac1k < \varepsilon\) for all \(b \in (a-\delta, a+\delta)\), so \(a \in C\). Hence \(C = \bigcap_{k=1}^{\infty} G_k\).

(c) Each \(G_k\) is open by (a), hence Borel by 2.29, and \(C\) is a countable intersection of Borel sets, hence Borel by 2.25(c).

Write \(f = c_1 \chi_{E_1} + \cdots + c_n \chi_{E_n}\).

(\(\Leftarrow\)) Suppose \(E_1, \ldots, E_n \in \mathcal{S}\). By 2.38 each \(\chi_{E_j}^{-1}(B)\) is one of \(E_j\), \(X \setminus E_j\), \(X\), \(\emptyset\), all in \(\mathcal{S}\) by hypothesis together with 2.23 and 2.25(a), so each \(\chi_{E_j}\) is \(\mathcal{S}\)-measurable. Constant functions are \(\mathcal{S}\)-measurable (inverse images are \(X\) or \(\emptyset\)), so 2.46(a) makes each \(c_j \chi_{E_j}\) measurable and then, inducting on the number of summands, makes \(f\) measurable.

(\(\Rightarrow\)) Suppose \(f\) is \(\mathcal{S}\)-measurable. Disjointness gives \(f(x) = c_i\) for \(x \in E_i\) and \(f(x) = 0\) off \(E_1 \cup \cdots \cup E_n\), so for each \(j\),

\begin{equation*} f^{-1}(\{c_j\}) = E_j : \end{equation*}

\(E_j \subseteq f^{-1}(\{c_j\})\) is clear, and if \(f(x) = c_j\) then \(x\) lies in some \(E_i\) (else \(f(x) = 0 \ne c_j\), the \(c\)s being nonzero) with \(c_i = c_j\), forcing \(i = j\) since the \(c\)s are distinct. As \(\{c_j\}\) is closed, hence Borel by 2.30, the definition 2.35 gives \(E_j = f^{-1}(\{c_j\}) \in \mathcal{S}\).

(a) The identity is the Cauchy criterion, written with a single anchor index. Indeed, by the definition 2.31 of inverse image, \(x\) lies in the right side \(R\) exactly when

\begin{equation*} \text{for every } n \text{ there is } j \text{ with } |f_j(x) - f_k(x)| < \tfrac{1}{n} \text{ for all } k \geq j . \end{equation*}

If \(f_1(x), f_2(x), \ldots\) converges it is Cauchy, and taking \(p = j\) in the Cauchy condition for \(\varepsilon = \tfrac1n\) puts \(x \in R\). Conversely, let \(x \in R\) and \(\varepsilon > 0\); choose \(n\) with \(\tfrac{2}{n} < \varepsilon\) and the corresponding \(j\), so that for all \(p, q \ge j\),

\begin{equation*} |f_p(x) - f_q(x)| \leq |f_p(x) - f_j(x)| + |f_j(x) - f_q(x)| < \tfrac{2}{n} < \varepsilon . \end{equation*}

Thus the sequence is Cauchy, hence convergent by completeness of \(\mathbb{R}\).

(b) Each \(f_j - f_k\) is \(\mathcal{S}\)-measurable by 2.46(a), and \(\bigl(-\tfrac1n, \tfrac1n\bigr)\) is open, hence Borel by 2.29, so the definition 2.35 gives

\begin{equation*} (f_j - f_k)^{-1}\Bigl( \bigl(-\tfrac{1}{n}, \tfrac{1}{n}\bigr) \Bigr) \in \mathcal{S} \end{equation*}

for all \(n, j, k\). The set of (a) is assembled from these by a countable intersection over \(k\), a countable union over \(j\), and a countable intersection over \(n\); \(\mathcal{S}\) is closed under the first and third by 2.25(c) and under the second by 2.23, so that set lies in \(\mathcal{S}\).

Since the \(E_k\) are disjoint with union \(X\), every \(x \in X\) lies in \(E_{k(x)}\) for exactly one index \(k(x)\), and \(x \in \bigcup_{k \in K} E_k \iff k(x) \in K\).

(a) Taking \(K = \emptyset\) gives \(\emptyset \in \mathcal{S}\). For \(E = \bigcup_{k \in K} E_k\), the equivalence above gives

\begin{equation*} X \setminus E = \bigcup_{k \in \mathbb{Z}^+ \setminus K} E_k \in \mathcal{S} , \end{equation*}

and for \(A_n = \bigcup_{k \in K_n} E_k\) with \(K = \bigcup_{n=1}^{\infty} K_n\),

\begin{equation*} \bigcup_{n=1}^{\infty} A_n = \bigcup_{k \in K} E_k \in \mathcal{S} . \end{equation*}

Hence \(\mathcal{S}\) is a \(\sigma\)-algebra on \(X\).

(b) (\(\Leftarrow\)) Let \(f\) be constant on each \(E_k\), let \(B\) be Borel, and put \(K = \{k : E_k \subseteq f^{-1}(B)\}\). Then \(f^{-1}(B) = \bigcup_{k \in K} E_k \in \mathcal{S}\): the inclusion \(\supseteq\) is the definition of \(K\), and if \(x \in f^{-1}(B)\) then constancy on \(E_{k(x)}\) gives \(E_{k(x)} \subseteq f^{-1}(B)\), so \(k(x) \in K\).

(\(\Rightarrow\)) Let \(f\) be \(\mathcal{S}\)-measurable and \(x, y \in E_k\). The singleton \(\{f(x)\}\) is closed, hence Borel, so \(f^{-1}(\{f(x)\}) = \bigcup_{j \in K} E_j\) for some \(K \subseteq \mathbb{Z}^+\); since \(x\) lies in it, \(k = k(x) \in K\), so \(E_k \subseteq f^{-1}(\{f(x)\})\) and \(f(y) = f(x)\).

(a) Since \(\emptyset \in \mathcal{S}\) and \(A \cap \emptyset = \emptyset\), we get \(\emptyset \in \mathcal{S}_A\). For \(E \in \mathcal{S}_A\) we have \(X \setminus E \in \mathcal{S}\), and \(A \subseteq E\) gives \(A \cap (X \setminus E) = \emptyset\) while \(A \cap E = \emptyset\) gives \(A \subseteq X \setminus E\); either way \(X \setminus E \in \mathcal{S}_A\). For \(E_1, E_2, \ldots \in \mathcal{S}_A\) and \(E = \bigcup_{k=1}^{\infty} E_k \in \mathcal{S}\): either \(A \subseteq E_k\) for some \(k\), whence \(A \subseteq E\), or \(A \cap E_k = \emptyset\) for every \(k\), whence

\begin{equation*} A \cap E = \bigcup_{k=1}^{\infty} (A \cap E_k) = \emptyset . \end{equation*}

Hence \(\mathcal{S}_A\) is a \(\sigma\)-algebra on \(X\).

(b) (\(\Rightarrow\)) If \(f\) is \(\mathcal{S}_A\)-measurable, then \(\mathcal{S}_A \subseteq \mathcal{S}\) makes \(f\) \(\mathcal{S}\)-measurable. Assuming \(A \ne \emptyset\), pick \(x \in A\); the singleton \(\{f(x)\}\) is closed, hence Borel, so \(f^{-1}(\{f(x)\}) \in \mathcal{S}_A\), and it meets \(A\) (at \(x\)), so the definition of \(\mathcal{S}_A\) forces \(A \subseteq f^{-1}(\{f(x)\})\); that is, \(f\) is constant on \(A\).

(\(\Leftarrow\)) Let \(f\) be \(\mathcal{S}\)-measurable and constant on \(A\). If \(A = \emptyset\) then \(\mathcal{S}_A = \mathcal{S}\); otherwise let \(c\) be the value of \(f\) on \(A\) and \(B\) Borel. Then \(f^{-1}(B) \in \mathcal{S}\), with \(A \subseteq f^{-1}(B)\) if \(c \in B\) and \(A \cap f^{-1}(B) = \emptyset\) if \(c \notin B\); either way \(f^{-1}(B) \in \mathcal{S}_A\).

By 2.39, applied with \(\mathcal{S}\) the \(\sigma\)-algebra of Borel sets contained in \(X\), it suffices to show \(f^{-1}\big((a,\infty)\big)\) is Borel for every \(a \in \mathbb{R}\). Write \(D\) for the countable set of discontinuity points; every countable set is Borel, being a countable union of closed singletons. Fix \(a\) and split

\begin{equation*} f^{-1}\big((a,\infty)\big) = U \cup V, \qquad U = \{x \in X \setminus D : f(x) > a\}, \qquad V = \{x \in D : f(x) > a\} , \end{equation*}

so \(V \subseteq D\) is countable, hence Borel. For \(x \in U\), continuity of \(f\) at \(x\) relative to \(X\), applied with \(\varepsilon = f(x) - a > 0\), yields \(\delta_x > 0\) with

\begin{equation*} f(y) > f(x) - \big(f(x) - a\big) = a \quad \text{for all } y \in (x - \delta_x, x + \delta_x) \cap X , \end{equation*}

points of \(D\) in that intersection included. Then \(G = \bigcup_{x \in U} (x - \delta_x, x + \delta_x)\) is open, hence Borel, so \(G \cap X\) is Borel, and

\begin{equation*} f^{-1}\big((a,\infty)\big) = (G \cap X) \cup V . \end{equation*}

Indeed \(\subseteq\) holds because a point of the left side lies in \(V\) if it is in \(D\) and otherwise lies in \(U \subseteq G \cap X\); and \(\supseteq\) holds because \(V\) is part of the left side while any \(y \in G \cap X\) lies in \((x-\delta_x, x+\delta_x) \cap X\) for some \(x \in U\), so \(f(y) > a\).

\(f^{\prime}\) is the pointwise limit of the difference quotients

\begin{equation*} g_k(x) = \frac{f\big(x + \tfrac{1}{k}\big) - f(x)}{1/k} , \qquad k \in \mathbb{Z}^+ . \end{equation*}

Each \(g_k\) is continuous on \(\mathbb{R}\), since \(f\) is continuous (being differentiable everywhere) and \(x \mapsto x + \tfrac1k\) is continuous; hence each \(g_k\) is Borel measurable by 2.41. Because \(f\) is differentiable at each \(x\), the limit

\begin{equation*} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = f^{\prime}(x) \end{equation*}

exists in \(\mathbb{R}\), so evaluating it along \(h = \tfrac1k \to 0\) gives \(g_k(x) \to f^{\prime}(x)\) for every \(x \in \mathbb{R}\). Thus \(f^{\prime}\) is Borel measurable by 2.48, applied with \(\mathcal{S}\) the Borel \(\sigma\)-algebra on \(\mathbb{R}\).

A function \(f : X \to \mathbb{R}\) is \(\mathcal{S}\)-measurable if and only if \(f\) is constant outside some countable subset of \(X\), that is, if and only if there is \(c \in \mathbb{R}\) with \(\{x \in X : f(x) \neq c\}\) countable.

Suppose \(C = \{x \in X : f(x) \neq c\}\) is countable and \(B \subseteq \mathbb{R}\) is Borel. If \(c \in B\), then \(X \setminus f^{-1}(B) \subseteq C\) is countable; if \(c \notin B\), then \(f^{-1}(B) \subseteq C\) is countable. Either way \(f^{-1}(B) \in \mathcal{S}\).

Conversely, suppose \(f\) is \(\mathcal{S}\)-measurable. If \(X\) is countable, take \(c = 0\); so assume \(X\) is uncountable, in which case no subset of \(X\) is both countable and cocountable. Thus each \(E_a := f^{-1}\big((a, \infty)\big) \in \mathcal{S}\) is countable or cocountable but not both, and

\begin{equation*} T = \{ a \in \mathbb{R} : E_a \text{ is countable} \} \end{equation*}

is closed upward, since \(b > a\) gives \(E_b \subseteq E_a\).

• (i) \(T \neq \emptyset\): otherwise every \(E_n\) with \(n \in \mathbb{Z}^+\) is cocountable, so \(\bigcap_{n=1}^{\infty} E_n\) has countable complement \(\bigcup_{n=1}^{\infty}(X \setminus E_n)\) and hence is nonempty as \(X\) is uncountable; but that intersection is empty because \(f\) is real-valued.
• (ii) \(T\) is bounded below: \(X = \bigcup_{n=1}^{\infty} E_{-n}\) is uncountable, so \(E_{-m}\) is uncountable for some \(m \in \mathbb{Z}^+\), i.e. \(-m \notin T\); upward closure then forces \(a > -m\) for every \(a \in T\).

Let \(c = \inf T \in \mathbb{R}\). For each \(n \in \mathbb{Z}^+\) there is \(a \in T\) with \(a < c + \tfrac1n\), so \(c + \tfrac1n \in T\) by upward closure, whence

\begin{equation*} \{ x \in X : f(x) > c \} = \bigcup_{n=1}^{\infty} E_{c + \frac{1}{n}} \end{equation*}

is countable. Also \(c - \tfrac1n \notin T\), so \(E_{c - 1/n}\) is cocountable and \(X \setminus E_{c-1/n} = \{f \le c - \tfrac1n\}\) is countable, whence

\begin{equation*} \{ x \in X : f(x) < c \} = \bigcup_{n=1}^{\infty} \Big\{ x \in X : f(x) \le c - \tfrac{1}{n} \Big\} \end{equation*}

is countable. Hence \(\{x \in X : f(x) \neq c\} = \{f > c\} \cup \{f < c\}\) is countable.

Because \(t^s = e^{s \ln t}\) for \(t > 0\) and \(f\) is positive everywhere,

\begin{equation*} f^g(x) = e^{\,g(x) \ln f(x)} \qquad \text{for all } x \in X , \end{equation*}

so \(f^g = \exp \circ h\) with \(h = g \cdot (\ln \circ f)\).

Now \(\ln\) is continuous on the Borel set \((0, \infty)\), hence Borel measurable by 2.41, and the range of \(f\) lies in \((0, \infty)\), so \(\ln \circ f\) is \(\mathcal{S}\)-measurable by 2.44. Hence \(h\) is \(\mathcal{S}\)-measurable by 2.46(a). Finally \(\exp\) is continuous on \(\mathbb{R}\), hence Borel measurable by 2.41, and its domain \(\mathbb{R}\) contains the range of \(h\), so \(f^g = \exp \circ h\) is \(\mathcal{S}\)-measurable by 2.44.

Let

\begin{equation*} \mathcal{T} = \{ A \subseteq [-\infty, \infty] : f^{-1}(A) \in \mathcal{S} \} ; \end{equation*}

it suffices to show \(\mathcal{T}\) contains every Borel subset of \([-\infty, \infty]\).

Because inverse images commute with complementation and countable unions, \(\mathcal{T}\) is a \(\sigma\)-algebra on \([-\infty, \infty]\) (the computation in the proof of 2.39, verbatim), hence closed also under countable intersections and set differences by 2.25. By hypothesis \((a, \infty] \in \mathcal{T}\) for all \(a \in \mathbb{R}\), so

\begin{equation*} \begin{aligned} \{\infty\} &= \bigcap_{n=1}^{\infty} (n, \infty] \in \mathcal{T} , \\ (-\infty, \infty] &= \bigcup_{n=1}^{\infty} (-n, \infty] \in \mathcal{T} , \end{aligned} \end{equation*}

the second giving \(\{-\infty\} \in \mathcal{T}\) by complementation and hence

\begin{equation*} \mathbb{R} = [-\infty, \infty] \setminus \big( \{\infty\} \cup \{-\infty\} \big) \in \mathcal{T} . \end{equation*}

Consequently \(\mathcal{T}_{\mathbb{R}} = \{ A \in \mathcal{T} : A \subseteq \mathbb{R} \}\) is a \(\sigma\)-algebra on \(\mathbb{R}\): it is closed under countable unions because \(\mathcal{T}\) is, and under complementation within \(\mathbb{R}\) because \(\mathbb{R} \setminus A = \mathbb{R} \cap ([-\infty, \infty] \setminus A) \in \mathcal{T}\), using \(\mathbb{R} \in \mathcal{T}\). It contains \((a, \infty) = (a, \infty] \setminus \{\infty\}\) for every \(a \in \mathbb{R}\), so by the second half of the proof of 2.39 it contains every Borel subset of \(\mathbb{R}\).

Now let \(C \subseteq [-\infty, \infty]\) be Borel. By 2.50 the set \(B = C \cap \mathbb{R}\) is a Borel subset of \(\mathbb{R}\), and \(C\) equals one of

\begin{equation*} B, \quad B \cup \{\infty\}, \quad B \cup \{-\infty\}, \quad B \cup \{\infty, -\infty\} . \end{equation*}

Each of these lies in \(\mathcal{T}\), since \(B \in \mathcal{T}_{\mathbb{R}} \subseteq \mathcal{T}\), both \(\{\infty\}\) and \(\{-\infty\}\) lie in \(\mathcal{T}\), and \(\mathcal{T}\) is closed under finite unions. Hence \(f^{-1}( C) \in \mathcal{S}\) for every Borel \(C\), so \(f\) is \(\mathcal{S}\)-measurable.

The set \(D\) of points of \(B\) at which \(f\) is discontinuous carries a pairwise disjoint family of nonempty open intervals, hence injects into \(\mathbb{Q}\) and is countable.

For \(x \in B\) put \(\alpha(x) = \sup\{f(t) : t \in B,\ t < x\}\) if \(B \cap (-\infty, x) \neq \emptyset\) and \(\alpha(x) = f(x)\) otherwise, and dually \(\beta(x) = \inf\{f(t) : t \in B,\ t > x\}\) if \(B \cap (x, \infty) \neq \emptyset\) and \(\beta(x) = f(x)\) otherwise; monotonicity bounds the first set above by \(f(x)\) and the second below by \(f(x)\), so \(\alpha(x), \beta(x) \in \mathbb{R}\) with \(\alpha(x) \le f(x) \le \beta(x)\). Call \(x\) a left (right) limit point of \(B\) when \(B \cap (x - \delta, x) \neq \emptyset\) (respectively \(B \cap (x, x+\delta) \neq \emptyset\)) for every \(\delta > 0\). Continuity is relative to the domain \(B\) throughout.

Then \(x \in D\) if and only if at least one of

• (i) \(x\) is a left limit point of \(B\) and \(\alpha(x) < f(x)\);
• (ii) \(x\) is a right limit point of \(B\) and \(f(x) < \beta(x)\)

holds. Indeed, suppose both fail and let \(\varepsilon > 0\). If \(x\) is not a left limit point, pick \(\delta_1 > 0\) with \(B \cap (x - \delta_1, x) = \emptyset\), so \(x\) is the only point of \(B \cap (x - \delta_1, x]\); otherwise \(\alpha(x) = f(x)\), and choosing \(t_0 \in B\) with \(t_0 < x\) and \(f(t_0) > f(x) - \varepsilon\) and setting \(\delta_1 = x - t_0\) gives \(f(x) - \varepsilon < f(t_0) \le f(t) \le f(x)\) for all \(t \in B \cap (x - \delta_1, x]\). The failure of (ii) yields \(\delta_2\) symmetrically, and \(\delta = \min\{\delta_1, \delta_2\}\) witnesses continuity at \(x\). Conversely, if (i) holds, take \(\varepsilon = f(x) - \alpha(x) > 0\); every \(\delta > 0\) admits \(t \in B \cap (x - \delta, x)\), and \(f(t) \le \alpha(x) = f(x) - \varepsilon\), so \(f\) is discontinuous at \(x\). The case (ii) is symmetric.

For \(x \in D\) set \(I_x = (\alpha(x), f(x))\) if (i) holds and \(I_x = (f(x), \beta(x))\) otherwise (so (ii) holds); either way \(I_x\) is a nonempty bounded open interval. For \(x < y\) in \(D\) we have \(\sup I_x \le \inf I_y\) in all four cases:

• \(I_x = (\alpha(x), f(x))\) and \(I_y = (\alpha(y), f(y))\): \(\alpha(y) \ge f(x)\), as \(x \in B\) and \(x < y\).
• \(I_x = (\alpha(x), f(x))\) and \(I_y = (f(y), \beta(y))\): \(f(y) \ge f(x)\), as \(f\) is increasing.
• \(I_x = (f(x), \beta(x))\) and \(I_y = (f(y), \beta(y))\): \(\beta(x) \le f(y)\), as \(y \in B\) and \(y > x\).
• \(I_x = (f(x), \beta(x))\) and \(I_y = (\alpha(y), f(y))\): here (ii) holds for \(x\), so \(x\) is a right limit point and some \(s \in B\) has \(x < s < y\); then \(\beta(x) \le f(s) \le \alpha(y)\).

So the \(I_x\) are pairwise disjoint, and choosing \(q_x \in I_x \cap \mathbb{Q}\) for each \(x \in D\) (density of \(\mathbb{Q}\)) gives an injection \(D \to \mathbb{Q}\). Hence \(D\) is countable.

Write \(g = f^{-1}\), which is defined on \(f(\mathbb{R})\) because a strictly increasing function is injective, and note that \(f(x) < f(u)\) forces \(x < u\) (otherwise \(x \ge u\) would give \(f(x) \ge f(u)\)).

Fix \(y_0 \in f(\mathbb{R})\), put \(x_0 = g(y_0)\), and let \(\varepsilon > 0\). Strict increase gives

\begin{equation*} f(x_0 - \varepsilon) < f(x_0) = y_0 < f(x_0 + \varepsilon) , \end{equation*}

the two outer values being defined since the domain of \(f\) is all of \(\mathbb{R}\), so

\begin{equation*} \delta = \min\bigl\{\, y_0 - f(x_0 - \varepsilon),\ f(x_0 + \varepsilon) - y_0 \,\bigr\} > 0 . \end{equation*}

If \(y \in f(\mathbb{R})\) with \(|y - y_0| < \delta\), then writing \(x = g(y)\),

\begin{equation*} f(x_0 - \varepsilon) \le y_0 - \delta < f(x) < y_0 + \delta \le f(x_0 + \varepsilon) , \end{equation*}

so \(x_0 - \varepsilon < x < x_0 + \varepsilon\), that is, \(|g(y) - g(y_0)| < \varepsilon\). Hence \(g\) is continuous at every point of \(f(\mathbb{R})\).

Let \(g = f^{-1} : f(\mathbb{R}) \to \mathbb{R}\), which is defined because \(f\) is injective and is continuous by Exercise 23 of this section, and observe that

\begin{equation*} g^{-1}(B) = \{\, y \in f(\mathbb{R}) : g(y) \in B \,\} = \{\, f(x) : x \in B \,\} = f(B), \end{equation*}

since every \(y \in f(\mathbb{R})\) equals \(f(x)\) for a unique \(x\), and then \(g(y) = x\). So by 2.41 it suffices to show that the domain \(f(\mathbb{R})\) of \(g\) is a Borel set: then \(g\) is Borel measurable and \(f(B) = g^{-1}(B)\) is Borel.

Put \(f(x^-) = \sup\{f(t) : t < x\}\) and \(f(x^+) = \inf\{f(t) : t > x\}\), which are real with \(f(x^-) \le f(x) \le f(x^+)\), and

\begin{equation*} L = \inf\{f(t) : t \in \mathbb{R}\}, \qquad M = \sup\{f(t) : t \in \mathbb{R}\}, \end{equation*}

with the conventions \((-\infty, L] = \emptyset\) if \(L = -\infty\) and \([M, \infty) = \emptyset\) if \(M = \infty\). Since every real number is both a left and a right limit point of \(\mathbb{R}\), the characterization in Exercise 22 of this section (with \(B = \mathbb{R}\), \(\alpha = f(\cdot^-)\), \(\beta = f(\cdot^+)\)) identifies

\begin{equation*} D = \{\, x \in \mathbb{R} : f(x^-) < f(x) \text{ or } f(x) < f(x^+) \,\} \end{equation*}

as the discontinuity set of \(f\), which that exercise shows is countable. We claim

\begin{equation*} \mathbb{R} \setminus f(\mathbb{R}) = (-\infty, L] \cup [M, \infty) \cup \bigcup_{x \in D} \Bigl( \bigl[f(x^-), f(x)\bigr) \cup \bigl(f(x), f(x^+)\bigr] \Bigr). \end{equation*}

No set on the right meets \(f(\mathbb{R})\). The infimum \(L\) is not attained (\(f(u) = L\) would give \(f(u-1) < L\)) and \(f(t) \ge L\) for all \(t\), so \((-\infty, L] \cap f(\mathbb{R}) = \emptyset\); symmetrically for \([M, \infty)\). If \(f(x) < f(x^+)\) and \(f(u) \in (f(x), f(x^+)]\), then \(u > x\) (as \(u \le x\) gives \(f(u) \le f(x)\)), so \(f(u) \ge f(x^+)\) and hence \(f(u) = f(x^+)\) — impossible, since \(f(u) = f(x^+)\) with \(u > x\) would give \(f(t) < f(x^+) = \inf\{f(s) : s > x\}\) for \(x < t < u\). So \((f(x), f(x^+)] \cap f(\mathbb{R}) = \emptyset\), and symmetrically \([f(x^-), f(x)) \cap f(\mathbb{R}) = \emptyset\).

Conversely, let \(y \notin f(\mathbb{R})\) with \(L < y < M\) (otherwise \(y\) lies in one of the first two sets). Then \(T = \{t \in \mathbb{R} : f(t) < y\}\) is nonempty because \(y > L\), and bounded above by any \(s\) with \(f(s) > y\) (such \(s\) exists as \(y < M\), and \(t \ge s\) would give \(f(t) \ge f(s) > y\)), so \(x = \sup T \in \mathbb{R}\). For \(t < x\) some \(t^{\prime} \in T\) has \(t < t^{\prime} \le x\), so \(f(t) \le f(t^{\prime}) < y\) and hence \(f(x^-) \le y\); for \(t > x\) we have \(t \notin T\), so \(f(t) \ge y\), indeed \(f(t) > y\) as \(y \notin f(\mathbb{R})\), giving \(f(x^+) \ge y\). Also \(f(x) \neq y\).

• (i) \(f(x) > y\): then \(y \in [f(x^-), f(x))\) and \(f(x^-) < f(x)\), so \(x \in D\).
• (ii) \(f(x) < y\): then \(y \in (f(x), f(x^+)]\) and \(f(x) < f(x^+)\), so \(x \in D\).

The right side is a countable union of intervals, hence Borel, so \(f(\mathbb{R})\) is Borel.

Take \(f_k : B \to \mathbb{R}\) defined by

\begin{equation*} f_k(x) = f(x) + \frac{x}{k} \qquad \text{for } k \in \mathbb{Z}^+ . \end{equation*}

If \(x, y \in B\) with \(x < y\), then \(f(x) \le f(y)\) because \(f\) is increasing and \(x/k < y/k\) because \(k > 0\), so

\begin{equation*} f_k(x) = f(x) + \frac{x}{k} < f(y) + \frac{y}{k} = f_k(y) ; \end{equation*}

thus each \(f_k\) is strictly increasing. For fixed \(x \in B\) we have \(x/k \to 0\), so \(f_k(x) \to f(x)\).

Take \(g = 0\) if \(B = \emptyset\), and otherwise

\begin{equation*} g(x) = \begin{cases} m & \text{if } B \cap (-\infty, x] = \emptyset, \\[2pt] \sup\{\, f(t) : t \in B,\ t \le x \,\} & \text{if } B \cap (-\infty, x] \neq \emptyset, \end{cases} \end{equation*}

where \(m = \inf f(B)\) and \(M = \sup f(B)\) are real because \(f\) is bounded and \(B \neq \emptyset\). In the second case the set is a nonempty subset of \(f(B)\), so its supremum lies in \([m, M] \subseteq \mathbb{R}\).

\(g\) is increasing: let \(x < y\). If \(B \cap (-\infty, x] = \emptyset\), then \(g(x) = m \le g(y)\), since \(g(y)\) is either \(m\) or a supremum of a subset of \(f(B)\), every element of which is at least \(m\). Otherwise \(B \cap (-\infty, y] \neq \emptyset\) as well and

\begin{equation*} \{\, f(t) : t \in B,\ t \le x \,\} \subseteq \{\, f(t) : t \in B,\ t \le y \,\} , \end{equation*}

so \(g(x) \le g(y)\).

\(g\) extends \(f\): for \(x \in B\) the second case applies, and \(f(x)\) both belongs to \(\{f(t) : t \in B,\ t \le x\}\) and bounds it above (as \(f\) is increasing), so \(g(x) = f(x)\).

False. Take \(X = \{1, 2\}\), \(\mathcal{S} = \{\emptyset, X\}\), and \(f : X \to [-\infty, \infty]\) given by

\begin{equation*} f(1) = \infty, \qquad f(2) = -\infty . \end{equation*}

Neither value lies in \(\mathbb{R}\), so \(f^{-1}\bigl((a, \infty)\bigr) = \emptyset \in \mathcal{S}\) for every \(a \in \mathbb{R}\). But \(\{\infty\}\) is a Borel subset of \([-\infty, \infty]\) by 2.50, since \(\{\infty\} \cap \mathbb{R} = \emptyset\) is Borel in \(\mathbb{R}\), while

\begin{equation*} f^{-1}\bigl(\{\infty\}\bigr) = \{1\} \notin \mathcal{S} . \end{equation*}

So \(f\) is not \(\mathcal{S}\)-measurable in the sense of 2.51.

Because \(g = f\) on \(B\) and \(g = 0\) off \(B\), every Borel set \(A \subseteq \mathbb{R}\) satisfies

\begin{equation*} g^{-1}(A) = \begin{cases} f^{-1}(A) \cup (\mathbb{R} \setminus B) & \text{if } 0 \in A, \\ f^{-1}(A) & \text{if } 0 \notin A, \end{cases} \end{equation*}

noting \(f^{-1}(A) \subseteq B\) since \(B\) is the domain of \(f\). Here \(f^{-1}(A)\) is Borel because \(f\) is Borel measurable, and \(\mathbb{R} \setminus B\) is Borel because \(B = f^{-1}(\mathbb{R})\) is Borel (as observed after 2.40) and the Borel sets form a \(\sigma\)-algebra. So \(g^{-1}(A)\) is Borel in both cases, and hence \(g\) is Borel measurable.

Take \(X = \mathbb{R}\) with

\begin{equation*} \mathcal{S} = \{E \subseteq \mathbb{R} : E \text{ or } \mathbb{R} \setminus E \text{ is countable}\} , \end{equation*}

a \(\sigma\)-algebra by the first bullet point of 2.28, and for \(t \in \mathbb{R}\) take

\begin{equation*} f_t = \chi_{\{t\}} \ \text{ if } t \in [0,1], \qquad f_t = 0 \ \text{ if } t \notin [0,1] , \end{equation*}

each mapping \(\mathbb{R}\) into \(\{0,1\} \subseteq [0,1]\). Each \(f_t\) is \(\mathcal{S}\)-measurable, since \(f_t^{-1}(B)\) is one of \(\emptyset\), \(\{t\}\), \(\mathbb{R} \setminus \{t\}\), \(\mathbb{R}\), all of which are countable or cocountable.

The pointwise supremum is \(f = \chi_{[0,1]}\): for \(x \in [0,1]\) the index \(t = x\) gives \(f_x(x) = 1\), the largest value available, while for \(x \notin [0,1]\) every \(f_t(x) = 0\) (either \(f_t \equiv 0\) or \(t \in [0,1]\), so \(t \neq x\)). But

\begin{equation*} f^{-1}\big((\tfrac{1}{2}, \infty)\big) = [0,1] \notin \mathcal{S} , \end{equation*}

since \([0,1]\) is uncountable and so is its complement, which contains \((1,2)\). Hence \(f\) is not \(\mathcal{S}\)-measurable.

Fix \(x \in \mathbb{R}\) and \(j \in \mathbb{Z}^{+}\), and set \(c = \cos(j!\,\pi x) \in [-1,1]\), so that \(c^{2k} = (c^2)^k\) with \(c^2 \in [0,1]\). Since \(1^k = 1\) for all \(k\) while \(0 \le r < 1\) forces \(r^k \to 0\), the inner limit

\begin{equation*} g_j(x) = \lim_{k \to \infty} \big(\cos(j!\,\pi x)\big)^{2k} = \begin{cases} 1 & \text{if } c^2 = 1,\\ 0 & \text{if } c^2 < 1 \end{cases} \end{equation*}

exists for every \(j\) and \(x\). Because \(|\cos\theta| = 1\) exactly when \(\theta\) is an integer multiple of \(\pi\),

\begin{equation*} |\cos(j!\,\pi x)| = 1 \iff j!\,\pi x \in \{m\pi : m \in \mathbb{Z}\} \iff j!\,x \in \mathbb{Z} , \end{equation*}

so \(g_j = \chi_{A_j}\) with \(A_j = \{ m/j! : m \in \mathbb{Z} \}\). Now take \(j \to \infty\).

• (i) \(x\) irrational: \(j!\,x \in \mathbb{Z}\) would make \(x = (j!\,x)/j!\) rational, so \(g_j(x) = 0\) for every \(j\) and the outer limit is \(0\).
• (ii) \(x = p/q\) with \(p \in \mathbb{Z}\), \(q \in \mathbb{Z}^{+}\): for \(j \ge q\) the integer \(q\) is a factor of \(j!\), so \(j!\,x = p \cdot (j!/q) \in \mathbb{Z}\) and \(g_j(x) = 1\); the outer limit is \(1\).

The set \(\{\mu(E) : E \in \mathcal{S}\}\) always contains its own upper bound \(\mu(X)\), whereas \([0,1)\) has no largest element.

Explicitly, if \(\{\mu(E) : E \in \mathcal{S}\} = [0,1)\), then \(\mu(X) < 1\) since \(X \in \mathcal{S}\), so there is \(t\) with \(\mu(X) < t < 1\), and \(t = \mu(E)\) for some \(E \in \mathcal{S}\). But \(E \subseteq X\), so 2.57(a) gives \[ t = \mu(E) \le \mu(X) < t, \] a contradiction.

Take \[ w_k = \mu(\{k\}) \in [0, \infty] \qquad \text{for } k \in \mathbb{Z}^+ . \] As in 2.55, \(\sum_{k \in E} w_k\) denotes the supremum of the subsums \(\sum_{k \in D} w_k\) over finite \(D \subseteq E\). Fix \(E \subseteq \mathbb{Z}^+\).

• (i) \(E\) finite, say \(E = \{k_1, \ldots, k_n\}\): finite additivity (from countable additivity by appending \(\emptyset, \emptyset, \ldots\), as noted after 2.54) applied to the disjoint singletons gives \(\mu(E) = \sum_{j=1}^{n} w_{k_j}\), and the supremum over finite \(D \subseteq E\) is attained at \(D = E\) since the \(w_k\) are nonnegative. (For \(E = \emptyset\) both sides are \(0\).)
• (ii) \(E\) infinite, listed increasingly as \(k_1 < k_2 < \cdots\): countable additivity applied to \(E = \bigcup_{j=1}^{\infty}\{k_j\}\) gives \(\mu(E) = \sum_{j=1}^{\infty} w_{k_j} = \sup_n \sum_{j=1}^{n} w_{k_j}\), the partial sums being increasing in \([0, \infty]\). That supremum equals the supremum over all finite \(D \subseteq E\): each \(\{k_1, \ldots, k_n\}\) is one such \(D\), and conversely each finite \(D \subseteq E\) lies inside some \(\{k_1, \ldots, k_n\}\), whose subsum is then at least that of \(D\).

Hence \(\mu(E) = \sum_{k \in E} w_k\) for every \(E \subseteq \mathbb{Z}^+\).

Take \[ \mu(E) = \sum_{k \in E} 2^{-k} \qquad \text{for } E \subseteq \mathbb{Z}^+ , \] a measure by the third bullet point of 2.55.

One inclusion: \(0 \le \mu(E) \le \mu(\mathbb{Z}^+) = \sum_{k=1}^{\infty} 2^{-k} = 1\), using 2.57(a).

For the other, let \(t \in [0,1]\); take \(E = \mathbb{Z}^+\) if \(t = 1\), so assume \(0 \le t < 1\). Define \(b_k \in \{0,1\}\) recursively by \(b_k = 1\) if \(s_{k-1} + 2^{-k} \le t\) and \(b_k = 0\) otherwise, where \(s_k = \sum_{j=1}^{k} b_j 2^{-j}\) and \(s_0 = 0\). Then \[ s_k \le t < s_k + 2^{-k} \qquad \text{for every } k \ge 0 , \] by induction: at \(k = 0\) this reads \(0 \le t < 1\); granting it at \(k-1\), if \(b_k = 1\) then \(s_k = s_{k-1} + 2^{-k} \le t\) and \(t < s_{k-1} + 2^{-(k-1)} = s_k + 2^{-k}\), while if \(b_k = 0\) then \(s_k = s_{k-1} \le t < s_{k-1} + 2^{-k} = s_k + 2^{-k}\). Hence \(0 \le t - s_k < 2^{-k}\), so \(s_k \to t\).

Now put \(E = \{k \in \mathbb{Z}^+ : b_k = 1\}\), so \(s_k\) is the subsum of \(\sum_{k \in E} 2^{-k}\) over the finite set \(E \cap \{1, \ldots, k\}\); thus \(s_k \le \mu(E)\), while every finite \(D \subseteq E\) satisfies \(D \subseteq E \cap \{1, \ldots, k\}\) for large \(k\) and hence \(\sum_{j \in D} 2^{-j} \le s_k\). Therefore \[ \mu(E) = \sup_{k} s_k = \lim_{k \to \infty} s_k = t , \] the suprema and limits agreeing because \(s_1 \le s_2 \le \cdots\).

Take \(X = \mathbb{Z}^+ \times \{1,2\}\) with \(\mathcal{S}\) all subsets of \(X\) and \(\mu(E) = \sum_{x \in E} w(x)\), where \[ w(k, 1) = 2^{-k}, \qquad w(k, 2) = 3 \qquad \text{for } k \in \mathbb{Z}^+ ; \] this is a measure by the third bullet point of 2.55 (the sums being suprema of finite subsums).

Writing \(E_1 = \{k : (k,1) \in E\}\) and \(E_2 = \{k : (k,2) \in E\}\), finite additivity applied to the disjoint pieces \(E_1 \times \{1\}\) and \(E_2 \times \{2\}\) gives \[ \mu(E) = s + b, \qquad s = \sum_{k \in E_1} 2^{-k}, \quad b = \sum_{k \in E_2} 3 , \] where \(b = 3n\) if \(E_2\) has exactly \(n\) elements and \(b = \infty\) if \(E_2\) is infinite, the finite subsums \(3, 6, 9, \ldots\) being unbounded. By Exercise 3 of this section, \(s\) ranges over exactly \([0,1]\) as \(E_1\) ranges over the subsets of \(\mathbb{Z}^+\); so \(b\) ranges over exactly \(\{0,3,6,\ldots\} \cup \{\infty\}\), and the two choices are independent because every pair \((E_1, E_2)\) arises from \(E = (E_1 \times \{1\}) \cup (E_2 \times \{2\})\). Since \(3k + s\) sweeps \([3k, 3k+1]\) as \(s\) sweeps \([0,1]\), while \(b = \infty\) forces \(\mu(E) = \infty\), \[ \{\mu(E) : E \in \mathcal{S}\} = \{\infty\} \cup \bigcup_{k=0}^{\infty} [3k, 3k+1] . \]

Write \(\mathcal{A} = \bigcup_{n=1}^{\infty} \mathcal{A}_n\), where \(\mathcal{A}_n = \{A \in \mathcal{A} : \mu(A) > \tfrac1n\}\); this is an equality because every \(A \in \mathcal{A}\) has \(\mu(A) > 0\) and hence \(\mu(A) > \tfrac1n\) for some \(n\) by the Archimedean property.

Each \(\mathcal{A}_n\) is finite: if \(A_1, \ldots, A_m\) are distinct elements of \(\mathcal{A}_n\), they are pairwise disjoint sets in \(\mathcal{S}\), so finite additivity (from countable additivity, as noted after 2.54) together with \(A_1 \cup \cdots \cup A_m \subseteq X\) and 2.57(a) gives \[ \frac{m}{n} < \sum_{j=1}^{m} \mu(A_j) = \mu(A_1 \cup \cdots \cup A_m) \le \mu(X) < \infty , \] so \(m < n\,\mu(X)\); were \(\mathcal{A}_n\) infinite it would supply \(m\) distinct elements with \(m \ge n\,\mu(X)\). Hence \(\mathcal{A}\) is a countable union of finite sets, so countable.

Only \(c = 4\).

Necessity. Suppose \(R := \{\mu(E) : E \in \mathcal{S}\} = [0,1] \cup [3,c]\). Since \(c \in R\), say \(c = \mu(E_0)\), and \(E_0 \subseteq X\) gives \(c \le \mu(X)\) by 2.57(a), while \(\mu(X) \in R\) gives \(\mu(X) \le c\), we get \(\mu(X) = c < \infty\). So for any \(t = \mu(E) \in R\) we have \(\mu(E) \le c < \infty\), and 2.57(b) applied to \(E \subseteq X\) yields \[ \mu(X \setminus E) = \mu(X) - \mu(E) = c - t , \] giving the reflection property \(t \in R \implies c - t \in R\).

• (i) \(t = 1\): \(c - 1 \in R\), and \(c - 1 \ge 2 > 1\) puts it in \([3,c]\), so \(c \ge 4\).
• (ii) \(c \ge 5\) is impossible: then \(c - 2 \in [3,c] \subseteq R\), so reflection gives \(2 \in R\), false.
• (iii) \(t = 3\): \(c - 3 \in R\), and \(4 \le c < 5\) gives \(1 \le c - 3 < 2\), whose only element in \(R\) is \(1\). So \(c = 4\).

Sufficiency. Take \(X = \{0,1,2,\ldots\}\) with \(\mathcal{S}\) all subsets of \(X\) and \(\mu(E) = \sum_{x \in E} w(x)\), where \(w(0) = 3\) and \(w(k) = 2^{-k}\) for \(k \in \mathbb{Z}^+\); this is a measure by the third bullet point of 2.55. Splitting \(E\) into \(E \cap \{0\}\) and \(E^{\prime} = E \cap \mathbb{Z}^+\), finite additivity gives \(\mu(E) = \sum_{k \in E^{\prime}} 2^{-k}\) if \(0 \notin E\) and \(\mu(E) = 3 + \sum_{k \in E^{\prime}} 2^{-k}\) if \(0 \in E\). By Exercise 3 of this section that sum ranges over exactly \([0,1]\), independently of whether \(0 \in E\), so \[ \{\mu(E) : E \in \mathcal{S}\} = [0,1] \cup \big(3 + [0,1]\big) = [0,1] \cup [3,4] . \]

Take \(X = \mathbb{Z}^+ \times \{1,2\}\) with \(\mathcal{S}\) all subsets of \(X\) and \(\mu(E) = \sum_{x \in E} w(x)\), where \[ w(k, 1) = 2^{-k}, \qquad w(k, 2) = k + 2 \qquad \text{for } k \in \mathbb{Z}^+ ; \] this is a measure by the third bullet point of 2.55.

With \(E_1 = \{k : (k,1) \in E\}\) and \(E_2 = \{k : (k,2) \in E\}\), finite additivity applied to the disjoint pieces \(E_1 \times \{1\}\) and \(E_2 \times \{2\}\) gives \(\mu(E) = s + b\), where \(s = \sum_{k \in E_1} 2^{-k}\) and \(b = \sum_{k \in E_2} (k+2)\); the two choices are independent, since every pair \((E_1, E_2)\) arises from \(E = (E_1 \times \{1\}) \cup (E_2 \times \{2\})\). By Exercise 3 of this section \(s\) ranges over exactly \([0,1]\), and the set \(B\) of possible values of \(b\) satisfies:

• (i) \(B \subseteq \{0\} \cup [3, \infty]\), since \(E_2 = \emptyset\) gives \(b = 0\) while any \(k \in E_2\) gives \(b \ge k + 2 \ge 3\);
• (ii) every integer \(n \ge 3\) lies in \(B\), via \(E_2 = \{n-2\}\);
• (iii) \(\infty \in B\), via \(E_2 = \mathbb{Z}^+\), whose finite subsums \(\sum_{k=1}^{m}(k+2)\) are unbounded.

So by (i) each \(\mu(E)\) is either \(s \in [0,1]\) or \(s + b \ge 3\), putting the range inside \([0,1] \cup [3,\infty]\). Conversely, \(t \in [0,1]\) arises from \(E_2 = \emptyset\) with \(E_1\) chosen by Exercise 3 so that \(s = t\); \(t = \infty\) arises from \(E_2 = \mathbb{Z}^+\) by (iii); and \(t \in [3,\infty)\) arises from \(E_2 = \{n-2\}\) with \(n = \lfloor t \rfloor \ge 3\), giving \(b = n\) by (ii), together with \(E_1\) chosen so that \(s = t - n \in [0,1)\). Hence \[ \{\mu(E) : E \in \mathcal{S}\} = [0, 1] \cup [3, \infty] . \]

Take

\begin{equation*} X = \{1, 2, 3, 4\}, \qquad \mathcal{S} = \{E : E \subseteq X\}, \qquad \mathcal{A} = \big\{\{1,2\}, \{2,3\}\big\}, \end{equation*}

and let \(\mu\) and \(\nu\) be the measures on \((X, \mathcal{S})\) defined by

\begin{equation*} \mu = \delta_2 + \delta_4, \qquad \nu = \delta_1 + \delta_3, \end{equation*}

where \(\delta_c\) is the Dirac measure at \(c\) (2.55), so that \(\mu\) and \(\nu\) are measures by Exercise 9 of this section. Explicitly, for \(E \subseteq X\),

\begin{equation*} \mu(E) = \#\big(E \cap \{2,4\}\big), \qquad \nu(E) = \#\big(E \cap \{1,3\}\big). \end{equation*}

Here \(\mathcal{S}\) is the smallest \(\sigma\)-algebra on \(X\) containing \(\mathcal{A}\): any \(\sigma\)-algebra \(\mathcal{T} \supseteq \mathcal{A}\) is closed under finite intersections and set differences, so

\begin{equation*} \begin{aligned} \{2\} &= \{1,2\} \cap \{2,3\}, \qquad \{1\} = \{1,2\} \setminus \{2\}, \\ \{3\} &= \{2,3\} \setminus \{2\} \end{aligned} \end{equation*}

lie in \(\mathcal{T}\), hence so does \(\{4\} = X \setminus (\{1\} \cup \{2\} \cup \{3\})\); as \(X\) is finite, every subset is a finite union of singletons and \(\mathcal{T} = \mathcal{S}\).

Finally \(\mu\) and \(\nu\) agree on \(\mathcal{A}\) and on \(X\) but differ at \(\{1\}\):

\begin{equation*} \mu(\{1,2\}) = \nu(\{1,2\}) = \mu(\{2,3\}) = \nu(\{2,3\}) = 1 , \qquad \mu(X) = \nu(X) = 2 , \end{equation*}

while \(\mu(\{1\}) = 0 \neq 1 = \nu(\{1\})\).

The map \(\mu + \nu\) sends \(\mathcal{S}\) into \([0,\infty]\), which is closed under addition with the convention \(a + \infty = \infty\), and \((\mu + \nu)(\varnothing) = 0 + 0 = 0\); only countable additivity needs work.

Series in \([0,\infty]\) add termwise: for sequences \((a_k)\), \((b_k)\) in \([0, \infty]\),

\begin{equation*} \sum_{k=1}^{\infty} (a_k + b_k) = \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} b_k , \end{equation*}

each series being the supremum of its increasing partial sums. Indeed, writing \(A_n\), \(B_n\), \(C_n\) for the \(n\)-th partial sums and \(A\), \(B\), \(C\) for their suprema, rearranging a finite sum gives \(C_n = A_n + B_n\). If \(A = \infty\), then \(C_n \geq A_n\) is unbounded, so \(C = \infty = A + B\), and symmetrically if \(B = \infty\). If \(A, B < \infty\), then \(C_n \leq A + B\) gives \(C \leq A + B\), while for \(\varepsilon > 0\) a single \(n\) with \(A_n > A - \varepsilon/2\) and \(B_n > B - \varepsilon/2\) (monotonicity) gives \(C_n > A + B - \varepsilon\); hence \(C = A + B\).

So for a disjoint sequence \(E_1, E_2, \dots\) in \(\mathcal{S}\), countable additivity of \(\mu\) and of \(\nu\) gives

\begin{equation*} \begin{aligned} (\mu + \nu)\Big(\bigcup_{k=1}^{\infty} E_k\Big) &= \sum_{k=1}^{\infty} \mu(E_k) + \sum_{k=1}^{\infty} \nu(E_k) \\ &= \sum_{k=1}^{\infty} \big(\mu(E_k) + \nu(E_k)\big) = \sum_{k=1}^{\infty} (\mu + \nu)(E_k) . \end{aligned} \end{equation*}

Hence \(\mu + \nu\) is a measure on \((X, \mathcal{S})\).

Take counting measure \(\mu\) on \(X = \mathbf{Z}^{+}\) with \(\mathcal{S}\) the \(\sigma\)-algebra of all subsets of \(X\) (a measure by 2.55), and

\begin{equation*} E_k = \{k, k+1, k+2, \dots\} \quad \text{for } k \in \mathbf{Z}^{+} , \end{equation*}

a decreasing sequence in \(\mathcal{S}\). Each \(E_k\) is infinite, so \(\mu(E_k) = \infty\) for every \(k\), while \(\bigcap_{k=1}^{\infty} E_k = \varnothing\) because \(n \notin E_{n+1}\) for every \(n \in \mathbf{Z}^{+}\). Hence

\begin{equation*} \mu\Big(\bigcap_{k=1}^{\infty} E_k\Big) = 0 \neq \infty = \lim_{k \to \infty} \mu(E_k) . \end{equation*}

The formula is inclusion-exclusion for three sets:

\begin{equation*} \begin{aligned} \mu(C \cup D \cup E) = {} & \mu( C) + \mu(D) + \mu(E) \\ & {} - \mu(C \cap D) - \mu(C \cap E) - \mu(D \cap E) + \mu(C \cap D \cap E) . \end{aligned} \end{equation*}

Write \(p = \mu(C \cap D)\), \(q = \mu(C \cap E)\), \(r = \mu(D \cap E)\), \(s = \mu(C \cap D \cap E)\). The first three are finite by hypothesis and \(s \leq p < \infty\) by 2.57(a), so only finite quantities are subtracted above (convention \(\infty - t = \infty\) for real \(t\)) and the formula is equivalent to the subtraction-free identity

\begin{equation*} \mu(C \cup D \cup E) + p + q + r = \mu( C) + \mu(D) + \mu(E) + s . \end{equation*}

Since \(p < \infty\), 2.61 applied to \(C\) and \(D\) gives \(\mu(C \cup D) + p = \mu( C) + \mu(D)\). Since \((C \cap E) \cap (D \cap E) = C \cap D \cap E\) has measure \(s < \infty\), 2.61 applied to \(C \cap E\) and \(D \cap E\), whose union is \((C \cup D) \cap E\), gives

\begin{equation*} \mu\big((C \cup D) \cap E\big) + s = q + r ; \end{equation*}

in particular \(\mu\big((C \cup D) \cap E\big) \leq q + r < \infty\), so 2.61 applied to \(C \cup D\) and \(E\) gives

\begin{equation*} \mu(C \cup D \cup E) + \mu\big((C \cup D) \cap E\big) = \mu(C \cup D) + \mu(E) . \end{equation*}

Adding \(s\) and then \(p\) to this last identity and substituting the previous two yields

\begin{equation*} \mu(C \cup D \cup E) + p + q + r = \mu( C) + \mu(D) + \mu(E) + s , \end{equation*}

every rearrangement being legitimate because addition on \([0, \infty]\) is commutative and associative. Subtracting the finite quantity \(p + q + r\) gives the formula.

The measures on \((X, \mathcal{S})\) are exactly the functions

\begin{equation*} \mu_{w,c}(E) = \sum_{x \in E} w(x) \; + \; c \cdot [E \text{ uncountable}] , \end{equation*}

with \(w : X \to [0, \infty]\) and \(c \in [0, \infty]\) arbitrary: a weighted counting measure with weights \(w(x) = \mu(\{x\})\), plus a lump of mass \(c\) carried by every uncountable set in \(\mathcal{S}\). Here \(\sum_{x \in A} w(x)\) is the supremum of the finite subsums, as in the third bullet of 2.55; \([E \text{ uncountable}]\) is \(1\) if \(E\) is uncountable and \(0\) otherwise; and \(0 \cdot \infty = 0\), so a countable \(E\) carries no lump even when \(c = \infty\).

Three facts carry the proof.

(L1) Unordered sums are countably additive: if \(A = \bigcup_{k=1}^{\infty} A_k\) with the \(A_k\) disjoint, then \(\sum_{x \in A} w(x) = \sum_{k=1}^{\infty} \sum_{x \in A_k} w(x)\). Indeed a finite \(D \subseteq A\) meets only finitely many \(A_k\), say those with \(k \le n\), and is the disjoint union of the \(D \cap A_k\), so \(\sum_{x \in D} w(x) \le \sum_{k=1}^{n} \sum_{x \in A_k} w(x)\); taking the supremum over \(D\) gives \(\le\). Conversely finite sets \(D_k \subseteq A_k\) for \(k \le n\) have union a finite subset of \(A\), so \(\sum_{k=1}^{n} \sum_{x \in D_k} w(x) \le \sum_{x \in A} w(x)\); taking the \(n\) suprema one at a time (legitimate since addition on \([0,\infty]\) is order preserving and commutes with suprema of increasing families in one variable) and letting \(n \to \infty\) gives \(\ge\).

(L2) If \(N\) is uncountable and \(w > 0\) on \(N\), then \(\sum_{x \in N} w(x) = \infty\): from \(N = \bigcup_{n=1}^{\infty} \{x \in N : w(x) > 1/n\}\) and the countability of a countable union of countable sets, some \(\{x \in N : w(x) > 1/n\}\) is uncountable, hence contains finite sets \(D\) of arbitrarily many elements, with \(\sum_{x \in D} w(x) > \#D/n\).

(L3) If \(E_1, E_2, \dots \in \mathcal{S}\) are disjoint with union \(E\), then \(E\) is uncountable if and only if exactly one \(E_k\) is. At most one can be, since an uncountable set of \(\mathcal{S}\) has countable complement while \(E_i \cap E_j = \varnothing\) forces \(E_i \subseteq X \setminus E_j\); and \(E\) is countable when all \(E_k\) are, uncountable when some \(E_k\) is.

Each \(\mu_{w,c}\) is a measure: it maps \(\mathcal{S}\) into \([0,\infty]\) with \(\mu_{w,c}(\varnothing) = 0\), and for disjoint \(E_1, E_2, \dots\) with union \(E\), (L3) gives \([E \text{ uncountable}] = \sum_{k=1}^{\infty} [E_k \text{ uncountable}]\) with at most one nonzero term, so this may be multiplied by \(c\) termwise unambiguously even when \(c = \infty\); combining that with (L1) and with termwise addition of series in \([0,\infty]\) (Exercise 9 of this section),

\begin{equation*} \mu_{w,c}(E) = \sum_{k=1}^{\infty} \Big( \sum_{x \in E_k} w(x) + c\,[E_k \text{ uncountable}] \Big) = \sum_{k=1}^{\infty} \mu_{w,c}(E_k) . \end{equation*}

Conversely, let \(\mu\) be a measure, put \(w(x) = \mu(\{x\})\) (singletons are countable, hence in \(\mathcal{S}\)) and \(N = \{x \in X : w(x) > 0\}\). For countable \(E \in \mathcal{S}\), countable additivity along the disjoint singletons \(\{x_1\}, \{x_2\}, \dots\) exhausting \(E\) gives \(\mu(E) = \sum_k w(x_k)\), which equals \(\sum_{x \in E} w(x)\) because the partial sums are the subsums over \(\{x_1, \dots, x_n\}\) and these are cofinal among finite subsets of \(E\). Also \(\mu(E) \ge \mu(D) = \sum_{x \in D} w(x)\) for finite \(D \subseteq E\) by 2.57(a) and finite additivity, so \(\mu(E) \ge \sum_{x \in E} w(x)\) for every \(E \in \mathcal{S}\).

• (i) \(N\) uncountable. If \(G \in \mathcal{S}\) is uncountable then \(X \setminus G\) is countable, so \(G \cap N = N \setminus (X \setminus G)\) is uncountable and (L2) gives \(\sum_{x \in G} w(x) = \infty\), whence \(\mu(G) = \infty\) by the inequality above. So \(\mu = \mu_{w,0}\).
• (ii) \(N\) countable. Put \(c = \mu(X \setminus N)\). For uncountable \(G \in \mathcal{S}\), the sets \(G \cap N\) and \(G \setminus N\) lie in \(\mathcal{S}\) (the first countable, the second with countable complement \((X \setminus G) \cup N\)) and partition \(G\), so \(\mu(G) = \mu(G \cap N) + \mu(G \setminus N)\). Here \(\mu(G \cap N) = \sum_{x \in G \cap N} w(x) = \sum_{x \in G} w(x)\), as \(w\) vanishes off \(N\); and splitting \(X \setminus N\) into \(G \setminus N\) and \((X \setminus N) \cap (X \setminus G)\), the latter countable with \(w \equiv 0\) on it and hence of measure \(0\), gives \(\mu(G \setminus N) = c\). So \(\mu = \mu_{w,c}\).

The set \(E\) of (a) is a countable union of closed intervals intersected with \((0,1)\), and \(|E| = 1\).

For a digit string \(s = (d_1,\ldots,d_m)\) put \(a_s = \sum_{j=1}^{m} d_j 10^{-j}\) and \(I_s = [a_s, a_s + 10^{-m}]\). Then \(x \in [0,1]\) has a decimal expansion beginning with \(s\) if and only if \(x \in I_s\), since the tail \(\sum_{j>m} d_j 10^{-j}\) ranges over exactly \([0,10^{-m}]\). (Check!)

(a). Let \(F\) be the string of one hundred 4s and let \(pF\) denote concatenation. A hundred consecutive 4s must begin in some position \(n\), preceded by some \(p \in \{0,\ldots,9\}^{n-1}\), so

\begin{equation*} E = (0,1) \cap \bigcup_{n=1}^{\infty} \ \bigcup_{p \in \{0,\ldots,9\}^{n-1}} I_{pF}. \end{equation*}

Only \(10^{\,n-1}\) strings \(p\) occur at each \(n\), so this is a countable union of closed intervals; hence \(E\) is Borel.

(b). Fix \(N\) and call \(s \in \{0,\ldots,9\}^{100N}\) good if none of its \(N\) blocks of length \(100\) is a hundred 4s; there are \((10^{100}-1)^{N}\) good strings, one choice per block. If \(x \in (0,1)\setminus E\), the first \(100N\) digits of any expansion of \(x\) form a good \(s\) (else \(x \in E\)), and then \(x \in I_s\); so \((0,1)\setminus E\) is covered by the good \(I_s\), whence by 2.5, 2.8, and \(|I_s| = 10^{-100N}\) (2.14),

\begin{equation*} |(0,1)\setminus E| \le (10^{100}-1)^{N} \cdot 10^{-100N} = \bigl(1 - 10^{-100}\bigr)^{N} . \end{equation*}

Letting \(N \to \infty\) gives \(|(0,1)\setminus E| = 0\). Now \(E \subseteq [0,1]\) gives \(|E| \le 1\) by 2.5 and 2.14, while \([0,1] = E \cup ((0,1)\setminus E) \cup \{0,1\}\) with \(|\{0,1\}| = 0\) (2.4) gives \(1 \le |E|\) by 2.8. Hence \(|E| = 1\).

Take \(A = \frac{1}{c}V\), where \(V \subseteq [-1,1]\) is the set constructed in the proof of 2.18 and \(c = |V|\).

Two facts from that proof: \(|V| > 0\), and \(V\) contains exactly one element of each \(\widetilde{a} = \{t \in [-1,1] : a - t \in \mathbb{Q}\}\), so that \((r+V) \cap (r^{\prime}+V) = \varnothing\) for distinct \(r, r^{\prime} \in \mathbb{Q}\) (if \(r + v_1 = r^{\prime} + v_2\) then \(v_1 - v_2 \in \mathbb{Q}\), placing \(v_1, v_2\) in one \(\widetilde{a}\), so \(v_1 = v_2\) and \(r = r^{\prime}\)). Note \(0 < c \le 2\) by 2.5 and 2.14.

Every closed \(F \subseteq V\) has \(|F| = 0\). Let \(r_1, r_2, \ldots\) be distinct rationals in \([0,1]\); the sets \(r_k + F\) are then disjoint closed, hence Borel, subsets of \([-1,2]\). Since outer measure is a measure on the Borel sets (2.68) and is translation invariant (2.7), while 2.5 and 2.14 bound the union,

\begin{equation*} 3 = |[-1,2]| \ \ge\ \Bigl| \bigcup_{k=1}^{\infty} (r_k + F) \Bigr| = \sum_{k=1}^{\infty} |F| , \end{equation*}

which forces \(|F| = 0\).

Now \(A = \frac{1}{c}V \subseteq [-\frac{1}{c}, \frac{1}{c}]\) is bounded, and the dilation formula \(|tB| = |t|\,|B|\) of Exercise 2 in Section 2A gives \(|A| = \frac{1}{c}|V| = 1\). If \(F \subseteq A\) is closed, then \(cF\) is a closed subset of \(cA = V\), so \(|cF| = 0\) by the previous paragraph and hence \(|F| = \frac{1}{c}|cF| = 0\). Thus

\begin{equation*} |F| = 0 = |A| - 1 \end{equation*}

for every closed \(F \subseteq A\).

Take \(A = \bigcup_{n=1}^{\infty} (n+V)\), where \(V \subseteq (0,1)\) contains exactly one element of each set \(\widetilde{a} = \{t \in (0,1) : a - t \in \mathbb{Q}\}\), \(a \in (0,1)\). Throughout, \(|(a,b)| = b-a\) is Exercise 6 in Section 2A.

Exactly as in the proof of 2.18, \((r+V) \cap (r^{\prime}+V) = \varnothing\) for distinct \(r, r^{\prime} \in \mathbb{Q}\). Each \(a \in (0,1)\) satisfies \(a - v \in \mathbb{Q} \cap (-1,1)\) for its representative \(v \in V\), so \((0,1) \subseteq \bigcup_k (q_k + V)\) with \(q_1, q_2, \ldots\) enumerating \(\mathbb{Q} \cap (-1,1)\); then 2.8 and 2.7 give \(1 \le \sum_k |V|\), forcing \(|V| > 0\).

\(V\) is not Lebesgue measurable. First, every closed \(F \subseteq V\) has \(|F| = 0\): for distinct rationals \(r_1, r_2, \ldots \in [0,1]\) the sets \(r_k + F\) are disjoint closed, hence Borel, subsets of \((0,2)\), so by 2.68, 2.7, 2.5, and 2.14,

\begin{equation*} 2 \ge \Bigl| \bigcup_{k=1}^{\infty} (r_k + F) \Bigr| = \sum_{k=1}^{\infty} |F| . \end{equation*}

Were \(V\) measurable, then since \(0 < |V| \le 1\), condition (b) of 2.71 with \(\varepsilon = |V|/2\) would give a closed \(F \subseteq V\) with \(|V \setminus F| < |V|/2\), whence \(|V| \le |F| + |V \setminus F| < |V|/2\) by 2.8, a contradiction.

The constant. Put \(c = |V| + |(0,1)\setminus V| - 1\), a real number since both terms lie in \([0,1]\) by 2.5. Subadditivity (2.8) gives \(c \ge 0\), and Exercise 12 of this section gives \(c \ne 0\) because \(V \subseteq (0,1)\) is not measurable. So \(c > 0\).

The estimate on one unit interval. Since \(V \subseteq (0,1)\), we have \(A \cap (n,n+1) = n+V\). Let \(G \supseteq A\) be open, fix \(n \in \mathbb{Z}^{+}\), and put \(W = G \cap (n,n+1)\) and \(E = (n,n+1) \setminus W\), a Borel (hence measurable) set disjoint from \(n + V \subseteq W\). Exercise 10 of this section, applied with \(E\) as the measurable set, and 2.5 give

\begin{equation*} |E| + |n+V| = \bigl| E \cup (n+V) \bigr| \le |(n,n+1)| = 1 , \end{equation*}

so \(|E| \le 1 - |V|\) by 2.7. Since \((n,n+1) \setminus (n+V) \subseteq E \cup \bigl( W \setminus (n+V) \bigr)\), subadditivity (2.8) and 2.7 give

\begin{equation*} |(0,1) \setminus V| \le |E| + \bigl| W \setminus (n+V) \bigr| \le 1 - |V| + \bigl| W \setminus (n+V) \bigr| , \end{equation*}

that is, \(\bigl| (G \setminus A) \cap (n,n+1) \bigr| = \bigl| W \setminus (n+V) \bigr| \ge c\).

Summing. Fix \(N \in \mathbb{Z}^{+}\) and put \(S = (G \setminus A) \cap (1, N+1)\). By the Lemma in Exercise 10 of this section there is a Borel set \(B \supseteq S\) with \(|B| = |S|\); the sets \(B \cap (n,n+1)\) for \(1 \le n \le N\) are disjoint Borel sets, the \(n\)-th containing \((G\setminus A) \cap (n,n+1)\), so by 2.5 and 2.68,

\begin{equation*} |G \setminus A| \ge |S| = |B| \ge \sum_{n=1}^{N} \bigl| B \cap (n,n+1) \bigr| \ge Nc . \end{equation*}

Letting \(N \to \infty\) gives \(|G \setminus A| = \infty\).

(a). Let \(\mathcal{C}\) be such a collection, \(U = \bigcup_{I \in \mathcal{C}} I\), and \(W = \bigcup_{I \in \mathcal{C}} I^{\circ}\); since each \(I \in \mathcal{C}\) has at least two points, \(I^{\circ}\) is a nonempty open interval and \(I \setminus I^{\circ}\) consists of at most the two endpoints of \(I\).

Let \(D\) be the countable set of rational pairs \((p,q)\), \(p<q\), with \((p,q) \subseteq I^{\circ}\) for at least one \(I \in \mathcal{C}\), and fix one such \(I_{p,q}\) for each. If \(x \in W\) then \(x \in I^{\circ}\) for some \(I\), and \(I^{\circ}\) is open, so \(p < x < q\) with \((p,q) \subseteq I^{\circ}\) for some rationals \(p, q\); hence

\begin{equation*} W \subseteq \bigcup_{(p,q) \in D} I_{p,q} \subseteq U . \end{equation*}

Next, \(U \setminus W\) is countable. If \(x \in U \setminus W\), say \(x \in I \in \mathcal{C}\), then \(x \notin I^{\circ}\), so \(x\) is an endpoint of \(I\) and \(I^{\circ}\) is a nonempty open interval with endpoint \(x\); thus \((x, x+\delta_x) \subseteq W\) or \((x - \delta_x, x) \subseteq W\) for some \(\delta_x > 0\). Write \(U \setminus W = U_{+} \cup U_{-}\) accordingly. For \(x \in U_{+}\) pick a rational \(r_x \in (x, x+\delta_x)\); the map \(x \mapsto r_x\) is injective on \(U_{+}\), since \(x < y\) and \(r_x = r_y\) would give \(y \in (x, x+\delta_x) \subseteq W\). So \(U_{+}\) is countable, and symmetrically so is \(U_{-}\).

Choosing \(J_x \in \mathcal{C}\) with \(x \in J_x\) for each \(x \in U \setminus W\), the collection \(\mathcal{D} = \{I_{p,q}\} \cup \{J_x\}\) is a countable subcollection of \(\mathcal{C}\) with

\begin{equation*} U = W \cup (U \setminus W) \subseteq \bigcup_{J \in \mathcal{D}} J \subseteq U . \end{equation*}

(b). By (a), \(U\) is a countable union of intervals, and every interval \(J \ne \varnothing\) is Borel: with \(a = \inf J\) and \(b = \sup J\), the set \(\{x \in \mathbb{R} : a < x < b\}\) is open and \(J\) differs from it by at most the points \(a, b\), each Borel because \(\{a\} = \bigcap_{m=1}^{\infty} (a - \frac{1}{m}, a+\frac{1}{m})\). As the Borel sets form a \(\sigma\)-algebra, \(U\) is Borel.

(c). Part (c) drops nontrivial, so singletons are allowed. By 2.67 there is a non-Borel set \(S \subseteq \mathbb{R}\); the collection \(\mathcal{C} = \{[x,x] : x \in S\}\) consists of closed intervals and has union \(S\).

Take \(F_k = E_1 \cup \cdots \cup E_k\), where (as \(A\) is Lebesgue measurable, so that (a) implies (b) in 2.71 applies) \(E_k \subseteq A\) is closed with \(|A \setminus E_k| < \frac{1}{k}\).

Each \(F_k\) is a finite union of closed subsets of \(A\), hence is closed and contained in \(A\); clearly \(F_1 \subseteq F_2 \subseteq \cdots\) and \(\bigcup_{k} F_k = \bigcup_{k} E_k\). Fixing \(n\) and using \(E_n \subseteq \bigcup_k F_k\) with 2.5,

\begin{equation*} \Bigl| A \setminus \bigcup_{k=1}^{\infty} F_k \Bigr| \le |A \setminus E_n| < \tfrac{1}{n} . \end{equation*}

Letting \(n \to \infty\) gives \(\bigl| A \setminus \bigcup_{k} F_k \bigr| = 0\).

Suppose \(A\) is Lebesgue measurable, and let \(\varepsilon > 0\). Because (a) implies (e) in 2.71, there is an open \(U \supseteq A\) with \(|U \setminus A| < \frac{\varepsilon}{2}\), and then \(|U| \le |A| + |U \setminus A| < \infty\) by 2.8 and the hypothesis \(|A| < \infty\). If \(U = \varnothing\) then \(A = \varnothing\) and \(G = (0, \varepsilon/3)\) serves; so assume \(U \ne \varnothing\) and write \(U = \bigcup_n I_n\) with \(I_1, I_2, \ldots\) disjoint nonempty open intervals (the connected components, as in the proof of 2.63), padding a finite list with copies of \(\varnothing\). Each \(I_n\) is bounded, since an unbounded one contains closed intervals of arbitrary length and would force \(|U| = \infty\) by 2.5 and 2.14. The \(I_n\) are disjoint Borel sets, so \(\sum_n |I_n| = |U| < \infty\) by 2.68; choose \(N\) with \(\sum_{n>N} |I_n| < \frac{\varepsilon}{2}\) and put \(G = I_1 \cup \cdots \cup I_N\). Since \(G \setminus A \subseteq U \setminus A\) and \(A \setminus G \subseteq U \setminus G = \bigcup_{n>N} I_n\), 2.5 and 2.8 give

\begin{equation*} |A \setminus G| + |G \setminus A| \le \sum_{n > N} |I_n| + |U \setminus A| < \varepsilon . \end{equation*}

Conversely, let \(\varepsilon > 0\) and choose such a \(G\) with \(|A \setminus G| + |G \setminus A| < \frac{\varepsilon}{2}\). As \(|A \setminus G| < \frac{\varepsilon}{2}\), the definition of outer measure (2.1) supplies open intervals \(J_k\) with \(A \setminus G \subseteq H = \bigcup_k J_k\) and \(\sum_k \ell(J_k) < \frac{\varepsilon}{2}\), so \(|H| < \frac{\varepsilon}{2}\) by 2.8 and 2.14. Then \(V = G \cup H\) is open, \(A = (A \cap G) \cup (A \setminus G) \subseteq V\), and \(V \setminus A \subseteq (G \setminus A) \cup H\), so by 2.5 and 2.8,

\begin{equation*} |V \setminus A| \le |G \setminus A| + |H| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

Thus \(A\) satisfies condition (e) of 2.71, and (e) implies (a) there, so \(A\) is Lebesgue measurable.

Take \(G_k = U_1 \cap \cdots \cap U_k\), where (as \(A\) is Lebesgue measurable, so that (a) implies (e) in 2.71 applies) \(U_k \supseteq A\) is open with \(|U_k \setminus A| < \frac{1}{k}\).

Each \(G_k\) is a finite intersection of open sets containing \(A\), hence is open and contains \(A\); clearly \(G_1 \supseteq G_2 \supseteq \cdots\) and \(\bigcap_k G_k = \bigcap_k U_k\). Fixing \(n\) and using \(\bigcap_k G_k \subseteq U_n\) with 2.5,

\begin{equation*} \Bigl| \Bigl( \bigcap_{k=1}^{\infty} G_k \Bigr) \setminus A \Bigr| \le |U_n \setminus A| < \tfrac{1}{n} . \end{equation*}

Letting \(n \to \infty\) gives the stated equality.

Translate the Borel witness of 2.70. Take a Borel set \(B \subseteq A\) with \(|A \setminus B| = 0\) (2.70). Then \(t + B\) is Borel by the paragraph below and \(t+B \subseteq t+A\); moreover \((t+A) \setminus (t+B) = t + (A \setminus B)\), since \(x\) belongs to the left side exactly when \(x - t \in A \setminus B\). Hence by translation invariance of outer measure (2.7),

\begin{equation*} \bigl| (t+A) \setminus (t+B) \bigr| = \bigl| t + (A \setminus B) \bigr| = |A \setminus B| = 0 , \end{equation*}

so \(t + A\) is Lebesgue measurable by 2.70.

The Borel sets are translation invariant. Fix \(t \in \mathbb{R}\) and put \(\mathcal{S} = \{E \subseteq \mathbb{R} : t+E \text{ is Borel}\}\). Since \(x \mapsto t+x\) is a bijection of \(\mathbb{R}\), we have \(t + \varnothing = \varnothing\), \(t + (\mathbb{R} \setminus E) = \mathbb{R} \setminus (t+E)\), and \(t + \bigcup_k E_k = \bigcup_k (t + E_k)\), so \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). It contains every open \(G\), because \(t + G\) is open: if \(x \in t+G\) then \((x-t-\delta, x-t+\delta) \subseteq G\) for some \(\delta > 0\), whence \((x-\delta, x+\delta) \subseteq t+G\). As the Borel sets form the smallest \(\sigma\)-algebra containing the open sets (2.29), every Borel set lies in \(\mathcal{S}\).

Dilate the Borel witness of 2.70, exactly as in Exercise 8 of this section, using the dilation formula \(|tA| = |t|\,|A|\) of Exercise 2 in Section 2A in place of 2.7.

(i) \(t = 0\): then \(0A\) is \(\varnothing\) or \(\{0\}\), in either case closed, hence Borel, hence Lebesgue measurable.

(ii) \(t \ne 0\): take a Borel set \(B \subseteq A\) with \(|A \setminus B| = 0\) (2.70). Then \(tB\) is Borel by the paragraph below and \(tB \subseteq tA\); also \(tA \setminus tB = t(A \setminus B)\), since \(x\) belongs to the left side exactly when \(\frac{x}{t} \in A \setminus B\). So by Exercise 2 in Section 2A,

\begin{equation*} |tA \setminus tB| = \bigl| t(A \setminus B) \bigr| = |t| \, |A \setminus B| = 0 , \end{equation*}

and \(tA\) is Lebesgue measurable by 2.70.

The Borel sets are dilation invariant. Fix \(t \ne 0\) and put \(\mathcal{S} = \{E \subseteq \mathbb{R} : tE \text{ is Borel}\}\). Since \(x \mapsto tx\) is a bijection of \(\mathbb{R}\) with inverse \(x \mapsto \frac{x}{t}\), we have \(t\varnothing = \varnothing\), \(t(\mathbb{R} \setminus E) = \mathbb{R} \setminus tE\), and \(t\bigcup_k E_k = \bigcup_k tE_k\), so \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). It contains every open \(G\), because \(tG\) is open: if \(x \in tG\) then \(\bigl(\frac{x}{t}-\delta, \frac{x}{t}+\delta\bigr) \subseteq G\) for some \(\delta > 0\), so the interval of radius \(|t|\delta\) centred at \(x\) lies in \(tG\). As the Borel sets form the smallest \(\sigma\)-algebra containing the open sets (2.29), every Borel set lies in \(\mathcal{S}\).

Subadditivity (2.8) gives \(|A \cup B| \le |A| + |B|\), so only the reverse inequality is at issue.

Lemma (Borel hull; used again in Exercises 3 and 12 of this section). Every \(S \subseteq \mathbb{R}\) lies in a Borel set \(E\) with \(|E| = |S|\). Indeed, if \(|S| = \infty\) take \(E = \mathbb{R}\). Otherwise, for each \(n \in \mathbb{Z}^{+}\) the infimum defining outer measure (2.2) supplies open intervals \(I_1, I_2, \ldots\) with \(S \subseteq G_n = \bigcup_k I_k\) and \(\sum_k \ell(I_k) < |S| + \frac{1}{n}\), whence \(|G_n| < |S| + \frac{1}{n}\) by 2.8 and \(|I_k| \le \ell(I_k)\). Then \(E = \bigcap_n G_n\) is Borel, contains \(S\), and by 2.5 satisfies \(|S| \le |E| \le |G_n| < |S| + \frac{1}{n}\) for every \(n\).

Now suppose \(A, B \subseteq \mathbb{R}\) are disjoint with \(B\) Lebesgue measurable, and take a Borel set \(D \subseteq B\) with \(|B \setminus D| = 0\) (2.70); then

\begin{equation*} |B| \le |D| + |B \setminus D| = |D| \le |B| \end{equation*}

by 2.8 and 2.5, so \(|D| = |B|\). By the Lemma take a Borel set \(E \supseteq A \cup B\) with \(|E| = |A \cup B|\). Since \(D \subseteq B \subseteq E\), the sets \(D\) and \(E \setminus D\) are disjoint Borel sets with union \(E\), so \(|E| = |D| + |E \setminus D|\) because outer measure is a measure on the Borel sets (2.68). Finally \(A \subseteq E \setminus D\), since \(A \subseteq E\) and \(A \cap D \subseteq A \cap B = \emptyset\); hence \(|A| \le |E \setminus D|\) by 2.5 and

\begin{equation*} |A \cup B| = |E| = |D| + |E \setminus D| \ge |B| + |A| . \end{equation*}

Suppose toward a contradiction that every subset of \(A\) is Lebesgue measurable.

The Vitali set. On \(\mathbb{R}\), the relation \(x \sim y\) when \(x - y \in \mathbb{Q}\) is an equivalence relation, and each class \(\tilde{x} = x + \mathbb{Q}\) meets \([0,1]\) (take a rational \(q \in [-x, -x+1]\)). So the Axiom of Choice, as in 2.18, provides \(V \subseteq [0,1]\) containing exactly one element of each class. Two consequences: (i) \(\mathbb{R} = \bigcup_{r \in \mathbb{Q}} (r+V)\), since \(x = (x-v)+v\) for the representative \(v\) of \(\tilde{x}\) and \(x - v \in \mathbb{Q}\); (ii) the sets \(r+V\), \(r \in \mathbb{Q}\), are pairwise disjoint, since \(r + v_1 = s + v_2\) gives \(v_1 - v_2 \in \mathbb{Q}\), hence \(v_1 = v_2\) and \(r = s\).

Put \(A_r = A \cap (r+V)\) for \(r \in \mathbb{Q}\). Each \(A_r \subseteq A\) is Lebesgue measurable by assumption, and by (i) and (ii) the countable union \(A = \bigcup_{r \in \mathbb{Q}} A_r\) is disjoint, so countable additivity on the Lebesgue measurable sets (2.72) gives

\begin{equation*} 0 < |A| = \sum_{r \in \mathbb{Q}} |A_r| . \end{equation*}

Fix \(r\) with \(|A_r| > 0\) and put \(W = (-r) + A_r\). Then \(W \subseteq V \subseteq [0,1]\), \(W\) is Lebesgue measurable by Exercise 8 in this section, and \(|W| = |A_r| > 0\) by 2.7.

Let \(q_1, q_2, \ldots\) enumerate \(\mathbb{Q} \cap [0,1]\). The sets \(q_k + W\) are Lebesgue measurable (Exercise 8) with \(|q_k + W| = |W|\) (2.7), are pairwise disjoint by (ii) since \(q_k + W \subseteq q_k + V\), and lie in \([0,2]\). Hence by 2.72, 2.5, and 2.14,

\begin{equation*} \sum_{k=1}^{\infty} |W| = \Bigl| \bigcup_{k=1}^{\infty} (q_k + W) \Bigr| \leq |[0,2]| = 2 , \end{equation*}

while \(|W| > 0\) makes the left side \(\infty\) – a contradiction. So some subset of \(A\) is not Lebesgue measurable.

Throughout, \(|(b,c)| = c-b\) (Exercise 6 in Section 2A), and both \(|A|\) and \(|(b,c)\setminus A|\) are at most \(c-b\) by 2.5, hence finite, so the arithmetic below is legitimate.

Suppose \(A\) is Lebesgue measurable. Exercise 10 of this section, applied to the disjoint sets \((b,c)\setminus A\) and \(A\) with \(A\) in the role of the measurable set, gives

\begin{equation*} |A| + |(b,c) \setminus A| = \bigl| A \cup \bigl( (b,c) \setminus A \bigr) \bigr| = |(b,c)| = c - b , \end{equation*}

the middle union being \((b,c)\) because \(A \subseteq (b,c)\).

Conversely, suppose \(|A| + |(b,c)\setminus A| = c-b\). By the Lemma in Exercise 10 of this section there is a Borel set \(E_0 \supseteq (b,c)\setminus A\) with \(|E_0| = |(b,c)\setminus A|\); replacing \(E_0\) by \(E = E_0 \cap (b,c)\) keeps both the inclusion and, by 2.5, the outer measure. Put \(B = (b,c) \setminus E\), a Borel set contained in \(A\) (if \(x \in B\) then \(x \in (b,c)\) and \(x \notin (b,c)\setminus A\)). The disjoint Borel sets \(B\) and \(E\) have union \((b,c)\), so \(|B| + |E| = c-b\) by 2.68; subtracting the finite \(|E| = |(b,c)\setminus A|\) and invoking the hypothesis gives \(|B| = |A|\). Now Exercise 10 again, applied to the disjoint sets \(A \setminus B\) and the Borel set \(B\), gives

\begin{equation*} |A| = \bigl| (A \setminus B) \cup B \bigr| = |A \setminus B| + |B| = |A \setminus B| + |A| , \end{equation*}

so cancelling the finite \(|A|\) yields \(|A \setminus B| = 0\). By 2.70, \(A\) is Lebesgue measurable.

Put \(A_n = (-n,n) \cap A \subseteq (-n,n)\). Since \((-n,n)\setminus A_n = (-n,n)\setminus A\), the displayed condition says exactly that

\begin{equation*} |A_n| + |(-n,n) \setminus A_n| = n - (-n) , \end{equation*}

which by Exercise 12 of this section (with \(b = -n\), \(c = n\)) holds if and only if \(A_n\) is Lebesgue measurable. So the exercise reduces to: \(A\) is Lebesgue measurable if and only if every \(A_n\) is.

(i) If \(A\) is Lebesgue measurable, then each \((-n,n)\) is open, hence Borel, hence Lebesgue measurable, and the Lebesgue measurable sets form a \(\sigma\)-algebra (2.72), which is closed under intersection (2.25); so each \(A_n\) is Lebesgue measurable.

(ii) Conversely, every real lies in \((-n,n)\) for all large \(n\), so \(A = \bigcup_{n=1}^{\infty} A_n\); that \(\sigma\)-algebra is closed under countable unions, so \(A\) is Lebesgue measurable.

Both numbers have a base \(3\) representation using only \(0\)s and \(2\)s, namely \(\frac{1}{4} = 0.\overline{02}_3\) and \(\frac{9}{13} = 0.\overline{200}_3\), so both lie in the Cantor set \(C\) by 2.75. (The repeating blocks come from the greedy digit algorithm, whose remainders return to \(\frac{1}{4}\) after two steps and to \(\frac{9}{13}\) after three.)

Summing the geometric series confirms the two representations:

\begin{equation*} \sum_{k=1}^{\infty} \frac{2}{3^{2k}} = 2 \cdot \frac{1/9}{1 - 1/9} = \frac{1}{4} , \end{equation*}

\begin{equation*} \sum_{k=0}^{\infty} \frac{2}{3^{3k+1}} = \frac{2}{3} \cdot \frac{1}{1 - 1/27} = \frac{18}{26} = \frac{9}{13} . \end{equation*}

The unique base \(3\) representation of \(\frac{13}{17}\) is \(0.2021\ldots_3\), whose fourth digit is \(1\), so \(\frac{13}{17} \notin C\) by 2.75.

Step 1: \(\frac{13}{17}\) has only one base \(3\) representation. Suppose \(x \in [0,1]\) has two different base \(3\) representations, say

\begin{equation*} x = \sum_{k=1}^{\infty} \frac{a_k}{3^k} = \sum_{k=1}^{\infty} \frac{b_k}{3^k}, \end{equation*}

with all \(a_k, b_k \in \{0,1,2\}\), and let \(n\) be the smallest index with \(a_n \neq b_n\); say \(a_n < b_n\). Then

\begin{equation*} 0 = \sum_{k=n}^{\infty} \frac{b_k - a_k}{3^k} \geq \frac{b_n - a_n}{3^n} - \sum_{k=n+1}^{\infty} \frac{2}{3^k} = \frac{b_n - a_n}{3^n} - \frac{1}{3^n} \geq 0 , \end{equation*}

because \(b_n - a_n \geq 1\). Hence equality holds throughout, which forces \(b_n - a_n = 1\) and \(b_k - a_k = -2\) for all \(k > n\), i.e. \(b_k = 0\) and \(a_k = 2\) for all \(k > n\). In particular

\begin{equation*} x = \sum_{k=1}^{n} \frac{b_k}{3^k} = \frac{m}{3^n} \quad \text{where } m = \sum_{k=1}^{n} b_k 3^{n-k} \in \mathbb{Z}. \end{equation*}

Thus a number with two base \(3\) representations must be of the form \(m/3^n\). But \(\frac{13}{17} = \frac{m}{3^n}\) would give \(17 \mid 13 \cdot 3^n\), impossible since the prime \(17\) divides neither \(13\) nor \(3\).

Step 2: that representation contains a \(1\). The digit algorithm \(x_0 = \frac{13}{17}\), \(a_k = \lfloor 3x_{k-1} \rfloor\), \(x_k = 3x_{k-1} - a_k\) gives

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{13}{17} &= \tfrac{39}{17} = 2 + \tfrac{5}{17}, &&a_1 = 2,\\ 3 \cdot \tfrac{5}{17} &= \tfrac{15}{17} = 0 + \tfrac{15}{17}, &&a_2 = 0,\\ 3 \cdot \tfrac{15}{17} &= \tfrac{45}{17} = 2 + \tfrac{11}{17}, &&a_3 = 2,\\ 3 \cdot \tfrac{11}{17} &= \tfrac{33}{17} = 1 + \tfrac{16}{17}, &&a_4 = 1 . \end{aligned} \end{equation*}

Every \(x_k\) lies in \([0,1)\), so these digits do represent \(\frac{13}{17}\); explicitly

\begin{equation*} \frac{13}{17} = \frac{61}{81} + \frac{1}{3^4} \cdot \frac{16}{17} . \qquad \text{(Check!)} \end{equation*}

By Step 1 this representation is the only one, and its fourth digit is \(1\); so no base \(3\) representation of \(\frac{13}{17}\) consists only of \(0\)s and \(2\)s.

The eight intervals are those \(\left(\frac{k}{81}, \frac{k+1}{81}\right)\) with \(k \in \{1,7,19,25,55,61,73,79\}\):

\begin{equation*} \begin{aligned} G_4 = &\left(\tfrac{1}{81},\tfrac{2}{81}\right) \cup \left(\tfrac{7}{81},\tfrac{8}{81}\right) \cup \left(\tfrac{19}{81},\tfrac{20}{81}\right) \cup \left(\tfrac{25}{81},\tfrac{26}{81}\right)\\ &\cup \left(\tfrac{55}{81},\tfrac{56}{81}\right) \cup \left(\tfrac{61}{81},\tfrac{62}{81}\right) \cup \left(\tfrac{73}{81},\tfrac{74}{81}\right) \cup \left(\tfrac{79}{81},\tfrac{80}{81}\right). \end{aligned} \end{equation*}

Indeed, \([0,1]\setminus(G_1 \cup G_2) = \left[0,\frac{1}{9}\right] \cup \left[\frac{2}{9},\frac{1}{3}\right] \cup \left[\frac{2}{3},\frac{7}{9}\right] \cup \left[\frac{8}{9},1\right]\) by 2.74, and deleting from each of these its middle third (the corresponding interval of \(G_3\)) leaves the eight closed intervals \(\left[\frac{3j}{81}, \frac{3j+3}{81}\right]\) with \(j \in \{0,2,6,8,18,20,24,26\}\). By 2.74, \(G_4\) is the union of their middle thirds \(\left(\frac{3j+1}{81}, \frac{3j+2}{81}\right)\), and \(3j+1\) runs over the eight values of \(k\) listed above.

Split the base \(3\) digits of \(t\) between two \(\{0,2\}\)-strings. Write \(S = \left\{\frac{1}{2}x+\frac{1}{2}y : x,y \in C\right\}\).

Since \(C \subseteq [0,1]\) (2.74) and \([0,1]\) is convex, \(S \subseteq [0,1]\). Conversely let \(t \in [0,1]\). If \(t = 1\), take \(x = y = 1 = 0.222\ldots_3 \in C\) (2.75). Otherwise fix a base \(3\) representation \(t = \sum_{k} t_k 3^{-k}\) and put

\begin{equation*} (a_k, b_k) = \begin{cases} (0,0) & \text{if } t_k = 0,\\ (0,2) & \text{if } t_k = 1,\\ (2,2) & \text{if } t_k = 2, \end{cases} \end{equation*}

so \(a_k, b_k \in \{0,2\}\) and \(a_k + b_k = 2t_k\) in every case. The series \(x = \sum_k a_k 3^{-k}\) and \(y = \sum_k b_k 3^{-k}\) converge, being dominated by \(\sum_k 2 \cdot 3^{-k} = 1\), and exhibit base \(3\) representations of \(x\) and \(y\) using only \(0\)s and \(2\)s, so \(x, y \in C\) by 2.75. Adding the two absolutely convergent series termwise,

\begin{equation*} \tfrac{1}{2}x + \tfrac{1}{2}y = \sum_{k=1}^{\infty} \frac{a_k + b_k}{2 \cdot 3^{k}} = \sum_{k=1}^{\infty} \frac{t_k}{3^k} = t , \end{equation*}

so \(t \in S\). Hence \(S = [0,1]\).

It suffices to show that the Cantor set \(C\) has no isolated points: if an open interval \(I\) contains some \(x \in C\), then \((x-\varepsilon, x+\varepsilon) \subseteq I\) for some \(\varepsilon > 0\), so \(I \cap C\) is infinite.

Fix \(x \in C\) and \(\varepsilon > 0\), and by 2.75 write \(x = \sum_{k} a_k 3^{-k}\) with every \(a_k \in \{0,2\}\). Flip the \(n\)-th digit:

\begin{equation*} x^{(n)} = \sum_{k \neq n} \frac{a_k}{3^k} + \frac{2 - a_n}{3^n} . \end{equation*}

Since \(2 - a_n \in \{0,2\}\), this display is a base \(3\) representation of \(x^{(n)}\) using only \(0\)s and \(2\)s, so \(x^{(n)} \in C\) by 2.75. The two strings differ only in position \(n\), so

\begin{equation*} \bigl| x - x^{(n)} \bigr| = \frac{|a_n - (2-a_n)|}{3^n} = \frac{2}{3^n} , \end{equation*}

whence \(x^{(1)}, x^{(2)}, \ldots\) are pairwise distinct and all differ from \(x\). Choosing \(N\) with \(\frac{2}{3^N} < \varepsilon\), the points \(x^{(N)}, x^{(N+1)}, \ldots\) are infinitely many elements of \(C \cap (x-\varepsilon, x+\varepsilon)\).

\(\int_0^1 \Lambda = \frac{1}{2}\), forced by the symmetry \(\Lambda(1-x) = 1 - \Lambda(x)\) on \([0,1]\).

By 2.79 the Cantor function \(\Lambda\) is continuous on \([0,1]\), hence Riemann integrable there; being bounded and continuous it is Lebesgue measurable with the same integral, so the value below settles either reading.

(i) \(x \in C\). By 2.75 write \(x = \sum_{k} a_k 3^{-k}\) with every \(a_k \in \{0,2\}\); this \(\{0,2\}\)-representation is unique, since by Step 1 of Exercise 15 two representations of one number differ by \(1\) at their first disagreement, whereas two digits from \(\{0,2\}\) differ by \(0\) or \(2\). As \(\sum_{k} 2 \cdot 3^{-k} = 1\),

\begin{equation*} 1 - x = \sum_{k=1}^{\infty} \frac{2 - a_k}{3^k} , \end{equation*}

again a \(\{0,2\}\)-representation, so \(1 - x \in C\); with \(e_k = a_k/2 \in \{0,1\}\), 2.77 gives

\begin{equation*} \Lambda(1-x) = \sum_{k=1}^{\infty} \frac{1 - e_k}{2^k} = 1 - \sum_{k=1}^{\infty} \frac{e_k}{2^k} = 1 - \Lambda(x) . \end{equation*}

(ii) \(x \in [0,1] \setminus C\). Then \(x \in G_n\) for some \(n\), since \([0,1] \setminus C = \bigcup_n G_n\) (2.74).

Subclaim. For every \(n \ge 1\), the set \([0,1] \setminus (G_1 \cup \cdots \cup G_n)\) is the union of the \(2^n\) closed intervals \(\left[c, c+\frac{1}{3^n}\right]\) with \(c = \sum_{i \le n} d_i 3^{-i}\) and \(d_i \in \{0,2\}\), while \(G_n\) is the union of the \(2^{n-1}\) open intervals \(\left(a, a+\frac{1}{3^n}\right)\) with \(a = \sum_{i<n} d_i 3^{-i} + 3^{-n}\) and \(d_i \in \{0,2\}\). This is induction on \(n\) from 2.74: the middle third of \(\left[c, c+3^{-(n-1)}\right]\) is \(\left(c+3^{-n}, c+2 \cdot 3^{-n}\right)\), which is the interval attached to \((d_1,\ldots,d_{n-1})\), and removing it leaves the two intervals attached to \((d_1,\ldots,d_{n-1},0)\) and \((d_1,\ldots,d_{n-1},2)\). (Check!)

Let \((a,b)\) be the interval of \(G_n\) containing \(x\), with \(a\) as in the subclaim, \(b = a + 3^{-n}\), and \(e_i = d_i/2\). Using \(3^{-n} = \sum_{k>n} 2 \cdot 3^{-k}\),

\begin{equation*} a = 0.d_1 \ldots d_{n-1} 0\,2\,2\,2\ldots_3 , \qquad b = 0.d_1 \ldots d_{n-1} 2\,0\,0\,0\ldots_3 , \end{equation*}

so \(a, b \in C\) and, by 2.77,

\begin{equation*} \Lambda(a) = \sum_{i=1}^{n-1}\frac{e_i}{2^i} + \sum_{k>n}\frac{1}{2^k} = \sum_{i=1}^{n-1}\frac{e_i}{2^i} + \frac{1}{2^n} = \Lambda(b) . \end{equation*}

Moreover every base \(3\) representation of \(x\) begins \(0.d_1 \ldots d_{n-1}1\): the numbers admitting a representation with a given prefix \(c_1 \ldots c_n\) form exactly the interval \(\left[\frac{k}{3^n}, \frac{k+1}{3^n}\right]\), where \(k = \sum_{i \le n} c_i 3^{n-i}\), because the tails \(\sum_{k>n} c_k 3^{-k}\) fill \([0, 3^{-n}]\); distinct prefixes give distinct \(k\), the prefix \(d_1 \ldots d_{n-1}1\) gives exactly \([a,b]\), and \(x\) lies in the interior of \([a,b]\), hence in no other prefix interval. So 2.77 gives \(\Lambda(x) = \sum_{i<n} e_i 2^{-i} + 2^{-n}\), which is \(\Lambda(a) = \Lambda(b)\).

Now \(1 - x \in (1-b, 1-a)\) with \(1-a, 1-b \in C\), so by (i) and monotonicity of \(\Lambda\) (2.79),

\begin{equation*} 1 - \Lambda(a) = \Lambda(1-b) \le \Lambda(1-x) \le \Lambda(1-a) = 1 - \Lambda(a) , \end{equation*}

whence \(\Lambda(1-x) = 1 - \Lambda(a) = 1 - \Lambda(x)\).

Substituting \(u = 1-x\) in the integral now gives

\begin{equation*} \int_0^1 \Lambda(x)\,dx = \int_0^1 \bigl( 1 - \Lambda(u) \bigr)\,du = 1 - \int_0^1 \Lambda(u)\,du , \end{equation*}

so \(\int_0^1 \Lambda = \frac{1}{2}\).

(a) \(\Lambda\!\left(\frac{9}{13}\right) = \frac{4}{7}\). (b) \(\Lambda(0.93) = \frac{7}{8}\).

(a). By Exercise 14, \(\frac{9}{13} = 0.\overline{200}_3 \in C\), so 2.77 replaces each \(2\) by \(1\) and reads the result in base \(2\):

\begin{equation*} \Lambda\!\left(\frac{9}{13}\right) = 0.\overline{100}_2 = \sum_{j=0}^{\infty} \frac{1}{2^{3j+1}} = \frac{1/2}{1 - 1/8} = \frac{4}{7}. \end{equation*}

(b). Here \(0.93 = \frac{93}{100}\), whose base \(3\) digits start

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{93}{100} &= \tfrac{279}{100} = 2 + \tfrac{79}{100}, &&\text{digit } 2,\\ 3 \cdot \tfrac{79}{100} &= \tfrac{237}{100} = 2 + \tfrac{37}{100}, &&\text{digit } 2,\\ 3 \cdot \tfrac{37}{100} &= \tfrac{111}{100} = 1 + \tfrac{11}{100}, &&\text{digit } 1 , \end{aligned} \end{equation*}

so \(\frac{93}{100} = 0.221\ldots_3\). This representation is unique by Step 1 of Exercise 15, since \(\frac{93}{100} = m/3^k\) would force \(100 \mid 93 \cdot 3^k\), impossible as \(\gcd(93 \cdot 3^k, 100) = 1\); it contains a \(1\), so \(\frac{93}{100} \notin C\) by 2.75. By 2.77, truncate after that first \(1\) and replace each earlier \(2\) by \(1\):

\begin{equation*} \Lambda(0.93) = 0.111_2 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}. \end{equation*}

\(\Lambda^{-1}\bigl(\{\tfrac{1}{3}\}\bigr) = \left\{\frac{1}{4}\right\}\) and \(\Lambda^{-1}\bigl(\{\tfrac{5}{16}\}\bigr) = \left[\frac{19}{81}, \frac{20}{81}\right]\).

Fact 1 (points of \(C\)). By 2.75 and the uniqueness of the \(\{0,2\}\)-representation noted in Exercise 19, the map \(x \mapsto (e_k)\), \(e_k = a_k/2\), is a bijection from \(C\) onto the set of binary digit strings, and 2.77 says \(\Lambda(x) = \sum_k e_k 2^{-k}\). So for \(y \in [0,1]\) the preimages of \(y\) lying in \(C\) correspond exactly to the base \(2\) representations of \(y\).

Fact 2 (points off \(C\)). By the subclaim in case (ii) of Exercise 19, together with \([0,1]\setminus C = \bigcup_n G_n\) (2.74) and the disjointness of \(G_1, G_2, \ldots\) noted in the proof of 2.76, the set \([0,1]\setminus C\) is the disjoint union, over \(n \ge 1\) and \((d_1,\ldots,d_{n-1}) \in \{0,2\}^{n-1}\), of the intervals \((a, a+3^{-n})\) with \(a = \sum_{i<n} d_i 3^{-i} + 3^{-n}\); and on such an interval \(\Lambda\) is constantly

\begin{equation*} \sum_{i=1}^{n-1} \frac{e_i}{2^i} + \frac{1}{2^n} = \frac{2m+1}{2^n}, \qquad e_i = \tfrac{d_i}{2}, \ \ m = \sum_{i<n} e_i 2^{n-1-i} . \end{equation*}

As \((d_1,\ldots,d_{n-1})\) ranges over \(\{0,2\}^{n-1}\), the integer \(m\) ranges bijectively over \(0,1,\ldots,2^{n-1}-1\). So the values of \(\Lambda\) off \(C\) are exactly the dyadic rationals in \((0,1)\), and such a value determines \(n\), then \(m\), then the interval.

Fact 3 (base \(2\) representations). Step 1 of Exercise 15, run with \(3\) replaced by \(2\) and the digit bound \(2\) by \(1\), shows that a number with two base \(2\) representations is dyadic. Conversely, if \(y = m/2^n \in (0,1)\) with \(m\) odd then \(y = 0.b_1 \ldots b_n 000\ldots_2\) with \(b_n = 1\), and also \(y = 0.b_1 \ldots b_{n-1} 0111\ldots_2\) since \(\sum_{k>n} 2^{-k} = 2^{-n}\). So a dyadic rational in \((0,1)\) has exactly two base \(2\) representations, every other number in \([0,1]\) exactly one.

(a). \(\frac{1}{3}\) is not dyadic, since \(\frac{1}{3} = \frac{m}{2^n}\) would give \(3 \mid 2^n\); so by Fact 2 no point off \(C\) maps to it, and by Fact 3 it has the single base \(2\) representation \(0.\overline{01}_2\) (indeed \(\sum_{j\ge1} 4^{-j} = \frac{1}{3}\)). By Fact 1 its only preimage is the point of \(C\) with \(\{0,2\}\)-representation \(0.\overline{02}_3\), namely

\begin{equation*} \sum_{j=1}^{\infty} \frac{2}{9^{j}} = \frac{2/9}{1 - 1/9} = \frac{1}{4} . \end{equation*}

(b). \(\frac{5}{16}\) is dyadic, so both cases contribute.

Off \(C\): \(\frac{5}{16} = \frac{2m+1}{2^n}\) forces \(n = 4\) and \(m = 2\), so \(e_1e_2e_3 = 010\) and \((d_1,d_2,d_3) = (0,2,0)\); by Fact 2 the interval is \((a, a + \frac{1}{81})\) with \(a = \frac{2}{9} + \frac{1}{81} = \frac{19}{81}\), one of the eight intervals of \(G_4\) found in Exercise 16.

On \(C\): by Fact 3 the two base \(2\) representations of \(\frac{5}{16}\) are \(0.0101\,000\ldots_2\) and \(0.0100\,111\ldots_2\), so by Fact 1 the two Cantor set preimages are

\begin{equation*} 0.0202\,000\ldots_3 = \frac{2}{9} + \frac{2}{81} = \frac{20}{81} , \end{equation*}

\begin{equation*} 0.0200\,222\ldots_3 = \frac{18}{81} + \frac{1}{81} = \frac{19}{81} . \end{equation*}

Combining the two cases,

\begin{equation*} \Lambda^{-1}\left(\left\{\tfrac{5}{16}\right\}\right) = \left\{\tfrac{19}{81}\right\} \cup \left(\tfrac{19}{81},\tfrac{20}{81}\right) \cup \left\{\tfrac{20}{81}\right\} = \left[\frac{19}{81}, \frac{20}{81}\right]. \end{equation*}

Both parts reduce to one fact: a base \(b\) representation of \(t \in [0,1]\) – digits \(d_j \in \{0,\ldots,b-1\}\) with \(t = \sum_{j\ge1} d_j b^{-j}\) – has eventually periodic digit string exactly when \(t\) is rational.

Lemma 1 (near-uniqueness). The argument of Step 1 in Exercise 15, run in base \(b\), gives: if \((c_j) \ne (c_j^{\prime})\) are base \(b\) representations of the same number, \(n\) is the first index where they differ, and \(c_n > c_n^{\prime}\), then \(c_n = c_n^{\prime}+1\) and \(c_j = 0\), \(c_j^{\prime} = b-1\) for every \(j > n\). In particular each string is eventually constant.

Lemma 2. Let \((d_j)\) be a base \(b\) representation of \(t \in [0,1]\). Then \(t\) is rational if and only if \((d_j)\) is eventually periodic.

Suppose \(d_{j+k} = d_j\) for all \(j > N\), and put \(z = \sum_{i \ge 1} d_{N+i} b^{-i}\), so that \(t = \sum_{j \le N} d_j b^{-j} + b^{-N} z\). Periodicity of \((d_{N+i})_{i \ge 1}\) gives

\begin{equation*} b^{k} z = \sum_{i=1}^{\infty} d_{N+i} b^{\,k-i} = m + z , \qquad m = \sum_{i=1}^{k} d_{N+i} b^{\,k-i} \in \mathbb{Z} , \end{equation*}

so \(z = m/(b^k - 1)\) and hence \(t\) is rational. Conversely let \(t = p/q\) with \(p, q\) integers, \(q \ge 1\), and put

\begin{equation*} s_n = \sum_{i=1}^{\infty} \frac{d_{n+i}}{b^{i}} = b^{n} t - \sum_{j=1}^{n} d_j b^{\,n-j} . \end{equation*}

Then \(s_n \in [0,1]\) and \(q s_n\) is an integer, hence \(q s_n \in \{0,1,\ldots,q\}\); by the pigeonhole principle \(s_n = s_m\) for some \(n < m\). Now \((d_{n+i})_{i \ge 1}\) and \((d_{m+i})_{i \ge 1}\) are base \(b\) representations of that one number: if they are equal then \(d_{j + (m-n)} = d_j\) for every \(j > n\), and if not then Lemma 1 makes each eventually constant, so \((d_j)\) is eventually constant. Either way \((d_j)\) is eventually periodic.

(a). Let \(x \in [0,1]\) be rational; the two bullet points of 2.77 give two cases.

(i) \(x \notin C\): by 2.75 every base \(3\) representation of \(x\) contains a digit \(1\), and 2.77 truncates after the first such \(1\), so \(\Lambda(x) = \sum_{j=1}^{n} e_j 2^{-j}\) is a finite sum of rationals.

(ii) \(x \in C\): let \((d_j)\) be the base \(3\) representation of \(x\) using only \(0\)s and \(2\)s, which is the one used in 2.77. Lemma 2 with \(b = 3\) makes \((d_j)\) eventually periodic, hence so is \((d_j/2)\), which by 2.77 is a base \(2\) representation of \(\Lambda(x)\); Lemma 2 with \(b = 2\) then makes \(\Lambda(x)\) rational.

(b). Let \(x \in C\) with \(\Lambda(x)\) rational, and let \((d_j)\) be its base \(3\) representation using only \(0\)s and \(2\)s; this one is unique, since by Lemma 1 two representations of one number differ by \(1\) at their first disagreement whereas digits in \(\{0,2\}\) differ by \(0\) or \(2\). By 2.77 the string \((d_j/2)\), with terms in \(\{0,1\}\), is a base \(2\) representation of \(\Lambda(x)\), so Lemma 2 with \(b = 2\) makes it eventually periodic; then \((d_j)\) is eventually periodic and Lemma 2 with \(b = 3\) makes \(x = \sum_{j} d_j 3^{-j}\) rational.

Take \(f = h \circ \varphi\), where \(\varphi\) reads off the limit superior of the decimal digit averages of the fractional part and \(h\) maps \([0,9]\) onto \(\mathbb{R}\):

\begin{equation*} \varphi(x) = \limsup_{n \to \infty} \frac{d_1(x) + d_2(x) + \cdots + d_n(x)}{n} \in [0,9] , \end{equation*}

\begin{equation*} h(t) = \tan\!\left( \pi \left( \frac{t}{9} - \frac12 \right) \right) \text{ for } t \in (0,9), \qquad h(0) = h(9) = 0 . \end{equation*}

Here \(d_j(x)\) is the \(j\)-th digit of the decimal representation of \(\langle x \rangle = x - \lfloor x \rfloor \in [0,1)\) that is not eventually \(9\); it exists by the greedy algorithm \(d_j = \lfloor 10^{j} \langle x \rangle \rfloor - 10 \lfloor 10^{j-1} \langle x \rangle \rfloor\) (Check!), and it is unique because by Lemma 1 of Exercise 22 with \(b = 10\), of two representations of one number one is eventually \(0\) and the other eventually \(9\). As \(t\) runs over \((0,9)\), \(\pi(t/9 - 1/2)\) runs over \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), on which \(\tan\) is a bijection onto \(\mathbb{R}\); so \(h([0,9]) = \mathbb{R}\), and it suffices to prove \(\varphi(I) = [0,9]\) for every nonempty open interval \(I\).

Step 1: \(I\) contains a full decimal cylinder. Shrink \(I\) to an interval strictly between consecutive integers: choose \(m \in \mathbb{Z}\) and \(u < v\) with \(m < u < v < m+1\) and \((u,v) \subseteq I\), so that \(\langle x \rangle = x - m\) on \((u,v)\). Choose \(k \in \mathbb{Z}^{+}\) with \(10^{-k} < (v-u)/2\) and put \(M = \lfloor (u-m)10^{k} \rfloor + 1\), so that \((u-m)10^k < M \le (u-m)10^k + 1\) and hence

\begin{equation*} u - m < \frac{M}{10^{k}} \quad\text{and}\quad \frac{M+1}{10^{k}} \le (u-m) + \frac{2}{10^{k}} < v - m . \end{equation*}

The second inequality gives \((M+1)10^{-k} < 1\), so \(1 \le M < 10^{k}\); let \(c_1 \ldots c_k\) be the \(k\)-digit form of \(M\), leading zeros allowed. For any digit string \((e_i)\) that is not eventually \(9\), put

\begin{equation*} x = m + \frac{M}{10^{k}} + \frac{1}{10^{k}} \sum_{i=1}^{\infty} \frac{e_i}{10^{i}} . \end{equation*}

Then \(\langle x \rangle\) lies in \(\left[ M 10^{-k}, (M+1)10^{-k} \right] \subseteq (u-m, v-m)\) by the two displayed inequalities, so \(x \in (u,v) \subseteq I\); and the string \(c_1, \ldots, c_k, e_1, e_2, \ldots\) is a decimal representation of \(\langle x \rangle\) that is not eventually \(9\), hence is the canonical one.

A fixed prefix does not move the limit superior: with \(K = c_1 + \cdots + c_k\) and \(S_j = e_1 + \cdots + e_j\), the average of the first \(j+k\) digits of \(\langle x \rangle\) is \((K+S_j)/(j+k)\), and

\begin{equation*} \frac{K + S_j}{j+k} - \frac{S_j}{j} = \frac{K}{j+k} - \frac{k\, S_j}{j (j+k)} , \end{equation*}

whose absolute value is at most \((K + 9k)/(j+k) \to 0\) because \(0 \le S_j \le 9j\). Hence \(\varphi(x) = \limsup_{j} S_j / j\).

Step 2: every \(t \in [0,9]\) is such a limit superior.

(i) \(t = 9\): take \(e_i = 0\) when \(i\) is a power of \(2\) and \(e_i = 9\) otherwise. At most \(1 + \log_2 n\) powers of \(2\) lie in \(\{1,\ldots,n\}\), so \(9 \ge S_n/n \ge 9 - 9(1 + \log_2 n)/n \to 9\). The string has infinitely many \(0\)s, so it is not eventually \(9\).

(ii) \(t \in [0,9)\): define \((e_i)\) greedily by \(e_n = 9\) if \((S_{n-1}+9)/n \le t\) and \(e_n = 0\) otherwise, and put \(A_n = S_n/n\). Then \(A_n \le t\) for every \(n\): if \(e_n = 9\) this is the defining rule, and if \(e_n = 0\) then \(A_n = S_{n-1}/n \le S_{n-1}/(n-1) = A_{n-1} \le t\) for \(n \ge 2\), while \(A_1 = 0 \le t\). So \(\limsup_n A_n \le t\). Also \(e_n = 0\) for infinitely many \(n\), since \(e_n = 9\) from some point on would force \(A_n \to 9 > t\); and each such \(n\) satisfies \(A_n = S_{n-1}/n > t - \frac{9}{n}\), so \(\limsup_n A_n \ge t\). Finally \((e_n)\) is not eventually \(9\).

Given \(I\) and \(t \in [0,9]\), Step 2 supplies a legal tail \((e_i)\) and Step 1 converts it into \(x \in I\) with \(\varphi(x) = t\). Thus \(\varphi(I) = [0,9]\) and \(f(I) = h([0,9]) = \mathbb{R}\).

Write \(\mu_{*}(A)\) for the inner measure and \(|A|\) for the outer measure. The empty set is an admissible \(F\) for every \(A\), so \(\mu_{*}(A) \in [0,\infty]\) is always defined; and every closed set is Borel, so each \(|F|\) occurring is a Lebesgue measure.

(a). The inequality \(\mu_{*}(A) \le |A|\) holds for every \(A\): each closed bounded \(F \subseteq A\) has \(|F| \le |A|\) by 2.5. For the reverse, let \(\varepsilon > 0\) and, \(A\) being Lebesgue measurable, use 2.71 ((a) implies (b)) to get a closed \(F \subseteq A\) with \(|A \setminus F| < \varepsilon\). The truncations \(F_n = F \cap [-n,n]\) are closed, bounded, contained in \(A\), and increase to \(F\), so \(\lim_n |F_n| = |F|\) by 2.59 applied to Lebesgue measure (a measure on the Lebesgue measurable sets by 2.72); as \(|F_n| \le \mu_{*}(A)\) for every \(n\), this gives \(\mu_{*}(A) \ge |F|\). Meanwhile \(A = F \cup (A \setminus F)\) and 2.8 give \(|A| \le |F| + \varepsilon\). If \(|A| = \infty\) this forces \(|F| = \infty\), hence \(\mu_{*}(A) = \infty = |A|\); otherwise \(\mu_{*}(A) \ge |F| > |A| - \varepsilon\) for every \(\varepsilon > 0\), so \(\mu_{*}(A) \ge |A|\).

(b). For \(r \in [0,1)\) and \(S \subseteq [0,1)\) write

\begin{equation*} S \oplus r = \bigl( (S \cap [0, 1-r)) + r \bigr) \cup \bigl( (S \cap [1-r, 1)) + r - 1 \bigr) \subseteq [0,1) , \end{equation*}

so that \(x \mapsto x \oplus r\) is a bijection of \([0,1)\) onto itself. If \(S\) is Lebesgue measurable then so is \(S \oplus r\), and since the two pieces are disjoint, additivity of Lebesgue measure (2.72) with translation invariance (2.7) gives \(|S \oplus r| = |S|\).

By the axiom of choice let \(V \subseteq [0,1)\) contain exactly one element of each class of the relation \(x \sim y\) when \(x - y \in \mathbb{Q}\) on \([0,1)\), let \(r_1, r_2, \ldots\) enumerate \(\mathbb{Q} \cap [0,1)\), and put \(V_k = V \oplus r_k\).

The \(V_k\) are disjoint with union \([0,1)\). If \(x \in V_j \cap V_k\), say \(x = v \oplus r_j = w \oplus r_k\) with \(v,w \in V\), then \(x - (v+r_j)\) and \(x - (w+r_k)\) lie in \(\{0,-1\}\), so \(v - w \in \mathbb{Q}\), forcing \(v = w\) and then \(r_j = r_k\) by injectivity of \(r \mapsto v \oplus r\). Conversely, given \(x \in [0,1)\) with representative \(v \in V\), the number \(r = x-v\) (or \(x-v+1\) if \(x < v\)) lies in \(\mathbb{Q} \cap [0,1)\) and \(x = v \oplus r\).

Each \(\mu_{*}(V_k) = 0\). Since \(x \oplus r_k \sim x\) for rational \(r_k\) and \(\oplus\, r_k\) is injective, \(V_k\) meets each class exactly once, so it suffices to show \(\mu_{*}(S) = 0\) whenever \(S \subseteq [0,1)\) meets each class at most once. Let \(F \subseteq S\) be closed and bounded, hence Borel and Lebesgue measurable. The sets \(F \oplus r\), \(r \in \mathbb{Q} \cap [0,1)\), are disjoint: \(x \oplus r = y \oplus r^{\prime}\) with \(x,y \in F\) and \(r \ne r^{\prime}\) forces \(x - y \in \mathbb{Q}\) and \(x \ne y\), two distinct points of \(S\) in one class. Each \(F \oplus r\) is a measurable subset of \([0,1)\) with \(|F \oplus r| = |F|\), so countable additivity (2.72) and 2.5 give

\begin{equation*} \sum_{r \in \mathbb{Q} \cap [0,1)} |F| = \Bigl| \bigcup_{r \in \mathbb{Q} \cap [0,1)} (F \oplus r) \Bigr| \le |[0,1)| = 1 , \end{equation*}

a sum of countably infinitely many copies of \(|F|\); hence \(|F| = 0\) and \(\mu_{*}(S) = 0\).

So \(V_1, V_2, \ldots\) are disjoint subsets of \(\mathbb{R}\) with \(\sum_{k} \mu_{*}(V_k) = 0\), whereas by (a)

\begin{equation*} \mu_{*}\Bigl( \bigcup_{k=1}^{\infty} V_k \Bigr) = \mu_{*}\bigl( [0,1) \bigr) = \bigl| [0,1) \bigr| = 1 . \end{equation*}

Thus \(\mu_{*}\) is not countably additive on the \(\sigma\)-algebra of all subsets of \(\mathbb{R}\).

Take \(n = \max\{n_1,\ldots,n_N\}\): a finite set of thresholds has a largest element, which is again in \(\mathbb{Z}^{+}\).

The case \(X = \emptyset\) is vacuous, so let \(X = \{x_1,\ldots,x_N\}\) with \(N \in \mathbb{Z}^{+}\), suppose \(f_1, f_2, \ldots\) converges pointwise on \(X\) to \(f\), and let \(\varepsilon > 0\). By pointwise convergence (2.82) there is for each \(j\) an \(n_j \in \mathbb{Z}^{+}\) with

\begin{equation*} |f_k(x_j) - f(x_j)| < \varepsilon \quad \text{for all integers } k \ge n_j . \end{equation*}

With \(n = \max\{n_1,\ldots,n_N\}\), every \(k \ge n\) and every \(x \in X\) satisfy \(x = x_j\) for some \(j\) and \(k \ge n_j\), so \(|f_k(x) - f(x)| < \varepsilon\). This single \(n\) works for all \(x \in X\) at once, so the convergence is uniform by 2.82.

Take \(f_k = \chi_{\{k\}}\).

For fixed \(m \in \mathbb{Z}^+\) we have \(f_k(m) = 0\) whenever \(k > m\), so \(f_k \to f \equiv 0\) pointwise. A uniform limit must agree with the pointwise limit, so uniform convergence would have to be to \(f\); but for \(\varepsilon = \tfrac12\) and any \(n \in \mathbb{Z}^+\), the index \(k = n\) and the point \(m = n\) give

\begin{equation*} |f_n(n) - f(n)| = 1 > \tfrac12 , \end{equation*}

so the condition in 2.82 fails. Hence the convergence is not uniform.

Take

\begin{equation*} f_k(x) = \begin{cases} k^2 x & \text{if } 0 \le x \le \frac{1}{k}, \\ \dfrac{1}{x} & \text{if } \frac{1}{k} < x \le 1, \end{cases} \qquad f(x) = \begin{cases} 0 & \text{if } x = 0, \\ \dfrac{1}{x} & \text{if } 0 < x \le 1. \end{cases} \end{equation*}

Each \(f_k\) is continuous on \([0,\frac1k]\) and on \([\frac1k,1]\) (where \(x \ge \frac1k > 0\)), and the two formulas agree at \(\frac1k\) because \(k^2 \cdot \frac1k = k\); hence \(f_k\) is continuous on \([0,1]\).

Pointwise convergence: \(f_k(0) = 0 = f(0)\) for every \(k\), while for \(0 < x \le 1\) any \(k > 1/x\) gives \(\frac1k < x\) and so \(f_k(x) = \frac1x = f(x)\), an eventually constant sequence. Finally \(f\bigl(\frac1k\bigr) = k\) for every \(k\), so \(f\) is unbounded.

The statement is true: \(f\) is uniformly continuous, by the \(\varepsilon/3\) argument of 2.84 with the \(\delta\) for \(f_n\) now independent of the point.

Suppose \(\varepsilon > 0\). Because \(f_1, f_2, \dots\) converges uniformly on \(A\) to \(f\), there exists \(n \in \mathbb{Z}^+\) such that

\begin{equation*} |f_n(z) - f(z)| < \frac{\varepsilon}{3} \quad \text{for all } z \in A . \end{equation*}

Because \(f_n\) is uniformly continuous on \(A\), there exists \(\delta > 0\) such that

\begin{equation*} |f_n(x) - f_n(y)| < \frac{\varepsilon}{3} \quad \text{for all } x, y \in A \text{ with } |x - y| < \delta . \end{equation*}

Now suppose \(x, y \in A\) and \(|x - y| < \delta\). Then

\begin{equation*} \begin{aligned} |f(x) - f(y)| &\le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ &= \varepsilon . \end{aligned} \end{equation*}

The \(\delta\) works for all pairs of points of \(A\) at once, so \(f\) is uniformly continuous on \(A\).

Take Lebesgue measure \(\mu\) on the Borel subsets of \(X = \mathbb{R}\) (so \(\mu(X) = \infty\)) and \(f_k = \chi_{[k,\,k+1]}\), with \(\varepsilon = 1\).

Each \(x \in \mathbb{R}\) has \(f_k(x) = 0\) for every \(k > x\), so \(f_k \to f \equiv 0\) pointwise. Now let \(E\) be any Borel set with \(\mu(\mathbb{R} \setminus E) < 1\). For each \(k \in \mathbb{Z}^+\), monotonicity applied to \([k,k+1] \setminus E \subseteq \mathbb{R} \setminus E\) gives

\begin{equation*} \bigl| [k, k+1] \setminus E \bigr| \le \bigl| \mathbb{R} \setminus E \bigr| < 1 = \bigl| [k, k+1] \bigr| , \end{equation*}

so \([k,k+1] \cap E \neq \emptyset\); choose \(x_k\) there. Then for \(\tfrac12\) and any \(n\), the index \(k = n\) and the point \(x_n \in E\) give \(|f_n(x_n) - f(x_n)| = 1 > \tfrac12\). Hence the convergence is uniform on no such \(E\), and the conclusion of Egorov’s Theorem (2.85) fails for \(\varepsilon = 1\) although all its other hypotheses hold.

Run the proof of Egorov’s Theorem (2.85) with \(\{x : |f_k(x) - f(x)| < \frac1n\}\) replaced by \(\{x : f_k(x) > n\}\).

Suppose \(\varepsilon > 0\) and fix \(n \in \mathbb{Z}^+\). For \(m \in \mathbb{Z}^+\) put

\begin{equation*} A_{m,n} = \bigcap_{k = m}^{\infty} \{ x \in X : f_k(x) > n \} \in \mathcal{S} . \end{equation*}

Then \(A_{1,n} \subseteq A_{2,n} \subseteq \cdots\), and \(\bigcup_{m} A_{m,n} = X\) because \(\lim_{k \to \infty} f_k(x) = \infty\) gives, for each \(x \in X\), an \(m\) with \(f_k(x) > n\) for all \(k \ge m\). Hence \(\lim_{m \to \infty} \mu(A_{m,n}) = \mu(X)\) by 2.59, and since \(\mu(X) < \infty\) we may choose \(m_n\) with \(\mu(X) - \mu(A_{m_n,n}) < \varepsilon/2^n\); as \(\mu(A_{m_n,n}) \le \mu(X) < \infty\), 2.57(b) turns this into \(\mu(X \setminus A_{m_n,n}) < \varepsilon/2^n\).

Let \(E = \bigcap_{n=1}^{\infty} A_{m_n,n} \in \mathcal{S}\). By countable subadditivity (2.58),

\begin{equation*} \begin{aligned} \mu(X \setminus E) &\le \sum_{n=1}^{\infty} \mu(X \setminus A_{m_n, n}) \\ &< \sum_{n=1}^{\infty} \frac{\varepsilon}{2^n} = \varepsilon . \end{aligned} \end{equation*}

Given \(t > 0\), choose \(n \ge t\); then \(E \subseteq A_{m_n,n}\) gives \(f_k(x) > n \ge t\) for all \(k \ge m_n\) and all \(x \in E\), which is uniform convergence to \(\infty\) on \(E\).

Each increasing bounded-above sequence \(g_1(x) \le g_2(x) \le \cdots\) converges to \(g(x) = \sup_k g_k(x) \in \mathbb{R}\), so \(g_k \le g\) on \(F\) for every \(k\).

(i) If \(g_1, g_2, \dots\) converges uniformly to \(g\) on \(F\), then \(g\) is continuous at each point of \(F\) by 2.84, since each \(g_k\) is.

(ii) Conversely, suppose \(g\) is continuous on \(F\) and set \(h_k = g - g_k\), a continuous real-valued function on \(F\) with \(h_k \ge 0\), \(h_1 \ge h_2 \ge \cdots\), and \(h_k(x) \to 0\) for each \(x \in F\). Uniform convergence means: for each \(\varepsilon > 0\) there is \(n\) with \(h_k < \varepsilon\) on \(F\) for all \(k \ge n\); by monotonicity in \(k\) it suffices to find one \(n\) with \(h_n < \varepsilon\) on \(F\).

If this fails for some \(\varepsilon > 0\), pick \(x_n \in F\) with \(h_n(x_n) \ge \varepsilon\) for every \(n\). As \(F\) is bounded, Bolzano-Weierstrass gives a subsequence \(x_{n_j} \to x\), and \(x \in F\) since \(F\) is closed. Fix \(m\); for all \(j\) with \(n_j \ge m\), monotonicity in the index gives \(h_m(x_{n_j}) \ge h_{n_j}(x_{n_j}) \ge \varepsilon\), so continuity of \(h_m\) at \(x\) yields

\begin{equation*} h_m(x) = \lim_{j \to \infty} h_m(x_{n_j}) \ge \varepsilon . \end{equation*}

This holds for every \(m\), contradicting \(h_m(x) \to 0\). Hence the convergence is uniform.

Given \(\varepsilon > 0\), choose \(N\) with \(\frac{1}{2^N} < \varepsilon\) and take \(E = \{1, \dots, N\}\), a set depending only on \(\varepsilon\). Summing the geometric series,

\begin{equation*} \mu(\mathbb{Z}^+ \setminus E) = \sum_{n=N+1}^{\infty} \frac{1}{2^n} = \frac{1}{2^N} < \varepsilon . \end{equation*}

Pointwise convergence on the finite set \(E\) is automatically uniform: if \(f_k \to f\) pointwise on \(\mathbb{Z}^+\) and \(\delta > 0\), pick \(M_n\) for each \(n \in E\) with \(|f_k(n) - f(n)| < \delta\) for all \(k \ge M_n\), and set \(M = \max\{M_1, \dots, M_N\}\), a maximum over finitely many indices. Then \(\sup_{n \in E} |f_k(n) - f(n)| \le \delta\) for all \(k \ge M\).

Near any \(x \in F = F_1 \cup \cdots \cup F_n\) the other pieces are a positive distance away, so continuity of \(g\) at \(x\) reduces to continuity of the single restriction \(g|_{F_j}\) with \(x \in F_j\).

Fix \(x \in F\) and \(\varepsilon > 0\), and pick \(j\) with \(x \in F_j\). The set \(C = \bigcup_{k \neq j} F_k\) is closed (a finite union of closed sets) and misses \(x\) (disjointness), so \(\mathbb{R} \setminus C\) is open about \(x\) and there is \(\delta_1 > 0\) with

\begin{equation*} (x - \delta_1, x + \delta_1) \cap C = \varnothing . \end{equation*}

Continuity of \(g|_{F_j}\) at \(x\) gives \(\delta_2 > 0\) with \(|g(y) - g(x)| < \varepsilon\) for all \(y \in F_j\) satisfying \(|y - x| < \delta_2\). Put \(\delta = \min\{\delta_1, \delta_2\}\). If \(y \in F\) and \(|y - x| < \delta\), then \(y \notin C\), so \(y \in F_j\), and hence \(|g(y) - g(x)| < \varepsilon\). Thus \(g\) is continuous at \(x\).

If \(F\) is not closed, then \(g(x) = \frac{1}{x - b}\) is a continuous function on \(F\) with no continuous extension to \(\mathbb{R}\); this contrapositive gives the result.

Not being closed, \(F\) omits some limit point \(b \notin F\) of itself, so there are \(x_j \in F\) with \(x_j \to b\) and \(x_j \neq b\). Since \(b \notin F\) we have \(F \subseteq \mathbb{R} \setminus \{b\}\), so \(g\) is the restriction to \(F\) of a function continuous on \(\mathbb{R} \setminus \{b\}\), hence continuous. If \(h : \mathbb{R} \to \mathbb{R}\) were continuous with \(h|_F = g\), then \(h(x_j) \to h(b) \in \mathbb{R}\), so \(h(x_1), h(x_2), \dots\) would be bounded; but

\begin{equation*} |h(x_j)| = |g(x_j)| = \frac{1}{|x_j - b|} \to \infty \end{equation*}

because \(0 < |x_j - b| \to 0\), a contradiction.

The statement is true: the unbounded witness of Exercise 10 can be replaced by a bounded function oscillating between \(0\) and \(1\) along a sequence tending to the missing limit point.

We again prove the contrapositive, so let \(b \notin F\) be a limit point of \(F\). Then \(F\) meets \((b, b+\delta)\) for every \(\delta > 0\) or meets \((b-\delta, b)\) for every \(\delta > 0\) (otherwise the smaller of the two failing radii would isolate \(b\) from \(F\), using \(b \notin F\)); replacing \(F, b\) by \(-F, -b\) if necessary — the reflection \(x \mapsto -x\) is a homeomorphism of \(\mathbb{R}\), and transports a non-extendable bounded continuous function back — we may assume the first.

Recursively choose \(x_1 \in F \cap (b, b+1)\) and \(x_{j+1} \in F \cap \bigl(b, b + \frac{x_j - b}{2}\bigr)\), nonempty at each stage. Then \(x_1 > x_2 > \cdots > b\) with \(0 < x_j - b \le 2^{1-j}\), so \(x_j \to b\). Let \(a_j = 1\) for \(j\) odd and \(a_j = 0\) for \(j\) even, and define \(\varphi : \mathbb{R} \setminus \{b\} \to [0,1]\) to be \(0\) on \((-\infty, b)\), equal to \(1\) on \([x_1, \infty)\), and affine on each \([x_{j+1}, x_j]\) with

\begin{equation*} \varphi(x_j) = a_j \quad \text{for every } j \in \mathbb{Z}^+ . \end{equation*}

These intervals together with \([x_1,\infty)\) cover \((b,\infty)\), since \(x_i \to b < x\) makes \(\{i : x \le x_i\}\) finite for each \(x > b\), and consecutive pieces agree at the shared endpoints; so \(\varphi\) is well defined. It is constant on the open set \((-\infty, b)\), and each \(x > b\) has a neighborhood \((x_j, \infty)\) on which \(\varphi\) is pasted from the finitely many closed affine pieces \([x_j, x_{j-1}], \dots, [x_1, \infty)\), continuous by Exercise 9. Hence \(\varphi\) is continuous on \(\mathbb{R} \setminus \{b\} \supseteq F\).

Set \(g = \varphi|_F\), a continuous function on \(F\) with \(0 \le g \le 1\). A continuous \(h : \mathbb{R} \to \mathbb{R}\) with \(h|_F = g\) would force

\begin{equation*} h(b) = \lim_{j \to \infty} h(x_j) = \lim_{j \to \infty} a_j , \end{equation*}

but \(1, 0, 1, 0, \dots\) does not converge. So \(g\) has no continuous extension, and \(F\) must be closed.

Take \(f = \chi_A\) for a Borel set \(A\) with \(|A \cap I| > 0\) and \(|I \setminus A| > 0\) for every nonempty open interval \(I\); we build such an \(A\) from nowhere dense sets of positive measure. (Note \(f = \chi_{\mathbb{Q}}\) fails, since \(B = \mathbb{R} \setminus \mathbb{Q}\) works for it.)

(i) Every nonempty bounded open interval \(J = (c,d)\) contains a closed set \(K\) with empty interior and \(|K| > 0\). Put \([c^{\prime},d^{\prime}] \subseteq J\) with \(d^{\prime} - c^{\prime} = \frac{d-c}{2}\), let \(\varepsilon = \frac{d^{\prime}-c^{\prime}}{2}\), enumerate \(\mathbb{Q}\) as \(q_1, q_2, \dots\), and let \(U = \bigcup_j (q_j - \varepsilon 2^{-j-1}, q_j + \varepsilon 2^{-j-1})\), so \(|U| \le \varepsilon\) by subadditivity (2.8) and 2.14. Then \(K = [c^{\prime},d^{\prime}] \setminus U\) is closed, contains no rational and so has empty interior, and \([c^{\prime},d^{\prime}] \subseteq K \cup U\) gives

\begin{equation*} d^{\prime} - c^{\prime} \le |K| + |U| \le |K| + \tfrac{d^{\prime}-c^{\prime}}{2} , \end{equation*}

whence \(|K| \ge \frac{d^{\prime}-c^{\prime}}{2} > 0\).

(ii) A finite union of closed sets with empty interior is closed with empty interior: for two such sets \(E, G\) and a nonempty open \(V \subseteq E \cup G\), the open set \(V \setminus E\) is nonempty (else \(V \subseteq E\)) and lies in \(G\), a contradiction; induct.

(iii) Enumerate the open intervals with rational endpoints as \(I_1, I_2, \dots\), and choose recursively disjoint closed sets \(K_n, L_n\) of positive measure and empty interior with \(K_n \cup L_n \subseteq I_n\): given \(C = K_1 \cup L_1 \cup \cdots \cup L_{n-1}\), which is closed with empty interior by (ii), the open set \(I_n \setminus C\) is nonempty, so it contains \((y-r, y+r)\) for some \(y, r\), and (i) supplies \(K_n \subseteq (y - \frac r2, y)\) and \(L_n \subseteq (y, y + \frac r2)\). Set \(A = \bigcup_n K_n\), a Borel set disjoint from every \(L_n\), and \(f = \chi_A\).

Each nonempty open interval \(I\) contains some \(I_n\), so \(|A \cap I| \ge |K_n| > 0\) and \(|I \setminus A| \ge |L_n| > 0\) by monotonicity. Now let \(|\mathbb{R} \setminus B| = 0\). From \(A \subseteq (A \cap B) \cup (\mathbb{R} \setminus B)\) and \(|A| \ge |K_1| > 0\), subadditivity gives \(|A \cap B| > 0\); pick \(x \in A \cap B\), so \(f(x) = 1\). If \(f|_B\) were continuous at \(x\), then taking \(\varepsilon = \frac12\) and using that \(f\) is \(\{0,1\}\)-valued, there is \(\delta > 0\) with \(B \cap I \subseteq A\) for \(I = (x-\delta, x+\delta)\); hence

\begin{equation*} I \setminus A \subseteq I \setminus (B \cap I) \subseteq \mathbb{R} \setminus B , \end{equation*}

forcing \(|I \setminus A| = 0\), contrary to \(|I \setminus A| > 0\).

The statement is false: take a non-Borel set \(A \subseteq \mathbb{R}\) (2.67) and set \(f_t = \chi_{\{t\}}\) if \(t \in A\), \(f_t = 0\) if \(t \notin A\). The countable analogue 2.53 holds only because \(\sigma\)-algebras are closed under countable unions.

Each \(f_t\) is Borel measurable, being constant or the characteristic function of the closed set \(\{t\}\). For \(x \in A\) the choice \(t = x\) gives \(f_x(x) = 1\), and all \(f_t\) are \(\{0,1\}\)-valued, so \(f(x) = 1\); for \(x \notin A\) every \(t \in A\) satisfies \(t \neq x\) and so \(f_t(x) = 0\), giving \(f(x) = 0\). Thus \(f = \chi_A\) and

\begin{equation*} f^{-1}\bigl( (\tfrac{1}{2}, \infty] \bigr) = A , \end{equation*}

which is not Borel; since \((\frac12, \infty]\) is a Borel subset of \([-\infty,\infty]\) by 2.50, the definition 2.51 shows \(f\) is not Borel measurable.

The set \(N = \{x : f(x) \ge 1\}\) is covered by the intervals \(A_k = [b_k - 2^{1-k}, b_k + 2^{1-k}]\), of total length \(4\), and a set whose complement has finite outer measure has infinite outer measure.

If \(x \notin \bigcup_k A_k\), then \(|x - b_k| > 2^{1-k} > 0\) for every \(k\), so \(x \notin \{b_1, b_2, \dots\}\) and \(f(x)\) is given by the series, with

\begin{equation*} \begin{aligned} f(x) = \sum_{k=1}^{\infty} \frac{1}{4^k |x - b_k|} &< \sum_{k=1}^{\infty} \frac{1}{2^{2k} \cdot 2^{1-k}} \\ &= \sum_{k=1}^{\infty} \frac{1}{2^{k+1}} = \frac12 . \end{aligned} \end{equation*}

Hence \(N \subseteq \bigcup_k A_k\), and countable subadditivity (2.8) with 2.14 gives

\begin{equation*} |N| \le \sum_{k=1}^{\infty} |A_k| = \sum_{k=1}^{\infty} \frac{4}{2^k} = 4 . \end{equation*}

Writing \(E = \{x : f(x) < 1\} = \mathbb{R} \setminus N\), for each \(n \in \mathbb{Z}^+\) the inclusion \([-n,n] \subseteq (E \cap [-n,n]) \cup N\) and subadditivity give

\begin{equation*} 2n = |[-n,n]| \le |E \cap [-n,n]| + |N| \le |E| + 4 . \end{equation*}

Letting \(n \to \infty\) yields \(|E| = \infty\).

Extend \(f\) by \(0\), apply 2.95 to the extension, and restrict the resulting Borel function back to \(B\).

Define \(\tilde{f} : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} \tilde{f}(x) = \begin{cases} f(x) & \text{if } x \in B, \\ 0 & \text{if } x \in \mathbb{R} \setminus B. \end{cases} \end{equation*}

Then \(\tilde{f}\) is Lebesgue measurable: for a Borel set \(A \subseteq \mathbb{R}\),

\begin{equation*} \tilde{f}^{-1}(A) = \begin{cases} f^{-1}(A) & \text{if } 0 \notin A, \\ f^{-1}(A) \cup (\mathbb{R} \setminus B) & \text{if } 0 \in A, \end{cases} \end{equation*}

a union of Lebesgue measurable sets, since \(f\) is Lebesgue measurable and the Borel set \(\mathbb{R} \setminus B\) is Lebesgue measurable.

By 2.95 there is a Borel measurable \(\tilde{g} : \mathbb{R} \to \mathbb{R}\) with \(|\{x \in \mathbb{R} : \tilde{g}(x) \neq \tilde{f}(x)\}| = 0\). Put \(g = \tilde{g}|_B\). For Borel \(A\) we have \(g^{-1}(A) = B \cap \tilde{g}^{-1}(A)\), Borel because \(B\) is; so \(g\) is Borel measurable. Finally \(\tilde{f} = f\) on \(B\) gives

\begin{equation*} \{x \in B : g(x) \neq f(x)\} \subseteq \{x \in \mathbb{R} : \tilde{g}(x) \neq \tilde{f}(x)\} , \end{equation*}

so \(|\{x \in B : g(x) \neq f(x)\}| = 0\) by monotonicity of outer measure (2.5).

A positive lower bound on a set of infinite measure would already make one lower Lebesgue sum infinite.

Suppose \(\mu(E) = \infty\) and \(c = \inf_E f > 0\) (note \(E \neq \varnothing\), since \(\mu(\varnothing) = 0\)). Taking the \(\mathcal{S}\)-partition \(\mathcal{P}\) of \(X\) given by \(E, X \setminus E\), the definition 3.2 and nonnegativity of the discarded term give

\begin{equation*} \mathcal{L}(f, \mathcal{P}) \geq \mu(E) \inf_E f = \infty \cdot c = \infty . \end{equation*}

Hence \(\int f \, d\mu = \infty\) by 3.3, contradicting \(\int f \, d\mu < \infty\). As \(\inf_E f \ge 0\) always, \(\inf_E f = 0\).

Every lower Lebesgue sum for \(\delta_c\) collapses to \(\inf_A f\) over the one partition piece \(A\) containing \(c\), and the level sets \(\{f > t\}\) supply pieces on which that infimum is nearly \(f( c)\).

(i) \(\int f \, d\delta_c \le f( c)\). If \(A_1, \dots, A_m\) is an \(\mathcal{S}\)-partition of \(X\), then \(c\) lies in exactly one piece \(A_{j_0}\), so \(\delta_c(A_{j_0}) = 1\) and the other terms vanish (with \(0 \cdot \infty = 0\)):

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \sum_{j=1}^m \delta_c(A_j) \inf_{A_j} f = \inf_{A_{j_0}} f \le f( c) . \end{equation*}

Take the supremum over \(\mathcal{P}\) and use 3.3.

(ii) \(\int f \, d\delta_c \ge f( c)\). Assume \(f( c) > 0\) and take any \(t \in [0, f( c))\). Since \((t,\infty]\) is a Borel subset of \([-\infty,\infty]\) by 2.50 and \(f\) is \(\mathcal{S}\)-measurable, the set \(A = \{x \in X : f(x) > t\}\) lies in \(\mathcal{S}\) by 2.51, and \(c \in A\); this replaces the partition by \(\{c\}, X \setminus \{c\}\), which is unavailable since \(\{c\}\) need not lie in \(\mathcal{S}\). For the partition \(A, X \setminus A\),

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \delta_c(A) \inf_A f = \inf_A f \ge t . \end{equation*}

Letting \(t \uparrow f( c)\) gives the inequality, also when \(f( c) = \infty\).

Let \(E = \{f > 0\}\) and \(E_k = \{f > \frac1k\}\), all in \(\mathcal{S}\) by 2.50 and 2.51.

(i) If \(\mu(E) > 0\): the sets \(E_k\) increase to \(E\) (any \(x\) with \(f(x) > 0\) has \(f(x) > \frac1k\) for some \(k\)), so \(\mu(E_k) \to \mu(E) > 0\) by 2.59 and \(\mu(E_k) > 0\) for some \(k\). Since \(\frac1k \chi_{E_k} \le f\) pointwise, 3.8 and 3.7 give

\begin{equation*} \int f \, d\mu \ge \int \tfrac{1}{k} \chi_{E_k} \, d\mu = \tfrac{1}{k} \mu(E_k) > 0 . \end{equation*}

(ii) If \(\mu(E) = 0\): every term \(\mu(A_j) \inf_{A_j} f\) of a lower Lebesgue sum vanishes. Either \(A_j \subseteq E\), whence \(\mu(A_j) = 0\) by 2.57(a), or \(A_j\) contains a point outside \(E\), whence \(\inf_{A_j} f = 0\); in both cases the convention \(0 \cdot \infty = 0\) gives the term \(0\). So \(\int f \, d\mu = 0\) by 3.3.

List \(\mathbb{Q} \cap [0,1]\) bijectively as \(r_1, r_2, \dots\) and take

\begin{equation*} f(x) = \begin{cases} \dfrac{1}{k} & \text{if } x = r_k , \\[2mm] 1 & \text{if } x \text{ is irrational,} \end{cases} \end{equation*}

whose values lie in \((0,1]\). It is Borel measurable, since for Borel \(B\) the set \(f^{-1}(B)\) is the union of the countable set \(\{r_k : \frac1k \in B\}\) with either \([0,1] \setminus \mathbb{Q}\) or \(\varnothing\).

Every nondegenerate subinterval \([x_{j-1}, x_j]\) contains \(r_k\) for infinitely many \(k\), so \(\inf_{[x_{j-1},x_j]} f = 0\). Hence every lower Riemann sum (1.3) of a partition \(P\) of \([0,1]\) is

\begin{equation*} L(f, P, [0,1]) = \sum_{j=1}^n (x_j - x_{j-1}) \cdot 0 = 0 , \end{equation*}

and taking the supremum over \(P\) gives \(L(f, [0,1]) = 0\) by 1.7.

The assertion is that \(\int b \, d\mu = \sum_{k=1}^{\infty} b_k\) for counting measure \(\mu\) on \(\mathbb{Z}^+\) (2.55) and \(b(k) = b_k \ge 0\); both inequalities come from 3.2 and 3.3, with \(\sum_{k=1}^{\infty} b_k = \sup\{\sum_{k \in F} b_k : F \text{ finite}\}\).

(i) \(\int b \, d\mu \ge \sum_{k=1}^{\infty} b_k\). Fix \(n\) and take the \(\mathcal{S}\)-partition \(\mathcal{P}_n\) of \(\mathbb{Z}^+\) given by

\begin{equation*} \{1\}, \{2\}, \ldots, \{n\}, \quad \{n+1, n+2, \ldots\} . \end{equation*}

Since \(\mu(\{k\}) = 1\) and the final term is nonnegative,

\begin{equation*} \mathcal{L}(b, \mathcal{P}_n) = \sum_{k=1}^n b_k + \mu(\{n+1, n+2, \ldots\}) \inf_{k > n} b_k \geq \sum_{k=1}^n b_k , \end{equation*}

so \(\int b \, d\mu \ge \sum_{k=1}^n b_k\) for every \(n\); let \(n \to \infty\).

(ii) \(\int b \, d\mu \le \sum_{k=1}^{\infty} b_k\). For an \(\mathcal{S}\)-partition \(A_1, \dots, A_m\) and \(c_j = \inf_{A_j} b\) we claim \(\mu(A_j) c_j \le \sum_{k \in A_j} b_k\): for \(A_j = \varnothing\) both sides are \(0\); for \(A_j\) finite with \(n_j\) elements each \(b_k \ge c_j\) gives \(n_j c_j \le \sum_{k \in A_j} b_k\); for \(A_j\) infinite with \(c_j > 0\), finite subsets of any size \(N\) give \(\sum_{k \in A_j} b_k \ge N c_j\), so the right side is \(\infty\) (and if \(c_j = 0\) the left side is \(\infty \cdot 0 = 0\)). Hence

\begin{equation*} \mathcal{L}(b, \mathcal{P}) \le \sum_{j=1}^m \sum_{k \in A_j} b_k \leq \sum_{k=1}^{\infty} b_k , \end{equation*}

the last step because approximating each \(\sum_{k \in A_j} b_k\) from below by a finite \(F_j \subseteq A_j\) and using disjointness makes \(F_1 \cup \cdots \cup F_m\) a finite subset of \(\mathbb{Z}^+\). Now apply 3.3.

Method (2): with \(b^{(n)} = \sum_{k=1}^n b_k \chi_{\{k\}}\) we have \(0 \le b^{(1)} \le b^{(2)} \le \cdots\) with \(b^{(n)}(k) = b_k\) once \(n \ge k\), so the Monotone Convergence Theorem 3.11 and 3.7 give

\begin{equation*} \int b \, d\mu = \lim_{n \to \infty} \sum_{k=1}^n b_k \mu(\{k\}) = \sum_{k=1}^{\infty} b_k . \end{equation*}

Group the pieces of \(\mathcal{P}^{\prime}\) by the piece of \(\mathcal{P}\) containing them; refining only raises the infima.

Write \(\mathcal{P} = A_1, \dots, A_m\) and \(\mathcal{P}^{\prime} = B_1, \dots, B_n\), and fix for each \(k\) an index \(j(k)\) with \(B_k \subseteq A_{j(k)}\). Then for each \(j\),

\begin{equation*} A_j = \bigcup_{\{k \, : \, j(k) = j\}} B_k , \end{equation*}

a disjoint union: \(\supseteq\) is clear, and if \(x \in A_j\) then the unique \(B_k\) containing \(x\) satisfies \(x \in A_j \cap A_{j(k)}\), forcing \(j(k) = j\) by disjointness of \(\mathcal{P}\). Finite additivity of \(\mu\) therefore gives \(\mu(A_j) = \sum_{\{k : j(k) = j\}} \mu(B_k)\). Grouping the terms of \(\mathcal{L}(f, \mathcal{P}^{\prime})\) by \(j\),

\begin{equation*} \begin{aligned} \mathcal{L}(f, \mathcal{P}^{\prime}) &= \sum_{k=1}^n \mu(B_k) \inf_{B_k} f \\ &= \sum_{j=1}^m \; \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \inf_{B_k} f \\ &\geq \sum_{j=1}^m \; \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \inf_{A_j} f \\ &= \sum_{j=1}^m \Bigl( \inf_{A_j} f \Bigr) \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \\ &= \sum_{j=1}^m \mu(A_j) \inf_{A_j} f \\ &= \mathcal{L}(f, \mathcal{P}) . \end{aligned} \end{equation*}

The third line uses \(\inf_{B_k} f \ge \inf_{A_j} f\) for \(B_k \subseteq A_j\) (multiplication by \(\mu(B_k) \in [0,\infty]\) preserving it, with \(0 \cdot \infty = 0\)); the fourth uses the distributive law \(\sum_i c \, t_i = c \sum_i t_i\) in \([0,\infty]\) (Check!), and the fifth the additivity above.

Write \(S = \sum_{x \in X} w(x) f(x)\), each such sum being the supremum of its finite subsums, and \(0 \cdot \infty = \infty \cdot 0 = 0\); note \(\mu(\{x\}) = w(x)\).

Two facts are used. (a) If \(A_1, \dots, A_m \subseteq A\) are disjoint and \(g : X \to [0,\infty]\), then \(\sum_{j} \sum_{x \in A_j} g(x) \le \sum_{x \in A} g(x)\): choosing finite \(F_j \subseteq A_j\) with \(\sum_{x \in F_j} g\) close to (or, if infinite, large below) \(\sum_{x \in A_j} g\) makes \(F_1 \cup \cdots \cup F_m\) a finite subset of \(A\), and taking suprema over such choices gives the claim. (b) If \(g(x) \ge c \, w(x)\) on \(A\) for some \(c \in [0,\infty]\), then \(\sum_{x \in A} g(x) \ge c \sum_{x \in A} w(x)\), by the distributive law on finite subsums and \(\sup_F (c \, t_F) = c \sup_F t_F\) (Check! in the cases \(c = 0, \infty\)).

(i) \(\int f \, d\mu \ge S\). For a finite \(F = \{x_1, \dots, x_p\} \subseteq X\) take the \(\mathcal{S}\)-partition \(\{x_1\}, \dots, \{x_p\}, X \setminus F\); discarding the last nonnegative term,

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \sum_{i=1}^p w(x_i) f(x_i) + \mu(X \setminus F) \inf_{X \setminus F} f \; \geq \; \sum_{x \in F} w(x) f(x) . \end{equation*}

Take suprema over \(F\) and apply 3.3.

(ii) \(\int f \, d\mu \le S\). For an \(\mathcal{S}\)-partition \(A_1, \dots, A_m\) and \(c_j = \inf_{A_j} f\), we have \(w(x)f(x) \ge c_j w(x)\) on \(A_j\), so (b) gives

\begin{equation*} \sum_{x \in A_j} w(x) f(x) \; \geq \; c_j \sum_{x \in A_j} w(x) = \mu(A_j) \inf_{A_j} f . \end{equation*}

Summing over \(j\) and applying (a) with \(A = X\) gives \(\mathcal{L}(f, \mathcal{P}) \le S\); now use 3.3.

Take \(f_k = \chi_{[k,\,k+1]}\), a simple Borel measurable function.

Each \(x \in \mathbb{R}\) has \(f_k(x) = 0\) for every \(k > x\), so \(f_k \to 0\) pointwise, while 3.4 gives

\begin{equation*} \int f_k \, d\lambda = \lambda\bigl([k, k+1]\bigr) = 1 \end{equation*}

for every \(k\).

Countable additivity of \(\nu\) is the Monotone Convergence Theorem 3.11 applied to the partial sums \(\chi_{A_1 \cup \cdots \cup A_n} f\); the rest is immediate.

Since \(\chi_\emptyset f\) is identically \(0\), \(\nu(\emptyset) = 0\). Let \(A_1, A_2, \dots \in \mathcal{S}\) be disjoint with union \(A\), and put

\begin{equation*} g_n = \sum_{k=1}^{n} \chi_{A_k} f = \chi_{A_1 \cup \cdots \cup A_n} f , \end{equation*}

the second equality by disjointness. Each \(g_n\) is \(\mathcal{S}\)-measurable, \(0 \le g_1 \le g_2 \le \cdots\), and \(g_n \to \chi_A f\) pointwise (each \(x \in A\) lies in exactly one \(A_k\), so \(g_n(x) = f(x)\) for \(n \ge k\); off \(A\) both sides are \(0\)). Hence 3.11 and additivity 3.16 give

\begin{equation*} \begin{aligned} \nu(A) = \lim_{n \to \infty} \int g_n \, d\mu &= \lim_{n \to \infty} \sum_{k=1}^{n} \nu(A_k) \\ &= \sum_{k=1}^{\infty} \nu(A_k) . \end{aligned} \end{equation*}

So \(\nu\) is a measure by 2.54.

Apply the Monotone Convergence Theorem 3.11 to the partial sums \(g_n = \sum_{k=1}^{n} f_k\).

Each \(g_n\) is \(\mathcal{S}\)-measurable, since for \(a \ge 0\) the sum of two measurable \(u, v : X \to [0,\infty]\) satisfies

\begin{equation*} \begin{aligned} (u + v)^{-1}\bigl((a, \infty]\bigr) &= \bigcup_{q \in \mathbb{Q}} \Bigl( u^{-1}\bigl((q, \infty]\bigr) \cap v^{-1}\bigl((a - q, \infty]\bigr) \Bigr) , \end{aligned} \end{equation*}

a countable union in \(\mathcal{S}\) (for \(\supseteq\) add the inequalities; for \(\subseteq\) take rational \(q\) between \(a - v(x)\) and \(u(x)\), or \(q > a\) if \(u(x) = \infty\) and \(q < 0\) if \(v(x) = \infty\)), so 2.52 applies; induct. Nonnegativity gives \(0 \le g_1 \le g_2 \le \cdots\) with \(g_n(x) \to f(x)\) for every \(x\), so \(f\) is \(\mathcal{S}\)-measurable by 2.53, and 3.11 together with additivity 3.16 gives

\begin{equation*} \begin{aligned} \int f \, d\mu = \lim_{n \to \infty} \int g_n \, d\mu &= \lim_{n \to \infty} \sum_{k=1}^{n} \int f_k \, d\mu \\ &= \sum_{k=1}^{\infty} \int f_k \, d\mu . \end{aligned} \end{equation*}

Take \(E = \{x \in X : g(x) < \infty\}\), where \(g(x) = \sum_{k=1}^{\infty} |f_k(x)|\).

Each \(|f_k|\) is \(\mathcal{S}\)-measurable, since for \(a \ge 0\) we have \(|f_k|^{-1}((a,\infty]) = f_k^{-1}((a,\infty)) \cup f_k^{-1}((-\infty,-a)) \in \mathcal{S}\), so 2.52 applies; hence \(g\) is \(\mathcal{S}\)-measurable as the pointwise supremum (2.53) of its partial sums, and \(E \in \mathcal{S}\). By Exercise 3A.10 applied to \(|f_1|, |f_2|, \dots\),

\begin{equation*} \int g \, d\mu = \sum_{k=1}^{\infty} \int |f_k| \, d\mu < \infty . \end{equation*}

For each \(t \in (0,\infty)\) we have \(g \ge t \chi_{X \setminus E}\) pointwise (the left side is \(\infty\) on \(X \setminus E\)), so 3.8 with 3.4 gives

\begin{equation*} t \, \mu(X \setminus E) \le \int g \, d\mu < \infty ; \end{equation*}

letting \(t \to \infty\) yields \(\mu(X \setminus E) = 0\). Finally, for \(x \in E\) the series \(\sum_k |f_k(x)|\) converges in \(\mathbb{R}\), so its terms tend to \(0\) and \(\lim_{k \to \infty} f_k(x) = 0\).

Take an enumeration \(r_1, r_2, \ldots\) of \(\mathbb{Q}\), put \(h_k(x) = 1/|x - r_k|\) with \(h_k(r_k) = \infty\), and set

\begin{equation*} g = \sum_{k=1}^{\infty} 8^{-k} h_k, \qquad S = g^{-1}(\{\infty\}), \qquad f = g \chi_{\mathbb{R} \setminus S} + \chi_S . \end{equation*}

Each \(h_k\) is Borel measurable (each \(h_k^{-1}((a, \infty])\) is an open interval or \(\mathbb{R}\)), so \(g\) is Borel measurable as the supremum of its partial sums (2.53) and \(g > 0\) everywhere; hence \(f : \mathbb{R} \to (0, \infty)\) is Borel measurable by 2.52.

\(\lambda(S) = 0\): with \(B = \bigcap_{n} \bigcup_{k \geq n} (r_k - 4^{-k}, r_k + 4^{-k})\), monotonicity and countable subadditivity (2.57, 2.58) give \(\lambda(B) \leq \sum_{k \geq n} 2 \cdot 4^{-k} = \frac{8}{3} 4^{-n}\) for every \(n\), so \(\lambda(B) = 0\); and if \(x \notin B \cup \mathbb{Q}\), choosing \(N\) with \(|x - r_k| \geq 4^{-k}\) for all \(k \geq N\) gives \(8^{-k} h_k(x) \leq 2^{-k}\) for \(k \geq N\) while the finitely many earlier terms are finite, so \(g(x) < \infty\). Thus \(S \subseteq B \cup \mathbb{Q}\) and \(\lambda(S) = 0\).

Fix a nonempty open interval \(I\); by density of \(\mathbb{Q}\) choose \(j\) with \(r_j \in I\), and \(\delta > 0\) with \((r_j - \delta, r_j + \delta) \subseteq I\). The disjoint sets \(A_m = (r_j + \delta 2^{-m-2}, \, r_j + \delta 2^{-m-1}]\) lie in \(I\), have \(\lambda(A_m) = \delta 2^{-m-2}\), and carry \(h_j \geq 2^{m+1}/\delta\), so 3.9 gives

\begin{equation*} \int \chi_I h_j \, d\lambda \;\geq\; \sum_{m=0}^{M} \frac{2^{m+1}}{\delta} \cdot \delta 2^{-m-2} = \frac{M+1}{2} \end{equation*}

for every \(M\), whence \(\int \chi_I h_j \, d\lambda = \infty\). Since \(\chi_I g = \sum_k 8^{-k} \chi_I h_k\) pointwise, 3A.10 and 3.20 give \(\int \chi_I g \, d\lambda \geq 8^{-j} \int \chi_I h_j \, d\lambda = \infty\). Also \(\int \chi_{I \cap S} \, g \, d\lambda = 0\): any \(\sum_i c_i \chi_{A_i} \leq \chi_{I \cap S} \, g\) with the \(A_i\) disjoint Borel sets and \(c_i > 0\) forces \(A_i \subseteq S\) and hence \(\lambda(A_i) = 0\), so every lower Lebesgue sum vanishes (3.3). Therefore additivity 3.16 gives

\begin{equation*} \infty = \int \chi_I g \, d\lambda = 0 + \int \chi_{I \setminus S} \, g \, d\lambda , \end{equation*}

and \(\chi_I f \geq \chi_{I \setminus S} \, g\) pointwise, so \(\int \chi_I f \, d\lambda = \infty\) by 3.8.

Take \(f_k = -\chi_{[k, \infty)}\) on \(\mathbb{R}\) with Lebesgue measure \(\lambda\) and the Borel \(\sigma\)-algebra.

Each \(f_k\) is a Borel measurable simple function, and \(f_1 \leq f_2 \leq \cdots\) because \([k+1,\infty) \subseteq [k,\infty)\); only the hypothesis \(f_k \geq 0\) fails. Given \(x\), we have \(f_k(x) = 0\) for all \(k > x\), so \(f_k \to 0\) pointwise and \(\int f \, d\lambda = 0\). But \(f_k^{\,+} = 0\) and \(f_k^{\,-} = \chi_{[k,\infty)}\), so \(\int f_k \, d\lambda\) is defined by 3.18 and 3.4 gives

\begin{equation*} \int f_k \, d\lambda = 0 - \lambda\bigl([k, \infty)\bigr) = -\infty \end{equation*}

for every \(k\). Hence \(\lim_{k \to \infty} \int f_k \, d\lambda = -\infty \neq 0 = \int f \, d\lambda\).

Take \(f_k = \chi_{[k, \infty)}\) on \(\mathbb{R}\) with Lebesgue measure \(\lambda\) and the Borel \(\sigma\)-algebra.

Each \(f_k\) is a nonnegative Borel measurable simple function, so every hypothesis of 3.11 holds except that \([k+1,\infty) \subseteq [k,\infty)\) makes the sequence decreasing: \(f_1 \geq f_2 \geq \cdots \geq 0\). Given \(x\), we have \(f_k(x) = 0\) for all \(k > x\), so \(f_k \to 0\) pointwise and \(\int f \, d\lambda = 0\). But 3.4 gives \(\int f_k \, d\lambda = \lambda([k,\infty)) = \infty\) for every \(k\), so

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\lambda = \infty \neq 0 = \int f \, d\lambda . \end{equation*}

Both parts follow from the fact that \(A \mapsto t + A\) and \(A \mapsto \frac{1}{t}A\) permute the Borel partitions of \(\mathbb{R}\), multiplying lower Lebesgue sums by \(1\) and by \(\frac{1}{|t|}\) respectively.

Write \(\mathcal{B}\) for the Borel \(\sigma\)-algebra. Since \(x \mapsto x - t\) and (for \(t \neq 0\)) \(x \mapsto tx\) are continuous, hence Borel measurable (2.41), the maps \(A \mapsto t + A\) and \(A \mapsto \frac{1}{t}A\) send \(\mathcal{B}\) into \(\mathcal{B}\); each is a bijection of \(\mathcal{B}\) onto \(\mathcal{B}\) (inverses \(A \mapsto -t + A\) and \(A \mapsto tA\)) preserving disjointness and unions, so each carries the Borel partitions of \(\mathbb{R}\) bijectively onto themselves. Moreover \(\lambda(t + A) = \lambda(A)\) by 2.7, and \(\lambda(\frac1t A) = \frac{1}{|t|} \lambda(A)\) by Exercise 2 in Section 2A together with 2.68. Since \(f_t^{-1}(B) = t + f^{-1}(B)\) in (a) and \(f_t^{-1}(B) = \frac1t f^{-1}(B)\) in (b), the function \(f_t\) is Borel measurable in both cases.

(a) Assume first \(f \geq 0\). For a Borel partition \(P = A_1, \ldots, A_m\) put \(P_t = t + A_1, \ldots, t + A_m\). As \(x \in t + A_j\) if and only if \(x - t \in A_j\), we get \(\inf_{t + A_j} f_t = \inf_{A_j} f\), so

\begin{equation*} L(f_t, P_t) = \sum_{j=1}^{m} \lambda(t + A_j) \inf_{t + A_j} f_t = \sum_{j=1}^{m} \lambda(A_j) \inf_{A_j} f = L(f, P) . \end{equation*}

Because \(P \mapsto P_t\) is a bijection of the Borel partitions, the two sets of lower Lebesgue sums coincide, and 3.3 gives \(\int f_t \, d\lambda = \int f \, d\lambda\).

For general \(f\) note \((f_t)^{\pm} = (f^{\pm})_t\) pointwise, so the nonnegative case gives \(\int (f_t)^{\pm} d\lambda = \int f^{\pm} d\lambda\); one of the latter is finite because \(\int f \, d\lambda\) is defined, hence \(\int f_t \, d\lambda\) is defined and 3.18 gives \(\int f_t \, d\lambda = \int f \, d\lambda\).

(b) Identically, with \(P^{(t)} = \frac1t A_1, \ldots, \frac1t A_m\) and \(\inf_{\frac1t A_j} f_t = \inf_{A_j} f\),

\begin{equation*} L\bigl(f_t, P^{(t)}\bigr) = \sum_{j=1}^{m} \frac{1}{|t|} \lambda(A_j) \inf_{A_j} f = \frac{1}{|t|} L(f, P) , \end{equation*}

so the lower Lebesgue sums for \(f_t\) are exactly \(\frac{1}{|t|}\) times those for \(f\); multiplication by the positive constant \(\frac{1}{|t|}\) commutes with suprema in \([0, \infty]\), so 3.3 gives \(\int f_t \, d\lambda = \frac{1}{|t|} \int f \, d\lambda\) for \(f \geq 0\). The passage to general \(f\) via \((f_t)^{\pm} = (f^{\pm})_t\) is as in (a), the final subtraction being legitimate because at most one of the two terms is infinite.

Both integrals are the limit of the same sequence of simple-function integrals.

Every \(\mathcal{S}\)-measurable function is \(\mathcal{T}\)-measurable, since its preimages lie in \(\mathcal{S} \subseteq \mathcal{T}\); in particular \(\int f \, d\mu_2\) makes sense. By 2.89 applied to \((X, \mathcal{S})\) there are simple \(\mathcal{S}\)-measurable \(u_1, u_2, \ldots\) with \(|u_k| \leq |u_{k+1}| \leq |f|\) pointwise and \(u_k \to f\) pointwise; put \(f_k = \max\{u_1^+, \ldots, u_k^+\}\). Each \(f_k\) takes finitely many values and \(f_k^{-1}((a,\infty]) = \bigcup_{j \leq k} (u_j^+)^{-1}((a,\infty]) \in \mathcal{S}\), so \(f_k\) is a nonnegative simple \(\mathcal{S}\)-measurable function (2.52); moreover \(0 \leq f_1 \leq f_2 \leq \cdots \leq f\) because \(u_j^+ \leq |u_j| \leq |f| = f\), and \(f_k \geq u_k^+ \to f^+ = f\), so \(f_k \to f\) pointwise.

If \(h = \sum_{k=1}^{n} c_k \chi_{E_k}\) is the standard representation of a nonnegative simple \(\mathcal{S}\)-measurable function, then \(E_1, \ldots, E_n\) are disjoint sets lying in \(\mathcal{S} \subseteq \mathcal{T}\), so 3.7 applies in each of the two measure spaces and gives

\begin{equation*} \int h \, d\mu_1 = \sum_{k=1}^{n} c_k \mu_1(E_k) = \sum_{k=1}^{n} c_k \mu_2(E_k) = \int h \, d\mu_2 , \end{equation*}

the middle equality because each \(E_k \in \mathcal{S}\). Applying this to each \(f_k\) and the Monotone Convergence Theorem 3.11 in \((X, \mathcal{S}, \mu_1)\) and in \((X, \mathcal{T}, \mu_2)\),

\begin{equation*} \int f \, d\mu_1 = \lim_{k \to \infty} \int f_k \, d\mu_1 = \lim_{k \to \infty} \int f_k \, d\mu_2 = \int f \, d\mu_2 . \end{equation*}

Put \(g_k = \inf\{f_j : j \geq k\}\), so that \(f = \lim_{k \to \infty} g_k\).

(a) Each \(g_k\) is \(\mathcal{S}\)-measurable by 2.53, and \(g_1 \leq g_2 \leq \cdots\) because the infimum is taken over a smaller set as \(k\) grows; hence \(f = \liminf_k f_k = \sup_k g_k\) is \(\mathcal{S}\)-measurable, again by 2.53.

(b) By (a), \(0 \leq g_1 \leq g_2 \leq \cdots\) are \(\mathcal{S}\)-measurable with \(g_k \to f\) pointwise, so 3.11 gives \(\int f \, d\mu = \lim_k \int g_k \, d\mu\). For every \(j \geq k\) we have \(g_k \leq f_j\), so 3.8 gives \(\int g_k \, d\mu \leq \inf_{j \geq k} \int f_j \, d\mu\); letting \(k \to \infty\),

\begin{equation*} \int f \, d\mu = \lim_{k \to \infty} \int g_k \, d\mu \leq \liminf_{k \to \infty} \int f_k \, d\mu . \end{equation*}

(c) On \(X = [0,1]\) with Borel sets and Lebesgue measure (so \(\mu(X) = 1\)), take \(f_k = \chi_{[0, 1/2]}\) for \(k\) odd and \(f_k = \chi_{(1/2, 1]}\) for \(k\) even; then \(0 \leq f_k \leq 1\), so the family is uniformly bounded. Each \(x \in [0,1]\) has \(f_k(x) = 0\) for infinitely many \(k\), so \(g_k = 0\) for every \(k\) and \(f = 0\), giving \(\int f \, d\lambda = 0\); while 3.4 gives \(\int f_k \, d\lambda = \frac12\) for every \(k\). Thus

\begin{equation*} 0 = \int f \, d\lambda < \liminf_{k \to \infty} \int f_k \, d\lambda = \tfrac12 . \end{equation*}

Take \(x_k = \dfrac{(-1)^k}{k}\).

The sequence \(1, \frac12, \frac13, \ldots\) decreases to \(0\), so by the alternating series test \(\lim_{n \to \infty} \sum_{k=1}^{n} x_k\) exists in \(\mathbb{R}\) (its value is \(-\ln 2\)). But integration with respect to counting measure on \(\mathbb{Z}^+\) is summation (Example 3.6), and every subset of \(\mathbb{Z}^+\) is measurable, so

\begin{equation*} \int x^+ \, d\mu = \sum_{m=1}^{\infty} \frac{1}{2m} = \infty , \qquad \int x^- \, d\mu = \sum_{m=1}^{\infty} \frac{1}{2m-1} = \infty , \end{equation*}

both by divergence of the harmonic series. Since neither is finite, 3.18 assigns no value to \(\int x \, d\mu\).

The one-set partition gives the left inequality, and \(f \leq (\sup_X f) \chi_X\) gives the right.

Assume \(X \neq \varnothing\) (otherwise \(\mu(X) = 0 = \int f \, d\mu\) and both inequalities are trivial), and recall the convention \(0 \cdot \infty = \infty \cdot 0 = 0\) stated with 3.2. The single set \(A_1 = X\) is an \(\mathcal{S}\)-partition \(P\) of \(X\), so by 3.2 and the definition 3.3 of the integral as a supremum of lower Lebesgue sums,

\begin{equation*} \mu(X) \inf_X f = L(f, P) \leq \int f \, d\mu . \end{equation*}

For the other inequality put \(c = \sup_X f \in [0, \infty]\), so \(f \leq c \chi_X\) pointwise; then 3.8 and 3.15 (with \(n = 1\), \(E_1 = X\), \(c_1 = c\)) give

\begin{equation*} \int f \, d\mu \leq \int c \, \chi_X \, d\mu = c \, \mu(X) = \mu(X) \sup_X f . \end{equation*}

The increasing case is 3.11 applied to \(f_k - f_1\); the decreasing case follows by passing to \(-f_k\).

Monotonicity makes \(f(x) = \lim_k f_k(x)\) exist in \([-\infty, \infty]\) for each \(x\), and \(f\) is \(\mathcal{S}\)-measurable by 2.53, being the pointwise supremum of the \(f_k\) in the increasing case and their pointwise infimum in the decreasing case.

Increasing case: assume \(f_1 \leq f_2 \leq \cdots\). Then \(f_k^- \leq f_1^- \leq |f_1|\) and likewise \(f^- \leq f_1^-\), so 3.8 gives \(\int f_k^- \, d\mu \leq \int |f_1| \, d\mu < \infty\) and \(\int f^- \, d\mu < \infty\); hence every \(\int f_k \, d\mu\) and \(\int f \, d\mu\) is defined, with values in \((-\infty, \infty]\).

Reduction to \(f_1\) real valued. Let \(N = f_1^{-1}(\{-\infty, \infty\}) \in \mathcal{S}\) and \(M = X \setminus N\). Since \(|f_1| \geq n \chi_N\) pointwise for every \(n\), 3.8 and 3.15 give \(n \mu(N) \leq \int |f_1| \, d\mu < \infty\), so \(\mu(N) = 0\). For nonnegative \(\mathcal{S}\)-measurable \(h\) we then have \(\int h \chi_N \, d\mu = 0\), because any \(\mathcal{S}\)-partition piece on which \(h \chi_N\) has positive infimum lies in \(N\) and so has measure \(0\), making every lower Lebesgue sum vanish (3.3). Hence for \(\mathcal{S}\)-measurable \(g\) with \(\int g \, d\mu\) defined, \(g \chi_M\) is \(\mathcal{S}\)-measurable (its set \((g\chi_M)^{-1}((a,\infty])\) is \(g^{-1}((a,\infty]) \cap M\) for \(a \geq 0\) and that set union \(N\) for \(a < 0\); apply 2.52), and \((g\chi_M)^{\pm} = g^{\pm}\chi_M\) with \(g^{\pm} = g^{\pm}\chi_M + g^{\pm}\chi_N\), so additivity 3.16 gives \(\int (g\chi_M)^{\pm} \, d\mu = \int g^{\pm} \, d\mu\) and therefore \(\int g \chi_M \, d\mu = \int g \, d\mu\). Replacing each \(f_k\) by \(f_k \chi_M\) and \(f\) by \(f \chi_M\) thus changes no integral while preserving monotonicity, the pointwise limit, and \(\int |f_1| \, d\mu < \infty\); so assume \(f_1\) is real valued.

Now \(g_k := f_k - f_1\) and \(g := f - f_1\) are well defined with values in \([0, \infty]\), and they are \(\mathcal{S}\)-measurable by 2.52, since for \(a \in \mathbb{R}\)

\begin{equation*} g_k^{-1}\bigl((a, \infty]\bigr) = \bigcup_{r \in \mathbb{Q}} \Bigl( f_k^{-1}\bigl((r, \infty]\bigr) \cap f_1^{-1}\bigl([-\infty, r - a)\bigr) \Bigr) \end{equation*}

(a rational \(r\) separates \(a + f_1(x) < r < f_k(x)\) exactly when \(g_k(x) > a\)). As \(0 \leq g_1 \leq g_2 \leq \cdots\) with \(g_k \to g\) pointwise, 3.11 gives \(\int g_k \, d\mu \to \int g \, d\mu\).

Next, if \(h : X \to (-\infty, \infty]\) is \(\mathcal{S}\)-measurable with \(h \geq f_1\), then

\begin{equation*} \int h \, d\mu = \int (h - f_1) \, d\mu + \int f_1 \, d\mu . \end{equation*}

Indeed \(h - f_1\) is \(\mathcal{S}\)-measurable by the display above with \(h\) in place of \(f_k\), and substituting \(h - f_1 = h^+ - h^- - f_1^+ + f_1^-\) (legitimate pointwise, since \(f_1\) is real valued and at most one of \(h^{\pm}(x)\) is nonzero; both sides are \(\infty\) where \(h = \infty\)) gives the identity between \([0,\infty]\)-valued functions \(h^+ + f_1^- = (h - f_1) + f_1^+ + h^-\), which integrates by 3.16 to

\begin{equation*} \int h^+ d\mu + \int f_1^- d\mu = \int (h - f_1) \, d\mu + \int f_1^+ d\mu + \int h^- d\mu . \end{equation*}

Here \(\int f_1^{\pm} d\mu \leq \int |f_1| \, d\mu < \infty\), and \(h \geq f_1\) forces \(h^- \leq f_1^-\) so \(\int h^- d\mu < \infty\); subtracting these two finite numbers and applying 3.18 yields the claim.

Taking \(h = f_k\) and \(h = f\) there, and using that adding the real number \(\int f_1 \, d\mu\) commutes with limits in \((-\infty, \infty]\),

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\mu = \int g \, d\mu + \int f_1 \, d\mu = \int f \, d\mu . \end{equation*}

Decreasing case: assume \(f_1 \geq f_2 \geq \cdots\). Each \(-f_k\) is \(\mathcal{S}\)-measurable (2.52), \(-f_1 \leq -f_2 \leq \cdots\) has pointwise limit \(-f\), and \(\int |-f_1| \, d\mu < \infty\); also \(f_k^+ \leq f_1^+ \leq |f_1|\) and \(f^+ \leq f_1^+\), so \(\int f_k \, d\mu\) and \(\int f \, d\mu\) are defined and 3.20 with \(c = -1\) applies. The increasing case gives \(\int (-f_k) \, d\mu \to \int (-f) \, d\mu\), that is \(-\int f_k \, d\mu \to -\int f \, d\mu\), which is the assertion.

Lebesgue’s heaps are the sets \(E_k = v^{-1}(\{c_k\})\) of 3.15, and his total is \(\sum_{k} c_k \mu(E_k)\): a sum over the finitely many values in the range, not over the many items in the domain.

Model the pocket by the finite set \(X\) of bills and coins, \(\mathcal{S}\) all subsets of \(X\), \(\mu\) counting measure, and \(v : X \to [0, \infty)\) the face value. The sum owed is \(\int v \, d\mu = \sum_{x \in X} v(x)\) (Example 3.6). Riemann’s method is “in the order I find them”: fix an enumeration of the domain and accumulate one item at a time. Lebesgue’s method is 3.15 read from left to right: let \(c_1, \ldots, c_n\) be the distinct denominations present and \(E_k = v^{-1}(\{c_k\})\) the heap of items of value \(c_k\), so that \(v = \sum_{k=1}^{n} c_k \chi_{E_k}\) and

\begin{equation*} \int v \, d\mu = \sum_{k=1}^{n} c_k \, \mu(E_k) = \sum_{k=1}^{n} (\text{denomination}) \times (\text{size of heap}) . \end{equation*}

That this is a statement about the range is visible in what 3.15 asks of \(E_1, \ldots, E_n\): only that they lie in \(\mathcal{S}\), so that \(\mu(E_k)\) makes sense. They need not be intervals, nor even disjoint. The \(c_k\) are values of the function; the \(E_k\) are whatever sets those values cut out. The same reading computes any \(\mathcal{S}\)-measurable \(f : X \to [0, \infty)\): chopping the range gives \(E_{j,k} = f^{-1}([k2^{-j}, (k+1)2^{-j}))\) for \(0 \leq k < j2^j\) and \(E_{j,\infty} = f^{-1}([j, \infty))\), all in \(\mathcal{S}\) precisely because \(f\) is \(\mathcal{S}\)-measurable, and the simple functions

\begin{equation*} s_j = \sum_{k=0}^{j2^j - 1} k 2^{-j} \chi_{E_{j,k}} + j \chi_{E_{j, \infty}} \end{equation*}

satisfy \(s_j = \min\{2^{-j}\lfloor 2^j f \rfloor, j\}\), hence increase to \(f\) pointwise; 3.15 evaluates each \(\int s_j \, d\mu\) as a sum over the range, and 3.11 delivers \(\int f \, d\mu\).

A Riemann sum, by contrast, starts from a partition \(a = t_0 < \cdots < t_n = b\) of the domain and forms \(\sum_j (\inf_{[t_{j-1},t_j]} f)(t_j - t_{j-1})\) (1.3); its accuracy is governed by the oscillation of \(f\) on each subinterval, and nothing about the range enters the choice of partition. So \(f = \chi_{\mathbb{Q} \cap [0,1]}\) has \(L(f, [0,1]) = 0 \neq 1 = U(f,[0,1])\), while sorting by value produces just two heaps and 3.15 gives at once

\begin{equation*} \int \chi_{\mathbb{Q} \cap [0,1]} \, d\lambda = 1 \cdot 0 + 0 \cdot 1 = 0 . \end{equation*}

The price is that the heaps must be measurable, which is why a theory of measure precedes the integral, and why the partitions in 3.3 range over arbitrary sets of \(\mathcal{S}\) rather than intervals: once they do, a domain partition can follow the level sets of \(f\), and 3.3 becomes the range-partition picture above.

Take \(f_k = \chi_{\{k\}}\).

Every function on \(\mathbb{Z}^+\) is measurable here. For fixed \(m\) we have \(f_k(m) = 0\) whenever \(k > m\), so \(\lim_{k \to \infty} f_k(m) = 0\); and 3.4 gives \(\int f_k \, d\mu = \mu(\{k\}) = 1\) for every \(k\), so \(\lim_{k \to \infty} \int f_k \, d\mu = 1\).

Take the tent functions \(f_k(x) = \max\bigl\{0, \, 1 - \frac{|x - k^2|}{k}\bigr\}\).

Each \(f_k\) is continuous with values in \([0,1]\) (a maximum of two continuous functions), hence Borel measurable by 2.41, and it vanishes outside \([k^2 - k, \, k^2 + k]\). Given \(x\), we have \(k^2 - k > |x|\) for all large \(k\), so \(f_k(x) = 0\) eventually and \(f_k \to 0\) pointwise. Being continuous, \(f_k\) is Riemann integrable on \([k^2-k, k^2+k]\), so 3.34 identifies its integral with the area of a triangle of base \(2k\) and height \(1\):

\begin{equation*} \int f_k \, d\lambda = \int_{[k^2 - k, \, k^2 + k]} f_k \, d\lambda = \tfrac12 \cdot 2k \cdot 1 = k , \end{equation*}

whence \(\lim_{k \to \infty} \int f_k \, d\lambda = \infty\).

Uniform continuity comes straight from 3.28: integrals of \(|f|\) over sets of small measure are small.

Each \(g(x)\) is a real number: by 3.24 it is \(\int \chi_{(-\infty,x)} f \, d\lambda\), and \(|\chi_{(-\infty,x)} f| \leq |f|\) with 3.8 makes both \((\chi_{(-\infty,x)}f)^{\pm}\) have finite integral, so 3.18 applies.

Let \(\varepsilon > 0\). Since \(|f|\) is nonnegative Borel measurable with \(\int |f| \, d\lambda < \infty\), 3.28 supplies \(\delta > 0\) such that \(\int_B |f| \, d\lambda < \varepsilon\) for every Borel set \(B\) with \(\lambda(B) < \delta\). Suppose \(x \leq y\) with \(y - x < \delta\). Then \(\chi_{(-\infty, y)} f = \chi_{(-\infty,x)} f + \chi_{[x,y)} f\), both summands having integrable absolute value, so 3.21 and then 3.23 and 3.8 give

\begin{equation*} |g(y) - g(x)| = \Bigl| \int_{[x, \, y)} f \, d\lambda \Bigr| \leq \int_{[x, \, y)} |f| \, d\lambda < \varepsilon , \end{equation*}

since \(\lambda([x,y)) = y - x < \delta\). As \(\delta\) depends only on \(\varepsilon\), \(g\) is uniformly continuous on \(\mathbb{R}\).

Write \(U(f, P) = \sum_{j} \mu(A_j) \sup_{A_j} f\) for an \(\mathcal{S}\)-partition \(P = A_1, \ldots, A_m\), and \(U(f) = \inf_P U(f,P)\); part (a) asserts \(\int f \, d\mu = U(f)\).

(a) Assume \(X \neq \varnothing\) (otherwise both sides are \(0\)), and discard the empty sets from every partition: their terms carry the factor \(\mu(\varnothing) = 0\), hence vanish by the convention \(0 \cdot \infty = \infty \cdot 0 = 0\) stated before 3.2. All infima and suprema below are then real numbers, because \(f\) is bounded; fix \(c \in [0,\infty)\) with \(0 \leq f \leq c\), so that every \(L(f,P)\) and \(U(f,P)\) lies in \([0, c\mu(X)] \subseteq \mathbb{R}\).

Refinement. Given \(\mathcal{S}\)-partitions \(P = A_1, \ldots, A_m\) and \(P’ = B_1, \ldots, B_n\), let \(Q\) consist of the nonempty sets \(A_i \cap B_j\), again an \(\mathcal{S}\)-partition of \(X\). Finite additivity of \(\mu\) gives \(\mu(A_i) = \sum_j \mu(A_i \cap B_j)\), and \(\inf_{A_i} f \leq \inf_{A_i \cap B_j} f\), so

\begin{equation*} \mu(A_i) \inf_{A_i} f \leq \sum_{j=1}^{n} \mu(A_i \cap B_j) \inf_{A_i \cap B_j} f ; \end{equation*}

summing over \(i\) gives \(L(f,P) \leq L(f,Q)\), and the same computation with \(B_j\) in place of \(A_i\) and suprema in place of infima gives \(U(f,Q) \leq U(f,P’)\). Since \(L(f,Q) \leq U(f,Q)\),

\begin{equation*} L(f, P) \leq L(f, Q) \leq U(f, Q) \leq U(f, P’) , \end{equation*}

so taking the supremum over \(P\) (3.3) and then the infimum over \(P’\) yields \(\int f \, d\mu \leq U(f)\).

Conversely let \(\varepsilon > 0\); we may assume \(c > 0\), since \(c = 0\) makes both quantities \(0\). Choose \(n\) with \(\frac{c}{n}\mu(X) < \varepsilon\) (possible as \(\mu(X) < \infty\)) and let \(P\) be the nonempty sets among

\begin{equation*} A_j = f^{-1}\Bigl( \Bigl[ \tfrac{(j-1)c}{n}, \tfrac{jc}{n} \Bigr) \Bigr) \ (j < n), \qquad A_n = f^{-1}\Bigl( \Bigl[ \tfrac{(n-1)c}{n}, c \Bigr] \Bigr) , \end{equation*}

which lie in \(\mathcal{S}\), are disjoint, and cover \(X\) because \(f(X) \subseteq [0,c]\). On each of them \(f\) varies by at most \(c/n\), so

\begin{equation*} U(f,P) - L(f,P) \leq \frac{c}{n} \sum_j \mu(A_j) = \frac{c}{n}\mu(X) < \varepsilon \end{equation*}

(the subtraction legitimate as all terms are finite), whence \(U(f) \leq U(f,P) < L(f,P) + \varepsilon \leq \int f \, d\mu + \varepsilon\) by 3.3. Let \(\varepsilon \to 0\).

(b) Take \(X = \mathbb{Z}^+\), \(\mathcal{S}\) all subsets, \(\mu(A) = \sum_{n \in A} 2^{-n}\) (countably additive, since nonnegative series may be summed in any order; \(\mu(X) = 1 < \infty\)), and \(f(n) = n\), which is measurable and unbounded. Applying 3.11 to \(f_N = \sum_{n \leq N} n \chi_{\{n\}} \uparrow f\) and 3.7,

\begin{equation*} \int f \, d\mu = \lim_{N \to \infty} \sum_{n=1}^{N} \frac{n}{2^n} = 2 < \infty . \end{equation*}

But any \(\mathcal{S}\)-partition of the infinite set \(X\) has an infinite member \(A_k\), for which \(\sup_{A_k} f = \infty\) and \(\mu(A_k) \geq 2^{-n_0} > 0\) for any \(n_0 \in A_k\); so \(U(f,P) = \infty\) for every \(P\) and \(U(f) = \infty \neq 2\).

(c) Same \(X\) and \(\mathcal{S}\), now with \(\mu\) counting measure (so \(\mu(X) = \infty\)) and \(f(n) = 2^{-n}\), which is measurable and bounded. As in (b), 3.11 and 3.7 give

\begin{equation*} \int f \, d\mu = \sum_{n=1}^{\infty} 2^{-n} = 1 . \end{equation*}

Again every \(\mathcal{S}\)-partition has an infinite member \(A_k\), now with \(\mu(A_k) = \infty\) and \(\sup_{A_k} f \geq f(n_0) > 0\) for any \(n_0 \in A_k\), so \(U(f) = \infty \neq 1\).

Apply the Dominated Convergence Theorem 3.31 to \(f_k = \chi_{[-k, \, k]} f\), with dominating function \(|f|\).

Each \(f_k\) is Borel measurable (2.46), \(|f_k| \leq |f|\) pointwise with \(\int |f| \, d\lambda < \infty\) by hypothesis, and for each \(x\) we have \(f_k(x) = f(x)\) once \(k \geq |x|\), so \(f_k \to f\) pointwise. Hence 3.31 and the definition 3.24 of integration over a subset give

\begin{equation*} \lim_{k \to \infty} \int_{[-k, \, k]} f \, d\lambda = \lim_{k \to \infty} \int f_k \, d\lambda = \int f \, d\lambda . \end{equation*}

Take \(f = \sum_{n=1}^{\infty} g_n\) with \(g_n(x) = \frac{(-1)^{n+1}}{n} \varphi(x - n)\), where \(\varphi(s) = \max\{0, 1 - |2s-1|\}\) is the tent function supported on \((0,1)\) with \(\int_0^1 \varphi = \frac12\).

The supports \((n, n+1)\) are disjoint, so at most one term is nonzero at each \(x\) and \(f\) is real valued; moreover \(f = \sum_{n=1}^{N} g_n\) on \([0, N+1)\), a finite sum of continuous functions, and these sets cover \([0,\infty)\), so \(f\) is continuous, hence Borel measurable (2.41). Since \(g_n\) is continuous and vanishes off \([n, n+1]\), 3.34 gives

\begin{equation*} \int g_n \, d\lambda = \frac{(-1)^{n+1}}{n} \int_0^1 \varphi = \frac{(-1)^{n+1}}{2n} , \qquad \int |g_n| \, d\lambda = \frac{1}{2n} . \end{equation*}

The limit exists. On \([0,N]\) we have \(f = \sum_{n=1}^{N-1} g_n\), so 3.21 (legitimate since each \(\int |g_n| \, d\lambda < \infty\)) gives \(\int_{[0,N]} f \, d\lambda = \sum_{n=1}^{N-1} \frac{(-1)^{n+1}}{2n}\), which converges to \(\frac{\ln 2}{2}\) by the alternating series test. For real \(t \geq 1\) put \(N = \lfloor t \rfloor\); then \((N, t] \subseteq (N, N+1]\), where \(|f| \leq \frac1N\), so 3.21 and 3.25 give

\begin{equation*} \Bigl| \int_{[0, \, t]} f \, d\lambda - \int_{[0, \, N]} f \, d\lambda \Bigr| \leq \lambda\bigl((N, t]\bigr) \sup_{(N, \, t]} |f| \leq \frac{1}{N} , \end{equation*}

and \(N \to \infty\) as \(t \to \infty\), so \(\lim_{t \to \infty} \int_{[0,t]} f \, d\lambda = \frac{\ln 2}{2} \in \mathbb{R}\).

The integral is not defined. The \(g_n\) have disjoint supports, with \(g_n \geq 0\) for \(n\) odd and \(g_n \leq 0\) for \(n\) even, so \(f^+ = \sum_{n \text{ odd}} g_n\) and \(f^- = -\sum_{n \text{ even}} g_n\); hence 3.8 and 3.16 give, for every \(K \in \mathbb{Z}^+\),

\begin{equation*} \int f^+ \, d\lambda \geq \sum_{k=1}^{K} \frac{1}{2(2k-1)} , \qquad \int f^- \, d\lambda \geq \sum_{k=1}^{K} \frac{1}{4k} . \end{equation*}

Both sums diverge as \(K \to \infty\) (comparison with the harmonic series), so \(\int f^+ \, d\lambda = \int f^- \, d\lambda = \infty\) and 3.18 assigns no value to \(\int_{[0,\infty)} f \, d\lambda\).

Take \(f = \sum_{n=1}^{\infty} h_n\) with \(h_n(x) = (-1)^{n+1} \, 2(n+1) \, \varphi\bigl(n(n+1)x - n\bigr)\), where \(\varphi(s) = \max\{0, 1 - |2s-1|\}\) is the tent function supported on \((0,1)\).

Since \(\varphi(n(n+1)x - n) \neq 0\) exactly when \(\frac{1}{n+1} < x < \frac1n\), the function \(h_n\) is continuous on \((0,1)\) and vanishes off \(I_n = (\frac{1}{n+1}, \frac1n)\), and the graph of \(|h_n|\) over \(I_n\) is a triangle of base \(\lambda(I_n) = \frac{1}{n(n+1)}\) and height \(2(n+1)\); as \(h_n\) is continuous on \(\overline{I_n}\), 3.34 gives

\begin{equation*} \int h_n \, d\lambda = (-1)^{n+1} \cdot \tfrac12 \cdot \tfrac{1}{n(n+1)} \cdot 2(n+1) = \frac{(-1)^{n+1}}{n} , \qquad \int |h_n| \, d\lambda = \frac1n . \end{equation*}

The \(I_n\) are disjoint, so at most one term is nonzero at each \(x\) and \(f\) is real valued; moreover \(f = \sum_{n=1}^{N} h_n\) on \((\frac{1}{N+1}, 1)\), a finite sum of continuous functions, and these sets cover \((0,1)\), so \(f\) is continuous, hence Borel measurable (2.41).

The limit exists. For \(n \geq 2\) we have \(f = \sum_{k=1}^{n-1} h_k\) on \((\frac1n, 1)\), and \(I_k \subseteq (\frac1n, 1)\) for \(k \leq n-1\), so \(\int_{(\frac1n, 1)} h_k \, d\lambda = \int h_k \, d\lambda\); since each \(\int |h_k| \, d\lambda < \infty\), additivity 3.21 gives

\begin{equation*} \int_{(\frac{1}{n}, \, 1)} f \, d\lambda = \sum_{k=1}^{n-1} \frac{(-1)^{k+1}}{k} \longrightarrow \ln 2 \in \mathbb{R} \end{equation*}

by the alternating series test.

The integral is not defined. The \(h_k\) have disjoint supports, with \(h_k \geq 0\) for \(k\) odd and \(h_k \leq 0\) for \(k\) even, so \(f^+ = \sum_{k \text{ odd}} h_k\) and \(f^- = -\sum_{k \text{ even}} h_k\); hence 3.8 and 3.16 give, for every \(K \in \mathbb{Z}^+\),

\begin{equation*} \int_{(0, \, 1)} f^+ \, d\lambda \geq \sum_{k=1}^{K} \frac{1}{2k-1} , \qquad \int_{(0, \, 1)} f^- \, d\lambda \geq \sum_{k=1}^{K} \frac{1}{2k} . \end{equation*}

Both sums diverge as \(K \to \infty\), so neither integral is finite and 3.18 assigns no value to \(\int_{(0,1)} f \, d\lambda\).

Both inclusions follow from the single formula \(g_n(x) = \inf_{J_n(x)} f\), \(h_n(x) = \sup_{J_n(x)} f\), where \(J_n(x)\) is the union of the one or two intervals of \(P_n\) that contain \(x\).

In the proof of 3.34, \(P_n\) cuts \([a,b]\) into \(2^n\) closed intervals of length \(\delta_n = (b-a)2^{-n}\); \(g_n\) and \(h_n\) are the corresponding lower and upper step functions, modified at the finitely many shared endpoints to take the infimum, respectively supremum, of \(f\) over the union of the two intervals meeting there; and \(f_L = \lim_n g_n\), \(f_U = \lim_n h_n\), these limits existing since \(f\) is bounded and \(g_1 \leq g_2 \leq \cdots\), \(h_1 \geq h_2 \geq \cdots\). With that modification the displayed formula holds at every \(x\), and \(J_n(x)\) is a closed interval containing \(x\) of length \(\delta_n\) or \(2\delta_n\). Hence \(g_n(x) \leq f(x) \leq h_n(x)\), so \(f_L \leq f \leq f_U\), and monotonicity of the two sequences gives

\begin{equation*} 0 \leq f_U(x) - f_L(x) \leq h_n(x) - g_n(x) = \sup_{J_n(x)} f - \inf_{J_n(x)} f \end{equation*}

for every \(n \in \mathbb{Z}^+\).

(i) Continuity at \(x\) gives \(f_L(x) = f_U(x)\). Let \(\varepsilon > 0\); choose \(\eta > 0\) with \(|f(y) - f(x)| < \varepsilon\) for all \(y \in [a,b]\) with \(|y - x| < \eta\), then \(n\) with \(2\delta_n < \eta\). Every \(y \in J_n(x)\) satisfies \(|y - x| < \eta\), so the display gives \(f_U(x) - f_L(x) \leq 2\varepsilon\); let \(\varepsilon \to 0\).

(ii) \(f_L(x) = f_U(x)\) gives continuity at \(x\). Then \(f_L(x) = f(x) = f_U(x)\), so for \(\varepsilon > 0\) there is \(n\) with \(h_n(x) - g_n(x) < \varepsilon\). Write \(J_n(x) = [c,d]\). If \(c > a\) then \(c\) is a partition point of \(P_n\) interior to \([a,b]\), hence lies in two of its intervals, so \(x = c\) would put the interval left of \(c\) inside \(J_n(x)\), contradicting \(\min J_n(x) = c\); thus \(c > a\) forces \(x > c\), and symmetrically \(d < b\) forces \(x < d\). So \(x\) is interior to \([c,d]\) relative to \([a,b]\), and there is \(\eta > 0\) with \(\{y \in [a,b] : |y - x| < \eta\} \subseteq J_n(x)\). For such \(y\) both \(f(y)\) and \(f(x)\) lie in \([g_n(x), h_n(x)]\), so \(|f(y) - f(x)| \leq h_n(x) - g_n(x) < \varepsilon\).

Both directions run through the identity \(E_k = f^{-1}(\{a_k\})\) and 3.7.

Put \(f = a_1 \chi_{E_1} + \cdots + a_n \chi_{E_n}\). Disjointness makes \(f = a_k\) on \(E_k\) and \(f = 0\) off \(E_1 \cup \cdots \cup E_n\); since the \(a_k\) are distinct and nonzero, this says exactly that

\begin{equation*} E_k = f^{-1}(\{a_k\}) \quad (1 \leq k \leq n), \qquad |f| = |a_1| \chi_{E_1} + \cdots + |a_n| \chi_{E_n} . \end{equation*}

(Note the \(E_k\) are not assumed measurable; that is part of what is proved.)

If every \(E_k \in \mathcal{S}\) with \(\mu(E_k) < \infty\), then \(f\) is \(\mathcal{S}\)-measurable by 2.46, and 3.7 applied to \(|f|\), whose sets \(E_k\) are disjoint and in \(\mathcal{S}\), gives

\begin{equation*} \|f\|_1 = \int |f| \, d\mu = |a_1| \mu(E_1) + \cdots + |a_n| \mu(E_n) < \infty , \end{equation*}

so \(f \in L^1(\mu)\) with the asserted norm.

Conversely if \(f \in L^1(\mu)\), then by 3.40 \(f\) is \(\mathcal{S}\)-measurable with \(\|f\|_1 < \infty\), so \(E_k = f^{-1}(\{a_k\}) \in \mathcal{S}\) by 2.35, the singleton \(\{a_k\}\) being Borel. The same application of 3.7 gives the displayed formula for \(\|f\|_1\); every term on the left lies in \([0,\infty]\) and their sum is finite, so \(|a_k| \mu(E_k) < \infty\), and \(a_k \neq 0\) forces \(\mu(E_k) < \infty\).

(a) The pointwise bound \(f^p \leq 1 + f^r\) does it; (b) take \(f(x) = 1/x\) on \((1,\infty)\).

(a) Both \(f^p\) and \(f^r\) are \(\mathcal{S}\)-measurable, being compositions of \(f\) with the continuous map \(s \mapsto s^t\) on \([0,\infty)\) (2.41 and 2.44). If \(f(x) \leq 1\) then \(f(x)^p \leq 1\); if \(f(x) > 1\) then \(f(x)^{r-p} \geq 1\), so \(f(x)^p \leq f(x)^r\). Either way \(f^p \leq 1 + f^r\) pointwise, so 3.8, additivity 3.16, and 3.4 applied to \(\chi_X\) give

\begin{equation*} \int f^p \, d\mu \leq \int (1 + f^r) \, d\mu = \mu(X) + \int f^r \, d\mu < \infty . \end{equation*}

(b) On \(X = (1,\infty)\) with the Borel sets and Lebesgue measure, so \(\mu(X) = \infty\), take \(p = 1\), \(r = 2\), and \(f(x) = 1/x\), which is continuous hence Borel measurable. The functions \(f^2 \chi_{(1,k)}\) and \(f \chi_{(1,k)}\) increase pointwise to \(f^2\) and \(f\), so 3.11 together with 3.34 gives

\begin{equation*} \int f^2 \, d\mu = \lim_{k \to \infty} \Bigl( 1 - \frac1k \Bigr) = 1 < \infty , \qquad \int f \, d\mu = \lim_{k \to \infty} \ln k = \infty . \end{equation*}

Take \(A_k = \{x \in X : |f(x)| > \frac1k\}\); then \(\{x : f(x) \neq 0\} = \bigcup_{k=1}^{\infty} A_k\), with every \(\mu(A_k) < \infty\).

Each \(A_k = |f|^{-1}\bigl((\frac1k, \infty)\bigr) \in \mathcal{S}\), since \(|f|\) is \(\mathcal{S}\)-measurable by 2.45. If \(f(x) \neq 0\) then \(|f(x)| > \frac1k\) for some \(k\), so \(x \in A_k\); conversely \(x \in A_k\) forces \(|f(x)| > 0\). This is the displayed union. Finally \(\frac1k \chi_{A_k} \leq |f|\) pointwise, so 3.8 and then 3.7 give

\begin{equation*} \tfrac{1}{k} \mu(A_k) = \int \tfrac{1}{k} \chi_{A_k} \, d\mu \leq \int |f| \, d\mu = \|f\|_1 , \end{equation*}

and \(\|f\|_1 < \infty\) because \(f \in L^1(\mu)\); hence \(\mu(A_k) \leq k \|f\|_1 < \infty\).

Dominate by \(g(x) = 1/\sqrt{x}\) and apply the Dominated Convergence Theorem 3.31 on \((0,1)\).

Each \(f_k\) is continuous on \((0,1)\), hence Borel measurable (2.41). For \(x \in (0,1)\) we have \(0 < (1-x)^k \leq 1\) and \(|\cos x^k| \leq 1\), so \(|f_k| \leq g\) for every \(k\); and \((1-x)^k \to 0\) for each fixed \(x\), so \(f_k \to 0\) pointwise. The functions \(g \chi_{(1/n, 1)}\) increase pointwise to \(g\), and \(g\) is continuous on \([1/n, 1]\), so 3.11 with 3.34 gives

\begin{equation*} \int_{(0,1)} g \, d\lambda = \lim_{n \to \infty} \int_{1/n}^{1} \frac{dx}{\sqrt{x}} = \lim_{n \to \infty} \Bigl( 2 - \frac{2}{\sqrt{n}} \Bigr) = 2 < \infty . \end{equation*}

Hence 3.31 applies, and with 3.39,

\begin{equation*} \lim_{k \to \infty} \int_0^1 f_k = \lim_{k \to \infty} \int_{(0,1)} f_k \, d\lambda = 0 . \end{equation*}

Take the typewriter sequence: writing \(k = \frac{(n-1)n}{2} + j\) with \(n \in \mathbb{Z}^+\) and \(j \in \{1, \ldots, n\}\), a decomposition that exists and is unique because \(\frac{(n-1)n}{2}\) is strictly increasing to \(\infty\) in \(n\), set

\begin{equation*} f_k = \sqrt{n} \; \chi_{[\frac{j-1}{n}, \, \frac{j}{n}]} . \end{equation*}

Each \(f_k\) is a nonnegative multiple of the characteristic function of a closed interval, hence a nonnegative Borel measurable function on \([0,1]\), and 3.7 gives

\begin{equation*} \int_0^1 f_k = \sqrt{n} \cdot \frac{1}{n} = \frac{1}{\sqrt{n}} . \end{equation*}

Only the \(n\) indices \(k \in \{\frac{(n-1)n}{2}+1, \ldots, \frac{n(n+1)}{2}\}\) carry a given \(n\), so \(n \to \infty\) as \(k \to \infty\), and the first condition holds.

For the second, fix \(x \in [0,1]\) and \(m \in \mathbb{Z}^+\), and let \(n \in \mathbb{Z}^+\) be arbitrary. The intervals \([\frac{j-1}{n}, \frac{j}{n}]\) for \(1 \leq j \leq n\) cover \([0,1]\), so some \(j\) has \(x \in [\frac{j-1}{n}, \frac{j}{n}]\), and then \(k_n = \frac{(n-1)n}{2} + j\) satisfies \(f_{k_n}(x) = \sqrt{n}\). Since \(k_n > \frac{(n-1)n}{2} \to \infty\), we have \(k_n \geq m\) for all large \(n\), so \(\sup_{k \geq m} f_k(x) \geq \sqrt{n}\) for all large \(n\), that is \(\sup_{k \geq m} f_k(x) = \infty\).

The values are \(2\) in (a) and \(\pi\) in (b); in (c) both \(\int f^{\pm} \, d\lambda = \infty\), while integration by parts makes the truncated integrals converge.

(a) Set \(f(0) = \infty\), which is immaterial since \(\lambda(\{0\}) = 0\) and makes the approximation below converge everywhere. Then \(f\) is Borel measurable by 2.52, since \(f^{-1}((a,\infty])\) is \([0,1]\) for \(a < 1\) and \([0, a^{-2})\) for \(a \geq 1\). The functions \(g_n = \min\{f, n\}\) are continuous on \([0,1]\) (the two formulas \(n\) on \([0, n^{-2}]\) and \(x^{-1/2}\) on \([n^{-2}, 1]\) agree at \(n^{-2}\)), increase, and converge pointwise to \(f\); as each is Riemann integrable, 3.34 gives

\begin{equation*} \int_{[0,1]} g_n \, d\lambda = n \cdot \frac{1}{n^2} + \int_{1/n^2}^{1} \frac{dx}{\sqrt{x}} = \frac1n + \Bigl( 2 - \frac2n \Bigr) = 2 - \frac1n , \end{equation*}

so 3.11 gives \(\int_{[0,1]} f \, d\lambda = 2\).

(b) Here \(f\) is continuous, hence Borel measurable (2.41), and \(g_n = f \chi_{[-n,n]}\) increases pointwise to \(f\), with \(\int_{\mathbb{R}} g_n \, d\lambda = \int_{-n}^{n} \frac{dx}{1+x^2} = 2 \arctan n\) by 3.34. Hence 3.11 gives

\begin{equation*} \int_{\mathbb{R}} f \, d\lambda = \lim_{n \to \infty} 2 \arctan n = \pi . \end{equation*}

(c) The function \(f(x) = (\sin x)/x\) is continuous on \((0,\infty)\), so \(f\) and hence \(f^{\pm}\) are Borel measurable (2.41, 2.44).

The integral is not defined. On \(A_k = (2k\pi, (2k+1)\pi)\) we have \(\sin x > 0\), so \(f^+ = f \geq \frac{\sin x}{(2k+1)\pi}\) there; and 3.34 with the fundamental theorem of calculus gives \(\int \chi_{A_k} \sin x \, d\lambda = [-\cos x]_{2k\pi}^{(2k+1)\pi} = 2\), the two endpoints forming a \(\lambda\)-null set on which the integrand is bounded. Hence \(\int f^+ \chi_{A_k} \, d\lambda \geq \frac{2}{(2k+1)\pi}\), and since the \(A_k\) are disjoint, 3.8 and 3.16 give

\begin{equation*} \int f^+ \, d\lambda \geq \sum_{k=0}^{n-1} \frac{2}{(2k+1)\pi} \qquad (n \in \mathbb{Z}^+) , \end{equation*}

whose right side diverges by comparison with the harmonic series; so \(\int f^+ \, d\lambda = \infty\). The same argument on \(B_k = ((2k+1)\pi, (2k+2)\pi)\), where \(f^- = -f \geq \frac{-\sin x}{(2k+2)\pi}\) and \(\int \chi_{B_k}(-\sin x) \, d\lambda = 2\), gives \(\int f^- \, d\lambda \geq \sum_{k=0}^{n-1} \frac{1}{(k+1)\pi}\) for every \(n\), so \(\int f^- \, d\lambda = \infty\) too, and 3.18 assigns no value to \(\int_{(0,\infty)} f \, d\lambda\).

The limit exists. Extend \(f\) continuously to \([0,\infty)\) by \(\tilde f(0) = 1\). For \(t > 0\) the function \(\tilde f\) is bounded on \([0,t]\) and \(\{0, t\}\) is \(\lambda\)-null, so by 3.8, 3.7 and the last assertion of 3.34

\begin{equation*} \int_{(0, \, t)} f \, d\lambda = \int_{[0, \, t]} \tilde f \, d\lambda = \int_0^t \frac{\sin x}{x} \, dx \in \mathbb{R} . \end{equation*}

For \(t \geq 1\), integration by parts for Riemann integrals gives

\begin{equation*} \int_1^t \frac{\sin x}{x} \, dx = \cos 1 - \frac{\cos t}{t} - \int_1^t \frac{\cos x}{x^2} \, dx . \end{equation*}

Now \(h(x) = (\cos x)/x^2\) is continuous with \(|h| \leq x^{-2}\), and 3.11 with 3.34 gives \(\int_{(1,\infty)} x^{-2} \, d\lambda = 1 < \infty\), so \(h \in L^1((1,\infty))\). For any sequence \(t_n \to \infty\) in \((1,\infty)\) we have \(h \chi_{(1,t_n)} \to h\) pointwise with \(|h \chi_{(1,t_n)}| \leq x^{-2}\), so 3.31 (with 3.34 identifying \(\int h \chi_{(1,t_n)} \, d\lambda = \int_1^{t_n} h\)) gives that

\begin{equation*} L = \lim_{t \to \infty} \int_1^t \frac{\cos x}{x^2} \, dx = \int_{(1,\infty)} h \, d\lambda \end{equation*}

exists in \(\mathbb{R}\). Since \(|\cos t / t| \leq 1/t \to 0\), letting \(t \to \infty\) in the two displays above gives

\begin{equation*} \lim_{t \to \infty} \int_{(0, \, t)} f \, d\lambda = \int_0^1 \frac{\sin x}{x} \, dx + \cos 1 - L \in \mathbb{R} . \end{equation*}

False: with \(r_1, r_2, \ldots\) an enumeration of \(\mathbb{Q} \cap (0,1)\), take

\begin{equation*} G = (0,1) \cap \bigcup_{k=1}^{\infty} \bigl( r_k - 2^{-k-2}, \; r_k + 2^{-k-2} \bigr) . \end{equation*}

Then \(G\) is open, and it contains every \(r_k\), hence is dense in \((0,1)\); being open it is Borel, so \(\lambda(G) = |G|\) (2.68, 2.69) and countable subadditivity of outer measure (2.8) with \(|I| \leq \ell(I)\) (2.2) gives

\begin{equation*} \lambda(G) \leq \sum_{k=1}^{\infty} 2 \cdot 2^{-k-2} = \tfrac12 . \end{equation*}

Put \(f = \chi_G\) on \([0,1]\); it is bounded, so \(L(f,[0,1])\) and \(U(f,[0,1])\) of 1.7 make sense. Let \(P : 0 = x_0 < \cdots < x_n = 1\) be any partition of \([0,1]\).

(i) Each \((x_{j-1}, x_j) \cap (0,1)\) is a nonempty open subset of \((0,1)\), so density gives \(y \in G\) with \(x_{j-1} < y < x_j\); hence \(\sup_{[x_{j-1}, x_j]} f = 1\) and, by 1.3,

\begin{equation*} U(f, P, [0,1]) = \sum_{j=1}^{n} (x_j - x_{j-1}) = 1 . \end{equation*}

(ii) Since \(f\) takes only the values \(0\) and \(1\), \(\inf_{[x_{j-1},x_j]} f = 1\) exactly when \([x_{j-1},x_j] \subseteq G\). Those intervals have pairwise disjoint interiors, so with \(J = \{j : [x_{j-1},x_j] \subseteq G\}\), finite additivity and monotonicity of \(\lambda\) (the overlaps being finitely many null points) give

\begin{equation*} L(f, P, [0,1]) = \sum_{j \in J} (x_j - x_{j-1}) = \lambda\Bigl( \bigcup_{j \in J} [x_{j-1}, x_j] \Bigr) \leq \lambda(G) \leq \tfrac12 . \end{equation*}

As \(P\) was arbitrary, \(L(f,[0,1]) \leq \frac12 < 1 = U(f,[0,1])\), so \(\chi_G\) is not Riemann integrable on \([0,1]\) by 1.9.

Both parts are the \(\varepsilon/3\) argument over the dense class supplied by 3.48, once we record how the \(\mathcal{L}^1\)-norm transforms.

Change of variables. If \(h \in \mathcal{L}^1(\mathbb{R})\), \(t \in \mathbb{R}\) and \(c > 0\), then \(x \mapsto h(x-t)\) and \(x \mapsto h(cx)\) lie in \(\mathcal{L}^1(\mathbb{R})\), with

\begin{equation*} \lVert h(\cdot - t) \rVert_1 = \lVert h \rVert_1 , \qquad \lVert h(c \, \cdot) \rVert_1 = \tfrac{1}{c} \lVert h \rVert_1 . \end{equation*}

Indeed the \(\sigma\)-algebra \(\mathcal{S}\) in question (Borel or Lebesgue, 3.45) is stable under \(A \mapsto t + A\) and \(A \mapsto cA\), since open sets are and null sets are, by 2.7 and 2.70; and \(\lambda(t+A) = \lambda(A)\) by 2.7 while \(\lambda(cA) = c\lambda(A)\) because scaling every interval of a cover in 2.2 by \(c\) scales its length by \(c\). So \(\{A_j\} \mapsto \{t + A_j\}\) and \(\{A_j\} \mapsto \{\frac{1}{c}A_j\}\) are bijections of the set of \(\mathcal{S}\)-partitions of \(\mathbb{R}\) (3.1) onto itself under which each lower Lebesgue sum (3.2) of \(|h|\) is reproduced exactly, respectively multiplied by \(\frac1c\); take suprema (3.3).

Part (a). Let \(\varepsilon > 0\). By 3.48 choose \(g\) continuous and vanishing outside \([-M, M]\) with \(\lVert f - g \rVert_1 < \frac{\varepsilon}{3}\); such a \(g\) is bounded and uniformly continuous on \(\mathbb{R}\), being uniformly continuous on the compact interval \([-M-1, M+1]\) and zero outside \([-M, M]\). Put \(g_t(x) = g(x-t)\). For \(|t| \le 1\) both \(g\) and \(g_t\) vanish outside \([-M-1, M+1]\), so 3.8 and 3.4 give

\begin{equation*} \lVert g - g_t \rVert_1 \le (2M+2) \sup_{x \in \mathbb{R}} |g(x) - g(x-t)| . \end{equation*}

Uniform continuity supplies \(\delta \in (0, 1]\) with that supremum at most \(\frac{\varepsilon}{3(2M+2)}\) whenever \(|t| < \delta\), whence \(\lVert g - g_t \rVert_1 \le \frac{\varepsilon}{3}\). Since \(f_t - g_t\) is the translate of \(f - g\), the change of variables gives \(\lVert f_t - g_t \rVert_1 = \lVert f - g \rVert_1 < \frac{\varepsilon}{3}\). Thus for \(|t| < \delta\) the triangle inequality (3.43) gives

\begin{equation*} \lVert f - f_t \rVert_1 \le \lVert f - g \rVert_1 + \lVert g - g_t \rVert_1 + \lVert g_t - f_t \rVert_1 < \varepsilon . \end{equation*}

Part (b). Let \(\varepsilon > 0\) and choose \(g\) as above with \(\lVert f - g \rVert_1 < \frac{\varepsilon}{4}\), and put \(g_t(x) = g(tx)\). By the change of variables, \(\lVert f_t - g_t \rVert_1 = \frac1t \lVert f - g \rVert_1 \le 2 \lVert f - g \rVert_1 < \frac{\varepsilon}{2}\) for every \(t \ge \frac12\). For \(\frac12 \le t \le 2\) and \(|x| > 2M\) we have \(|x| > M\) and \(|tx| > M\), so \(g - g_t\) vanishes outside \([-2M, 2M]\), where \(|x - tx| \le 2M|t-1|\). Uniform continuity supplies \(\eta > 0\) with \(|g(u) - g(v)| < \frac{\varepsilon}{16M}\) whenever \(|u - v| < \eta\); set \(\rho = \min\{\frac12, \frac{\eta}{2M}\}\). If \(|t - 1| < \rho\) then \(|g - g_t| \le \frac{\varepsilon}{16M}\chi_{[-2M, 2M]}\), so \(\lVert g - g_t \rVert_1 \le \frac{\varepsilon}{4}\) by 3.8 and 3.4. Hence \(\lVert f - f_t \rVert_1 < \varepsilon\) for \(|t-1| < \rho\) by 3.43.

Everything comes from the pointwise inequality \(c^p \chi_E \le |h|^p\) on \(X\), where \(E = \{x \in X : |h(x)| \ge c\}\): off \(E\) the left side is \(0\), while on \(E\) we have \(|h(x)|^p \ge c^p\) because \(t \mapsto t^p\) is increasing on \([0, \infty)\). Both sides are nonnegative and \(\mathcal{S}\)-measurable, so 3.8 and 3.15 give

\begin{equation*} c^p \, \mu\bigl(\{x \in X : |h(x)| \ge c\}\bigr) = \int c^p \chi_E \, d\mu \le \int |h|^p \, d\mu . \end{equation*}

Divide by \(c^p > 0\).

Write \(m = \int h \, d\mu\), a real number because \(h \in \mathcal{L}^1(\mu)\). Applying Exercise 1 in this section to the \(\mathcal{S}\)-measurable function \(h - m\) with \(p = 2\) gives

\begin{equation*} \mu\bigl(\{x \in X : |h(x) - m| \ge c\}\bigr) \le \frac{1}{c^2} \int (h-m)^2 \, d\mu , \end{equation*}

so it suffices to show \(\int (h-m)^2 \, d\mu = \int h^2 \, d\mu - m^2\). If \(\int h^2 \, d\mu = \infty\) the desired right side is \(\infty\) and there is nothing to prove, so assume \(h^2 \in \mathcal{L}^1(\mu)\). Then the constant function \(m\) is in \(\mathcal{L}^1(\mu)\) too, since \(\int |m| \, d\mu = |m| \mu(X) = |m| < \infty\), and so is \(-2mh\); hence additivity and homogeneity of integration (3.21, 3.20) applied to the pointwise identity \((h-m)^2 = h^2 - 2mh + m^2\) yield

\begin{equation*} \begin{aligned} \int (h-m)^2 \, d\mu &= \int h^2 \, d\mu - 2m \int h \, d\mu + m^2 \mu(X) \\ &= \int h^2 \, d\mu - m^2 , \end{aligned} \end{equation*}

using \(\mu(X) = 1\) again.

Call \(c \in (0,\infty)\) an equality point if \(\mu(E_c) = \frac{1}{c}\lVert h \rVert_1\), where \(E_c = \{x \in X : |h(x)| \ge c\}\). The whole point is that an equality point pins down \(|h|\) on \(E_c\): we show \(0 < \mu(E_c) < \infty\) and \(|h| = c\) almost everywhere on \(E_c\).

Since \(|h| = \chi_{E_c}|h| + \chi_{X \setminus E_c}|h|\) pointwise and \(\chi_{E_c}|h| \ge c\chi_{E_c}\), additivity for nonnegative functions (3.16) and 3.8 give

\begin{equation*} \lVert h \rVert_1 = \int_{E_c} |h| \, d\mu + \int_{X \setminus E_c} |h| \, d\mu \ge c\,\mu(E_c) + \int_{X \setminus E_c} |h| \, d\mu , \end{equation*}

whose first term is \(\lVert h \rVert_1\); as \(\lVert h \rVert_1 < \infty\) we may cancel it, leaving

\begin{equation*} \int_{E_c} |h| \, d\mu = c \, \mu(E_c) = \lVert h \rVert_1 \in (0, \infty) , \end{equation*}

so \(0 < \mu(E_c) < \infty\). Applying 3.16 to \(\chi_{E_c}|h| = \chi_{E_c}(|h| - c) + c\chi_{E_c}\) and subtracting the finite number \(c\,\mu(E_c)\) gives \(\int \chi_{E_c}(|h| - c) \, d\mu = 0\). A nonnegative \(\mathcal{S}\)-measurable \(g\) with \(\int g \, d\mu = 0\) vanishes almost everywhere, since Exercise 1 in this section with \(p = 1\) gives \(\mu(\{g \ge \frac1n\}) = 0\) for every \(n \in \mathbb{Z}^+\) and \(\{g > 0\} = \bigcup_n \{g \ge \frac1n\}\) is then null by 2.58. Hence \(|h| = c\) almost everywhere on \(E_c\).

Now suppose \(c_1 < c_2\) were both equality points. Then \(E_{c_2} \subseteq E_{c_1}\), and no \(x\) can have \(|h(x)|\) equal to both \(c_1\) and \(c_2\), so

\begin{equation*} E_{c_2} \subseteq \{x \in E_{c_1} : |h(x)| \ne c_1\} \cup \{x \in E_{c_2} : |h(x)| \ne c_2\} , \end{equation*}

a union of two sets of measure \(0\); thus \(\mu(E_{c_2}) = 0\) by 2.58, contradicting \(\mu(E_{c_2}) = \lVert h \rVert_1 / c_2 > 0\).

Given \(\alpha \in (0,3)\), choose \(\varepsilon\) with \(0 < \varepsilon < \min\{2, 3-\alpha\}\) and take the two intervals

\begin{equation*} I_1 = (0, 2), \qquad I_2 = (2 - \varepsilon, \, 4 - \varepsilon) , \end{equation*}

writing \(\alpha * I\) for the open interval with the same center as \(I\) and \(\alpha\) times its length (4.2 is the case \(\alpha = 3\)). Since \(I_1 \cap I_2 = (2-\varepsilon, 2) \ne \emptyset\), the only disjoint sublists are the empty one and the two singletons, and \(I_1 \cup I_2 = (0, 4-\varepsilon)\). The empty sublist fails because \(I_1 \cup I_2 \ne \emptyset\). For the singletons,

\begin{equation*} \alpha * I_1 = (1 - \alpha, \, 1 + \alpha), \qquad \alpha * I_2 = (3 - \varepsilon - \alpha, \, 3 - \varepsilon + \alpha) , \end{equation*}

so \(I_1 \cup I_2 \subseteq \alpha * I_1\) would force \(4 - \varepsilon \le 1 + \alpha\) and \(I_1 \cup I_2 \subseteq \alpha * I_2\) would force \(3 - \varepsilon - \alpha \le 0\), each saying \(\alpha \ge 3 - \varepsilon\), contrary to \(\varepsilon < 3 - \alpha\). Hence no disjoint sublist has the covering property, so the conclusion of 4.4 fails with \(\alpha\) in place of \(3\).

Write \(I = (a-r, a+r)\) and \(J = (b-s, b+s)\) with \(r, s > 0\), so that the hypothesis \(|I| \ge |J|\) reads \(s \le r\), and \(3 * I = (a - 3r, \, a + 3r)\) by 4.2. Choosing \(x \in I \cap J\) gives \(|a - b| \le |a - x| + |x - b| < r + s\), so every \(y \in J\) satisfies

\begin{equation*} |y - a| \le |y - b| + |b - a| < s + (r + s) = r + 2s \le 3r . \end{equation*}

Thus \(J \subseteq 3 * I\).

Write \(h = \chi_{[0,1]}\) and, following 4.6,

\begin{equation*} h^*(b) = \sup_{t > 0} \psi_b(t), \qquad \psi_b(t) = \frac{\bigl| [0,1] \cap [b-t, \, b+t] \bigr|}{2t} , \end{equation*}

the numerator being \(\int_{b-t}^{b+t} |h|\). Since that intersection lies inside \([b-t, b+t]\), we always have \(\psi_b(t) \le 1\).

(i) \(0 < b < 1\). For \(0 < t \le \min\{b, 1-b\}\) we have \([b-t, b+t] \subseteq [0,1]\), so \(\psi_b(t) = 1\), and therefore \(h^*(b) = 1\).

(ii) \(b \le 0\). Here \(b - t < 0\) for every \(t > 0\), so the overlap is \(\max\{0, \min\{b+t, 1\}\}\) and

\begin{equation*} \psi_b(t) = \begin{cases} 0 & \text{if } 0 < t \le -b, \\ \frac{1}{2} + \frac{b}{2t} & \text{if } -b < t \le 1-b, \\ \frac{1}{2t} & \text{if } t > 1 - b . \end{cases} \end{equation*}

The middle branch is nondecreasing in \(t\) because \(b \le 0\), with value \(\frac{1}{2(1-b)}\) at \(t = 1-b\), while the third branch is smaller than \(\frac{1}{2(1-b)}\) throughout. Hence \(h^*(b) = \frac{1}{2(1-b)}\).

(iii) \(b \ge 1\). Here \(b + t > 1\) for every \(t > 0\), so the overlap is \(\max\{0, 1 - \max\{b-t, 0\}\}\) and

\begin{equation*} \psi_b(t) = \begin{cases} 0 & \text{if } 0 < t \le b-1, \\ \frac{1}{2} + \frac{1-b}{2t} & \text{if } b - 1 < t \le b, \\ \frac{1}{2t} & \text{if } t > b . \end{cases} \end{equation*}

The middle branch is nondecreasing because \(1 - b \le 0\), with value \(\frac{1}{2b}\) at \(t = b\), and the third branch stays below that. Hence \(h^*(b) = \frac{1}{2b}\).

These three values are exactly the formula in Example 4.7.

With \(E = [0,1] \cup [2,3]\) and \(h = \chi_E\),

\begin{equation*} h^*(b) = \begin{cases} \dfrac{1}{3-b} & \text{if } b \le -1, \\[1mm] \dfrac{1}{2(1-b)} & \text{if } -1 \le b \le 0, \\[1mm] 1 & \text{if } 0 < b < 1, \\[1mm] \dfrac{2b-1}{2b} & \text{if } 1 \le b \le \tfrac{3}{2}, \\[1mm] \dfrac{5-2b}{2(3-b)} & \text{if } \tfrac{3}{2} \le b \le 2, \\[1mm] 1 & \text{if } 2 < b < 3, \\[1mm] \dfrac{1}{2(b-2)} & \text{if } 3 \le b \le 4, \\[1mm] \dfrac{1}{b} & \text{if } b \ge 4 \end{cases} \end{equation*}

(the two formulas agree at \(b = -1, \frac32, 4\)). Put \(g(y) = |E \cap (-\infty, y]|\), which is continuous, nondecreasing, and equal to \(0, y, 1, y-1, 2\) on \((-\infty, 0], [0,1], [1,2], [2,3], [3,\infty)\) respectively. By additivity of Lebesgue measure and 4.6,

\begin{equation*} h^*(b) = \sup_{t>0} \psi_b(t), \qquad \psi_b(t) = \frac{g(b+t) - g(b-t)}{2t} \le 1 , \end{equation*}

the bound because \(E \cap [b-t, b+t] \subseteq [b-t, b+t]\).

Reduction to finitely many \(t\). Let \(t_1 < \cdots < t_N\) be the positive numbers with \(b+t\) or \(b-t\) in \(\{0,1,2,3\}\). Across any \(t\)-interval meeting none of them, both \(t \mapsto g(b \pm t)\) are affine, so \(\psi_b(t) = \frac{\alpha}{2} + \frac{\beta}{2t}\) is monotone there; \(\psi_b\) is continuous, \(\lim_{t \downarrow 0}\psi_b(t)\) exists in \([0,1]\) by monotonicity and the bound \(\psi_b \le 1\), and \(\psi_b(t) \to 0\) as \(t \to \infty\) since \(g(b+t) - g(b-t) \le 2\), so on \([t_N, \infty)\) the monotone \(\psi_b \ge 0\) is maximal at \(t_N\). Hence

\begin{equation*} h^*(b) = \max\bigl\{ \lim_{t \downarrow 0} \psi_b(t), \ \psi_b(t_1), \ \ldots, \ \psi_b(t_N) \bigr\} . \end{equation*}

Also \(y \mapsto 3-y\) maps \(E\) onto \(E\) and preserves Lebesgue measure, so \(\psi_{3-b} = \psi_b\) and \(h^*(3-b) = h^*(b)\); it therefore suffices to treat \(b \le \frac32\).

(i) \(0 < b < 1\). For \(0 < t \le \min\{b, 1-b\}\) we have \([b-t, b+t] \subseteq [0,1]\), so \(\psi_b(t) = 1\) and \(h^*(b) = 1\).

(ii) \(b \le 0\). Then \(g(b-t) = 0\) for all \(t > 0\), the candidates are \(t = -b, 1-b, 2-b, 3-b\), and

\begin{equation*} \psi_b(-b) = 0, \quad \psi_b(1-b) = \frac{1}{2(1-b)}, \quad \psi_b(2-b) = \frac{1}{2(2-b)}, \quad \psi_b(3-b) = \frac{1}{3-b} , \end{equation*}

while \(\lim_{t \downarrow 0}\psi_b(t)\) is \(0\) if \(b < 0\) and \(\frac12 = \frac{1}{2(1-b)}\) if \(b = 0\). As \(\frac{1}{2(2-b)} < \frac{1}{2(1-b)}\), we get \(h^*(b) = \max\{\frac{1}{2(1-b)}, \frac{1}{3-b}\}\), and \(\frac{1}{2(1-b)} \ge \frac{1}{3-b}\) exactly when \(b \ge -1\). This gives the first two branches.

(iii) \(1 \le b \le \frac32\). The candidates are \(t = b-1, b, 2-b, 3-b\), and since then \(2b-1 \in [1,2]\), \(2b \in [2,3]\), \(2b-2 \in [0,1]\), \(2b-3 \le 0\),

\begin{equation*} \begin{aligned} \psi_b(b-1) &= \frac{1-1}{2(b-1)} = 0, \\ \psi_b(b) &= \frac{(2b-1) - 0}{2b} =: A, \\ \psi_b(2-b) &= \frac{1 - (2b-2)}{2(2-b)} =: B, \\ \psi_b(3-b) &= \frac{2 - 0}{2(3-b)} =: C, \end{aligned} \end{equation*}

while \(\lim_{t \downarrow 0}\psi_b(t)\) is \(0\) for \(b > 1\) and \(\frac12 = A\) at \(b = 1\). Now \(A \ge C\) is equivalent to \((2b-3)(b-1) \le 0\), which holds on \([1, \frac32]\); and \(A = 1 - \frac{1}{2b}\) is increasing while \(B = \frac{3-2b}{4-2b}\) is decreasing, with \(A = B = \frac12\) at \(b = 1\), so \(B \le A\). Hence \(h^*(b) = A = \frac{2b-1}{2b}\).

(iv) \(b \ge \frac32\). By the symmetry \(h^*(b) = h^*(3-b)\), cases (iii), (i), (ii) applied at \(3 - b\) give \(\frac{5-2b}{2(3-b)}\) on \([\frac32, 2]\), then \(1\) on \((2,3)\), then \(\frac{1}{2(b-2)}\) on \([3,4]\), then \(\frac1b\) for \(b \ge 4\).

The answer is

\begin{equation*} h^{*}(b) = \begin{cases} \dfrac{1}{4(1-b)} & \text{if } b \le \frac12, \\[2mm] b & \text{if } \frac12 \le b < 1, \\[2mm] \dfrac{b - \sqrt{b^2-1}}{2} = \dfrac{1}{2(b + \sqrt{b^2-1})} & \text{if } b \ge 1 \end{cases} \end{equation*}

(the first two branches agree at \(b = \frac12\)). Since \(|h| = h\), writing \(F(x) = \int_{-\infty}^{x} h\), so that \(F = 0, \frac{x^2}{2}, \frac12\) on \((-\infty, 0], [0,1], [1,\infty)\), Definition 4.6 reads

\begin{equation*} h^{*}(b) = \sup_{t > 0} V_b(t), \qquad V_b(t) = \frac{F(b+t) - F(b-t)}{2t} . \end{equation*}

(i) \(b \le 0\). Then \(F(b-t) = 0\) for every \(t > 0\). For \(0 < t \le -b\), \(V_b(t) = 0\). For \(-b \le t \le 1-b\), substituting \(v = b + t \in [0,1]\),

\begin{equation*} V_b(t) = \frac{v^2}{4(v-b)}, \qquad \frac{d}{dv}\Bigl(\frac{v^2}{v-b}\Bigr) = \frac{v(v-2b)}{(v-b)^2} \ge 0 , \end{equation*}

so \(V_b\) increases to \(\frac{1}{4(1-b)}\) at \(t = 1-b\); for \(t \ge 1-b\), \(V_b(t) = \frac{1}{4t}\) decreases from that same value. Hence \(h^{*}(b) = \frac{1}{4(1-b)}\).

(ii) \(0 < b < 1\). For \(0 < t \le \min\{b, 1-b\}\) both \(b \pm t\) lie in \([0,1]\), so \(V_b(t) = \frac{(b+t)^2 - (b-t)^2}{4t} = b\). If \(b \le \frac12\), then for \(b \le t \le 1-b\) we get \(V_b(t) = \frac{(b+t)^2}{4t}\) with derivative \(\frac{(b+t)(t-b)}{4t^2} \ge 0\), so \(V_b\) increases to \(\frac{1}{4(1-b)}\) at \(t = 1-b\), beyond which \(V_b(t) = \frac{1}{4t}\) decreases; since \(1 - 4b(1-b) = (2b-1)^2 \ge 0\) gives \(\frac{1}{4(1-b)} \ge b\), we get \(h^{*}(b) = \frac{1}{4(1-b)}\). If \(b \ge \frac12\), then for \(1-b \le t \le b\), substituting \(u = b - t \in [0, 2b-1]\),

\begin{equation*} V_b(t) = \frac{1-u^2}{4(b-u)}, \qquad \frac{d}{du}\Bigl(\frac{1-u^2}{b-u}\Bigr) = \frac{u^2 - 2bu + 1}{(b-u)^2} > 0 \end{equation*}

because that quadratic has discriminant \(4b^2 - 4 < 0\); so \(V_b\) decreases in \(t\) from its value \(b\) at \(t = 1-b\), while for \(t \ge b\) we have \(V_b(t) = \frac{1}{4t} \le \frac{1}{4b} \le b\). Hence \(h^{*}(b) = b\).

(iii) \(b \ge 1\). Then \(F(b+t) = \frac12\) for every \(t > 0\), and \(V_b(t) = 0\) for \(0 < t \le b-1\). For \(b - 1 \le t < b\), substituting \(u = b - t\), the function \(g(u) = \frac{1-u^2}{4(b-u)}\) has \(g^{\prime}(u) = \frac{u^2 - 2bu + 1}{4(b-u)^2}\), whose numerator has roots \(b \pm \sqrt{b^2-1}\) with \(u_0 = b - \sqrt{b^2-1} \in (0,1]\) (as \((b-1)^2 \le (b-1)(b+1)\)) and the other root at least \(1\); so \(g\) increases on \([0, u_0)\) and decreases on \((u_0, 1]\), and since \(b - u_0 = \sqrt{b^2-1}\) and \(1 - u_0^2 = 2\sqrt{b^2-1}\,(b - \sqrt{b^2-1})\),

\begin{equation*} g(u_0) = \frac{b - \sqrt{b^2-1}}{2} . \end{equation*}

When \(b > 1\) this value is attained at \(t = \sqrt{b^2-1} > 0\); when \(b = 1\) we have \(u_0 = 1\), i.e. the excluded \(t = 0\), but \(g\) increases on \([0,1)\) with \(\lim_{u \uparrow 1} \frac{1+u}{4} = \frac12\), again the same number. Finally \(t \ge b\) gives \(V_b(t) = \frac{1}{4t} \le \frac{1}{4b} \le g(u_0)\), because \(b - \sqrt{b^2-1} = \frac{1}{b + \sqrt{b^2-1}} \ge \frac{1}{2b}\). Hence \(h^{*}(b) = \frac{b - \sqrt{b^2-1}}{2}\).

Fix \(c \in \mathbb{R}\) and \(b\) with \(h^{*}(b) > c\); it suffices to produce \(\delta > 0\) with \(h^{*} > c\) on \((b-\delta, b+\delta)\). By 4.6 there is \(t > 0\) with \(\alpha > 2tc\), where \(\alpha = \int_{b-t}^{b+t} |h| \in [0,\infty]\). Choose \(t^{\prime} > t\) with \(2t^{\prime}c < \alpha\): for \(c \le 0\) any \(t^{\prime} > t\) serves (when \(c = 0\) note \(\alpha > 0\)), while for \(c > 0\) take \(t^{\prime} = t+1\) if \(\alpha = \infty\) and any \(t^{\prime} \in (t, \frac{\alpha}{2c})\) otherwise, that interval being nonempty because \(\frac{\alpha}{2c} > t\). Put \(\delta = t^{\prime} - t > 0\). If \(|b^{\prime} - b| < \delta\), then \(b^{\prime} - t^{\prime} < b - t\) and \(b^{\prime} + t^{\prime} > b + t\), so \((b-t, b+t) \subseteq (b^{\prime}-t^{\prime}, b^{\prime}+t^{\prime})\) and monotonicity of the integral of \(|h| \ge 0\) gives

\begin{equation*} h^{*}(b^{\prime}) \ge \frac{1}{2t^{\prime}} \int_{b^{\prime}-t^{\prime}}^{b^{\prime}+t^{\prime}} |h| \ge \frac{\alpha}{2t^{\prime}} > c . \end{equation*}

Thus every point of \(\{h^{*} > c\}\) is an interior point.

True: \(h^{*}\) is increasing. Suppose \(b_1 < b_2\) and put \(s = b_2 - b_1 > 0\) (note \(h\) is Borel measurable, each \(\{h > c\}\) being an interval, so \(h^{*}\) is defined, and \(|h| = h\)). Translation invariance of Lebesgue measure (2.7) passes from characteristic functions through simple functions to all nonnegative Borel functions, giving \(\int_{b_1-t}^{b_1+t} h(x+s) \, dx = \int_{b_2-t}^{b_2+t} h\) for every \(t > 0\). Since \(h\) is increasing and \(s > 0\), we have \(h(x) \le h(x+s)\) pointwise, so monotonicity of the integral gives

\begin{equation*} \frac{1}{2t}\int_{b_1-t}^{b_1+t} |h| \le \frac{1}{2t}\int_{b_2-t}^{b_2+t} |h| \le h^{*}(b_2) \end{equation*}

for every \(t > 0\). Take the supremum over \(t > 0\) on the left.

Take

\begin{equation*} h = \sum_{n=1}^{\infty} n \, \chi_{[n, \, n + n^{-3}]} , \end{equation*}

a tall thin spike at each positive integer. The intervals are disjoint because \(n + n^{-3} < n+1\), so \(h\) is \([0,\infty)\)-valued, and it is Borel measurable as a pointwise limit of its partial sums (2.48). By the Monotone Convergence Theorem (3.11),

\begin{equation*} \lVert h \rVert_1 = \sum_{n=1}^{\infty} n \cdot n^{-3} = \sum_{n=1}^{\infty} \frac{1}{n^2} < \infty . \end{equation*}

Finiteness of \(h^{*}\): fix \(b\), put \(R = |b| + 1\), and note \(M_R := \sup_{[-R,R]} h \le \lceil R \rceil < \infty\), only the spikes with \(n \le R\) meeting \([-R, R]\). If \(0 < t \le 1\) then \([b-t, b+t] \subseteq [-R, R]\), so the average of \(h\) over it is at most \(M_R\); if \(t \ge 1\) that average is at most \(\frac{1}{2t}\lVert h \rVert_1 \le \frac12 \lVert h \rVert_1\). Hence \(h^{*}(b) \le \max\{M_R, \frac12 \lVert h \rVert_1\} < \infty\).

Unboundedness: taking \(b = n\) and \(t = n^{-3}\) gives

\begin{equation*} h^{*}(n) \ge \frac{n^3}{2} \int_{n - n^{-3}}^{n + n^{-3}} h \ge \frac{n^3}{2} \cdot n \cdot n^{-3} = \frac{n}{2} \longrightarrow \infty . \end{equation*}

Write \(E = \{b \in \mathbb{R} : h^{*}(b) = \infty\}\). For every positive integer \(n\) we have \(E \subseteq \{b \in \mathbb{R} : h^{*}(b) > n\}\), so monotonicity of outer measure (2.5) and the Hardy–Littlewood maximal inequality (4.8), applicable since \(h \in \mathcal{L}^1(\mathbb{R})\), give

\begin{equation*} |E| \le \bigl| \{b \in \mathbb{R} : h^{*}(b) > n\} \bigr| \le \frac{3}{n} \lVert h \rVert_1 . \end{equation*}

As \(\lVert h \rVert_1 < \infty\), the right side tends to \(0\), so \(|E| = 0\). (In fact \(E = \bigcap_n \{h^{*} > n\}\) is Borel, each of those sets being open by Exercise 9 in this section.)

Superimpose translated copies of one function with a single bad point. Define \(g \colon \mathbb{R} \to [0,\infty)\) by

\begin{equation*} g(x) = \begin{cases} \dfrac{1}{x (\log x)^2} & \text{if } 0 < x < \frac12, \\[2mm] 0 & \text{otherwise,} \end{cases} \end{equation*}

which is Borel measurable, being continuous on \((0, \frac12)\) and zero elsewhere. Since \(\frac{d}{dx}\bigl(-\frac{1}{\log x}\bigr) = \frac{1}{x(\log x)^2}\) on \((0,1)\) and \(-\frac{1}{\log x} \to 0\) as \(x \downarrow 0\), we get \(\int_0^t g = \frac{1}{|\log t|}\) for \(0 < t < \frac12\); in particular \(\lVert g \rVert_1 = \frac{1}{\log 2} < \infty\), and

\begin{equation*} \frac{1}{2t}\int_{-t}^{t} g = \frac{1}{2t\,|\log t|} \longrightarrow \infty \qquad (t \downarrow 0) , \end{equation*}

because \(t|\log t| \to 0\); hence \(g^{*}(0) = \infty\).

Let \(r_1, r_2, \ldots\) enumerate \(\mathbb{Q}\) and put \(\tilde h = \sum_{k=1}^{\infty} 2^{-k} g(\cdot - r_k)\). Each summand is Borel measurable (a translate of a Borel set is Borel), so \(\tilde h\) is Borel measurable as the supremum of the partial sums (2.53), and 3.11 together with translation invariance (2.7) gives

\begin{equation*} \int_{-\infty}^{\infty} \tilde h = \sum_{k=1}^{\infty} 2^{-k} \lVert g \rVert_1 = \frac{1}{\log 2} < \infty . \end{equation*}

Hence \(N = \{\tilde h = \infty\}\) is a null set (otherwise \(\int \tilde h \ge c|N| \to \infty\)), so \(h = \tilde h \chi_{\mathbb{R} \setminus N}\) is a real-valued Borel function in \(\mathcal{L}^1(\mathbb{R})\) agreeing with \(\tilde h\) off \(N\). The integral of a nonnegative function over a null set is \(0\), so \(\int_{b-t}^{b+t} |h| = \int_{b-t}^{b+t} \tilde h\) for all \(b\) and \(t > 0\), giving \(h^{*} = \tilde h^{*}\).

Now fix \(b \in \mathbb{Q}\), say \(b = r_k\). Since \(\tilde h \ge 2^{-k}g(\cdot - r_k)\) pointwise, and since \(f_1 \le f_2\) implies \(f_1^{*} \le f_2^{*}\), while \((\lambda f)^{*} = \lambda f^{*}\) for \(\lambda > 0\) and \((g(\cdot - r))^{*}(b) = g^{*}(b - r)\) by 2.7 (all immediate from 4.6),

\begin{equation*} h^{*}(b) = \tilde h^{*}(b) \ge 2^{-k} g^{*}(b - r_k) = 2^{-k} g^{*}(0) = \infty . \end{equation*}

Approach \(c\) from below. For every integer \(n > \frac1c\) the number \(c - \frac1n\) is positive and \(\{b : h^{*}(b) \ge c\} \subseteq \{b : h^{*}(b) > c - \frac1n\}\), so monotonicity of outer measure (2.5) and the Hardy–Littlewood maximal inequality (4.8), applicable because \(h \in \mathcal{L}^1(\mathbb{R})\), give

\begin{equation*} \bigl| \{b \in \mathbb{R} : h^{*}(b) \ge c\} \bigr| \le \frac{3}{c - \frac1n} \lVert h \rVert_1 . \end{equation*}

Since \(\lVert h \rVert_1 < \infty\), letting \(n \to \infty\) gives the bound \(\frac{3}{c}\lVert h \rVert_1\).

The limit holds at every Lebesgue point of \(f\), of which almost every \(b \in \mathbb{R}\) is one by the first version of the Lebesgue Differentiation Theorem (4.10). So let \(E\) be the set of \(b\) with \(\lim_{t \downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)| = 0\), fix \(b \in E\) and \(t > 0\), and write \(I = [b-t, b+t]\). Since \(f\) and the constant \(f(b)\) are both integrable on \(I\), 3.21 and 3.23 give

\begin{equation*} |f_I - f(b)| = \Bigl| \frac{1}{2t}\int_{b-t}^{b+t}\bigl(f - f(b)\bigr) \Bigr| \le \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)| , \end{equation*}

so integrating the pointwise bound \(|f - f_I| \le |f - f(b)| + |f(b) - f_I|\) over \(I\) (3.21, 3.22) and dividing by \(2t\) yields

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t} |f - f_I| \le 2 \cdot \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)| \xrightarrow[\ t \downarrow 0\ ]{} 0 . \end{equation*}

Again the limit holds at every \(b\) in the full-measure set \(E\) of 4.10, that is, where \(\lim_{t \downarrow 0}\frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)| = 0\). Fix such a \(b\) and \(t > 0\), and let \(I\) be any interval of length \(t\) containing \(b\); then every \(x \in I\) has \(|x - b| \le t\), so \(I \subseteq [b-t, b+t]\). Exactly the estimate of Exercise 1 in this section, carried out on \(I\), gives \(\frac{1}{|I|}\int_I |f - f_I| \le \frac{2}{|I|}\int_I |f - f(b)|\), whence

\begin{equation*} \frac{1}{|I|}\int_I |f - f_I| \le \frac{2}{t}\int_{b-t}^{b+t}|f - f(b)| = 4 \cdot \frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)| . \end{equation*}

The right side is independent of \(I\), hence bounds the supremum over all such \(I\), and it tends to \(0\) as \(t \downarrow 0\) because \(b \in E\).

Approximate \(f(b)\) by rationals, applying a localized form of 4.10 to each \((f-q)^2\).

Localization. If \(h \colon \mathbb{R} \to \mathbb{R}\) is Lebesgue measurable with \(\int_{-n}^{n}|h| < \infty\) for every \(n \in \mathbb{Z}^+\), then \(\frac{1}{2t}\int_{b-t}^{b+t} h \to h(b)\) as \(t \downarrow 0\) for almost every \(b\). Indeed \(h_n = h\chi_{(-n,n)}\) lies in \(L^1(\mathbb{R})\), so 4.10 supplies a null set \(N_n\) off which \(\frac{1}{2t}\int_{b-t}^{b+t}|h_n - h_n(b)| \to 0\); if \(b \notin N = \bigcup_n N_n\) (null by 2.8) and \(n > |b|\), then \(h = h_n\) on \((b-t, b+t)\) for \(0 < t < n - |b|\), and \(|\frac{1}{2t}\int_{b-t}^{b+t} h - h(b)| \le \frac{1}{2t}\int_{b-t}^{b+t}|h - h(b)|\) by 3.23.

For \(q \in \mathbb{Q}\) put \(h_q = (f - q)^2\), Lebesgue measurable by 2.46. Since \((\alpha - \beta)^2 \le 2\alpha^2 + 2\beta^2\) and \(f^2 \in L^1(\mathbb{R})\),

\begin{equation*} \int_{-n}^{n} |h_q| \le 2 \int_{-\infty}^{\infty} f^2 + 4nq^2 < \infty \qquad (n \in \mathbb{Z}^+) , \end{equation*}

so localization gives a null set \(N_q\) with \(\frac{1}{2t}\int_{b-t}^{b+t} h_q \to (f(b)-q)^2\) for \(b \notin N_q\). Let \(N = \bigcup_{q \in \mathbb{Q}} N_q\), null because \(\mathbb{Q}\) is countable.

Fix \(b \notin N\) and \(\varepsilon > 0\), and choose \(q \in \mathbb{Q}\) with \(|f(b) - q| < \varepsilon\). Then \(|f(x) - f(b)|^2 \le 2(f(x)-q)^2 + 2(q - f(b))^2\) for every \(x\), so averaging over \((b-t, b+t)\) and letting \(t \downarrow 0\) gives

\begin{equation*} \limsup_{t \downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)|^2 \le 4\bigl(f(b) - q\bigr)^2 < 4\varepsilon^2 . \end{equation*}

As \(\varepsilon > 0\) was arbitrary and the averages are nonnegative, the limit is \(0\).

Truncate on the right and use that differentiability at \(b\) is a local property. The hypothesis makes \(g(x) = \int_{-\infty}^{x} f\) real valued, and for \(n \in \mathbb{Z}^+\) the function \(f_n = f\chi_{(-\infty, n)}\) satisfies \(\int_{-\infty}^{\infty}|f_n| = \int_{-\infty}^{n}|f| < \infty\), so \(f_n \in L^1(\mathbb{R})\). Hence 4.19 applied to \(f_n\) gives a set \(N_n\) with \(|N_n| = 0\) such that \(g_n^{\prime}(b) = f_n(b)\) for \(b \notin N_n\), where \(g_n(x) = \int_{-\infty}^{x} f_n\).

For \(x < n\) we have \(f = f_n\) on \((-\infty, x)\), so \(g_n = g\) on \((-\infty, n)\). Thus if \(b < n\), the difference quotients of \(g\) and of \(g_n\) at \(b\) coincide for all \(t \ne 0\) with \(b + t < n\), so 4.16 gives \(g^{\prime}(b) = g_n^{\prime}(b)\) whenever the latter exists; and \(f_n(b) = f(b)\). Therefore

\begin{equation*} b \in (-\infty, n) \setminus N_n \ \Longrightarrow \ g^{\prime}(b) = f_n(b) = f(b) . \end{equation*}

Now \(N = \bigcup_{n=1}^{\infty} N_n\) is null by 2.8, and every \(b \notin N\) satisfies \(b < n\) for some \(n \in \mathbb{Z}^+\), so \(g^{\prime}(b) = f(b)\) for almost every \(b \in \mathbb{R}\).

Split \(\mathbb{R}\) according to local integrability of \(|f|\): let

\begin{equation*} \mathcal{F} = \Bigl\{ (p,q) : p, q \in \mathbb{Q}, \ p < q, \ \int_p^q |f| < \infty \Bigr\} , \qquad L = \bigcup_{(p,q) \in \mathcal{F}} (p,q) , \end{equation*}

so that \(\mathcal{F}\) is countable.

(i) \(b \notin L\). Then \(\int_{b-t}^{b+t}|f| = \infty\) for every \(t > 0\): were it finite, rationals \(p, q\) with \(b - t < p < b < q < b+t\) would give \((p,q) \in \mathcal{F}\) and hence \(b \in L\). So \(f^{*}(b) = \infty > |f(b)|\), with no exceptional set.

(ii) \(b \in L\). Fix \((p,q) \in \mathcal{F}\) and put \(h = |f|\chi_{(p,q)}\), which lies in \(L^1(\mathbb{R})\) since \(\int h = \int_p^q |f| < \infty\). By 4.10 and 3.23 there is a null set \(N_{p,q}\) off which \(\frac{1}{2t}\int_{b-t}^{b+t} h \to h(b)\). If \(b \in (p,q) \setminus N_{p,q}\) and \(0 < t < \min\{b-p, q-b\}\), then \((b-t, b+t) \subseteq (p,q)\), so \(h = |f|\) there and \(h(b) = |f(b)|\); since every such average is at most \(f^{*}(b)\) by 4.6,

\begin{equation*} |f(b)| = \lim_{t \downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t}|f| \le f^{*}(b) . \end{equation*}

Since \(\mathcal{F}\) is countable, \(N = \bigcup_{(p,q) \in \mathcal{F}} N_{p,q}\) is null by 2.8, and every \(b \notin N\) falls under (i) or (ii).

The hypothesis says exactly that \(g(x) = \int_{-\infty}^{x} h\) is the zero function, so \(g^{\prime}(b) = 0\) for every \(b \in \mathbb{R}\). The Lebesgue Differentiation Theorem in its second version (4.19), applicable because \(h \in L^1(\mathbb{R})\), gives \(g^{\prime}(b) = h(b)\) for almost every \(b\). Hence \(h(s) = 0\) for almost every \(s \in \mathbb{R}\).

Take

\begin{equation*} E = \bigcup_{k=0}^{\infty} \Bigl( \bigl[-4^{-2k}, \, -4^{-2k-1}\bigr] \cup \bigl[4^{-2k-1}, \, 4^{-2k}\bigr] \Bigr) , \end{equation*}

a Borel set symmetric about \(0\), for which the ratio \(\frac{|E \cap (-t,t)|}{2t}\) oscillates between \(\frac45\) and \(\frac15\) as \(t \downarrow 0\). Its intervals in \((0,\infty)\) are the disjoint \([4^{-2j-1}, 4^{-2j}]\), \(j \ge 0\), of length \(\frac34 16^{-j}\), and by symmetry the ratio equals \(\frac{|E \cap (0,t)|}{t}\).

(i) \(t_m = 4^{-2m}\). The intervals meeting \((0, t_m)\) are exactly those with \(j \ge m\), the one with \(j = m\) having right endpoint \(t_m\); so countable additivity gives

\begin{equation*} |E \cap (0, t_m)| = \sum_{j \ge m} \tfrac34 16^{-j} = \tfrac34 \cdot \tfrac{16}{15} \cdot 16^{-m} = \tfrac45 \, t_m . \end{equation*}

(ii) \(s_m = 4^{-2m-1}\). This is the left endpoint of the interval with \(j = m\), so only the intervals with \(j \ge m+1\) meet \((0, s_m)\), and they lie inside it; by (i) with \(m+1\) in place of \(m\),

\begin{equation*} |E \cap (0, s_m)| = \tfrac45 \, 4^{-2m-2} = \tfrac15 \, s_m . \end{equation*}

Both \(t_m \downarrow 0\) and \(s_m \downarrow 0\), yet the ratios are constantly \(\frac45\) and \(\frac15\), so the limit in 4.22 does not exist.

Take \(E = F \cup (-F)\), where \(F = \bigcup_{k=1}^{\infty} J_k\) and

\begin{equation*} J_k = \Bigl[ \tfrac{1}{k+1}, \ \tfrac{1}{k+1} + \tfrac{1}{3k(k+1)} \Bigr) \end{equation*}

is the left third of \(\bigl[\frac{1}{k+1}, \frac1k\bigr)\), since \(\frac1k - \frac{1}{k+1} = \frac{1}{k(k+1)}\). These blocks sit on the harmonic scale rather than a geometric one, which is what makes the limit exist. As a countable union of intervals together with its reflection, \(E\) is Borel, and \(F \subseteq (0,1)\), \(-F \subseteq (-1,0)\).

Because Lebesgue measure is invariant under \(x \mapsto -x\) (reflection carries a cover by open intervals to a cover of the same total length, as in 2.7), and \(F \cap (0,t)\), \((-F) \cap (-t, 0)\) are disjoint Borel sets with union \(E \cap (-t,t)\),

\begin{equation*} \frac{|E \cap (-t,t)|}{2t} = \frac{|F \cap (0,t)|}{t} , \end{equation*}

so it suffices to show the right side tends to \(\frac13\). Since \(J_k \subseteq \bigl[\frac{1}{k+1}, \frac1k\bigr)\), we have \(F \cap (0, \frac1n) = \bigcup_{k \ge n} J_k\), a disjoint union, so 2.68 and 2.14 give

\begin{equation*} \Bigl| F \cap \bigl(0, \tfrac1n\bigr) \Bigr| = \frac13 \sum_{k=n}^{\infty} \Bigl( \frac1k - \frac{1}{k+1} \Bigr) = \frac{1}{3n} \end{equation*}

by telescoping. For \(t \in (0,1]\) put \(n = \lfloor 1/t \rfloor\), so \(\frac{1}{n+1} < t \le \frac1n\); then \((0, \frac{1}{n+1}) \subseteq (0,t) \subseteq (0, \frac1n]\) and monotonicity of outer measure give

\begin{equation*} \frac{n}{3(n+1)} \le \frac{|F \cap (0,t)|}{t} \le \frac{n+1}{3n} . \end{equation*}

As \(t \downarrow 0\) we have \(n \to \infty\) and both bounds tend to \(\frac13\), so the density of \(E\) at \(0\) is \(\frac13\).

Write the given number as \(c \in [0,1]\), to free the letter \(t\) for 4.22, and run the construction of Exercise 8 in this section with \(c\) in place of \(\frac13\): let \(E = F \cup (-F)\), where \(F = \bigcup_{k=1}^{\infty} J_k\) and

\begin{equation*} J_k = \Bigl[ \tfrac{1}{k+1}, \ \tfrac{1}{k+1} + \tfrac{c}{k(k+1)} \Bigr) \end{equation*}

is the left-hand fraction \(c\) of \(\bigl[\frac{1}{k+1}, \frac1k\bigr)\), of length \(\frac{c}{k(k+1)}\) (empty if \(c = 0\), the whole block if \(c = 1\); both are allowed below). As before \(E\) is Borel, the blocks are disjoint, \(F \subseteq (0,1)\), \(-F \subseteq (-1,0)\), and reflection invariance of Lebesgue measure gives

\begin{equation*} \frac{|E \cap (-s,s)|}{2s} = \frac{|F \cap (0,s)|}{s} . \end{equation*}

Exactly as in Exercise 8, \(F \cap (0, \frac1n) = \bigcup_{k \ge n} J_k\) with

\begin{equation*} \Bigl| F \cap \bigl(0, \tfrac1n\bigr) \Bigr| = c \sum_{k=n}^{\infty} \Bigl( \frac1k - \frac{1}{k+1} \Bigr) = \frac{c}{n} \end{equation*}

by 2.68, 2.14 and telescoping, so taking \(n = \lfloor 1/s \rfloor\) for \(s \in (0,1]\), whence \(\frac{1}{n+1} < s \le \frac1n\), monotonicity of outer measure gives

\begin{equation*} \frac{cn}{n+1} \le \frac{|F \cap (0,s)|}{s} \le \frac{c(n+1)}{n} . \end{equation*}

Both bounds tend to \(c\) as \(s \downarrow 0\), so the density of \(E\) at \(0\) equals \(c\).

Build from \(E\) an everywhere-differentiable function with derivative exactly \(\chi_E\); Darboux’s argument then forbids that derivative from taking both the values \(1\) and \(0\). Define

\begin{equation*} g(x) = |E \cap [0,x]| \quad \text{for } x \ge 0, \qquad g(x) = -|E \cap [x,0]| \quad \text{for } x < 0 . \end{equation*}

Each \(E \cap [u,v]\) is Lebesgue measurable with \(|E \cap [u,v]| \le v - u\) by 2.72 and 2.14, so \(g\) is real valued, and additivity of the measure supplied by 2.72 (endpoints being null) gives

\begin{equation*} g(v) - g(u) = |E \cap [u,v]| \in [0, \, v-u] \qquad (u < v) , \end{equation*}

so \(g\) is nondecreasing and \(1\)-Lipschitz, hence continuous.

\(g^{\prime} = \chi_E\) at every point. Fix \(b\) and \(t > 0\); the last display gives \(\frac{g(b+t)-g(b)}{t} = \frac{|E \cap [b,b+t]|}{t}\) and \(\frac{g(b)-g(b-t)}{t} = \frac{|E \cap [b-t,b]|}{t}\).

(i) \(b \in E\), so \(E\) has density \(1\) at \(b\). Since \([b, b+t] \subseteq [b-t,b+t]\), additivity and monotonicity give

\begin{equation*} 0 \le 1 - \frac{|E \cap [b,b+t]|}{t} = \frac{|[b,b+t] \setminus E|}{t} \le \frac{|[b-t,b+t] \setminus E|}{t} = 2\Bigl(1 - \frac{|E \cap (b-t,b+t)|}{2t}\Bigr) , \end{equation*}

which tends to \(0\); the same bound holds with \([b-t,b]\) in place of \([b,b+t]\). Hence \(g^{\prime}(b) = 1\).

(ii) \(b \notin E\), so \(E\) has density \(0\) at \(b\). Then

\begin{equation*} 0 \le \frac{|E \cap [b,b+t]|}{t} \le 2 \cdot \frac{|E \cap (b-t,b+t)|}{2t} \longrightarrow 0 , \end{equation*}

and likewise for \([b-t,b]\), so \(g^{\prime}(b) = 0\).

Now suppose \(p \in E\) and \(q \in \mathbb{R}\setminus E\). Replacing \(E\) by \(-E\) if necessary, we may assume \(p < q\): reflection invariance of Lebesgue measure makes the density of \(-E\) at \(-b\) equal to that of \(E\) at \(b\), so \(-E\) satisfies the same hypotheses and is empty (all of \(\mathbb{R}\)) exactly when \(E\) is. Put \(\varphi(x) = g(x) - \frac{x}{2}\), continuous and everywhere differentiable with \(\varphi^{\prime} = \chi_E - \frac12\), so \(\varphi^{\prime}(p) = \frac12 > 0\) and \(\varphi^{\prime}(q) = -\frac12 < 0\). The maximum of \(\varphi\) over the compact interval \([p,q]\) is attained at some \(c\), not at \(p\) (as \(\varphi^{\prime}(p) > 0\) gives \(\varphi(p+t) > \varphi(p)\) for some small \(t > 0\)) and not at \(q\) (as \(\varphi^{\prime}(q) < 0\) gives \(\varphi(q+t) > \varphi(q)\) for some small \(t < 0\)); so \(c \in (p,q)\), where the one-sided difference quotients force \(\varphi^{\prime}( c) \le 0\) and \(\varphi^{\prime}( c) \ge 0\). Then \(g^{\prime}( c) = \frac12\), contradicting \(g^{\prime}( c) = \chi_E( c) \in \{0,1\}\). Hence no such pair \(p, q\) exists, that is, \(E = \varnothing\) or \(E = \mathbb{R}\).

\(A \in \mathcal{S}\) and \(B \in \mathcal{T}\), because each is a cross section of \(A \times B\).

Because \(A\) and \(B\) are nonempty, pick \(a \in A\) and \(b \in B\). As \(A \times B \in \mathcal{S} \otimes \mathcal{T}\), 5.6 gives

\begin{equation*} [A \times B]^b \in \mathcal{S} \qquad\text{and}\qquad [A \times B]_a \in \mathcal{T} . \end{equation*}

By Example 5.5, \([A \times B]^b = A\) (since \(b \in B\)) and \([A \times B]_a = B\) (since \(a \in A\)). Hence \(A \in \mathcal{S}\) and \(B \in \mathcal{T}\).

Write \(\Delta(E) = \{x \in X : (x,x) \in E\}\) and let

\begin{equation*} \mathcal{E} = \{E \subseteq X \times X : \Delta(E) \in \mathcal{S}\} ; \end{equation*}

the assertion is that \(\mathcal{S} \otimes \mathcal{S} \subseteq \mathcal{E}\).

Every measurable rectangle lies in \(\mathcal{E}\), since for \(A, B \in \mathcal{S}\),

\begin{equation*} \Delta(A \times B) = A \cap B \in \mathcal{S} . \end{equation*}

Moreover \(\Delta(\varnothing) = \varnothing\) and

\begin{equation*} \Delta\bigl((X \times X) \setminus E\bigr) = X \setminus \Delta(E), \end{equation*}

\begin{equation*} \Delta\Bigl(\bigcup_{k=1}^{\infty} E_k\Bigr) = \bigcup_{k=1}^{\infty} \Delta(E_k) \end{equation*}

(Check!), so \(\mathcal{E}\) is a \(\sigma\)-algebra on \(X \times X\) because \(\mathcal{S}\) is one on \(X\). As \(\mathcal{S} \otimes \mathcal{S}\) is the smallest \(\sigma\)-algebra containing the measurable rectangles, \(\mathcal{S} \otimes \mathcal{S} \subseteq \mathcal{E}\).

Take \(E = \{(x,x) : x \in A\}\), where \(A \subseteq \mathbb{R}\) is a set that is not Borel (2.67).

For \(a \in \mathbb{R}\), the relation \((a,y) \in E\) forces \(y = a \in A\), and \((x,a) \in E\) forces \(x = a \in A\); hence

\begin{equation*} [E]_a = [E]^a = \begin{cases} \{a\} & \text{if } a \in A,\\ \varnothing & \text{if } a \notin A, \end{cases} \end{equation*}

which is closed and therefore in \(\mathcal{B}\). But \(\{x \in \mathbb{R} : (x,x) \in E\} = A \notin \mathcal{B}\), so \(E \notin \mathcal{B} \otimes \mathcal{B}\) by Exercise 5A.2 (applied with \(\mathcal{S} = \mathcal{B}\)).

\(h = FG\), where \(F(x,y) = f(x)\) and \(G(x,y) = g(y)\), and both factors are \((\mathcal{S} \otimes \mathcal{T})\)-measurable.

Indeed, for every Borel set \(D \subseteq \mathbb{R}\),

\begin{equation*} F^{-1}(D) = f^{-1}(D) \times Y, \qquad G^{-1}(D) = X \times g^{-1}(D), \end{equation*}

which are measurable rectangles in \(\mathcal{S} \otimes \mathcal{T}\) (5.2), because \(f^{-1}(D) \in \mathcal{S}\) and \(g^{-1}(D) \in \mathcal{T}\) by the measurability of \(f\) and of \(g\). Hence \(FG\) is \((\mathcal{S} \otimes \mathcal{T})\)-measurable by 2.46(a), and \((FG)(x,y) = f(x)g(y) = h(x,y)\).

A subset of \(\mathbb{R}\) is an interval exactly when it is order convex (meaning that \(x, z \in I\) and \(x < y < z\) imply \(y \in I\)), and this makes the complement of an interval a union of two intervals; De Morgan’s Laws then finish the proof.

Order convexity characterizes intervals. Each form \((\alpha,\beta)\), \([\alpha,\beta)\), \((\alpha,\beta]\), \([\alpha,\beta]\) (with \(\alpha = -\infty\) or \(\beta = \infty\) allowed at an open end), together with singletons and \(\varnothing\), is cut out by inequalities that an intermediate point inherits. Conversely, let \(I \ne \varnothing\) be order convex, \(\alpha = \inf I\), \(\beta = \sup I\). Then

\begin{equation*} (\alpha,\beta) \subseteq I \subseteq [\alpha,\beta] \cap \mathbb{R} : \end{equation*}

the right inclusion is the definition of \(\inf\) and \(\sup\), and if \(\alpha < y < \beta\) then \(y\) is neither a lower nor an upper bound of \(I\), so \(x < y < z\) for some \(x, z \in I\), whence \(y \in I\). So \(I\) is \((\alpha,\beta)\) with possibly its finite endpoints adjoined, hence an interval (a single point when \(\alpha = \beta\)); and \(\varnothing = (0,0)\) is an interval.

Let \(\mathcal{A}\) be the collection of finite unions of intervals.

(i) \(\mathcal{A}\) is closed under finite intersections. The intersection of two order convex sets is order convex, hence an interval, and

\begin{equation*} \Bigl(\bigcup_{j=1}^{n} I_j\Bigr) \cap \Bigl(\bigcup_{k=1}^{m} J_k\Bigr) = \bigcup_{j=1}^{n} \bigcup_{k=1}^{m} (I_j \cap J_k) \in \mathcal{A} ; \end{equation*}

induction extends this to finitely many elements of \(\mathcal{A}\).

(ii) The complement of an interval \(I\) is a union of two intervals. Put

\begin{equation*} L = \{x \in \mathbb{R} : x < y \text{ for every } y \in I\}, \end{equation*}

\begin{equation*} U = \{x \in \mathbb{R} : x > y \text{ for every } y \in I\} . \end{equation*}

Then \(\mathbb{R} \setminus I = L \cup U\): a point of \(I \cap L\) would satisfy \(x < x\), and similarly for \(U\), while if \(x\) lies in none of \(I\), \(L\), \(U\), then some \(y, z \in I\) satisfy \(y < x < z\), forcing \(x \in I\). Also \(L\) is closed downward and \(U\) closed upward, so both are order convex, hence intervals.

Now let \(E = I_1 \cup \cdots \cup I_n \in \mathcal{A}\). By De Morgan’s Laws,

\begin{equation*} \mathbb{R} \setminus E = \bigcap_{j=1}^{n} (\mathbb{R} \setminus I_j), \end{equation*}

an intersection of finitely many elements of \(\mathcal{A}\) by (ii), hence in \(\mathcal{A}\) by (i).

\(\mathbb{Q}\) is a countable union of intervals but \(\mathbb{R} \setminus \mathbb{Q}\) is not.

Let \(\mathcal{A}\) be the collection of countable unions of intervals. Then \(\mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{q\} \in \mathcal{A}\), each \(\{q\}\) counting as an interval in Example 5.11.

An interval \(I \subseteq \mathbb{R} \setminus \mathbb{Q}\) has at most one point: if \(x, z \in I\) with \(x < z\), density of \(\mathbb{Q}\) gives \(q \in \mathbb{Q}\) with \(x < q < z\), and every interval contains the points lying between two of its points, so \(q \in I \cap \mathbb{Q}\), a contradiction. So if \(\mathbb{R} \setminus \mathbb{Q} = \bigcup_{k=1}^{\infty} I_k\) with each \(I_k\) an interval, then \(\mathbb{R} \setminus \mathbb{Q}\) is a countable union of sets with at most one element, hence countable, making \(\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})\) countable. This contradiction gives \(\mathbb{R} \setminus \mathbb{Q} \notin \mathcal{A}\), so \(\mathcal{A}\) is not closed under complementation.

Both directions are De Morgan’s Laws.

Suppose \(\mathcal{A}\) is an algebra on \(W\). Closure under complementation is the second condition of 5.10, and for \(E, F \in \mathcal{A}\),

\begin{equation*} E \cap F = W \setminus \bigl((W \setminus E) \cup (W \setminus F)\bigr) \in \mathcal{A} \end{equation*}

by closure under complementation and under unions of two elements; induction gives \(E_1 \cap \cdots \cap E_n \in \mathcal{A}\).

Conversely, suppose \(\mathcal{A}\) is nonempty and closed under finite intersections and under complementation. Choosing any \(E \in \mathcal{A}\) (this is the only use of nonemptiness) gives

\begin{equation*} \varnothing = E \cap (W \setminus E) \in \mathcal{A}, \end{equation*}

and for \(E, F \in \mathcal{A}\),

\begin{equation*} E \cup F = W \setminus \bigl((W \setminus E) \cap (W \setminus F)\bigr) \in \mathcal{A} . \end{equation*}

The three conditions of 5.10 hold, so \(\mathcal{A}\) is an algebra on \(W\).

(a) \(\Rightarrow\) (b). By 5.18 there are \(E_1, E_2, \ldots \in \mathcal{S}\) with \(X = \bigcup_{k=1}^{\infty} E_k\) and \(\mu(E_k) < \infty\). Put \(X_k = E_1 \cup \cdots \cup E_k \in \mathcal{S}\). These increase, satisfy \(\bigcup_{k=1}^{\infty} X_k = \bigcup_{k=1}^{\infty} E_k = X\), and by finite subadditivity (2.58 applied to \(E_1, \ldots, E_k, \varnothing, \varnothing, \ldots\)),

\begin{equation*} \mu(X_k) \le \mu(E_1) + \cdots + \mu(E_k) < \infty . \end{equation*}

(b) \(\Rightarrow\) (c). Put \(Y_1 = X_1\) and \(Y_k = X_k \setminus X_{k-1}\) for \(k \ge 2\), all in \(\mathcal{S}\). If \(j < k\) then \(Y_j \subseteq X_j \subseteq X_{k-1}\) while \(Y_k \cap X_{k-1} = \varnothing\), so the sequence is disjoint. Each \(x \in X\) lies in \(Y_m\) for the smallest \(m\) with \(x \in X_m\), so \(X = \bigcup_{k=1}^{\infty} Y_k\). Finally \(Y_k \subseteq X_k\) gives \(\mu(Y_k) \le \mu(X_k) < \infty\) by 2.57(a).

(c) \(\Rightarrow\) (a). Such a sequence is in particular a sequence in \(\mathcal{S}\) with union \(X\) and with every term of finite measure, which is the definition 5.18 of \(\sigma\)-finite.

The sets \(A_j \times B_k\) form a countable cover of \(X \times Y\) by sets of finite \(\mu \times \nu\) measure.

Say \(\mu\) is a measure on \((X, \mathcal{S})\) and \(\nu\) on \((Y, \mathcal{T})\); since both are \(\sigma\)-finite, \(\mu \times \nu\) is defined on \(\mathcal{S} \otimes \mathcal{T}\) by 5.25 and is a measure by 5.27. Choose \(A_j \in \mathcal{S}\) with \(X = \bigcup_{j=1}^{\infty} A_j\) and \(\mu(A_j) < \infty\), and \(B_k \in \mathcal{T}\) with \(Y = \bigcup_{k=1}^{\infty} B_k\) and \(\nu(B_k) < \infty\). Each \(A_j \times B_k\) is a measurable rectangle (5.2), the index set \(\mathbf{Z}^+ \times \mathbf{Z}^+\) is countable,

\begin{equation*} X \times Y = \bigcup_{j=1}^{\infty} \bigcup_{k=1}^{\infty} (A_j \times B_k), \end{equation*}

and by 5.26,

\begin{equation*} (\mu \times \nu)(A_j \times B_k) = \mu(A_j)\, \nu(B_k) < \infty . \end{equation*}

Thus \(\mu \times \nu\) is \(\sigma\)-finite by 5.18.

The two measures agree on the algebra of finite unions of measurable rectangles, and the Monotone Class Theorem (5.17) propagates this to \(\mathcal{S} \otimes \mathcal{T}\).

By Exercise 5A.8 choose increasing sequences \(X_n \in \mathcal{S}\) and \(Y_n \in \mathcal{T}\) with \(X = \bigcup_n X_n\), \(Y = \bigcup_n Y_n\), \(\mu(X_n) < \infty\), \(\nu(Y_n) < \infty\), and set \(W_n = X_n \times Y_n\). Then \(W_1 \subseteq W_2 \subseteq \cdots\) and \(\bigcup_n W_n = X \times Y\) (for \((x,y) \in X_j \times Y_k\) take \(n = \max\{j,k\}\)), while the hypothesis on \(\omega\) and 5.26 give

\begin{equation*} \omega(W_n) = \mu(X_n)\, \nu(Y_n) = (\mu \times \nu)(W_n) < \infty \end{equation*}

(here \(\mu \times \nu\) is defined by 5.25 and is a measure by 5.27, both spaces being \(\sigma\)-finite). Fix \(n\) and set

\begin{equation*} \mathcal{M}_n = \{ E \in \mathcal{S} \otimes \mathcal{T} : \omega(E \cap W_n) = (\mu \times \nu)(E \cap W_n) \} . \end{equation*}

(i) \(\mathcal{M}_n\) contains the collection \(\mathcal{A}\) of finite unions of measurable rectangles, an algebra by 5.13(a). For \(A \in \mathcal{S}\) and \(B \in \mathcal{T}\),

\begin{equation*} (A \times B) \cap W_n = (A \cap X_n) \times (B \cap Y_n), \end{equation*}

so the hypothesis on \(\omega\) and 5.26 give \(A \times B \in \mathcal{M}_n\); a general \(E \in \mathcal{A}\) is a disjoint union \(E_1 \cup \cdots \cup E_m\) of measurable rectangles by 5.13(b), and finite additivity of both measures then gives \(E \in \mathcal{M}_n\).

(ii) \(\mathcal{M}_n\) is a monotone class. For \(E_k \uparrow E\) in \(\mathcal{M}_n\), apply 2.59 to \(E_k \cap W_n \uparrow E \cap W_n\) for each measure. For \(E_k \downarrow E\), we have \(\omega(E_1 \cap W_n) \le \omega(W_n) < \infty\) and likewise for \(\mu \times \nu\) by 2.57(a), so 2.60 applies to both and

\begin{equation*} \omega(E \cap W_n) = \lim_{k \to \infty} \omega(E_k \cap W_n) = \lim_{k \to \infty} (\mu \times \nu)(E_k \cap W_n) = (\mu \times \nu)(E \cap W_n). \end{equation*}

By 5.17, \(\mathcal{M}_n\) contains the smallest \(\sigma\)-algebra containing \(\mathcal{A}\); that \(\sigma\)-algebra is \(\mathcal{S} \otimes \mathcal{T}\), since \(\mathcal{A}\) contains every measurable rectangle (5.2) and \(\mathcal{A} \subseteq \mathcal{S} \otimes \mathcal{T}\). As \(\mathcal{M}_n \subseteq \mathcal{S} \otimes \mathcal{T}\), we get \(\mathcal{M}_n = \mathcal{S} \otimes \mathcal{T}\). Finally, for \(E \in \mathcal{S} \otimes \mathcal{T}\) we have \(E \cap W_n \uparrow E\), so 2.59 gives

\begin{equation*} \omega(E) = \lim_{n \to \infty} \omega(E \cap W_n) = \lim_{n \to \infty} (\mu \times \nu)(E \cap W_n) = (\mu \times \nu)(E) . \end{equation*}

Hence \(\omega = \mu \times \nu\).

(a) Everything comes from the pair of antiderivatives

\begin{equation*} \frac{\partial}{\partial y}\Bigl(\frac{y}{x^2+y^2}\Bigr) =\frac{\partial}{\partial x}\Bigl(\frac{-x}{x^2+y^2}\Bigr) =\frac{x^2-y^2}{(x^2+y^2)^2}, \end{equation*}

valid wherever \((x,y)\neq(0,0)\). Write \(f(x,y)=(x^2-y^2)/(x^2+y^2)^2\) off the origin and \(f(0,0)=0\); note \(f(y,x)=-f(x,y)\).

For \(x\in(0,1]\) the map \(y\mapsto f(x,y)\) is continuous on \([0,1]\) (the denominator is at least \(x^4>0\)), so 3.34 and the Fundamental Theorem of Calculus give

\begin{equation*} \int_{[0,1]}f(x,y)\,d\lambda(y)=\left[\frac{y}{x^2+y^2}\right]_{y=0}^{y=1} =\frac{1}{1+x^2}. \end{equation*}

For \(x=0\) we have \(f(0,\cdot)^+=0\) and \(f(0,\cdot)^-=y^{-2}\chi_{(0,1]}\), whose integral is at least \(\int_{[1/k,1]}y^{-2}\,dy=k-1\) for every \(k\in\mathbb{Z}^+\) (3.34), hence is \(\infty\); so by 3.18 the inner integral is \(0-\infty=-\infty\). Thus the inner integral is the function \(g\) with \(g(0)=-\infty\) and \(g(x)=1/(1+x^2)\) for \(x>0\). Since \(g^-=\infty\cdot\chi_{\{0\}}\) has integral \(\infty\cdot\lambda(\{0\})=0\), and \(g^+\) differs from the continuous function \(x\mapsto 1/(1+x^2)\) only on the null set \(\{0\}\), 3.18 and 3.34 give

\begin{equation*} \int_{[0,1]}\int_{[0,1]}f(x,y)\,d\lambda(y)\,d\lambda(x) =\int_0^1\frac{dx}{1+x^2}=\frac{\pi}{4}. \end{equation*}

Interchanging the two variables (which by \(f(y,x)=-f(x,y)\) changes the sign) gives

\begin{equation*} \int_{[0,1]}\int_{[0,1]}f(x,y)\,d\lambda(x)\,d\lambda(y)=-\frac{\pi}{4}. \end{equation*}

(b) Tonelli’s Theorem (5.28) requires a nonnegative integrand, and \(f\) takes both signs; its other hypothesis holds, since \(\lambda\) on \([0,1]\) is finite, hence \(\sigma\)-finite. Fubini’s Theorem (5.32) requires \(f\in\mathcal{L}^1(\lambda\times\lambda)\), and this fails. Here \(f\) is \(\mathcal{B}\otimes\mathcal{B}\)-measurable: the coordinate maps are \(\mathcal{B}\otimes\mathcal{B}\)-measurable because their level sets are measurable rectangles (2.39), so each

\begin{equation*} f_k(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2+\frac{1}{k}} \end{equation*}

is measurable by 2.46, and \(f_k\to f\) pointwise, so 2.48 applies. Applying 5.28 to \(|f|\) and using \(f(x,y)\ge 0\) for \(0\le y\le x\),

\begin{equation*} \int_{[0,1]}|f(x,y)|\,d\lambda(y) \ge\left[\frac{y}{x^2+y^2}\right]_{y=0}^{y=x}=\frac{1}{2x}, \end{equation*}

whence for every \(k\in\mathbb{Z}^+\)

\begin{equation*} \int_{[0,1]^2}|f|\,d(\lambda\times\lambda)\ge\frac{1}{2}\int_{[1/k,1]}\frac{dx}{x} =\frac{\ln k}{2}, \end{equation*}

so \(\int_{[0,1]^2}|f|\,d(\lambda\times\lambda)=\infty\).

(a) Take

\begin{equation*} x_{m,n}= \begin{cases} 1 & \text{if }n=m,\\ -1 & \text{if }n=2m,\\ 0 & \text{otherwise.} \end{cases} \end{equation*}

Row \(m\) has exactly the two nonzero terms \(x_{m,m}=1\) and \(x_{m,2m}=-1\), distinct because \(m\ge 1\), so every row sums to \(0\) and

\begin{equation*} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}x_{m,n}=0 . \end{equation*}

Column \(n\) receives \(+1\) from \(m=n\) and, when \(n\) is even, \(-1\) from \(m=n/2\) (a distinct index); so the column sums are \(1\) for \(n\) odd and \(0\) for \(n\) even, and the partial sums of \(\sum_n\sum_m x_{m,n}\) count the odd integers up to \(n\), giving

\begin{equation*} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}x_{m,n}=\infty . \end{equation*}

(b) Let \(\mu=\nu\) be counting measure on all subsets of \(\mathbb{Z}^+\), which is \(\sigma\)-finite since \(\mathbb{Z}^+=\bigcup_{k=1}^{\infty}\{1,\dots,k\}\), and let \(f(m,n)=x_{m,n}\). Every subset of the countable set \(\mathbb{Z}^+\times\mathbb{Z}^+\) is a countable union of the measurable rectangles \(\{m\}\times\{n\}\), so \(f\) is \(\mathcal{S}\otimes\mathcal{T}\)-measurable; integration against counting measure is summation (3.6, and 3.42 in the real-valued case), so the two sums in (a) are the two iterated integrals of 5.28 and 5.32 (each inner function has at most two nonzero values, hence lies in \(\mathcal{L}^1\)).

Tonelli’s Theorem (5.28), and likewise its corollary 5.31 on double sums, requires a nonnegative integrand, and \(f\) takes the value \(-1\); the \(\sigma\)-finiteness hypothesis does hold. Fubini’s Theorem (5.32) requires \(f\in\mathcal{L}^1(\mu\times\nu)\), which fails: applying 5.28 to \(|f|\), whose rows each have two entries of absolute value \(1\),

\begin{equation*} \int_{\mathbb{Z}^+\times\mathbb{Z}^+}|f|\,d(\mu\times\nu) =\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|x_{m,n}|=\sum_{m=1}^{\infty}2=\infty . \end{equation*}

Both implications come from a rectangle decomposition of \(U_f=\{(x,t):0<t<f(x)\}\) and from 5.6.

Suppose \(f\) is \(\mathcal{S}\)-measurable. Then

\begin{equation*} U_f=\bigcup_{r\in\mathbb{Q}\cap(0,\infty)}f^{-1}\bigl((r,\infty]\bigr)\times(0,r] : \end{equation*}

for \(\supseteq\), such a point satisfies \(0<t\le r<f(x)\); for \(\subseteq\), density of \(\mathbb{Q}\) supplies a rational \(r\) with \(t<r<f(x)\) (any rational \(r>t\) if \(f(x)=\infty\)), and then \(r>0\). Each \(f^{-1}((r,\infty])\) is in \(\mathcal{S}\) by 2.51, since \((r,\infty]\) is Borel (2.50), and \((0,r]\in\mathcal{B}\), so each term is a measurable rectangle; the union is countable, so \(U_f\in\mathcal{S}\otimes\mathcal{B}\).

Conversely, suppose \(U_f\in\mathcal{S}\otimes\mathcal{B}\). By 5.6 each cross section \([U_f]^t\) lies in \(\mathcal{S}\), and for \(t\in(0,\infty)\) the condition \((x,t)\in U_f\) says exactly \(f(x)>t\), so

\begin{equation*} f^{-1}\bigl((t,\infty]\bigr)=\{x\in X:f(x)>t\}=[U_f]^t\in\mathcal{S}. \end{equation*}

Now let \(a\in\mathbb{R}\). (i) If \(a>0\), then \(f^{-1}((a,\infty])=[U_f]^a\in\mathcal{S}\). (ii) If \(a<0\), then \(f^{-1}((a,\infty])=X\in\mathcal{S}\), as \(f\) takes values in \([0,\infty]\). (iii) If \(a=0\), then \(f(x)>0\) exactly when \(f(x)>1/k\) for some \(k\), so

\begin{equation*} f^{-1}\bigl((0,\infty]\bigr)=\bigcup_{k=1}^{\infty}[U_f]^{1/k}\in\mathcal{S}. \end{equation*}

Hence \(f\) is \(\mathcal{S}\)-measurable by 2.52.

\(\operatorname{graph}(f)=(g-h)^{-1}\bigl(\{0\}\bigr)\), where \(g(x,y)=f(x)\) and \(h(x,y)=y\).

For every Borel set \(B\subseteq\mathbb{R}\),

\begin{equation*} g^{-1}(B)=f^{-1}(B)\times\mathbb{R} \qquad\text{and}\qquad h^{-1}(B)=X\times B , \end{equation*}

which are measurable rectangles in \(\mathcal{S}\otimes\mathcal{B}\) because \(f^{-1}(B)\in\mathcal{S}\) by the measurability of \(f\). So \(g\) and \(h\) are \(\mathcal{S}\otimes\mathcal{B}\)-measurable, hence so is \(g-h\) by 2.46(a), and \(\{0\}\) is a Borel subset of \(\mathbb{R}\). Thus

\begin{equation*} \operatorname{graph}(f)=\{(x,y):f(x)-y=0\}=(g-h)^{-1}\bigl(\{0\}\bigr) \in\mathcal{S}\otimes\mathcal{B}. \end{equation*}

Method (2): build the graph from rectangles directly,

\begin{equation*} \operatorname{graph}(f)=\bigcap_{k=1}^{\infty}\ \bigcup_{j\in\mathbb{Z}} \Bigl(f^{-1}\bigl([\tfrac{j}{k},\tfrac{j+1}{k})\bigr) \times[\tfrac{j}{k},\tfrac{j+1}{k})\Bigr). \end{equation*}

Indeed if \(y=f(x)\), then for each \(k\) the integer \(j\) with \(j\le kf(x)<j+1\) puts both \(f(x)\) and \(y\) in \([\tfrac{j}{k},\tfrac{j+1}{k})\); conversely a point of the right side has \(f(x)\) and \(y\) in a common interval of length \(1/k\) for every \(k\), so \(|y-f(x)|<1/k\) for all \(k\) and \(y=f(x)\). Each term is a measurable rectangle, and a \(\sigma\)-algebra is closed under countable unions and countable intersections.

The two conditions are equivalent because \(\|x\|_\infty \le \|x\|_2 \le \sqrt{n}\,\|x\|_\infty\).

Write \(\|x\|_2=(x_1^2+\cdots+x_n^2)^{1/2}\) and \(D(b,r)=\{a\in\mathbb{R}^n:\|a-b\|_2<r\}\) for the set displayed in the exercise, and recall that \(G\) is open means every \(x\in G\) has an open cube \(B(x,\delta)=\{y:\|y-x\|_\infty<\delta\}\) inside \(G\). The first inequality holds because \(\|x\|_\infty^2=x_j^2\le\|x\|_2^2\) for an index \(j\) attaining the maximum; the second because \(\|x\|_2^2\le n\|x\|_\infty^2\). They say exactly that

\begin{equation*} D(b,\delta)\subseteq B(b,\delta) \quad\text{and}\quad B(b,r/\sqrt{n}\,)\subseteq D(b,r) \end{equation*}

for all \(b\in\mathbb{R}^n\) and \(\delta,r>0\). So if \(G\) is open and \(b\in G\), then choosing \(\delta>0\) with \(B(b,\delta)\subseteq G\) gives \(D(b,\delta)\subseteq G\); and conversely, if \(D(b,r)\subseteq G\), then \(B(b,r/\sqrt{n}\,)\subseteq G\), so \(G\) is open.

Take \(E=E_A=\mathbb{R}^2\setminus\Gamma_A\) for a suitable \(A\subseteq(0,\infty)\), where \(\Gamma_A\) is the graph of

\begin{equation*} f_A(x)=\begin{cases} -x & \text{if } |x|\in A,\\ x & \text{if } |x|\notin A; \end{cases} \end{equation*}

there are \(2^{\mathfrak{c}}\) such sets but only \(\mathfrak{c}\) Borel sets, so some \(E_A\) is not Borel. (Here \(\mathfrak{c}=2^{\aleph_0}\).)

(i) Every \(E_A\) has all cross sections open, and \(A\mapsto E_A\) is injective. Since \(|f_A(x)|=|x|\), we get \(f_A(f_A(x))=x\), so \(f_A\) is a bijection of \(\mathbb{R}\) equal to its own inverse; hence \((a,y)\in\Gamma_A\) exactly when \(y=f_A(a)\), and \((x,a)\in\Gamma_A\) exactly when \(x=f_A(a)\), giving

\begin{equation*} [E_A]_a=[E_A]^a=\mathbb{R}\setminus\{f_A(a)\}, \end{equation*}

an open subset of \(\mathbb{R}\). Injectivity holds because \(f_A(t)=-t\) for \(t\in A\) and \(f_A(t)=t\) for \(t\notin A\), with \(-t\ne t\) when \(t>0\), so

\begin{equation*} A=\{t>0:(t,-t)\in\mathbb{R}^2\setminus E_A\}. \end{equation*}

(ii) \(|\mathcal{B}_2|\le\mathfrak{c}\). By the proof of 5.38(a), every open subset of \(\mathbb{R}^2\) is the union of a subcollection of the countable collection of open cubes with rational center and rational side length, so the collection \(\Sigma_0\) of open sets satisfies \(|\Sigma_0|\le 2^{\aleph_0}=\mathfrak{c}\). For \(0<\alpha<\omega_1\) (the first uncountable ordinal) let \(\Sigma_\alpha\) be the collection of countable unions \(\bigcup_k F_k\) in which each \(F_k\) or its complement lies in \(\bigcup_{\beta<\alpha}\Sigma_\beta\), and put \(\Sigma=\bigcup_{\alpha<\omega_1}\Sigma_\alpha\). Taking all \(F_k\) equal shows \(\Sigma_\beta\subseteq\Sigma_\alpha\) for \(\beta<\alpha\); \(\Sigma\) is closed under complementation, and under countable unions because a countable set of countable ordinals has countable supremum. So \(\Sigma\) is a \(\sigma\)-algebra containing the open sets, whence \(\mathcal{B}_2\subseteq\Sigma\). By transfinite induction \(|\Sigma_\alpha|\le\mathfrak{c}\) for every \(\alpha<\omega_1\): the sets available at stage \(\alpha\) number at most \(\aleph_0\cdot\mathfrak{c}=\mathfrak{c}\), and each element of \(\Sigma_\alpha\) is the union of a sequence of them, so

\begin{equation*} |\Sigma_\alpha|\le\mathfrak{c}^{\aleph_0}=\bigl(2^{\aleph_0}\bigr)^{\aleph_0} =2^{\aleph_0}=\mathfrak{c}. \end{equation*}

Hence \(|\mathcal{B}_2|\le|\Sigma|\le\aleph_1\cdot\mathfrak{c}=\mathfrak{c}\), using \(\aleph_1\le\mathfrak{c}\).

By Cantor’s Theorem \(2^{\mathfrak{c}}>\mathfrak{c}\ge|\mathcal{B}_2|\), so by (i) the sets \(E_A\) cannot all be Borel; choose \(A\) with \(E=E_A\notin\mathcal{B}_2\).

Both equalities follow from the fact that the generators of each side lie in the other, the reverse inclusion being proved by freezing one factor.

Identify \(((x,y),z)\) and \((x,(y,z))\) with \((x,y,z)\), so that \((A\times B)\times C\), \(A\times(B\times C)\), and \(A\times B\times C\) are the same set, and write \(\mathcal{P}=\mathcal{S}\otimes\mathcal{T}\otimes\mathcal{U}\).

\(\mathcal{P}\subseteq(\mathcal{S}\otimes\mathcal{T})\otimes\mathcal{U}\): for \(A\in\mathcal{S}\), \(B\in\mathcal{T}\), \(C\in\mathcal{U}\), the set \(A\times B\) lies in \(\mathcal{S}\otimes\mathcal{T}\), so \(A\times B\times C=(A\times B)\times C\) is a measurable rectangle in \((\mathcal{S}\otimes\mathcal{T})\otimes\mathcal{U}\); thus that \(\sigma\)-algebra contains all generators of \(\mathcal{P}\), which is the smallest one that does.

For the reverse inclusion, fix \(C\in\mathcal{U}\) and set

\begin{equation*} \mathcal{E}=\{D\subseteq X\times Y : D\times C\in\mathcal{P}\}. \end{equation*}

Then \(X\times Y\in\mathcal{E}\) (as \(X\times Y\times C\) is a generator of \(\mathcal{P}\)); \(\mathcal{E}\) is closed under countable unions because \((\bigcup_k D_k)\times C=\bigcup_k(D_k\times C)\); and \(\mathcal{E}\) is closed under complementation because

\begin{equation*} \bigl((X\times Y)\setminus D\bigr)\times C =\bigl((X\times Y\times Z)\setminus(D\times C)\bigr) \cap\bigl(X\times Y\times C\bigr), \end{equation*}

both sides consisting of the points \((x,y,z)\) with \(z\in C\) and \((x,y)\notin D\). So \(\mathcal{E}\) is a \(\sigma\)-algebra on \(X\times Y\) containing every measurable rectangle \(A\times B\), whence \(\mathcal{S}\otimes\mathcal{T}\subseteq\mathcal{E}\); that is, \(D\times C\in\mathcal{P}\) for all \(D\in\mathcal{S}\otimes\mathcal{T}\) and \(C\in\mathcal{U}\). These sets generate \((\mathcal{S}\otimes\mathcal{T})\otimes\mathcal{U}\), so \((\mathcal{S}\otimes\mathcal{T})\otimes\mathcal{U}\subseteq\mathcal{P}\).

The proof of \(\mathcal{P}=\mathcal{S}\otimes(\mathcal{T}\otimes\mathcal{U})\) is symmetric: fix \(A\in\mathcal{S}\) and let \(\mathcal{F}=\{F\subseteq Y\times Z : A\times F\in\mathcal{P}\}\), which is a \(\sigma\)-algebra by the same three verifications (for complementation use \(A\times((Y\times Z)\setminus F)=((X\times Y\times Z)\setminus(A\times F))\cap(A\times Y\times Z)\)) and contains every \(B\times C\); hence \(\mathcal{T}\otimes\mathcal{U}\subseteq\mathcal{F}\), so the generators \(A\times F\) of \(\mathcal{S}\otimes(\mathcal{T}\otimes\mathcal{U})\) all lie in \(\mathcal{P}\).

Fix \(a\in\mathbb{R}^n\).

(i) \(a+E\in\mathcal{B}_n\) for every \(E\in\mathcal{B}_n\). The collection \(\{E\in\mathcal{B}_n : a+E\in\mathcal{B}_n\}\) contains every open set, because \(a+B(x,\delta)=B(a+x,\delta)\), and it is closed under complementation and countable unions because \(x\mapsto a+x\) is a bijection of \(\mathbb{R}^n\):

\begin{equation*} a+(\mathbb{R}^n\setminus E)=\mathbb{R}^n\setminus(a+E), \qquad a+\bigcup_{k=1}^{\infty}E_k=\bigcup_{k=1}^{\infty}(a+E_k). \end{equation*}

So it is a \(\sigma\)-algebra containing the open sets, hence equals \(\mathcal{B}_n\).

(ii) Translation invariance of \(\lambda_m\) forces translation invariance of the integral: if it holds in dimension \(m\) and \(h:\mathbb{R}^m\to[0,\infty]\) is \(\mathcal{B}_m\)-measurable and \(b\in\mathbb{R}^m\), then \(x\mapsto h(x-b)\) is \(\mathcal{B}_m\)-measurable (its inverse images are \(b+h^{-1}(B)\), measurable by (i)) and

\begin{equation*} \int_{\mathbb{R}^m}h(x-b)\,d\lambda_m(x)=\int_{\mathbb{R}^m}h\,d\lambda_m . \end{equation*}

For \(h=\chi_F\) this is \(\lambda_m(b+F)=\lambda_m(F)\), since \(\chi_F(x-b)=\chi_{b+F}(x)\); for nonnegative simple \(h\) it follows by 3.16 and 3.20; for general \(h\) take simple \(s_k\uparrow h\) (2.89), note \(s_k(\cdot-b)\uparrow h(\cdot-b)\), and apply the Monotone Convergence Theorem (3.11) twice.

(iii) Induction on \(n\). For \(n=1\), \(\lambda_1\) is the restriction of outer measure to \(\mathcal{B}_1\), and outer measure is translation invariant by 2.7. Let \(n>1\) and assume the result in dimension \(n-1\). View \(\mathbb{R}^n=\mathbb{R}^{n-1}\times\mathbb{R}\), so \(\mathcal{B}_n=\mathcal{B}_{n-1}\otimes\mathcal{B}_1\) by 5.39 and \(\lambda_n=\lambda_{n-1}\times\lambda_1\); each \(\lambda_k\) is \(\sigma\)-finite, since \(\mathbb{R}^k\) is the union of the cubes \(C_m\) centered at the origin with side \(m\) and \(\lambda_k(C_m)=m^k<\infty\), so 5.20 and 5.25 apply. Write \(a=(a^{\prime},a_n)\). Since \((x,y)\in a+E\) exactly when \((x-a^{\prime},y-a_n)\in E\),

\begin{equation*} [a+E]_x=a_n+[E]_{x-a^{\prime}} , \end{equation*}

so by the case \(n=1\) we get \(\lambda_1([a+E]_x)=g(x-a^{\prime})\), where \(g(x)=\lambda_1([E]_x)\) is \(\mathcal{B}_{n-1}\)-measurable by 5.20(a). Hence, by 5.25 and then (ii) with \(m=n-1\) and \(b=a^{\prime}\),

\begin{equation*} \begin{aligned} \lambda_n(a+E) &=\int_{\mathbb{R}^{n-1}}\lambda_1\bigl([a+E]_x\bigr)\,d\lambda_{n-1}(x)\\ &=\int_{\mathbb{R}^{n-1}}g(x-a^{\prime})\,d\lambda_{n-1}(x)\\ &=\int_{\mathbb{R}^{n-1}}g\,d\lambda_{n-1}=\lambda_n(E). \end{aligned} \end{equation*}

(a) \(f_t^{-1}(B)=\tfrac{1}{t}f^{-1}(B)\in\mathcal{B}_n\) for every Borel \(B\subseteq\mathbb{R}\), so \(f_t\) is \(\mathcal{B}_n\)-measurable by 2.35.

Indeed \(f(tx)\in B\) exactly when \(tx\in f^{-1}(B)\), that is, when \(x\in\tfrac{1}{t}f^{-1}(B)\); and \(f^{-1}(B)\in\mathcal{B}_n\) by the measurability of \(f\), so the following lemma (applied to the nonzero number \(1/t\)) puts \(\tfrac{1}{t}f^{-1}(B)\) in \(\mathcal{B}_n\). The same argument applies to \([0,\infty]\)-valued measurable functions.

Lemma. If \(t\in\mathbb{R}\setminus\{0\}\) and \(E\in\mathcal{B}_n\), then \(tE\in\mathcal{B}_n\) and \(\lambda_n(tE)=|t|^n\lambda_n(E)\).

For \(t>0\) this is 5.41, and \(tE=-(|t|E)\) for \(t<0\), so only \(x\mapsto -x\) needs treatment. The collection \(\{F\in\mathcal{B}_n : -F\in\mathcal{B}_n\}\) contains the open sets (\(-B(x,\delta)=B(-x,\delta)\)) and is a \(\sigma\)-algebra (\(x\mapsto -x\) is a bijection), hence equals \(\mathcal{B}_n\). That \(\lambda_n(-F)=\lambda_n(F)\) goes by induction on \(n\). For \(n=1\), negating a cover of \(F\) by open intervals gives a cover of \(-F\) by open intervals of the same lengths, so \(|-F|=|F|\). For \(n>1\), view \(\mathbb{R}^n=\mathbb{R}^{n-1}\times\mathbb{R}\), so \(\mathcal{B}_n=\mathcal{B}_{n-1}\otimes\mathcal{B}_1\) (5.39) and \(\lambda_n=\lambda_{n-1}\times\lambda_1\), each \(\lambda_k\) being \(\sigma\)-finite (\(\mathbb{R}^k\) is a union of cubes of finite measure), so 5.20 and 5.25 apply. The induction hypothesis gives

\begin{equation*} \int_{\mathbb{R}^{n-1}}h(-x)\,d\lambda_{n-1}(x) =\int_{\mathbb{R}^{n-1}}h\,d\lambda_{n-1} \end{equation*}

for \(\mathcal{B}_{n-1}\)-measurable \(h:\mathbb{R}^{n-1}\to[0,\infty]\), via \(\chi_F(-x)=\chi_{-F}(x)\), then 3.16 and 3.20 for simple functions, then 2.89 and 3.11 in general. Since \([-F]_x=-[F]_{-x}\), the function \(g(x)=\lambda_1([F]_x)\) (\(\mathcal{B}_{n-1}\)-measurable by 5.20(a)) satisfies \(\lambda_1([-F]_x)=g(-x)\), so by 5.25

\begin{equation*} \lambda_n(-F)=\int_{\mathbb{R}^{n-1}}g(-x)\,d\lambda_{n-1}(x) =\int_{\mathbb{R}^{n-1}}g\,d\lambda_{n-1}=\lambda_n(F). \end{equation*}

Finally, for \(t<0\), \(\lambda_n(tE)=\lambda_n(|t|E)=|t|^n\lambda_n(E)\) by 5.41.

(b) Write \(h_t(x)=h(tx)\). For \(E\in\mathcal{B}_n\) we have \((\chi_E)_t=\chi_{(1/t)E}\), so the Lemma gives

\begin{equation*} \int_{\mathbb{R}^n}(\chi_E)_t\,d\lambda_n=\lambda_n\Bigl(\tfrac{1}{t}E\Bigr) =\frac{1}{|t|^n}\int_{\mathbb{R}^n}\chi_E\,d\lambda_n . \end{equation*}

This extends to nonnegative simple functions by 3.16 and 3.20, and then to every \(\mathcal{B}_n\)-measurable \(h:\mathbb{R}^n\to[0,\infty]\) by taking simple \(s_k\uparrow h\) (2.89), so that \((s_k)_t\uparrow h_t\), and applying 3.11 twice. Since \((f_t)^{\pm}=(f^{\pm})_t\),

\begin{equation*} \int_{\mathbb{R}^n}(f_t)^{\pm}\,d\lambda_n =\frac{1}{|t|^n}\int_{\mathbb{R}^n}f^{\pm}\,d\lambda_n . \end{equation*}

Because \(0<|t|^{-n}<\infty\) and \(\int_{\mathbb{R}^n}f\,d\lambda_n\) is defined, at least one of these is finite, so \(\int_{\mathbb{R}^n}f_t\,d\lambda_n\) is defined (3.18) and subtracting gives

\begin{equation*} \int_{\mathbb{R}^n}f_t\,d\lambda_n=\frac{1}{|t|^n}\int_{\mathbb{R}^n}f\,d\lambda_n . \end{equation*}

Take \(F=\{0\}\times\mathbb{R}\) and \(E=\{0\}\times V\), where \(V\subseteq\mathbb{R}\) is not Lebesgue measurable.

Such a \(V\) exists: 2.18 gives disjoint \(A,B\subseteq\mathbb{R}\) with \(|A\cup B|\ne|A|+|B|\), while outer measure is a measure on \((\mathbb{R},\mathcal{L})\) by 2.72(b) and so is additive on disjoint sets in \(\mathcal{L}\); hence \(A\) and \(B\) are not both in \(\mathcal{L}\). Also \(\lambda\) is \(\sigma\)-finite, since \(\mathbb{R}=\bigcup_{k=1}^{\infty}(-k,k)\) with \(\lambda((-k,k))=2k<\infty\), so \(\lambda\times\lambda\) is defined by 5.25 and is a measure by 5.27.

Clearly \(E\subseteq F\). Since \(\{0\}\) and \(\mathbb{R}\) lie in \(\mathcal{L}\), the set \(F\) is a measurable rectangle, so \(F\in\mathcal{L}\otimes\mathcal{L}\) and, by 5.26,

\begin{equation*} (\lambda\times\lambda)(F)=\lambda(\{0\})\,\lambda(\mathbb{R})=0\cdot\infty=0 . \end{equation*}

But \(E\notin\mathcal{L}\otimes\mathcal{L}\): otherwise 5.6 would give \([E]_0=V\in\mathcal{L}\), contradicting the choice of \(V\).

\(\mathcal{E}_m\) is closed under countable increasing unions and countable decreasing intersections, which by 5.15 is what must be checked.

In the proof of 5.41, \(t>0\) is fixed, \(C_m\) is the open cube in \(\mathbb{R}^n\) centered at the origin with side length \(m\), and

\begin{equation*} \mathcal{E}_m=\{E\in\mathcal{B}_n : E\subseteq C_m \text{ and } \lambda_n(tE)=t^n\lambda_n(E)\}, \end{equation*}

the condition making sense because \(tE\in\mathcal{B}_n\) (first paragraph of that proof). Since \(x\mapsto tx\) is a bijection of \(\mathbb{R}^n\), it carries unions to unions and intersections to intersections and preserves inclusions.

(i) Increasing unions. Let \(E_k\uparrow E\) with \(E_k\in\mathcal{E}_m\). Then \(E\in\mathcal{B}_n\), \(E\subseteq C_m\), and \(tE_k\uparrow tE\), so two applications of 2.59 give

\begin{equation*} \lambda_n(tE)=\lim_{k\to\infty}\lambda_n(tE_k) =t^n\lim_{k\to\infty}\lambda_n(E_k)=t^n\lambda_n(E). \end{equation*}

(ii) Decreasing intersections. Let \(E_k\downarrow E\) with \(E_k\in\mathcal{E}_m\). Then \(E\in\mathcal{B}_n\), \(E\subseteq C_m\), and \(tE_k\downarrow tE\). The finiteness needed for 2.60 holds for both sequences, since by 2.57(a)

\begin{equation*} \lambda_n(E_1)\le\lambda_n(C_m)=m^n<\infty, \qquad \lambda_n(tE_1)\le\lambda_n(tC_m)=( tm)^n<\infty \end{equation*}

(here \(\lambda_n(C_m)=m^n\) by induction on \(n\) from \(\lambda_n=\lambda_{n-1}\times\lambda_1\) and 5.26, the case \(n=1\) being the length of an interval; and \(tC_m\) is the cube of side \(tm\)). So 2.60 gives the same chain of equalities as in (i).

In both cases \(E\in\mathcal{E}_m\), so \(\mathcal{E}_m\) is a monotone class.

Given \(x\) in the open unit ball \(B_n\), the open cube \(B(x,\delta)\) with \(\delta=(1-\|x\|_2)/\sqrt{n}>0\) is contained in \(B_n\), where \(\|y\|_2=(y_1{}^2+\cdots+y_n{}^2)^{1/2}\).

Indeed \(\|y\|_2\le\sqrt{n}\,\|y\|_\infty\), because \(|y_j|\le\|y\|_\infty\) for each \(j\); so \(\|y-x\|_\infty<\delta\) gives

\begin{equation*} \|y-x\|_2\le\sqrt{n}\,\|y-x\|_\infty<\sqrt{n}\,\delta=1-\|x\|_2 . \end{equation*}

The triangle inequality for \(\|\cdot\|_2\), which follows from the Cauchy–Schwarz inequality since

\begin{equation*} \|u+v\|_2{}^2=\|u\|_2{}^2+2\sum_{j=1}^{n}u_jv_j+\|v\|_2{}^2 \le(\|u\|_2+\|v\|_2)^2 , \end{equation*}

then gives \(\|y\|_2\le\|x\|_2+\|y-x\|_2<1\), so \(y\in B_n\). Hence \(B_n\) is open.

If \(G_1\) and \(G_2\) are open, then \(G_1\times G_2\) is open by 5.36 (nonemptiness is not needed here).

Conversely, suppose \(G_1\times G_2\) is open in \(\mathbb{R}^{m+n}\), and recall \(B(x,\delta)\times B(y,\delta)=B((x,y),\delta)\), which holds because \(\|(u,v)\|_\infty=\max\{\|u\|_\infty,\|v\|_\infty\}\). Fix \(y_0\in G_2\), possible since \(G_2\ne\varnothing\). Given \(x\in G_1\), openness supplies \(\delta>0\) with \(B((x,y_0),\delta)\subseteq G_1\times G_2\); then for every \(u\in B(x,\delta)\),

\begin{equation*} (u,y_0)\in B(x,\delta)\times B(y_0,\delta) =B\bigl((x,y_0),\delta\bigr)\subseteq G_1\times G_2 , \end{equation*}

so \(u\in G_1\). Thus \(B(x,\delta)\subseteq G_1\) and \(G_1\) is open. The argument for \(G_2\) is symmetric, using a fixed \(x_0\in G_1\).

If \(F_1\) and \(F_2\) are closed, then

\begin{equation*} \mathbb{R}^{m+n}\setminus(F_1\times F_2) =\bigl[(\mathbb{R}^m\setminus F_1)\times\mathbb{R}^n\bigr] \cup\bigl[\mathbb{R}^m\times(\mathbb{R}^n\setminus F_2)\bigr], \end{equation*}

since \((x,y)\notin F_1\times F_2\) exactly when \(x\notin F_1\) or \(y\notin F_2\); each bracket is open by 5.36, so the complement of \(F_1\times F_2\) is open and \(F_1\times F_2\) is closed. (Nonemptiness is not needed here.)

Conversely, suppose \(F_1\times F_2\) is closed, and fix \(y_0\in F_2\), possible since \(F_2\ne\varnothing\). Given \(x\in\mathbb{R}^m\setminus F_1\), we have \((x,y_0)\notin F_1\times F_2\), so openness of the complement supplies \(\delta>0\) with

\begin{equation*} B\bigl((x,y_0),\delta\bigr)\cap(F_1\times F_2)=\varnothing . \end{equation*}

For \(u\in B(x,\delta)\), the identity \(B(x,\delta)\times B(y_0,\delta)=B((x,y_0),\delta)\) puts \((u,y_0)\) outside \(F_1\times F_2\), and \(y_0\in F_2\) then forces \(u\notin F_1\). Hence \(B(x,\delta)\subseteq\mathbb{R}^m\setminus F_1\), so that complement is open and \(F_1\) is closed. The argument for \(F_2\) is symmetric, using a fixed \(x_0\in F_1\).

(a) Yes. (b) False: for \(m=n=1\) the hyperbola \(E=\{(x,y)\in\mathbb{R}^2 : xy=1\}\) is closed while \(A=\mathbb{R}\setminus\{0\}\) is not.

Throughout we identify \(\mathbb{R}^m\times\mathbb{R}^n\) with \(\mathbb{R}^{m+n}\) and use \(B(x,\delta)\times B(y,\delta)=B((x,y),\delta)\).

(a) Let \(x\in A\), so \((x,y)\in E\) for some \(y\). Openness of \(E\) supplies \(\delta>0\) with

\begin{equation*} B\bigl((x,y),\delta\bigr)=B(x,\delta)\times B(y,\delta)\subseteq E , \end{equation*}

so every \(u\in B(x,\delta)\) has \((u,y)\in E\) and hence lies in \(A\). Thus \(B(x,\delta)\subseteq A\), and \(A\) is open.

(b) The hyperbola is closed: if \(x_0y_0\ne 1\), put \(\varepsilon=|x_0y_0-1|>0\) and choose \(\delta>0\) with \(\delta\le 1\) and \(\delta(|x_0|+|y_0|+1)<\varepsilon\); then \(\|(x,y)-(x_0,y_0)\|_\infty<\delta\) gives \(|y|<|y_0|+1\) and

\begin{equation*} |xy-x_0y_0|\le|y|\,|x-x_0|+|x_0|\,|y-y_0|<\delta(|x_0|+|y_0|+1)<\varepsilon , \end{equation*}

so \(xy\ne 1\). Every \(x\ne 0\) has \((x,1/x)\in E\), while \(0\cdot y=0\ne 1\), so \(A=\mathbb{R}\setminus\{0\}\), whose complement \(\{0\}\) is not open.

For general \(m\) and \(n\), take

\begin{equation*} E=\{(x,y) : x_1y_1=1,\ x_2=\cdots=x_m=0,\ y_2=\cdots=y_n=0\}, \end{equation*}

a finite intersection of closed sets: \(\{(x,y):x_1y_1=1\}\) is closed by the estimate above (since \(\|(x,y)-(x^0,y^0)\|_\infty<\delta\) controls \(|x_1-x_1{}^0|\) and \(|y_1-y_1{}^0|\)), and each \(\{(x,y):x_j=0\}\) is closed because a point with \(x_j\ne 0\) has \(B((x,y),|x_j|)\) in the complement, similarly for \(y_k\). Then

\begin{equation*} A=\{x\in\mathbb{R}^m : x_1\ne 0,\ x_2=\cdots=x_m=0\}, \end{equation*}

which is not closed: the origin is in \(\mathbb{R}^m\setminus A\), yet every \(B(0,\delta)\) contains \((\delta/2,0,\dots,0)\in A\).

(a) The limit is \(0\); (b) the maximum is at \(n = 5\), where \(\lambda_5(B_5) = 8\pi^2/15\).

Both parts run on the recursion established in the proof of 5.44, valid for every integer \(n > 2\), together with the values 5.44 supplies:

\begin{equation*} \lambda_n(B_n) = \frac{2\pi}{n}\,\lambda_{n-2}(B_{n-2}), \end{equation*}

\begin{equation*} \lambda_1(B_1) = 2,\quad \lambda_2(B_2) = \pi,\quad \lambda_3(B_3) = \frac{4\pi}{3}, \end{equation*}

\begin{equation*} \lambda_4(B_4) = \frac{\pi^2}{2},\quad \lambda_5(B_5) = \frac{8\pi^2}{15}, \quad \lambda_6(B_6) = \frac{\pi^3}{6}. \end{equation*}

(a) Put \(r = 2\pi/7\), so \(0 < r < 1\) because \(\pi < 3.5\). If \(n \ge 5\) then \(n + 2 \ge 7\), so

\begin{equation*} \lambda_{n+2}(B_{n+2}) = \frac{2\pi}{n+2}\,\lambda_n(B_n) \le r\,\lambda_n(B_n). \end{equation*}

Iterating from \(n = 5\) and from \(n = 6\), and setting \(M = \max\{\lambda_5(B_5), \lambda_6(B_6)\}\), we get for all \(n \ge 5\)

\begin{equation*} 0 \le \lambda_n(B_n) \le M\,r^{\,(n-6)/2}, \end{equation*}

since \(n = 5 + 2j\) or \(n = 6 + 2j\) with \(j = \lfloor (n-5)/2 \rfloor \ge (n-6)/2\). As \(0 < r < 1\), the squeeze theorem gives \(\lim_{n \to \infty} \lambda_n(B_n) = 0\).

(b) For \(n \ge 5\) we have \(n + 2 \ge 7 > 2\pi\), so the recursion and \(\lambda_n(B_n) > 0\) give \(\lambda_{n+2}(B_{n+2}) < \lambda_n(B_n)\); each parity chain therefore decreases strictly from its start at \(n = 5\) and \(n = 6\), and the supremum is the largest of the first six values. Using \(3 < \pi < 3.2\), and abbreviating \(\lambda_j = \lambda_j(B_j)\):

\begin{equation*} \frac{\lambda_2}{\lambda_1} = \frac{\pi}{2} > 1,\qquad \frac{\lambda_3}{\lambda_2} = \frac43 > 1,\qquad \frac{\lambda_4}{\lambda_3} = \frac{3\pi}{8} > \frac98 > 1, \end{equation*}

\begin{equation*} \frac{\lambda_5}{\lambda_4} = \frac{16}{15} > 1,\qquad \frac{\lambda_5}{\lambda_6} = \frac{48}{15\pi} = \frac{16}{5\pi} > 1 \ \text{ since } 5\pi < 16 . \end{equation*}

Thus \(\lambda_5(B_5) = 8\pi^2/15\) is the unique maximum.

Induct in steps of \(2\), using the recursion \(\lambda_n(B_n) = \frac{2\pi}{n}\lambda_{n-2}(B_{n-2})\) for \(n > 2\) from the proof of 5.44 and the gamma identities \(\Gamma(x+1) = x\Gamma(x)\), \(\Gamma(1) = 1\), \(\Gamma(\frac12) = \sqrt{\pi}\).

The two base cases, \(n = 1\) and \(n = 2\), start the two parity chains that together reach every positive integer:

\begin{equation*} \frac{\pi^{1/2}}{\Gamma\bigl(\frac32\bigr)} = \frac{\sqrt{\pi}}{\frac12\sqrt{\pi}} = 2 = \lambda_1(B_1), \qquad \frac{\pi}{\Gamma(2)} = \frac{\pi}{1} = \pi = \lambda_2(B_2). \end{equation*}

For \(n > 2\), the recursion, the induction hypothesis for \(n-2\), and \(\Gamma(\frac{n}{2}+1) = \frac{n}{2}\Gamma(\frac{n}{2})\) give

\begin{equation*} \lambda_n(B_n) = \frac{2\pi}{n} \cdot \frac{\pi^{(n-2)/2}}{\Gamma\bigl(\frac{n}{2}\bigr)} = \frac{2\,\pi^{n/2}}{n\,\Gamma\bigl(\frac{n}{2}\bigr)} = \frac{\pi^{n/2}}{\Gamma\bigl(\frac{n}{2} + 1\bigr)}, \end{equation*}

completing the induction.

(a) On the open set \(G = \mathbb{R}^2 \setminus \{(0,0)\}\) the formula \(f(x,y) = (x^3y - xy^3)/(x^2+y^2)\) is a quotient of polynomials with nonvanishing denominator, hence infinitely differentiable there; so both mixed partials exist at every point of \(G\), and the quotient rule gives

\begin{equation*} (D_1 f)(x,y) = \frac{x^4 y + 4x^2 y^3 - y^5}{(x^2+y^2)^2}, \qquad (D_2 f)(x,y) = \frac{x^5 - 4x^3 y^2 - x y^4}{(x^2+y^2)^2} \end{equation*}

on \(G\). (Check!) Since \(f(t,0) = f(0,t) = 0\) for all \(t\), the difference quotients at the origin give \((D_1 f)(0,0) = (D_2 f)(0,0) = 0\), so \(D_1 f\) and \(D_2 f\) exist on all of \(\mathbb{R}^2\), and the displayed formulas evaluated on the axes yield

\begin{equation*} (D_1 f)(0,y) = -y \quad (y \in \mathbb{R}), \qquad (D_2 f)(x,0) = x \quad (x \in \mathbb{R}), \end{equation*}

the value \(0\) at the origin being consistent with both. Hence the two mixed partials also exist at the origin:

\begin{equation*} \bigl(D_1(D_2 f)\bigr)(0,0) = \lim_{t \to 0} \frac{(D_2 f)(t,0) - 0}{t} = \lim_{t \to 0} \frac{t}{t} = 1, \end{equation*}

\begin{equation*} \bigl(D_2(D_1 f)\bigr)(0,0) = \lim_{t \to 0} \frac{(D_1 f)(0,t) - 0}{t} = \lim_{t \to 0} \frac{-t}{t} = -1 . \end{equation*}

(b) Those two limits are \(1\) and \(-1\), so the mixed partials differ at the origin.

(c) Result 5.48 requires the four functions \(D_1 f\), \(D_2 f\), \(D_1(D_2 f)\), \(D_2(D_1 f)\) to be continuous on the open set, not merely to exist there. All four are continuous on \(G\), where \(f\) is smooth, so 5.48 does apply on \(G\) and gives \(D_1(D_2 f) = D_2(D_1 f)\) there; if both mixed partials were also continuous at \((0,0)\), letting \((x,y) \to (0,0)\) within \(G\) in that identity would force equality of the two values at the origin, contradicting (b). So the continuity hypothesis of 5.48 fails on every open set containing \((0,0)\).

In each of the five examples the first three conditions of 6.1 follow at once from the corresponding property of \(|\cdot|\), coordinatewise or termwise. (Check!) What remains is finiteness of \(d\) and the triangle inequality.

(i) Discrete metric on a nonempty set \(V\). The values lie in \(\{0,1\}\). If \(f = h\) then \(d(f,h) = 0 \le d(f,g) + d(g,h)\); if \(f \ne h\) then \(g\) differs from at least one of \(f, h\), so \(d(f,g) + d(g,h) \ge 1 = d(f,h)\).

(ii) \(\mathbb{R}\) with \(d(x,y) = |x-y|\).

\begin{equation*} d(x,z) = |(x-y) + (y-z)| \le |x-y| + |y-z| = d(x,y) + d(y,z). \end{equation*}

(iii) \(\mathbb{R}^n\) with the max metric. Finite, as a maximum of \(n\) nonnegative reals. For each \(j\),

\begin{equation*} |x_j - z_j| \le |x_j - y_j| + |y_j - z_j| \le d(x,y) + d(y,z), \end{equation*}

and the right side is independent of \(j\), so the maximum of the left side over \(j\) obeys the same bound.

(iv) \(C([0,1])\) with the sup metric. Finite because \(|f-g|\) is continuous on the closed bounded interval \([0,1]\), hence bounded. For each \(t \in [0,1]\),

\begin{equation*} |f(t) - h(t)| \le |f(t) - g(t)| + |g(t) - h(t)| \le d(f,g) + d(g,h), \end{equation*}

so \(d(f,g) + d(g,h)\) is an upper bound for the set whose supremum is \(d(f,h)\).

(v) \(\ell^1\). Finite because \(|a_k - b_k| \le |a_k| + |b_k|\) termwise gives \(d(a,b) \le \sum_k |a_k| + \sum_k |b_k| < \infty\). For every \(n \in \mathbb{Z}^+\),

\begin{equation*} \sum_{k=1}^{n} |a_k - c_k| \le \sum_{k=1}^{n} |a_k - b_k| + \sum_{k=1}^{n} |b_k - c_k| \le d(a,b) + d(b,c), \end{equation*}

the last step because a nonnegative series dominates its partial sums; let \(n \to \infty\) to get \(d(a,c) \le d(a,b) + d(b,c)\).

For \(E = \{f_1,\dots,f_n\}\) finite in a metric space \((V,d)\) and \(g \in V \setminus E\), the witness is

\begin{equation*} r = \min\{d(g,f_1),\dots,d(g,f_n)\}, \end{equation*}

which is positive because \(g \ne f_j\) forces \(d(g,f_j) > 0\) for each \(j\) (second bullet of 6.1, contrapositive, with \(d \ge 0\)) and the minimum is over finitely many terms. If some \(h \in B(g,r)\) lay in \(E\), say \(h = f_j\), then \(d(g,f_j) = d(g,h) < r \le d(g,f_j)\), a contradiction; hence \(B(g,r) \subseteq V \setminus E\). So \(V \setminus E\) is open by 6.4 (for \(E = \emptyset\) take \(r = 1\)), and \(E\) is closed by 6.6.

For \(h\) outside the closed ball \(\overline{B}(f,r) = \{g : d(f,g) \le r\}\) of 6.3, the witness is \(s = d(f,h) - r > 0\). If \(g \in B(h,s)\), then the triangle inequality and the symmetry of \(d\) (6.1) give

\begin{equation*} d(f,g) \ge d(f,h) - d(g,h) > d(f,h) - s = r, \end{equation*}

so \(g \notin \overline{B}(f,r)\). Thus \(B(h,s) \subseteq V \setminus \overline{B}(f,r)\), the complement is open by 6.4, and \(\overline{B}(f,r)\) is closed by 6.6.

Both parts produce the ball required by 6.4 at an arbitrary point of the set.

(a) Let \(G = \bigcup_{A \in \mathcal{A}} A\) with each \(A \in \mathcal{A}\) open, and let \(f \in G\). Choose \(A \in \mathcal{A}\) with \(f \in A\); openness of \(A\) gives \(r > 0\) with

\begin{equation*} B(f,r) \subseteq A \subseteq G . \end{equation*}

Hence \(G\) is open.

(b) Let \(G = G_1 \cap \cdots \cap G_n\) with each \(G_j\) open, and let \(f \in G\). Openness of \(G_j\) gives \(r_j > 0\) with \(B(f,r_j) \subseteq G_j\), and \(r = \min\{r_1,\dots,r_n\} > 0\) because the collection is finite. If \(g \in B(f,r)\) and \(j \in \{1,\dots,n\}\), then \(d(f,g) < r \le r_j\), so

\begin{equation*} B(f,r) \subseteq B(f,r_j) \subseteq G_j \quad \text{for every } j, \end{equation*}

whence \(B(f,r) \subseteq G\) and \(G\) is open.

Both parts are Exercise 6A.4 read through complements, using 6.6 (a set is closed exactly when its complement is open) and De Morgan’s laws.

(a) For \(\mathcal{F}\) a collection of closed subsets of \(V\) and \(F = \bigcap_{A \in \mathcal{F}} A\),

\begin{equation*} V \setminus F = \bigcup_{A \in \mathcal{F}} (V \setminus A), \end{equation*}

a union of open sets, hence open by 6A.4(a). So \(F\) is closed.

(b) For closed \(F_1,\dots,F_n\) and \(F = F_1 \cup \cdots \cup F_n\),

\begin{equation*} V \setminus F = \bigcap_{j=1}^{n} (V \setminus F_j), \end{equation*}

an intersection of finitely many open sets, hence open by 6A.4(b). So \(F\) is closed.

(a) The closed ball \(\overline{B}(f,r)\) is a closed set containing \(B(f,r)\) (closed by 6A.3, containing it because \(d(f,g) < r\) implies \(d(f,g) \le r\)), and by 6.9(b) the closure \(\overline{B(f,r)}\) is the intersection of all such sets; hence \(\overline{B(f,r)} \subseteq \overline{B}(f,r)\).

Method (2): If \(g \in \overline{B(f,r)}\) and \(\varepsilon > 0\), pick \(h \in B(g,\varepsilon) \cap B(f,r)\) (6.7); then the triangle inequality and symmetry (6.1) give

\begin{equation*} d(f,g) \le d(f,h) + d(h,g) < r + \varepsilon . \end{equation*}

Letting \(\varepsilon \downarrow 0\) gives \(d(f,g) \le r\).

(b) Take \(V = \{0,1\}\) with the discrete metric of 6.2, \(f = 0\), \(r = 1\). Then \(d(0,1) = 1\) is not less than \(1\) but is at most \(1\), so

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

The finite set \(\{0\}\) is closed by 6A.2, hence equals its own closure by 6.9(d), so \(\overline{B(0,1)} = \{0\} \ne \overline{B}(0,1)\).

Two limits of the same sequence are at distance \(0\), hence equal. Suppose \(f_k \to f\) and \(f_k \to g\) in a metric space \((V,d)\), and let \(\varepsilon > 0\). By 6.8 both \(d(f_k,f) \to 0\) and \(d(f_k,g) \to 0\), so some single index \(n\) (the larger of the two the definitions supply) satisfies \(d(f_n,f) < \varepsilon/2\) and \(d(f_n,g) < \varepsilon/2\); the triangle inequality with the symmetry of \(d\) (6.1) then gives

\begin{equation*} d(f,g) \le d(f,f_n) + d(f_n,g) < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

Thus \(0 \le d(f,g) < \varepsilon\) for every \(\varepsilon > 0\), forcing \(d(f,g) = 0\) and hence \(f = g\) by the second bullet of 6.1.

Throughout, \(\overline{E} = \{g \in V : B(g,\varepsilon) \cap E \neq \emptyset \text{ for every } \varepsilon > 0\}\) is the closure as defined in 6.7, and \(E \subseteq \overline{E}\) because \(f \in B(f,\varepsilon) \cap E\) for every \(f \in E\) and \(\varepsilon > 0\). The parts are proved in the order (a), (c), (b), (d), (e).

(a) If \(g \in \overline{E}\), choose \(f_k \in B(g,1/k) \cap E\) for each \(k \in \mathbb{Z}^+\); then \(d(f_k,g) < 1/k \to 0\), so \(f_k \to g\) by 6.8. Conversely, if \(f_1, f_2, \ldots\) lie in \(E\) with \(f_k \to g\), then each \(\varepsilon > 0\) admits a \(k\) with \(d(f_k,g) < \varepsilon\), so \(f_k \in B(g,\varepsilon) \cap E\) and \(g \in \overline{E}\).

(c) Suppose \(g \in V \setminus \overline{E}\), and take \(\varepsilon > 0\) with \(B(g,\varepsilon) \cap E = \emptyset\). For \(h \in B(g,\varepsilon)\) put \(\delta = \varepsilon - d(g,h) > 0\); the triangle inequality gives

\begin{equation*} d(g,u) \leq d(g,h) + d(h,u) < d(g,h) + \delta = \varepsilon \qquad (u \in B(h,\delta)), \end{equation*}

so \(B(h,\delta) \subseteq B(g,\varepsilon)\) misses \(E\) and \(h \notin \overline{E}\). Hence \(B(g,\varepsilon) \subseteq V \setminus \overline{E}\), so \(V \setminus \overline{E}\) is open and \(\overline{E}\) is closed by 6.6.

(b) First, \((\ast)\): every closed \(F \supseteq E\) contains \(\overline{E}\). Indeed, for \(g \in V \setminus F\) the openness of \(V \setminus F\) gives \(\varepsilon > 0\) with \(B(g,\varepsilon) \subseteq V \setminus F \subseteq V \setminus E\), so \(g \notin \overline{E}\). Thus \(\overline{E}\) is contained in the intersection \(C\) of all closed sets containing \(E\); and \(\overline{E}\) is itself one of those sets, by (c) and \(E \subseteq \overline{E}\), so \(C \subseteq \overline{E}\). Hence \(C = \overline{E}\).

(d) If \(E = \overline{E}\), then \(E\) is closed by (c). If \(E\) is closed, then \((\ast)\) with \(F = E\) gives \(\overline{E} \subseteq E \subseteq \overline{E}\).

(e) If \(E\) is closed and \(f_1, f_2, \ldots \in E\) converge to \(g\), then \(g \in \overline{E} = E\) by (a) and (d). Conversely, if \(E\) contains every such limit, then (a) gives \(\overline{E} \subseteq E\), so \(E = \overline{E}\) and \(E\) is closed by (d).

For \(G\) open in \(V\), take the sets of points at distance at least \(1/k\) from the complement:

\begin{equation*} F_k = V \setminus \bigcup_{h \in V \setminus G} B\!\left(h, \tfrac{1}{k}\right) = \left\{ f \in V : d(f,h) \geq \tfrac{1}{k} \text{ for all } h \in V \setminus G \right\} \end{equation*}

for \(k \in \mathbb{Z}^+\) (so \(F_k = V\) when \(G = V\), the union being empty).

Each \(F_k\) is closed, since each \(B(h,1/k)\) is open by 6.5, their union is open by 6A.4(a), and 6.6 applies. Each \(F_k \subseteq G\), since \(f \notin G\) gives \(f \in B(f,1/k)\) and hence \(f \notin F_k\). Conversely, if \(f \in G\), choose \(r > 0\) with \(B(f,r) \subseteq G\) and then \(k\) with \(1/k \le r\): were \(f \in B(h,1/k)\) for some \(h \in V \setminus G\), then \(d(f,h) < 1/k \le r\) would put \(h \in B(f,r) \subseteq G\), a contradiction, so \(f \in F_k\). Hence \(G = \bigcup_{k=1}^{\infty} F_k\).

The statement is true.

Closure is order preserving: if \(E_1 \subseteq E_2\) and \(g \in \overline{E_1}\), then \(\emptyset \neq B(g,\varepsilon) \cap E_1 \subseteq B(g,\varepsilon) \cap E_2\) for every \(\varepsilon > 0\), so \(g \in \overline{E_2}\) by 6.7. Applying this to \(U \subseteq U \cup W\) and \(W \subseteq U \cup W\) gives \(\overline{U} \cup \overline{W} \subseteq \overline{U \cup W}\).

For the reverse inclusion, suppose \(g \notin \overline{U} \cup \overline{W}\) and pick \(\varepsilon_1, \varepsilon_2 > 0\) with \(B(g,\varepsilon_1) \cap U = \emptyset\) and \(B(g,\varepsilon_2) \cap W = \emptyset\). With \(\varepsilon = \min\{\varepsilon_1,\varepsilon_2\} > 0\),

\begin{equation*} B(g,\varepsilon) \cap (U \cup W) = \bigl(B(g,\varepsilon) \cap U\bigr) \cup \bigl(B(g,\varepsilon) \cap W\bigr) = \emptyset , \end{equation*}

so \(g \notin \overline{U \cup W}\). Hence \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W}\), and the two inclusions give equality.

Method (2) for the reverse inclusion: \(\overline{U} \cup \overline{W}\) is closed, being a union of two closed sets (6.9(c) and 6A.5(b)), and it contains \(U \cup W\); now apply 6.9(b).

False: in \(V = \mathbb{R}\) with \(d(x,y) = |x-y|\), take

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

Here \(U \cap W = \emptyset\), and \(\overline{\emptyset} = \emptyset\) because no ball meets \(\emptyset\) (6.7), so \(\overline{U \cap W} = \emptyset\). But \(1 \in \overline{U} \cap \overline{W}\): given \(\varepsilon > 0\), the points

\begin{equation*} t = \max\left\{\tfrac12,\, 1 - \tfrac{\varepsilon}{2}\right\} \in U, \qquad s = \min\left\{\tfrac32,\, 1 + \tfrac{\varepsilon}{2}\right\} \in W \end{equation*}

both lie in \(B(1,\varepsilon)\), since \(|1-t| \le \varepsilon/2 < \varepsilon\) and \(|s-1| \le \varepsilon/2 < \varepsilon\). (Check!) Hence \(1 \in \overline{U} \cap \overline{W}\) while \(1 \notin \overline{U \cap W}\).

(a) Fix \(f \in U\) and \(\varepsilon > 0\). Continuity of \(S\) at \(T(f)\) supplies \(\gamma > 0\) with \(d_W(S(T(f)),S(v)) < \varepsilon\) whenever \(d_V(T(f),v) < \gamma\), and continuity of \(T\) at \(f\) then supplies (with \(\gamma\) as the tolerance) \(\delta > 0\) with \(d_V(T(f),T(g)) < \gamma\) whenever \(d_U(f,g) < \delta\). Taking \(v = T(g)\), we get for all \(g \in U\) with \(d_U(f,g) < \delta\)

\begin{equation*} d_W\bigl((S \circ T)(f), (S \circ T)(g)\bigr) = d_W\bigl(S(T(f)), S(T(g))\bigr) < \varepsilon , \end{equation*}

so \(S \circ T\) is continuous at \(f\) by 6.10, and \(f\) was arbitrary.

(b) If \(f_k \to f\) in \(U\), then 6.11(a) \(\Rightarrow\) (b) applied to the continuous \(T\) gives \(T(f_k) \to T(f)\) in \(V\), and the same implication applied to the continuous \(S\) with the convergent sequence \(T(f_1), T(f_2), \ldots\) gives

\begin{equation*} \lim_{k \to \infty} (S \circ T)(f_k) = \lim_{k \to \infty} S\bigl(T(f_k)\bigr) = S\bigl(T(f)\bigr) = (S \circ T)(f). \end{equation*}

So \(S \circ T\) satisfies 6.11(b), hence is continuous by 6.11(b) \(\Rightarrow\) (a).

(c) For every \(G \subseteq W\) we have \((S \circ T)^{-1}(G) = T^{-1}(S^{-1}(G))\), since \(S(T(f)) \in G\) exactly when \(T(f) \in S^{-1}(G)\). If \(G\) is open, then 6.11(a) \(\Rightarrow\) (c) applied to \(S\) makes \(S^{-1}(G)\) open in \(V\), and the same implication applied to \(T\) with that open set makes

\begin{equation*} (S \circ T)^{-1}(G) = T^{-1}\bigl(S^{-1}(G)\bigr) \end{equation*}

open in \(U\). So \(S \circ T\) satisfies 6.11(c), hence is continuous by 6.11(c) \(\Rightarrow\) (a).

Since the text gives (b) \(\Rightarrow\) (d) and (c) \(\Leftrightarrow\) (d), only (a) \(\Rightarrow\) (b) and (c) \(\Rightarrow\) (a) are missing; with them the cycle (a) \(\Rightarrow\) (b) \(\Rightarrow\) (d) \(\Rightarrow\) (c) \(\Rightarrow\) (a) makes all four equivalent. Write \(d_V, d_W\) for the two metrics.

(a) \(\Rightarrow\) (b). Suppose \(T\) is continuous, \(f_k \to f\) in \(V\), and \(\varepsilon > 0\). Continuity at \(f\) gives \(\delta > 0\) with \(d_W(T(f),T(g)) < \varepsilon\) whenever \(d_V(f,g) < \delta\) (6.10), and \(d_V(f_k,f) \to 0\) (6.8) gives \(n \in \mathbb{Z}^+\) with \(d_V(f_k,f) < \delta\) for all \(k \ge n\). Hence

\begin{equation*} d_W\bigl(T(f_k), T(f)\bigr) < \varepsilon \quad \text{for all } k \ge n , \end{equation*}

so \(d_W(T(f_k),T(f)) \to 0\), that is, \(T(f_k) \to T(f)\).

(c) \(\Rightarrow\) (a). Let \(f \in V\) and \(\varepsilon > 0\). The ball \(B(T(f),\varepsilon)\) is open in \(W\) by 6.5, so

\begin{equation*} G = T^{-1}\bigl(B(T(f),\varepsilon)\bigr) \end{equation*}

is open in \(V\) by (c), and \(f \in G\). By 6.4 there is \(\delta > 0\) with \(B(f,\delta) \subseteq G\); then \(d_V(f,g) < \delta\) forces \(T(g) \in B(T(f),\varepsilon)\), that is, \(d_W(T(f),T(g)) < \varepsilon\). So \(T\) is continuous at each \(f \in V\).

The Cauchy sequence converges to the limit \(f\) of its subsequence \(f_{k_1}, f_{k_2}, \ldots\), where \(k_1 < k_2 < \cdots\); note \(k_m \ge m\) for every \(m\) (induction: \(k_1 \ge 1\), and \(k_{m+1} > k_m \ge m\) forces \(k_{m+1} \ge m+1\)).

Let \(\varepsilon > 0\). Being Cauchy (6.12) gives \(n \in \mathbb{Z}^+\) with \(d(f_p,f_q) < \varepsilon/2\) for all \(p,q \ge n\), and \(d(f_{k_m},f) \to 0\) (6.8) gives \(M\) with \(d(f_{k_m},f) < \varepsilon/2\) for all \(m \ge M\). Fix an integer \(m \ge \max\{M,n\}\) and put \(j = k_m\), so that \(j \ge m \ge n\) and \(d(f_j,f) < \varepsilon/2\). Then for every \(k \ge n\),

\begin{equation*} d(f_k, f) \leq d(f_k, f_j) + d(f_j, f) < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon , \end{equation*}

the first term being bounded by the Cauchy estimate since \(k, j \ge n\). Hence \(d(f_k,f) \to 0\), which by 6.8 says \(f_k \to f\).

Each space is complete; (ii) is proved from the least-upper-bound property of \(\mathbb{R}\) and then drives (iii), (iv), (v).

(i) Discrete metric. Taking \(\varepsilon = 1\) in 6.12 gives \(n\) with \(d(f_j,f_k) < 1\), hence \(d(f_j,f_k) = 0\) and \(f_j = f_k\), for all \(j,k \ge n\). The sequence is eventually constant, so \(f_k \to f_n \in V\).

(ii) \(\mathbb{R}\). A Cauchy sequence \(x_1, x_2, \dots\) is bounded: \(\varepsilon = 1\) in 6.12 gives \(n\) with \(|x_k| \le |x_n| + 1\) for \(k \ge n\), so \(|x_k| \le M\) for all \(k\), where \(M = \max\{|x_1|,\dots,|x_{n-1}|,|x_n|+1\}\). Hence \(s_m = \sup\{x_k : k \ge m\}\) exists for each \(m\), with \(s_1 \ge s_2 \ge \cdots \ge -M\), so \(L = \inf_m s_m\) exists. Given \(\varepsilon > 0\), take \(n\) with \(|x_p - x_q| < \varepsilon/2\) for \(p,q \ge n\). Then \(x_k \le x_n + \varepsilon/2\) for \(k \ge n\) gives \(L \le s_n \le x_n + \varepsilon/2\), while \(s_m \ge x_m > x_n - \varepsilon/2\) for \(m \ge n\) together with the monotonicity of \((s_m)\) gives \(L = \inf_{m \ge n} s_m \ge x_n - \varepsilon/2\); so \(|x_n - L| \le \varepsilon/2\) and

\begin{equation*} |x_j - L| \leq |x_j - x_n| + |x_n - L| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon \qquad (j \ge n). \end{equation*}

(iii) \(\mathbb{R}^n\), max metric. Each coordinate sequence is Cauchy, since \(|x^{(j)}_i - x^{(k)}_i| \le d(x^{(j)},x^{(k)})\), so converges to some \(y_i\) by (ii). Given \(\varepsilon > 0\), let \(N\) be the largest of the \(n\) thresholds \(m_i\) beyond which \(|x^{(k)}_i - y_i| < \varepsilon\); then for \(k \ge N\),

\begin{equation*} d\big(x^{(k)}, y\big) = \max_{1 \leq i \leq n} |x^{(k)}_i - y_i| < \varepsilon . \end{equation*}

(iv) \(C([0,1])\), sup metric. Since \(|f_j(t) - f_k(t)| \le d(f_j,f_k)\), each \(f_1(t), f_2(t), \dots\) is Cauchy in \(\mathbb{R}\); put \(f(t) = \lim_{k} f_k(t)\), which exists by (ii). Given \(\varepsilon > 0\), take \(n\) with \(d(f_j,f_k) < \varepsilon/3\) for \(j,k \ge n\); letting \(j \to \infty\) in \(|f_j(t) - f_k(t)| < \varepsilon/3\) gives

\begin{equation*} |f(t) - f_k(t)| \leq \tfrac{\varepsilon}{3} \qquad (t \in [0,1],\ k \ge n). \end{equation*}

Moreover \(f\) is continuous: given \(t\), continuity of \(f_n\) at \(t\) gives \(\delta > 0\) with \(|f_n(t)-f_n(s)| < \varepsilon/3\) whenever \(|s-t| < \delta\), and then

\begin{equation*} |f(t) - f(s)| \leq |f(t) - f_n(t)| + |f_n(t) - f_n(s)| + |f_n(s) - f(s)| < \varepsilon . \end{equation*}

So \(f \in C([0,1])\), and the previous display gives \(d(f_k,f) \le \varepsilon/3 < \varepsilon\) for \(k \ge n\).

(v) \(\ell^1\). Since \(|a^{(j)}_k - a^{(m)}_k| \le d(a^{(j)},a^{(m)})\), each coordinate sequence is Cauchy and converges by (ii) to some \(b_k\); set \(b = (b_1,b_2,\dots)\). Given \(\varepsilon > 0\), take \(n\) with \(d(a^{(j)},a^{(m)}) < \varepsilon/2\) for \(j,m \ge n\). For \(m \ge n\) and \(N \in \mathbb{Z}^+\), letting \(j \to \infty\) in the finite sum \(\sum_{k=1}^{N} |a^{(j)}_k - a^{(m)}_k| \le \varepsilon/2\) gives \(\sum_{k=1}^{N} |b_k - a^{(m)}_k| \le \varepsilon/2\), and taking the supremum over \(N\) gives

\begin{equation*} d\big(a^{(m)}, b\big) = \sum_{k=1}^{\infty} |b_k - a^{(m)}_k| \leq \tfrac{\varepsilon}{2} < \varepsilon \qquad (m \ge n). \end{equation*}

With \(m = n\) this puts \(b - a^{(n)}\) in \(\ell^1\), so \(b \in \ell^1\) because \(a^{(n)} \in \ell^1\); the display then says \(a^{(m)} \to b\) in \(\ell^1\).

Completeness of \(\mathbb{R}\) (6.15, verified in 6A.15) is used throughout.

(a) The three properties hold termwise, from the first, third, and fourth bullets of 6.1: \(d(f_k,f_k) = 0\); \(d(g_k,f_k) = d(f_k,g_k)\); and

\begin{equation*} 0 \leq d(f_k,h_k) \leq d(f_k,g_k) + d(g_k,h_k) \to 0 \end{equation*}

when \((f_k) \equiv (g_k) \equiv (h_k)\).

(b) Everything follows from the quadrilateral inequality, which two applications of the triangle inequality give for \(f, f^{\prime}, g, g^{\prime} \in U\):

\begin{equation*} \big| d(f,g) - d(f^{\prime},g^{\prime}) \big| \leq d(f,f^{\prime}) + d(g,g^{\prime}) . \end{equation*}

Existence. Taking \(f = f_j\), \(f^{\prime} = f_k\), \(g = g_j\), \(g^{\prime} = g_k\) and choosing \(n\) beyond which \(d(f_j,f_k) < \varepsilon/2\) and \(d(g_j,g_k) < \varepsilon/2\) (both sequences being Cauchy) shows \(d(f_1,g_1), d(f_2,g_2), \dots\) is Cauchy in \(\mathbb{R}\), hence convergent, with limit in \([0,\infty)\).

Representatives. Taking \(f = f_k\), \(f^{\prime} = f^{\prime}_k\), \(g = g_k\), \(g^{\prime} = g^{\prime}_k\) makes the right side tend to \(0\) when \((f_k) \equiv (f^{\prime}_k)\) and \((g_k) \equiv (g^{\prime}_k)\), so the two limits agree and \(d_V\) is well defined.

Metric axioms. Write \(F, G, H\) for the classes of \((f_k), (g_k), (h_k)\). Then \(d_V(F,F) = \lim_k d(f_k,f_k) = 0\); \(d_V(F,G) = 0\) says precisely \((f_k) \equiv (g_k)\), that is, \(F = G\); symmetry is termwise; and since all three limits exist, letting \(k \to \infty\) in \(d(f_k,h_k) \le d(f_k,g_k) + d(g_k,h_k)\) gives

\begin{equation*} d_V(F,H) \leq d_V(F,G) + d_V(G,H) . \end{equation*}

(c) Write \(\iota(f) = \widehat{(f,f,f,\dots)}\), so \(d_V(\iota(f),\iota(g)) = d(f,g)\) by (d). Let \(F^{(1)}, F^{(2)}, \dots\) be Cauchy in \(V\), with representatives \((f^{(m)}_1, f^{(m)}_2, \dots) \in W\). For each \(m\) pick \(k_m\) with \(d(f^{(m)}_j, f^{(m)}_k) < 1/m\) for all \(j,k \ge k_m\) and set \(h_m = f^{(m)}_{k_m}\); since \(d(f^{(m)}_k, h_m) < 1/m\) for \(k \ge k_m\),

\begin{equation*} d_V\big(F^{(m)}, \iota(h_m)\big) = \lim_{k \to \infty} d\big(f^{(m)}_k, h_m\big) \leq \tfrac{1}{m} . \end{equation*}

Then \(h_1, h_2, \dots\) is Cauchy in \(U\): by (d) and the triangle inequality in \(V\),

\begin{equation*} d(h_j, h_m) = d_V\big(\iota(h_j), \iota(h_m)\big) \leq \tfrac{1}{j} + d_V\big(F^{(j)}, F^{(m)}\big) + \tfrac{1}{m}, \end{equation*}

which is \(< \varepsilon\) for \(j,m \ge n\) once \(n \ge n_0\) (where \(d_V(F^{(j)},F^{(m)}) < \varepsilon/3\) for \(j,m \ge n_0\)) and \(n > 3/\varepsilon\). Put \(H = \widehat{(h_1,h_2,\dots)}\). Given \(\varepsilon > 0\), take \(n_1\) with \(d(h_m,h_k) < \varepsilon/3\) for \(m,k \ge n_1\); letting \(k \to \infty\) gives \(d_V(\iota(h_m),H) \le \varepsilon/3\) for \(m \ge n_1\), so for every integer \(m \ge n_1\) with \(m > 3/\varepsilon\),

\begin{equation*} d_V\big(F^{(m)}, H\big) \leq d_V\big(F^{(m)}, \iota(h_m)\big) + d_V\big(\iota(h_m), H\big) \leq \tfrac{1}{m} + \tfrac{\varepsilon}{3} < \varepsilon . \end{equation*}

Hence \(F^{(m)} \to H\), and \((V,d_V)\) is complete by 6.14.

(d) The constant sequence \((f,f,f,\dots)\) is Cauchy, so \(\iota\) is defined on all of \(U\), and the real sequence with \(k\)-th term \(d(f,g)\) is constant, so

\begin{equation*} d_V\big(\widehat{(f,f,f,\dots)}, \widehat{(g,g,g,\dots)}\big) = \lim_{k \to \infty} d(f,g) = d(f,g) . \end{equation*}

(e) Distance preservation makes \(\iota\) injective (if \(\iota(f) = \iota(g)\) then \(d(f,g) = 0\), so \(f = g\) by the second bullet of 6.1), hence a bijection of \(U\) onto \(\iota(U) \subseteq V\) carrying \(d\) to the metric \(\iota(U)\) inherits from \(d_V\). Identifying each \(f\) with \(\iota(f)\) exhibits \(U\), with its own metric, as a subset of the complete space \(V\) of (c).

Set \(w = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}\,i\), which makes sense because \(a + bi \ne 0\) forces \(a^2 + b^2 > 0\). The multiplication rule of 6.17 gives

\begin{equation*} (a + bi)w = \frac{a^2 + b^2}{a^2 + b^2} + \frac{-ab + ab}{a^2 + b^2}\, i = 1 , \end{equation*}

and the multiplicative inverse of a nonzero element of the field \(\mathbb{C}\) is unique, so \(1/(a+bi) = w\).

Method (2): By 6.24, \(z\bar z = |z|^2\); taking \(z = a + bi\), so \(|z|^2 = a^2 + b^2 \ne 0\), and dividing by \(z|z|^2\) gives

\begin{equation*} \frac{1}{a + bi} = \frac{\overline{a + bi}}{|a + bi|^2} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}\, i . \end{equation*}

Both inequalities are one squared chain. Write \(z = a + bi\) with \(a, b \in \mathbb{R}\), so \(|z| = \sqrt{a^2+b^2}\) by 6.18, and put \(M = \max\{|a|,|b|\}\). Then

\begin{equation*} M^2 = \max\{a^2, b^2\} \leq a^2 + b^2 = |z|^2 \leq 2M^2 = \bigl(\sqrt{2}\,M\bigr)^2 , \end{equation*}

the first equality because \(t \mapsto t^2\) is increasing on \([0,\infty)\) and the last inequality because \(a^2, b^2 \le M^2\). All the quantities squared are nonnegative, so taking square roots preserves the order and gives \(M \le |z| \le \sqrt{2}\,M\).

Write \(z = a + bi\) with \(a, b \in \mathbb{R}\), so \(|z| = \sqrt{a^2+b^2}\) by 6.18. Since \(2|a|\,|b| \ge 0\), and since \((|a|-|b|)^2 \ge 0\) expands to \(2|a|\,|b| \le a^2 + b^2\),

\begin{equation*} |z|^2 = a^2 + b^2 \leq \bigl(|a| + |b|\bigr)^2 = a^2 + 2|a|\,|b| + b^2 \leq 2\bigl(a^2+b^2\bigr) = \bigl(\sqrt{2}\,|z|\bigr)^2 . \end{equation*}

All the quantities squared are nonnegative, so taking square roots gives \(|z| \le |a| + |b| \le \sqrt{2}\,|z|\), which is the assertion.

Write \(w = a + bi\) and \(z = c + di\) with \(a,b,c,d \in \mathbb{R}\). By the multiplication rule 6.17 and the formula \(|u+vi|^2 = u^2 + v^2\) of 6.18,

\begin{equation*} \begin{aligned} |wz|^2 &= (ac - bd)^2 + (ad + bc)^2 \\ &= a^2 c^2 + b^2 d^2 + a^2 d^2 + b^2 c^2 \\ &= \left( a^2 + b^2 \right) \left( c^2 + d^2 \right) = \left( |w| \, |z| \right)^2 , \end{aligned} \end{equation*}

the cross terms \(\mp 2abcd\) cancelling; both sides are nonnegative, so \(|wz| = |w|\,|z|\).

For the triangle inequality, 6.24 supplies \(u\bar u = |u|^2\), \(u + \bar u = 2\operatorname{Re} u\), \(|\bar u| = |u|\), and the additivity and multiplicativity of conjugation, so

\begin{equation*} \begin{aligned} |w + z|^2 &= (w + z)(\bar{w} + \bar{z}) = |w|^2 + |z|^2 + w\bar{z} + \overline{w \bar{z}} \\ &= |w|^2 + |z|^2 + 2\operatorname{Re}(w \bar z) \\ &\leq |w|^2 + 2|w \bar z| + |z|^2 = \left( |w| + |z| \right)^2 , \end{aligned} \end{equation*}

using \(\overline{w \bar z} = \bar w z\), then \(\operatorname{Re} u \le |\operatorname{Re} u| \le |u|\) (6B.2), then \(|w\bar z| = |w|\,|\bar z| = |w|\,|z|\) by the first part. Taking square roots of nonnegative quantities gives \(|w+z| \le |w| + |z|\).

Write \(u = \operatorname{Re} f\) and \(v = \operatorname{Im} f\), so that \(f(x) = (u(x),v(x))\) under the identification and, by 6.19, condition (a) says exactly that \(u\) and \(v\) are \(\mathcal{S}\)-measurable. We prove (a) \(\Rightarrow\) (c) \(\Rightarrow\) (b) \(\Rightarrow\) (a).

(a) \(\Rightarrow\) (c). The collection \(\mathcal{T} = \{B \subseteq \mathbb{R}^2 : f^{-1}(B) \in \mathcal{S}\}\) is a \(\sigma\)-algebra on \(\mathbb{R}^2\), since \(f^{-1}(\varnothing) = \varnothing\) and inverse images commute with complements and countable unions (2.33). It contains every open cube \(C = I_1 \times I_2\) (two bounded open intervals of equal length), because

\begin{equation*} f^{-1}( C) = u^{-1}(I_1) \cap v^{-1}(I_2) \in \mathcal{S} : \end{equation*}

the intervals are Borel sets, so the two inverse images lie in \(\mathcal{S}\) by 2.35, and \(\mathcal{S}\) is closed under finite intersections (2.25(b)). Since \(\mathcal{B}_2\) is the smallest \(\sigma\)-algebra containing the open cubes (5.38(b)), \(\mathcal{B}_2 \subseteq \mathcal{T}\).

(c) \(\Rightarrow\) (b). Every open subset of \(\mathbb{R}^2\) lies in \(\mathcal{B}_2\) by 5.37.

(b) \(\Rightarrow\) (a). For each \(a \in \mathbb{R}\) the sets \((a,\infty) \times \mathbb{R}\) and \(\mathbb{R} \times (a,\infty)\) are open in \(\mathbb{R}^2\), and

\begin{equation*} u^{-1}\bigl((a,\infty)\bigr) = f^{-1}\bigl((a,\infty) \times \mathbb{R}\bigr), \qquad v^{-1}\bigl((a,\infty)\bigr) = f^{-1}\bigl(\mathbb{R} \times (a,\infty)\bigr), \end{equation*}

both in \(\mathcal{S}\) by (b). Hence \(u\) and \(v\) are \(\mathcal{S}\)-measurable by 2.39, and \(f\) is \(\mathcal{S}\)-measurable by 6.19.

Write \(f = u_1 + i v_1\) and \(g = u_2 + i v_2\) with real and imaginary parts as named; by 6.19 the hypothesis says \(u_1, v_1, u_2, v_2\) are \(\mathcal{S}\)-measurable, and the same result reduces each claim to the measurability of two real-valued functions. Sums, differences, products, and quotients with nowhere-vanishing denominators of such functions are \(\mathcal{S}\)-measurable by 2.46.

(a) The definitions of addition and multiplication in \(\mathbb{C}\) (6.17) give, pointwise on \(X\),

\begin{equation*} \operatorname{Re}(f \pm g) = u_1 \pm u_2 , \qquad \operatorname{Im}(f \pm g) = v_1 \pm v_2 , \end{equation*}

\begin{equation*} \operatorname{Re}(fg) = u_1 u_2 - v_1 v_2 , \qquad \operatorname{Im}(fg) = u_1 v_2 + v_1 u_2 , \end{equation*}

and every right side is \(\mathcal{S}\)-measurable by 2.46(a).

(b) Put \(h = |g|^2 = u_2^{\,2} + v_2^{\,2}\), which is \(\mathcal{S}\)-measurable by 2.45 and 2.46(a) and never \(0\), since \(h(x) = 0\) would force \(u_2(x) = v_2(x) = 0\) and hence \(g(x) = 0\). By 6.24 we have \(g\bar g = h\), so \(f/g = f\bar g/h\) pointwise, and \(\bar g = u_2 - iv_2\) (6.23) with the rule 6.17 gives \(f\bar g = (u_1u_2 + v_1v_2) + (v_1u_2 - u_1v_2)i\). Dividing a complex number by the nonzero real \(h\) divides both parts by \(h\), so

\begin{equation*} \operatorname{Re}\frac{f}{g} = \frac{u_1 u_2 + v_1 v_2}{u_2^{\,2} + v_2^{\,2}} , \qquad \operatorname{Im}\frac{f}{g} = \frac{v_1 u_2 - u_1 v_2}{u_2^{\,2} + v_2^{\,2}} , \end{equation*}

both \(\mathcal{S}\)-measurable by 2.46(a) for the numerators and 2.46(b) for the quotients.

Split into real and imaginary parts: each \(\operatorname{Re} f_k\) and \(\operatorname{Im} f_k\) is \(\mathcal{S}\)-measurable by 6.19. Fix \(x \in X\) and put \(z_k = f_k(x)\), \(L = f(x)\), so \(|z_k - L| \to 0\) by 6.18. Since \(\operatorname{Re}\) and \(\operatorname{Im}\) are additive, Exercise 6B.2 gives

\begin{equation*} \bigl| \operatorname{Re} z_k - \operatorname{Re} L \bigr| \le |z_k - L| \quad\text{and}\quad \bigl| \operatorname{Im} z_k - \operatorname{Im} L \bigr| \le |z_k - L| , \end{equation*}

so \((\operatorname{Re} f_k)(x) \to (\operatorname{Re} f)(x)\) and \((\operatorname{Im} f_k)(x) \to (\operatorname{Im} f)(x)\) for every \(x \in X\). Thus \(\operatorname{Re} f\) and \(\operatorname{Im} f\) are everywhere-existing pointwise limits of \(\mathcal{S}\)-measurable real-valued functions, hence \(\mathcal{S}\)-measurable by 2.48. Therefore \(f\) is \(\mathcal{S}\)-measurable by 6.19.

Both sides equal \((a I_u - b I_v) + i (a I_v + b I_u)\), where \(f = u + iv\) and \(\alpha = a + bi\) with \(u = \operatorname{Re} f\), \(v = \operatorname{Im} f\) \(\mathcal{S}\)-measurable by 6.19, and \(I_u = \int u \, d\mu\), \(I_v = \int v \, d\mu\).

Because \(|u| \le |f|\) and \(|v| \le |f|\) pointwise, 3.8 gives \(\int |u| \, d\mu < \infty\) and \(\int |v| \, d\mu < \infty\), so \(I_u\) and \(I_v\) are real numbers and \(\int f \, d\mu = I_u + i I_v\) by 6.21. Multiplying out with 6.17,

\begin{equation*} \operatorname{Re}(\alpha f) = au - bv \quad\text{and}\quad \operatorname{Im}(\alpha f) = av + bu , \end{equation*}

while \(|\alpha f| = |\alpha| \, |f|\) (Exercise 6B.4) and 3.20 give \(\int |\alpha f| \, d\mu = |\alpha| \int |f| \, d\mu < \infty\), so \(\int \alpha f \, d\mu\) is defined. Each of \(au, bv, av, bu\) has absolute value with finite integral by 3.20, so additivity 3.21 and homogeneity 3.20 apply termwise:

\begin{equation*} \begin{aligned} \int \alpha f \, d\mu &= \int (au - bv) \, d\mu + i \int (av + bu) \, d\mu \\ &= (a I_u - b I_v) + i (a I_v + b I_u), \end{aligned} \end{equation*}

the first equality by 6.21. Expanding with 6.17,

\begin{equation*} \alpha \int f \, d\mu = (a + bi)(I_u + i I_v) = (a I_u - b I_v) + i (a I_v + b I_u) . \end{equation*}

The intersection inherits each of the three conditions of 6.31 memberwise. Let \(\mathcal{A}\) be a collection of subspaces of \(V\) and set

\begin{equation*} U = \bigcap_{U^{\prime} \in \mathcal{A}} U^{\prime} , \end{equation*}

so \(U \subseteq V\) (with the convention that the empty intersection inside \(V\) is \(V\) itself). Every \(U^{\prime} \in \mathcal{A}\) contains \(0\) by 6.31, so \(0 \in U\). If \(f, g \in U\) and \(\alpha \in \mathbf{F}\), then \(f, g \in U^{\prime}\) for each \(U^{\prime} \in \mathcal{A}\), whence \(f + g \in U^{\prime}\) and \(\alpha f \in U^{\prime}\) by 6.31; as \(U^{\prime}\) was arbitrary, \(f + g \in U\) and \(\alpha f \in U\). Hence \(U\) is a subspace of \(V\) by 6.31.

Every axiom of 6.27 holds coordinatewise, inherited from \(V\) and \(W\) (vector spaces over the same \(\mathbf{F}\)), because two ordered pairs are equal exactly when both pairs of coordinates are.

The operations land in \(V \times W\), since \(f_1 + f_2 \in V\), \(g_1 + g_2 \in W\), \(\alpha f \in V\), \(\alpha g \in W\). A typical axiom check reads

\begin{equation*} \begin{aligned} \alpha \bigl( (f_1, g_1) + (f_2, g_2) \bigr) &= \bigl( \alpha(f_1 + f_2),\ \alpha(g_1 + g_2) \bigr) \\ &= (\alpha f_1 + \alpha f_2,\ \alpha g_1 + \alpha g_2) \\ &= \alpha(f_1, g_1) + \alpha(f_2, g_2), \end{aligned} \end{equation*}

the middle equality being distributivity in \(V\) and in \(W\); commutativity, associativity of addition, \((\alpha\beta)(f,g) = \alpha(\beta(f,g))\), \(1(f,g) = (f,g)\), and \((\alpha + \beta)(f,g) = \alpha(f,g) + \beta(f,g)\) go identically. (Check!) The additive identity is \((0_V, 0_W)\), and \((f,g)\) has additive inverse \((f^{\prime}, g^{\prime})\), where \(f + f^{\prime} = 0_V\) in \(V\) and \(g + g^{\prime} = 0_W\) in \(W\). Hence \(V \times W\) is a vector space over \(\mathbf{F}\).

The map \(N(f) = \|f\|\) is Lipschitz with constant \(1\), hence continuous. Indeed, the triangle inequality of 6.33 gives

\begin{equation*} \|f\| \le \|f - g\| + \|g\| \quad\text{and}\quad \|g\| \le \|g - f\| + \|f\| , \end{equation*}

while homogeneity (6.33, with \(\alpha = -1\)) gives \(\|g - f\| = \|f - g\|\); since \(\bigl| \|f\| - \|g\| \bigr|\) is one of \(\|f\| - \|g\|\), \(\|g\| - \|f\|\), the two inequalities yield

\begin{equation*} \bigl| N(f) - N(g) \bigr| = \bigl| \|f\| - \|g\| \bigr| \le \|f - g\| \end{equation*}

for all \(f, g \in V\). Given \(f \in V\) and \(\varepsilon > 0\), take \(\delta = \varepsilon\) in 6.10 (the metrics being \(\|f - g\|\) on \(V\) and \(|s - t|\) on \(\mathbf{F}\), by 6.36): \(\|g - f\| < \delta\) forces \(|N(g) - N(f)| < \varepsilon\).

Both inclusions follow from the sequential description 6.9(a) of the closure, taken in the metric \(d(g,h) = \|g - h\|\) of 6.36.

(i) \(\overline{B(f,r)} \subseteq \overline{B}(f,r)\). If \(g_k \in B(f,r)\) and \(g_k \to g\), then for each \(k\)

\begin{equation*} \|f - g\| \le \|f - g_k\| + \|g_k - g\| < r + \|g_k - g\| , \end{equation*}

and letting \(k \to \infty\) gives \(\|f - g\| \le r\).

(ii) \(\overline{B}(f,r) \subseteq \overline{B(f,r)}\). Given \(\|f - g\| \le r\), put \(g_k = f + \bigl(1 - \tfrac{1}{k}\bigr)(g - f)\). Homogeneity of the norm gives

\begin{equation*} \begin{aligned} \|g_k - f\| &= (1 - \tfrac1k) \|g - f\| \le (1 - \tfrac1k) r < r , \\ \|g_k - g\| &= \tfrac1k \|g - f\| \le \tfrac{r}{k} , \end{aligned} \end{equation*}

so each \(g_k \in B(f,r)\) and \(g_k \to g\).

One-point vectors in \(\mathbf{F}^n\) break a condition of 6.33 in each case.

(i) For \(\|(a_1, \ldots, a_n)\| = |a_1|^{1/2} + \cdots + |a_n|^{1/2}\), homogeneity fails: with \(f = (1, 0, \ldots, 0)\) and \(\alpha = 4\),

\begin{equation*} \|\alpha f\| = \|(4, 0, \ldots, 0)\| = |4|^{1/2} = 2 \ne 4 = |\alpha| \, \|f\| . \end{equation*}

(ii) For \(\|(a_1, \ldots, a_n)\|_{1/2} = \bigl( |a_1|^{1/2} + \cdots + |a_n|^{1/2} \bigr)^2\), the triangle inequality fails whenever \(n > 1\) (for \(n = 1\) this is the absolute value): with \(f = (1, 0, \ldots, 0)\) and \(g = (0, 1, 0, \ldots, 0)\), so that \(f + g = (1, 1, 0, \ldots, 0)\),

\begin{equation*} \begin{aligned} \|f + g\|_{1/2} &= \bigl( |1|^{1/2} + |1|^{1/2} \bigr)^2 = 4 \\ &> 2 = \|f\|_{1/2} + \|g\|_{1/2} . \end{aligned} \end{equation*}

Take \(M = \max\bigl\{ \|f_1\|, \ldots, \|f_{n-1}\|, \, 1 + \|f_n\| \bigr\} + 1\), where \(n \in \mathbf{Z}^+\) is supplied by the Cauchy condition 6.12 with \(\varepsilon = 1\), so that \(\|f_j - f_k\| < 1\) for all \(j, k \ge n\). (The maximum is over a finite nonempty set of reals, so \(M \in \mathbf{R}\).)

If \(k \ge n\), then the triangle inequality gives

\begin{equation*} \|f_k\| = \|(f_k - f_n) + f_n\| \le \|f_k - f_n\| + \|f_n\| < 1 + \|f_n\| \le M - 1 , \end{equation*}

and if \(k < n\), then \(\|f_k\| \le M - 1\) by the definition of \(M\). Hence \(\|f_k\| < M\) for every \(k \in \mathbf{Z}^+\).

Both \(\|\cdot\|_1\) and \(\|\cdot\|_\infty\) reduce convergence in \(\mathbf{F}^n\) to convergence of each coordinate in the complete space \(\mathbf{F}\), so \(\mathbf{F}^n\) is a Banach space for each.

Each is a norm: positive definiteness and homogeneity follow from \(|\alpha a_m| = |\alpha| \, |a_m|\), and \(|a_m + b_m| \le |a_m| + |b_m|\) gives the triangle inequality on summing over \(m\), respectively on maximizing over \(m\). (Check!) They are comparable: for \(a = (a_1, \ldots, a_n) \in \mathbf{F}^n\) and each \(m\),

\begin{equation*} |a_m| \le \|a\|_\infty \le \|a\|_1 \le n \|a\|_\infty , \end{equation*}

since the maximum is one of the \(n\) summands of \(\|a\|_1\), each of which is at most \(\|a\|_\infty\).

Now let \(\|\cdot\|\) be either norm and let \(f_k = (a_{k,1}, \ldots, a_{k,n})\) be Cauchy for \(\|\cdot\|\). Fix \(m\); the comparison gives

\begin{equation*} |a_{j,m} - a_{k,m}| \le \|f_j - f_k\|_\infty \le \|f_j - f_k\| \end{equation*}

(the last step an equality when \(\|\cdot\| = \|\cdot\|_\infty\)), so \(a_{1,m}, a_{2,m}, \ldots\) is Cauchy in \(\mathbf{F}\) and hence converges to some \(b_m \in \mathbf{F}\) by completeness of \(\mathbf{F}\). Put \(b = (b_1, \ldots, b_n)\). Given \(\varepsilon > 0\), pick \(N_m\) with \(|a_{k,m} - b_m| < \varepsilon/n\) for all \(k \ge N_m\) and set \(N = \max\{N_1, \ldots, N_n\}\); then for every \(k \ge N\),

\begin{equation*} \|f_k - b\|_\infty \le \|f_k - b\|_1 = \sum_{m=1}^{n} |a_{k,m} - b_m| < \varepsilon . \end{equation*}

Hence \(f_k \to b\) in \((\mathbf{F}^n, \|\cdot\|)\) for both choices of \(\|\cdot\|\), so each is a Banach space by 6.37.

The limit of a \(\|\cdot\|\)-Cauchy sequence is its pointwise limit, which exists because \(\mathbf{F}\) is complete and is bounded because Cauchy sequences are bounded.

That \(\|\cdot\|\) is a norm on \(b(X)\) is the third bullet point of 6.34: \(\|f\| = 0\) forces \(|f(x)| \le 0\) hence \(f = 0\); \(\|\alpha f\| = \sup_X |\alpha| \, |f| = |\alpha| \, \|f\|\); and \(|(f+g)(x)| \le \|f\| + \|g\|\) for each \(x \in X\) gives \(\|f + g\| \le \|f\| + \|g\|\) on taking the supremum. (Check!)

Let \(f_1, f_2, \ldots\) be Cauchy in \(b(X)\). For each \(x \in X\) we have \(|f_j(x) - f_k(x)| \le \|f_j - f_k\|\), so \(f_1(x), f_2(x), \ldots\) is Cauchy in \(\mathbf{F}\); set \(f(x) = \lim_{k \to \infty} f_k(x)\), which exists by completeness of \(\mathbf{F}\). By 6C.4 there is \(M \in \mathbf{R}\) with \(\|f_k\| < M\) for every \(k\), so \(|f(x)| = \lim_{k \to \infty} |f_k(x)| \le M\) for each \(x \in X\) and hence \(f \in b(X)\).

Given \(\varepsilon > 0\), pick \(n\) with \(\|f_j - f_k\| < \varepsilon/2\) for all \(j, k \ge n\). Fixing \(k \ge n\) and \(x \in X\) and letting \(j \to \infty\) in \(|f_j(x) - f_k(x)| < \varepsilon/2\),

\begin{equation*} |f(x) - f_k(x)| = \lim_{j \to \infty} |f_j(x) - f_k(x)| \le \frac{\varepsilon}{2} , \end{equation*}

so the supremum over \(x \in X\) gives \(\|f - f_k\| \le \varepsilon/2 < \varepsilon\) for all \(k \ge n\). Thus \(f_k \to f\) in \(b(X)\), which is therefore a Banach space by 6.37.

The truncated harmonic sequences

\begin{equation*} f_k = \bigl( 1, \tfrac12, \tfrac13, \ldots, \tfrac1k, 0, 0, \ldots \bigr) \in \ell^1 \end{equation*}

form a \(\|\cdot\|_\infty\)-Cauchy sequence with no limit in \(\ell^1\). (Each \(f_k\) has finitely many nonzero coordinates, so lies in \(\ell^1\); write \(h(m)\) for the \(m^{\text{th}}\) coordinate of \(h\).)

Cauchy: for \(j > k\) the coordinates of \(f_j - f_k\) vanish except in positions \(k+1, \ldots, j\), where they are \(\frac{1}{k+1}, \ldots, \frac{1}{j}\), so \(\|f_j - f_k\|_\infty = \frac{1}{k+1}\); hence \(\|f_j - f_k\|_\infty \le \frac{1}{n+1} < \varepsilon\) for all \(j, k \ge n\) once \(n > 1/\varepsilon\) (the cases \(j = k\) and \(j < k\) being trivial and symmetric).

No limit: suppose \(\|f_k - g\|_\infty \to 0\) for some \(g = (b_1, b_2, \ldots) \in \ell^1\). Fix \(m\); for every \(k \ge m\) we have \(f_k(m) = \frac{1}{m}\), so

\begin{equation*} \bigl| \tfrac1m - b_m \bigr| = |f_k(m) - b_m| \le \|f_k - g\|_\infty \to 0 , \end{equation*}

and the left side is independent of \(k\), forcing \(b_m = \frac{1}{m}\) for every \(m\). Then \(\sum_{m=1}^{\infty} |b_m| = \sum_{m=1}^{\infty} \frac{1}{m} = \infty\), so \(g \notin \ell^1\), a contradiction. Hence \((\ell^1, \|\cdot\|_\infty)\) is not complete.

The coordinatewise limit of an \(\|\cdot\|_1\)-Cauchy sequence lies in \(\ell^1\) and is its limit there.

First, \(\|\cdot\|_1\) is a norm: \(\sum_{k=1}^{\infty} |a_k| = 0\) forces every \(a_k = 0\), homogeneity is \(\sum_k |\alpha a_k| = |\alpha| \sum_k |a_k|\), and for \(a = (a_1, a_2, \ldots)\), \(b = (b_1, b_2, \ldots) \in \ell^1\) and each \(n \in \mathbf{Z}^+\),

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

so \(n \to \infty\) gives \(\|a + b\|_1 \le \|a\|_1 + \|b\|_1\) (and \(a + b \in \ell^1\)).

Now let \(f^n = (a_1^n, a_2^n, \ldots)\) be Cauchy in \(\ell^1\). For fixed \(k\) we have \(|a_k^n - a_k^m| \le \|f^n - f^m\|_1\), so \(a_k^1, a_k^2, \ldots\) is Cauchy in the complete space \(\mathbf{F}\) and converges to some \(a_k\); put \(f = (a_1, a_2, \ldots)\). Given \(\varepsilon > 0\), choose \(N\) with \(\|f^n - f^m\|_1 < \varepsilon\) for all \(m, n \ge N\). Fix \(n \ge N\) and \(M \in \mathbf{Z}^+\); the finite sum \(\sum_{k=1}^{M} |a_k^n - a_k^m| \le \|f^n - f^m\|_1 < \varepsilon\) may be passed to the limit \(m \to \infty\) term by term, giving \(\sum_{k=1}^{M} |a_k^n - a_k| \le \varepsilon\), and the partial sums increase in \(M\), so

\begin{equation*} \sum_{k=1}^{\infty} |a_k^n - a_k| \le \varepsilon \quad \text{for all } n \ge N . \end{equation*}

Hence \(f^N - f \in \ell^1\), so \(f = f^N - (f^N - f) \in \ell^1\), and the display reads \(\|f^n - f\|_1 \le \varepsilon\) for all \(n \ge N\). Thus \(f^n \to f\) in \(\ell^1\), which is therefore a Banach space by 6.37.

The ramps \(f_k \in C([0,1])\), \(k \ge 3\), defined by

\begin{equation*} f_k(x) = \begin{cases} 0 & \text{if } 0 \le x \le \tfrac12, \\[2pt] k(x - \tfrac12) & \text{if } \tfrac12 \le x \le \tfrac12 + \tfrac1k, \\[2pt] 1 & \text{if } \tfrac12 + \tfrac1k \le x \le 1, \end{cases} \end{equation*}

are Cauchy for \(\|f\| = \int_0^1 |f|\) but have no limit in \(C([0,1])\). (The formulas agree at both junctions, so each \(f_k\) is continuous, and \(0 \le f_k \le 1\).)

Used twice below: if \(g\) is continuous on \([a,b]\) with \(a < b\) and \(\int_a^b |g| = 0\), then \(g = 0\) on \([a,b]\), since otherwise continuity puts \(|g| > c/2\) on an interval of positive length \(\delta\) about a point where \(|g| = c > 0\), forcing \(\int_a^b |g| \ge c\delta/2 > 0\).

Cauchy: for \(j, k \ge 3\) with \(m = \min\{j,k\}\), both \(f_j\) and \(f_k\) vanish on \([0, \tfrac12]\) and both equal \(1\) on \([\tfrac12 + \tfrac1m, 1]\), so \(f_j - f_k\) vanishes off an interval of length \(\tfrac1m\) on which \(|f_j - f_k| \le 1\); hence

\begin{equation*} \|f_j - f_k\| \le \frac{1}{\min\{j,k\}} < \varepsilon \quad \text{for all } j, k \ge n > 1/\varepsilon . \end{equation*}

No limit: suppose \(f \in C([0,1])\) with \(\|f_k - f\| \to 0\). Since \(f_k = 0\) on \([0, \tfrac12]\),

\begin{equation*} \int_0^{1/2} |f| = \int_0^{1/2} |f - f_k| \le \|f_k - f\| \longrightarrow 0 , \end{equation*}

and the left side is independent of \(k\), so \(f = 0\) on \([0, \tfrac12]\). Fixing \(\delta \in (0, \tfrac12)\) and using \(f_k = 1\) on \([\tfrac12 + \delta, 1]\) for \(k > 1/\delta\),

\begin{equation*} \int_{1/2 + \delta}^{1} |f - 1| = \int_{1/2+\delta}^{1} |f - f_k| \le \|f_k - f\| \longrightarrow 0 , \end{equation*}

so \(f = 1\) on \([\tfrac12 + \delta, 1]\) and hence, \(\delta\) being arbitrary, on \((\tfrac12, 1]\). Then \(\lim_{x \downarrow 1/2} f(x) = 1 \ne 0 = f(\tfrac12)\), contradicting continuity at \(\tfrac12\).

Shrink each \(f \in V\) into the ball, then scale back. Suppose \(B(h,r) = \{g \in V : \|g - h\| < r\} \subseteq U\). If \(\|g\| < r\), then \(h + g\) and \(h\) both lie in \(B(h,r) \subseteq U\), so \(g = (h + g) - h \in U\); thus \(B(0,r) \subseteq U\).

Now let \(f \in V\) with \(f \ne 0\) (the case \(f = 0\) holding since \(U\) is a subspace), so \(\|f\| > 0\) by positive definiteness, and set \(\alpha = r/(2\|f\|) > 0\). Homogeneity gives \(\|\alpha f\| = r/2 < r\), so \(\alpha f \in B(0,r) \subseteq U\) and therefore

\begin{equation*} f = \frac{1}{\alpha} (\alpha f) \in U . \end{equation*}

Hence \(V \subseteq U\), and \(U = V\).

The segment joining a point of \(E\) to a point outside \(E\) cannot leave \(E\). Both \(\emptyset\) and \(V\) are open, and each is the complement of the other, so both are also closed.

Suppose \(E \subseteq V\) is open and closed with \(f \in E\) and \(g \in V \setminus E\), so \(\|g - f\| > 0\). Put \(\gamma(t) = f + t(g - f)\) for \(t \in [0,1]\), so homogeneity of the norm gives

\begin{equation*} \|\gamma(s) - \gamma(t)\| = \|(s - t)(g - f)\| = |s - t| \, \|g - f\| , \end{equation*}

and let \(c = \sup A\), where \(A = \{t \in [0,1] : \gamma(t) \in E\}\) contains \(0\) and is bounded above by \(1\).

Then \(c \in A\): choosing \(t_k \in A\) with \(t_k \to c\), the display gives \(\gamma(t_k) \to \gamma( c)\), and \(\gamma( c) \in E\) by 6.9(e) since \(E\) is closed. Also \(c < 1\), because \(\gamma(1) = g \notin E\). But \(E\) is open, so \(B(\gamma( c), \varepsilon) \subseteq E\) for some \(\varepsilon > 0\), and any

\begin{equation*} t \in \Bigl( c, \ \min\Bigl\{ 1, \ c + \frac{\varepsilon}{\|g - f\|} \Bigr\} \Bigr) \end{equation*}

(a nonempty interval, as \(c < 1\)) satisfies \(\|\gamma(t) - \gamma( c)\| = (t - c) \|g - f\| < \varepsilon\), so \(\gamma(t) \in E\) and \(t \in A\) with \(t > \sup A\), a contradiction.

Limits of sums are sums of limits. Let \(U\) be a subspace of \(V\); by 6.9(a), \(\overline{U}\) is exactly the set of limits of sequences in \(U\), and \(0 \in U \subseteq \overline{U}\).

Suppose \(f, g \in \overline{U}\) and \(\alpha \in \mathbf{F}\); choose \(f_k \to f\) and \(g_k \to g\) with all \(f_k, g_k \in U\), so \(f_k + g_k \in U\) and \(\alpha f_k \in U\). The triangle inequality and homogeneity give

\begin{equation*} \|(f_k + g_k) - (f + g)\| \le \|f_k - f\| + \|g_k - g\| \to 0 \quad\text{and}\quad \|\alpha f_k - \alpha f\| = |\alpha| \, \|f_k - f\| \to 0 , \end{equation*}

so \(f + g \in \overline{U}\) and \(\alpha f \in \overline{U}\) by 6.9(a). Hence \(\overline{U}\) is a subspace of \(V\) by 6.31.

Everything is defined termwise on Cauchy sequences:

\begin{equation*} \begin{aligned} (f_k)^{\wedge} + (g_k)^{\wedge} &= (f_k + g_k)^{\wedge}, \qquad \alpha (f_k)^{\wedge} = (\alpha f_k)^{\wedge}, \\ \bigl\| (f_k)^{\wedge} \bigr\|_V &= \lim_{k \to \infty} \|f_k\| , \end{aligned} \end{equation*}

where \((f_k)^{\wedge}\) denotes the class of the Cauchy sequence \((f_1, f_2, \ldots)\) in the construction of 6A.16, equivalence meaning \(\lim_{k \to \infty} d(f_k, g_k) = 0\) with \(d(f,g) = \|f - g\|\).

(a) The two operations produce Cauchy sequences and respect equivalence, by

\begin{equation*} \|(f_j + g_j) - (f_k + g_k)\| \le \|f_j - f_k\| + \|g_j - g_k\| , \qquad \|\alpha f_j - \alpha f_k\| = |\alpha| \, \|f_j - f_k\| , \end{equation*}

read first for large \(j, k\), then with \((f_k^{\prime}), (g_k^{\prime})\) equivalent to \((f_k), (g_k)\) in place of the second entries. Each axiom of 6.27 then holds because it holds at every index in \(U\); the zero is \((0, 0, \ldots)^{\wedge}\) and \((f_k)^{\wedge}\) has additive inverse \((-f_k)^{\wedge}\). (Check!)

(b) The limit defining \(\|\cdot\|_V\) exists because \(\bigl| \|f_j\| - \|f_k\| \bigr| \le \|f_j - f_k\|\) makes \(\|f_1\|, \|f_2\|, \ldots\) Cauchy in the complete space \(\mathbf{R}\), and the same inequality applied to equivalent sequences shows it depends only on the class. It is a norm: \(\lim_k \|f_k\| = 0\) says exactly that \((f_k) \equiv (0,0,\ldots)\), i.e. \((f_k)^{\wedge}\) is the zero of \(V\), while homogeneity and the triangle inequality pass to the limit from \(U\). Moreover

\begin{equation*} \begin{aligned} \bigl\| (f_k)^{\wedge} - (g_k)^{\wedge} \bigr\|_V &= \bigl\| (f_k - g_k)^{\wedge} \bigr\|_V = \lim_{k \to \infty} \|f_k - g_k\| \\ &= d_V\bigl( (f_k)^{\wedge}, (g_k)^{\wedge} \bigr) , \end{aligned} \end{equation*}

so the metric that 6.36 attaches to \(\|\cdot\|_V\) is \(d_V\), which is complete by 6A.16(c). Hence \(V\) is a Banach space by 6.37.

(c) The map \(\varphi(f) = (f, f, f, \ldots)^{\wedge}\) is linear by the definitions in (a) and norm preserving, since \(\|\varphi(f)\|_V = \lim_{k \to \infty} \|f\| = \|f\|\); in particular \(\varphi\) is injective, and \(\varphi(U)\) is a subspace of \(V\). Thus \(U\) is isometrically isomorphic to a subspace of the Banach space \(V\).

The extension is \(Tf = \lim_{k \to \infty} S f_k\) for any \(f_k \in U\) with \(f_k \to f \in \overline{U}\). Throughout, \(\overline{U}\) is a subspace of \(V\) by 6C.12, and \(\|Sh\| \le \|S\| \, \|h\|\) for \(h \in U\) by 6C.16(c), with \(\|S\| < \infty\) as \(S\) is bounded.

(a) Uniqueness: if \(T_1, T_2\) are continuous on \(\overline{U}\) with \(T_i|_U = S\), pick \(f_k \in U\) with \(f_k \to f\) (6.9(a)); then 6.11 gives

\begin{equation*} T_1 f = \lim_{k \to \infty} S f_k = T_2 f . \end{equation*}

Existence: such a sequence is Cauchy (6.13), and

\begin{equation*} \|S f_j - S f_k\| = \|S(f_j - f_k)\| \le \|S\| \, \|f_j - f_k\| \end{equation*}

makes \(Sf_1, Sf_2, \ldots\) Cauchy in the Banach space \(W\), hence convergent. The limit does not depend on the sequence: for another \(g_k \to f\) in \(U\),

\begin{equation*} \|S f_k - S g_k\| \le \|S\| \bigl( \|f_k - f\| + \|f - g_k\| \bigr) \to 0 . \end{equation*}

Constant sequences give \(T|_U = S\). Finally, for \(f, g \in \overline{U}\) with \(f_k \to f\) and \(g_k \to g\) in \(U\) we have \(Sf_k - Sg_k \to Tf - Tg\) (triangle inequality), so letting \(k \to \infty\) in \(\|Sf_k - Sg_k\| \le \|S\| \, \|f_k - g_k\|\) and using continuity of the norm (6C.1),

\begin{equation*} \|Tf - Tg\| \le \|S\| \, \|f - g\| ; \end{equation*}

thus \(T\) is Lipschitz, hence continuous.

(b) Linearity: \(f_k + g_k \to f + g\) and \(\alpha f_k \to \alpha f\) in \(U\) (as in 6C.12), so evaluating \(T\) along these sequences and using linearity of \(S\),

\begin{equation*} T(f + g) = \lim_{k \to \infty} (S f_k + S g_k) = Tf + Tg , \qquad T(\alpha f) = \lim_{k \to \infty} \alpha S f_k = \alpha T f . \end{equation*}

Putting \(g = 0\) in the Lipschitz estimate (with \(T0 = S0 = 0\)) gives \(\|Tf\| \le \|S\| \, \|f\|\) on \(\overline{U}\), so \(\|T\| \le \|S\| < \infty\); conversely \(\|S\| = \sup\{\|Tf\| : f \in U, \ \|f\| \le 1\} \le \|T\|\), the supremum for \(\|T\|\) being over the larger set \(\overline{U}\). Hence \(\|T\| = \|S\|\).

(c) Take \(V = \ell^1\) with \(\|\cdot\|_1\), let \(U\) be the subspace of sequences with only finitely many nonzero coordinates, let \(W = U\) with the inherited norm, and let \(S : U \to W\) be the identity, so \(\|S\| = 1\). Truncations \(a^{(n)} = (a_1, \ldots, a_n, 0, 0, \ldots) \in U\) satisfy

\begin{equation*} \|a - a^{(n)}\|_1 = \sum_{k = n+1}^{\infty} |a_k| \to 0 , \end{equation*}

so \(\overline{U} = \ell^1\) by 6.9(a), while \(f = (1/2, 1/4, 1/8, \ldots) \in \ell^1 \setminus U\). If \(T : \ell^1 \to W\) were continuous with \(T|_U = S\), then 6.11 applied to the truncations \(f_k \to f\) would give \(Tf = \lim_{k \to \infty} f_k\) in \(W\), hence in \(\ell^1\) since the norm of \(W\) is the restriction of \(\|\cdot\|_1\), so \(Tf = f\) by uniqueness of limits. That is impossible: \(Tf \in W = U\) but \(f \notin U\).

The quotient norm is the distance to \(U\): replacing \(g\) by \(-g\) in the infimum gives \(\|f + U\| = \operatorname{dist}(f, U)\), which depends only on the coset and satisfies \(0 \le \|f + U\| \le \|f\|\) (take \(g = 0\)).

(a) Homogeneity and the triangle inequality hold for any subspace \(U\), so positive definiteness carries the whole statement. For \(\alpha \ne 0\) the map \(g \mapsto \alpha g\) is a bijection of \(U\), so

\begin{equation*} \|\alpha(f + U)\| = \inf_{h \in U} \|\alpha f + \alpha h\| = |\alpha| \, \|f + U\| , \end{equation*}

while \(\|0(f + U)\| = \|0 + U\| = 0\); and for \(f, h \in V\), \(g_1, g_2 \in U\),

\begin{equation*} \|(f + h) + U\| \le \|(f + h) + (g_1 + g_2)\| \le \|f + g_1\| + \|h + g_2\| , \end{equation*}

whose infimum over \(g_1\) and then \(g_2\) is the triangle inequality. Since \(\|0 + U\| = 0\) always, \(\|\cdot\|\) is a norm exactly when \(\operatorname{dist}(f, U) = 0\) forces \(f \in U\); and \(\operatorname{dist}(f, U) = 0\) says precisely that \(f\) is a limit of a sequence in \(U\), i.e. \(f \in \overline{U}\). So \(\|\cdot\|\) is a norm if and only if \(\overline{U} \subseteq U\), i.e. if and only if \(U\) is closed.

(b) By (a), \(V/U\) is a normed vector space; apply the series criterion 6.41. Suppose \(\sum_{k=1}^{\infty} \|f_k + U\| < \infty\) and choose \(g_k \in U\) with \(\|f_k + g_k\| \le \|f_k + U\| + 2^{-k}\), so \(\sum_{k=1}^{\infty} \|f_k + g_k\| < \infty\). Since \(V\) is a Banach space, 6.41 gives \(f \in V\) with \(\sum_{k=1}^{n} (f_k + g_k) \to f\); as \(\sum_{k=1}^{n} g_k \in U\), the cosets agree and \(\|h + U\| \le \|h\|\) yields

\begin{equation*} \Bigl\| \sum_{k=1}^{n} (f_k + U) - (f + U) \Bigr\| \le \Bigl\| \sum_{k=1}^{n} (f_k + g_k) - f \Bigr\| \longrightarrow 0 . \end{equation*}

Thus every absolutely convergent series in \(V/U\) converges, so \(V/U\) is a Banach space by 6.41.

(c) First \(U\) is closed: if \(g_k \in U\) and \(g_k \to f\) in \(V\), then \((g_k)\) is Cauchy by 6.13, so completeness of \(U\) gives \(g_k \to g \in U\), and \(f = g \in U\) by uniqueness of limits; hence \(\|\cdot\|\) is a norm on \(V/U\) by (a). Now let \(f_1, f_2, \ldots\) be Cauchy in \(V\). Since

\begin{equation*} \|(f_j + U) - (f_k + U)\| = \|(f_j - f_k) + U\| \le \|f_j - f_k\| , \end{equation*}

the cosets are Cauchy in \(V/U\) and so converge to some \(f + U\), that is, \(\|(f_k - f) + U\| \to 0\). Choose \(h_k \in U\) with \(\|f_k - f + h_k\| \le \|(f_k - f) + U\| + \tfrac{1}{k}\) and put \(u_k = f_k + h_k\), so \(u_k \to f\) in \(V\). Then \(h_k = u_k - f_k\) obeys \(\|h_j - h_k\| \le \|u_j - u_k\| + \|f_j - f_k\|\), and both sequences on the right are Cauchy (6.13 for \(u_k\)), so \((h_k)\) is Cauchy in \(U\) and converges to some \(h \in U\). Therefore

\begin{equation*} \|f_k - (f - h)\| = \|(u_k - h_k) - (f - h)\| \le \|u_k - f\| + \|h_k - h\| \longrightarrow 0 , \end{equation*}

so \(f_k \to f - h\) in \(V\), and \(V\) is a Banach space.

Rescaling moves each defining set onto the closed unit ball. Write \(A = \|T\| = \sup\{\|Tf\| : \|f\| \le 1\}\) (6.43) and let \(B, C, D, E\) be the quantities on the right in (a), (b), (c), (d); no boundedness is assumed, so all lie in \([0, \infty]\), with \(\inf \emptyset = \infty\). Used throughout: \(T0 = 0\) and \(\|T(\alpha f)\| = |\alpha| \, \|Tf\|\), and \(V \ne \{0\}\) makes the sets for \(C\) and \(E\) nonempty.

(a) \(B \le A\) since \(\{\|f\| < 1\} \subseteq \{\|f\| \le 1\}\). Conversely, if \(\|f\| \le 1\) and \(t \in (0,1)\), then \(\|tf\| \le t < 1\), so \(t \|Tf\| = \|T(tf)\| \le B\); this gives \(\|Tf\| \le B\) trivially when \(B = \infty\), and when \(B < \infty\) by letting \(t \uparrow 1\) in \(\|Tf\| \le B/t\). Hence \(A \le B\).

(b) \(C \le A\) since \(\{\|f\| = 1\} \subseteq \{\|f\| \le 1\}\). Conversely, let \(\|f\| \le 1\); then \(\|Tf\| = 0 \le C\) if \(f = 0\), and otherwise \(u = f/\|f\|\) has \(\|u\| = 1\) and

\begin{equation*} \|Tf\| = \bigl\| T(\|f\| u) \bigr\| = \|f\| \, \|Tu\| \le \|Tu\| \le C . \end{equation*}

Hence \(A \le C\).

(c) Put \(S = \{c \in [0, \infty) : \|Tf\| \le c\|f\| \text{ for all } f \in V\}\), so \(D = \inf S\). If \(c \in S\) and \(\|f\| \le 1\), then \(\|Tf\| \le c\|f\| \le c\), so \(A \le c\) and \(S \subseteq [A, \infty)\). Also \(\|Tf\| \le A\|f\|\) for every \(f \in V\), both sides vanishing at \(f = 0\) and otherwise

\begin{equation*} \|Tf\| = \|f\| \, \|T(f/\|f\|)\| \le A \|f\| . \end{equation*}

So if \(A < \infty\), then \(A \in S\) and \(D = \inf S = A\); and if \(A = \infty\), then \(S \subseteq [\infty, \infty)\) is empty, so \(D = \inf \emptyset = \infty = A\).

(d) The two sets \(\{\|Tf\|/\|f\| : f \ne 0\}\) and \(\{\|Tu\| : \|u\| = 1\}\) are equal: \(\|u\| = 1\) gives \(\|Tu\| = \|Tu\|/\|u\|\), and \(f \ne 0\) gives \(u = f/\|f\|\) of norm \(1\) with \(\|Tu\| = \|Tf\|/\|f\|\). Hence \(E = C = A\) by (b).

Apply \(\|Rf\| \le \|R\| \, \|f\|\) (6C.16(c)) twice, to \(S\) at \(Tf\) and to \(T\) at \(f\). The composition \(S \circ T\) is linear, so \(\|S \circ T\|\) is defined by 6.43.

(i) \(\|S\| \ne 0\) and \(\|T\| \ne 0\). If either is \(\infty\), then \(\|S\| \, \|T\| = \infty\) and there is nothing to prove; so let \(\|S\|, \|T\| \in (0, \infty)\) and \(f \in U\) with \(\|f\| \le 1\). Then

\begin{equation*} \|(S \circ T)f\| = \|S(Tf)\| \le \|S\| \, \|Tf\| \le \|S\| \, \|T\| \, \|f\| \le \|S\| \, \|T\| , \end{equation*}

and the supremum over such \(f\) gives \(\|S \circ T\| \le \|S\| \, \|T\|\).

(ii) \(\|S\| = 0\) or \(\|T\| = 0\), where the convention \(0 \cdot \infty = 0\) makes the claim \(\|S \circ T\| = 0\). If \(\|T\| = 0\), the cited inequality forces \(Tf = 0\) for every \(f \in U\), so \((S \circ T)f = S0 = 0\); if \(\|S\| = 0\), it forces \(Sg = 0\) for every \(g \in V\), so again \((S \circ T)f = 0\). Either way \(\|S \circ T\| = 0 \le \|S\| \, \|T\|\).

We run (a) \(\Rightarrow\) (c) \(\Rightarrow\) (b) \(\Rightarrow\) (a) and (a) \(\Rightarrow\) (d) \(\Rightarrow\) (a), with \(V\) and \(W\) metrized by 6.36 and \(T^{-1}\bigl(B(0,r)\bigr) = \{f \in V : \|Tf\| < r\}\). The engine for both returns to (a): if \(\|Th\| \le M < \infty\) for all \(\|h\| < \delta\), then \(\|T\| \le 2M/\delta\), since \(\|f\| \le 1\) gives \(\|(\delta/2) f\| < \delta\) and hence \((\delta/2) \|Tf\| = \|T((\delta/2) f)\| \le M\).

(a) \(\Rightarrow\) (c). By 6C.16(c), \(\|Th\| \le \|T\| \, \|h\|\); given \(\varepsilon > 0\), set \(\delta = \varepsilon/(1 + \|T\|) > 0\). If \(\|f - g\| < \delta\), then

\begin{equation*} \|Tf - Tg\| = \|T(f - g)\| \le \|T\| \, \|f - g\| \le \|T\| \, \delta = \frac{\|T\|}{1 + \|T\|} \, \varepsilon < \varepsilon \end{equation*}

(the second \(\le\) rather than \(<\) because \(\|T\|\) may be \(0\)).

(c) \(\Rightarrow\) (b). Uniform continuity gives, for each \(\varepsilon > 0\), a \(\delta > 0\) with \(\|Tg - T0\| < \varepsilon\) whenever \(\|g - 0\| < \delta\); by 6.10 this is continuity at \(f = 0\).

(b) \(\Rightarrow\) (a). Continuity at \(f\) with \(\varepsilon = 1\) in 6.10 gives \(\delta > 0\) with \(\|Tg - Tf\| < 1\) whenever \(\|g - f\| < \delta\). For \(\|h\| < \delta\), take \(g = f + h\):

\begin{equation*} \|Th\| = \|T(g - f)\| = \|Tg - Tf\| < 1 , \end{equation*}

so the engine with \(M = 1\) gives \(\|T\| \le 2/\delta < \infty\).

(a) \(\Rightarrow\) (d). A bounded \(T\) is continuous by 6.48, and \(B(0,1)\) is open in \(W\) by 6.5, so \(T^{-1}\bigl(B(0,1)\bigr)\) is open in \(V\) by 6.11(a) \(\Leftrightarrow\) 6.11(c); thus (d) holds with \(r = 1\).

(d) \(\Rightarrow\) (a). Since \(T0 = 0\), the open set \(T^{-1}\bigl(B(0,r)\bigr)\) contains \(0\), so 6.4 supplies \(\delta > 0\) with \(\|Th\| < r\) whenever \(\|h\| < \delta\); the engine with \(M = r\) gives \(\|T\| \le 2r/\delta < \infty\).

Write \(S = \varphi^{-1}(\alpha)\), a translate of \(\operatorname{null} \varphi\). Translation \(\tau_h(f) = h + f\) is a bijection of \(V\) with \(\|\tau_h(f) - \tau_h(g)\| = \|f - g\|\), hence a homeomorphism, so \(\overline{h + A} = h + \overline{A}\) for every \(A \subseteq V\).

(i) \(\varphi\) identically \(0\). Then \(S = \varnothing\) because \(\alpha \ne 0\), and all three statements hold: \(\|\varphi\| = 0\), \(\varnothing\) is closed, and \(\overline{\varnothing} = \varnothing \ne V\).

(ii) \(\varphi\) not identically \(0\), so 6.52 applies. Pick \(g\) with \(\varphi(g) = \beta \ne 0\) and set \(h = (\alpha/\beta) g\), so \(\varphi(h) = \alpha\) and, since \(\varphi(f) = \alpha\) exactly when \(f - h \in \operatorname{null} \varphi\),

\begin{equation*} S = h + \operatorname{null} \varphi . \end{equation*}

(a) \(\Rightarrow\) (b): a bounded \(\varphi\) is continuous by 6.48, and \(\{\alpha\}\) is closed in \(\mathbb{F}\), so \(S = \varphi^{-1}(\{\alpha\})\) is closed by 6.11(d).

(b) \(\Rightarrow\) (c): \(\varphi(0) = 0 \ne \alpha\), so \(0 \notin S = \overline{S}\) and hence \(\overline{S} \ne V\).

(c) \(\Rightarrow\) (a), contrapositively: if \(\varphi\) is unbounded, then statement (a) of 6.52 fails, hence so does its equivalent statement (d), giving \(\overline{\operatorname{null} \varphi} = V\) and

\begin{equation*} \overline{S} = h + \overline{\operatorname{null} \varphi} = h + V = V . \end{equation*}

If \(U \ne \operatorname{null} \varphi\), a single vector of \(U\) outside \(\operatorname{null} \varphi\) sweeps out \(V\). Pick \(u \in U\) with \(\beta = \varphi(u) \ne 0\), and let \(g \in V\). Then

\begin{equation*} \varphi\Bigl(g - \frac{\varphi(g)}{\beta} u\Bigr) = \varphi(g) - \frac{\varphi(g)}{\beta}\,\varphi(u) = 0 , \end{equation*}

so \(g - \frac{\varphi(g)}{\beta} u \in \operatorname{null} \varphi \subseteq U\), while \(\frac{\varphi(g)}{\beta} u \in U\) because \(U\) is a subspace containing \(u\); adding gives \(g \in U\). Hence \(V \subseteq U\), so \(U = V\).

The scalar is \(\alpha = \psi(h)\) for any \(h\) with \(\varphi(h) = 1\). Let \(V\) be the common domain.

If \(\psi = \alpha \varphi\) and \(\varphi(f) = 0\), then \(\psi(f) = \alpha \cdot 0 = 0\), so \(\operatorname{null} \varphi \subseteq \operatorname{null} \psi\).

Conversely, suppose \(\operatorname{null} \varphi \subseteq \operatorname{null} \psi\).

(i) \(\varphi\) identically \(0\). Then \(V = \operatorname{null} \varphi \subseteq \operatorname{null} \psi\), so \(\psi\) is identically \(0\) and \(\psi = 0 \cdot \varphi\).

(ii) \(\varphi(g) \ne 0\) for some \(g \in V\). Put \(h = g/\varphi(g)\), so \(\varphi(h) = 1\), and set \(\alpha = \psi(h)\). For \(f \in V\),

\begin{equation*} \varphi\bigl(f - \varphi(f) h\bigr) = \varphi(f) - \varphi(f)\varphi(h) = 0 , \end{equation*}

so \(f - \varphi(f) h \in \operatorname{null} \varphi \subseteq \operatorname{null} \psi\), and linearity of \(\psi\) gives

\begin{equation*} 0 = \psi\bigl(f - \varphi(f) h\bigr) = \psi(f) - \varphi(f) \psi(h) = \psi(f) - \alpha \varphi(f) . \end{equation*}

Hence \(\psi = \alpha \varphi\).

Every linear map \(T : \mathbb{F}^n \to V\) is bounded, hence continuous by 6.48.

Let \(e_1, \ldots, e_n\) be the standard basis of \(\mathbb{F}^n\) and put \(c = \sum_{k=1}^{n} \|T e_k\| < \infty\). For \(x = (a_1, \ldots, a_n) \in \mathbb{F}^n\) we have \(x = \sum_{k=1}^{n} a_k e_k\) and \(|a_k| \leq \|x\|_\infty\), so

\begin{equation*} \begin{aligned} \|T x\| &= \Bigl\| \sum_{k=1}^{n} a_k \, T e_k \Bigr\| \leq \sum_{k=1}^{n} |a_k| \, \|T e_k\| \\ &\leq \Bigl( \max_{1 \leq k \leq n} |a_k| \Bigr) \sum_{k=1}^{n} \|T e_k\| = c \, \|x\|_\infty . \end{aligned} \end{equation*}

Thus \(\|T\| \leq c < \infty\), so \(T\) is bounded and therefore continuous by 6.48.

(a) Were the infimum \(0\), a Bolzano-Weierstrass extraction would produce a nonzero element of \(\operatorname{null} T\).

Write \(S = \{x \in \mathbb{F}^n : \|x\|_\infty = 1\}\) and \(c = \inf\{\|Tx\| : x \in S\} \geq 0\), and suppose \(c = 0\). Choose \(x_k \in S\) with \(\|T x_k\| < \frac{1}{k}\). Every coordinate of every \(x_k\) has absolute value at most \(1\), so applying Bolzano-Weierstrass successively in each of the \(n\) coordinates (to real and imaginary parts when \(\mathbb{F} = \mathbb{C}\)) yields a subsequence converging coordinatewise to some \(x \in \mathbb{F}^n\); coordinatewise convergence of finitely many coordinates is convergence in \(\|\cdot\|_\infty\), so \(\|x_{k_m} - x\|_\infty \to 0\), and the reverse triangle inequality then forces \(\|x\|_\infty = 1\). By 6D.4 the map \(T\) is continuous, so

\begin{equation*} \|Tx\| = \lim_{m \to \infty} \|T x_{k_m}\| \leq \lim_{m \to \infty} \tfrac{1}{k_m} = 0 . \end{equation*}

Thus \(x\) is a nonzero element of \(\operatorname{null} T\), contradicting that \(T\) is one-to-one. Hence \(c > 0\).

(b) \(T^{-1}\) is bounded with \(\|T^{-1}\| \leq 1/c\), where \(c > 0\) is the infimum from (a).

Linearity of \(T^{-1}\): applying \(T^{-1}\) to \(T(T^{-1} g + \alpha \, T^{-1} h) = g + \alpha h\) gives \(T^{-1}(g + \alpha h) = T^{-1} g + \alpha \, T^{-1} h\). Next, \(\|Tx\| \geq c \|x\|_\infty\) for every \(x \in \mathbb{F}^n\): this is trivial for \(x = 0\), and for \(x \neq 0\) the vector \(x / \|x\|_\infty\) lies in \(S\), so homogeneity gives \(c \leq \|Tx\| / \|x\|_\infty\). Hence for \(g \in V\), taking \(x = T^{-1} g\),

\begin{equation*} \|T^{-1} g\|_\infty \leq \tfrac{1}{c} \|T x\| = \tfrac{1}{c} \|g\| , \end{equation*}

so \(T^{-1}\) is a bounded linear map with \(\|T^{-1}\| \leq 1/c\).

Both parts follow from the comparison: for every norm \(\|\cdot\|\) on \(\mathbb{F}^n\) there are \(c, C \in (0,\infty)\) with

\begin{equation*} c \, \|x\|_\infty \leq \|x\| \leq C \, \|x\|_\infty \quad \text{for all } x \in \mathbb{F}^n . \end{equation*}

For the upper bound, take \(C = 1 + \sum_{k=1}^{n} \|e_k\|\) with \(e_1, \ldots, e_n\) the standard basis; the computation of 6D.4, applied to the identity map \(I\) from \((\mathbb{F}^n, \|\cdot\|_\infty)\) to \((\mathbb{F}^n, \|\cdot\|)\), gives \(\|x\| \leq C \|x\|_\infty\). For the lower bound, \(I\) is linear, one-to-one, and onto the normed vector space \((\mathbb{F}^n, \|\cdot\|)\), so 6D.5(a) gives

\begin{equation*} c := \inf \{ \|x\| : \|x\|_\infty = 1 \} > 0 ; \end{equation*}

then \(c \|x\|_\infty \leq \|x\|\) for all \(x\), trivially at \(x = 0\) and by applying the definition of \(c\) to \(x / \|x\|_\infty\) otherwise.

Consequently any two norms \(\|\cdot\|_a, \|\cdot\|_b\) on \(\mathbb{F}^n\) satisfy \(\lambda \|x\|_b \leq \|x\|_a \leq \Lambda \|x\|_b\) for all \(x\), with \(\Lambda = C_a / c_b\) and \(\lambda = c_a / C_b\).

(a) Because \(\lambda \|x_k - x\|_b \leq \|x_k - x\|_a \leq \Lambda \|x_k - x\|_b\), one side tends to \(0\) exactly when the other does, so the two norms have the same convergent sequences with the same limits. For open sets, the same bounds give

\begin{equation*} B_b(y, r/\Lambda) \subseteq B_a(y, r) \quad\text{and}\quad B_a(y, \lambda r) \subseteq B_b(y, r) , \end{equation*}

so a set open for one norm is open for the other. Complements then give the same closed sets.

(b) Let \(\|\cdot\|\) be a norm on \(\mathbb{F}^n\) with \(c, C\) as above, and let \(x_1, x_2, \ldots\) be \(\|\cdot\|\)-Cauchy. From \(\|x_j - x_k\|_\infty \leq \frac{1}{c} \|x_j - x_k\|\) the sequence is \(\|\cdot\|_\infty\)-Cauchy, hence each of its \(n\) coordinate sequences \((a_{k,m})_{k \in \mathbb{Z}^+}\) is Cauchy in \(\mathbb{F}\) and so converges to some \(a_m\), the field \(\mathbb{F}\) being complete. With \(x = (a_1, \ldots, a_n)\),

\begin{equation*} \begin{aligned} \|x_k - x\| &\leq C \, \|x_k - x\|_\infty \\ &= C \max_{1 \leq m \leq n} |a_{k,m} - a_m| \longrightarrow 0 . \end{aligned} \end{equation*}

Thus \((\mathbb{F}^n, \|\cdot\|)\) is a Banach space.

Every linear map \(T : V \to W\) is bounded, hence continuous by 6.48.

Assume \(V \neq \{0\}\) (otherwise \(T = 0\)), let \(e_1, \ldots, e_n\) be a basis of \(V\), and define \(S : \mathbb{F}^n \to V\) by \(S(a_1, \ldots, a_n) = a_1 e_1 + \cdots + a_n e_n\), where \(\mathbb{F}^n\) carries \(\|\cdot\|_\infty\). Because \(e_1, \ldots, e_n\) spans \(V\) and is linearly independent, \(S\) is linear, one-to-one, and onto \(V\); hence \(S^{-1}\) is bounded by 6D.5(b), and the linear map \(T \circ S\) is bounded by 6D.4 together with 6.48. Set \(M = \|S^{-1}\|\) and \(K = \|T \circ S\|\). For \(f \in V\), applying \(\|Rg\| \leq \|R\| \, \|g\|\) (immediate from 6.43) twice,

\begin{equation*} \|T f\| = \|(T \circ S)(S^{-1} f)\| \leq K \, \|S^{-1} f\|_\infty \leq K M \, \|f\| . \end{equation*}

Hence \(\|T\| \leq KM < \infty\), so \(T\) is bounded and therefore continuous by 6.48.

Suppose \(V\) is a finite-dimensional normed vector space; completeness transfers to \(V\) from \(\mathbf{F}^n\) through a coordinate map.

If \(V = \{0\}\) then \(V\) is trivially complete. Otherwise 6.54, after discarding from a finite spanning family each element lying in the span of the others, supplies a basis \(e_1, \ldots, e_n\) of \(V\); define \(T \colon \mathbf{F}^n \to V\) by \(T(a_1, \ldots, a_n) = a_1 e_1 + \cdots + a_n e_n\), with \(\mathbf{F}^n\) carrying \(\|\cdot\|_\infty\). Then \(T\) is linear, and one-to-one and onto \(V\) because \(e_1, \ldots, e_n\) is a basis. By 6D.4 the map \(T\) is bounded; set \(M = \|T\|\). By 6D.5(b) the map \(T^{-1}\) is bounded; set \(N = \|T^{-1}\|\).

Let \(f_1, f_2, \ldots\) be Cauchy in \(V\) and put \(x_j = T^{-1} f_j\). Then

\begin{equation*} \|x_j - x_k\|_\infty \le N \|f_j - f_k\| , \end{equation*}

so \(x_1, x_2, \ldots\) is \(\|\cdot\|_\infty\)-Cauchy; each of its \(n\) coordinate sequences is then Cauchy in the complete field \(\mathbf{F}\), so \(\|x_j - x\|_\infty \to 0\) for the coordinatewise limit \(x\). Hence

\begin{equation*} \|f_j - Tx\| = \|T(x_j - x)\| \le M \|x_j - x\|_\infty \to 0 , \end{equation*}

so \(f_j \to Tx \in V\) and \(V\) is a Banach space.

Suppose \(U\) is a finite-dimensional subspace of a normed vector space \(V\); with the inherited norm, \(U\) is a Banach space by Exercise 8 in this section.

Suppose \(f \in \overline{U}\). By 6.7 the ball of radius \(\frac{1}{j}\) about \(f\) meets \(U\), so there is a sequence \(f_1, f_2, \ldots\) in \(U\) with \(\|f_j - f\| \to 0\). A convergent sequence is Cauchy, so completeness of \(U\) supplies \(g \in U\) with \(\|f_j - g\| \to 0\), whence

\begin{equation*} \|f - g\| \le \|f - f_j\| + \|f_j - g\| \to 0 . \end{equation*}

Thus \(f = g \in U\), so \(\overline{U} = U\) and \(U\) is closed in \(V\).

Take

\begin{equation*} V = \{(a_1, a_2, \ldots) \in \ell^\infty : a_k = 0 \text{ for all but finitely many } k\} , \end{equation*}

with the norm \(\|\cdot\|_\infty\) inherited from \(\ell^\infty\), and let \(e_k \in V\) be the sequence with \(1\) in slot \(k\) and \(0\) elsewhere; then \(\{e_k\}_{k \in \mathbf{Z}^+}\) is a basis of \(V\) in the sense of 6.54.

\(V\) is a subspace of \(\ell^\infty\), since the coordinates of \(a + \alpha b\) vanish off the union of the two finite sets where those of \(a\) and \(b\) are nonzero, so \(V\) is a normed vector space. The family \(\{e_k\}\) spans \(V\): if \(a \in V\) has \(a_k = 0\) for \(k > n\), then \(a = \sum_{k=1}^{n} a_k e_k\). It is linearly independent: if \(\Omega\) is finite and \(\sum_{k \in \Omega} \alpha_k e_k = 0\), reading off the \(j^{\text{th}}\) coordinate gives \(\alpha_j = 0\) for each \(j \in \Omega\).

Finally \(V\) is infinite-dimensional: given \(f_1, \ldots, f_m \in V\), all their coordinates past some index \(N\) vanish, as then do those of every element of \(\operatorname{span}(f_1, \ldots, f_m)\), whereas \(e_{N+1} \in V\) has \((N+1)^{\text{st}}\) coordinate \(1\).

The union of a chain \(\mathcal{C} \subseteq \mathcal{A}\) is its largest member \(m\mathbf{Z} \in \mathcal{A}\), where \(m\) is the smallest \(k\) with \(k\mathbf{Z} \in \mathcal{C}\). (Chains are understood nonempty, the only reading under which 6.60 can apply to a collection omitting \(\emptyset\).)

For \(j, k \in \{2, 3, 4, \ldots\}\) we have the divisibility criterion

\begin{equation*} k\mathbf{Z} \subseteq j\mathbf{Z} \iff j \text{ divides } k : \end{equation*}

if \(k\mathbf{Z} \subseteq j\mathbf{Z}\) then \(k \in j\mathbf{Z}\), so \(k = jm\) for some \(m \in \mathbf{Z}\); conversely \(k = jm\) gives \(kn = j(mn) \in j\mathbf{Z}\).

Now let \(\mathcal{C} \subseteq \mathcal{A}\) be a chain, put \(S = \{k \ge 2 : k\mathbf{Z} \in \mathcal{C}\}\), and let \(m = \min S\), which exists by the well-ordering of \(\mathbf{Z}^+\). Then \(m\mathbf{Z} \subseteq \bigcup_{E \in \mathcal{C}} E\). Conversely let \(k \in S\); since \(\mathcal{C}\) is a chain, either \(k\mathbf{Z} \subseteq m\mathbf{Z}\), or \(m\mathbf{Z} \subseteq k\mathbf{Z}\), in which case \(k\) divides \(m\), so \(k \le m\) and minimality forces \(k = m\). Either way \(k\mathbf{Z} \subseteq m\mathbf{Z}\), so

\begin{equation*} \bigcup_{E \in \mathcal{C}} E = \bigcup_{k \in S} k\mathbf{Z} = m\mathbf{Z} \in \mathcal{A} , \end{equation*}

which is the hypothesis of 6.60.

Apply Zorn’s Lemma (6.60) to the collection \(\mathcal{A}\) of all linearly independent subsets of \(V\) containing \(A = \{f_j : j \in \Lambda\}\), where \(\{f_j\}_{j \in \Lambda}\) is the given linearly independent family.

Linear independence forces \(j \mapsto f_j\) to be one-to-one (otherwise \(f_i - f_j = 0\) is a nontrivial vanishing combination), so, as in the convention preceding 6.57, the family may be identified with the set \(A\); note \(A \in \mathcal{A}\).

Hypothesis of 6.60: let \(\mathcal{C} \subseteq \mathcal{A}\) be a chain (nonempty, as in Exercise 11; when \(A = \emptyset\) the empty union \(\emptyset\) lies in \(\mathcal{A}\) too) and let \(U = \bigcup_{E \in \mathcal{C}} E\), so \(A \subseteq U\). Every finite subset of \(U\) lies in a single element of \(\mathcal{C}\): choosing \(E_k \in \mathcal{C}\) with \(u_k \in E_k\), an induction on \(n\) using the chain condition makes one of \(E_1, \ldots, E_n\) contain the others. Since linear independence involves only finitely many elements at a time, a relation \(\sum_{u \in \Omega} \alpha_u u = 0\) over a finite \(\Omega \subseteq U\) takes place inside one linearly independent \(E \in \mathcal{C}\), so every \(\alpha_u = 0\). Hence \(U \in \mathcal{A}\).

By 6.60 there is a maximal \(\Gamma \in \mathcal{A}\), and \(\Gamma\) is maximal among all linearly independent subsets of \(V\): any linearly independent \(\Gamma^{\prime} \supsetneq \Gamma\) contains \(A\) and so lies in \(\mathcal{A}\). Therefore 6.57 makes \(\Gamma\) a basis of \(V\), and \(A \subseteq \Gamma\) says it extends the given family.

The bridge is that the norm-preserving extensions of \(\psi\) to \(U + \mathbf{R}h\) are exactly the \(\varphi_c\) with \(s(h) \le c \le t(h)\), so uniqueness on \(V\) is equivalent to \(s(h) = t(h)\) throughout \(V \setminus U\). Here \(\mathbf{F} = \mathbf{R}\), since \(\psi\) is real-valued and over \(\mathbf{C}\) the identity \(i \psi(f) = \psi(if) \in \mathbf{R}\) would force \(\psi = 0\); for \(h \in V \setminus U\) write

\begin{equation*} s(h) = \sup_{f \in U} \big( -\|\psi\| \, \|f + h\| - \psi(f) \big), \qquad t(h) = \inf_{g \in U} \big( \|\psi\| \, \|g + h\| - \psi(g) \big) . \end{equation*}

Taking \(f = g = 0\) and invoking 6.66 gives \(-\|\psi\| \, \|h\| \le s(h) \le t(h) \le \|\psi\| \, \|h\|\), so \(s(h)\) and \(t(h)\) are real and the displayed condition says exactly \(s(h) = t(h)\).

Claim. For \(c \in \mathbf{R}\), the linear functional \(\varphi_c(f + \alpha h) = \psi(f) + \alpha c\) on \(U + \mathbf{R}h\) (well defined since \(h \notin U\)) extends \(\psi\) with \(\|\varphi_c\| = \|\psi\|\) if and only if \(s(h) \le c \le t(h)\). Indeed \(\varphi_c|_U = \psi\) gives \(\|\varphi_c\| \ge \|\psi\|\), so \(\|\varphi_c\| = \|\psi\|\) is exactly 6.64; as in the proof of 6.63, 6.64 is equivalent to 6.65, since 6.64 at \(\alpha = 1\) is 6.65 while 6.64 is trivial at \(\alpha = 0\) and follows for \(\alpha \ne 0\) from 6.65 applied to \(f/\alpha\) after multiplying by \(|\alpha|\). Finally 6.65 reads

\begin{equation*} \bigl( -\|\psi\| \, \|f + h\| - \psi(f) \bigr) \le c \le \bigl( \|\psi\| \, \|f + h\| - \psi(f) \bigr) \end{equation*}

for all \(f \in U\), whose supremum and infimum over \(U\) are \(s(h) \le c \le t(h)\).

(i) Suppose \(s(h) = t(h)\) for all \(h \in V \setminus U\). At least one norm-preserving extension exists by the Hahn-Banach Theorem (6.69). If \(\varphi_1, \varphi_2\) are two of them and \(h \in V \setminus U\), then \(c_j = \varphi_j(h)\) satisfies

\begin{equation*} |\psi(f) + c_j| = |\varphi_j(f + h)| \le \|\psi\| \, \|f + h\| \quad (f \in U) , \end{equation*}

which is 6.65, so the claim’s computation gives \(s(h) \le c_j \le t(h)\) and hence \(c_1 = c_2\). As \(\varphi_1 = \psi = \varphi_2\) on \(U\), we get \(\varphi_1 = \varphi_2\).

(ii) Conversely, suppose \(s(h) \neq t(h)\) for some \(h \in V \setminus U\), so \(s(h) < t(h)\) by 6.66. By the claim, \(\varphi_{s(h)}\) and \(\varphi_{t(h)}\) are norm-\(\|\psi\|\) extensions of \(\psi\) to \(U + \mathbf{R}h\) taking different values at \(h\); applying 6.69 to each yields \(\Phi_1, \Phi_2\) on \(V\) with \(\|\Phi_j\| = \|\psi\|\), both restricting to \(\psi\) on \(U\), and \(\Phi_1(h) = s(h) \neq t(h) = \Phi_2(h)\). So the extension is not unique.

Take for \(\varphi\) a norm-preserving Hahn-Banach extension of the limit functional on the subspace of convergent sequences.

Let \(U = \{a \in \ell^\infty : \lim_{k \to \infty} a_k \text{ exists}\}\) and \(\psi(a) = \lim_{k \to \infty} a_k\). Every convergent sequence is bounded and the limit laws are exactly additivity and homogeneity, so \(U\) is a subspace of \(\ell^\infty\) and \(\psi\) is a linear functional on it. Continuity of the absolute value gives

\begin{equation*} |\psi(a)| = \lim_{k \to \infty} |a_k| \le \sup_{k \in \mathbf{Z}^+} |a_k| = \|a\|_\infty , \end{equation*}

so \(\|\psi\| \le 1\), with equality because \(\psi(1,1,1,\ldots) = 1\) and \(\|(1,1,1,\ldots)\|_\infty = 1\).

By the Hahn-Banach Theorem (6.69), applied to the bounded linear functional \(\psi\) on the subspace \(U\) of \(\ell^\infty\), there is a bounded linear functional \(\varphi\) on \(\ell^\infty\) with \(\varphi|_U = \psi\) and \(\|\varphi\| = 1\). Hence \(|\varphi(a)| \le \|\varphi\| \, \|a\|_\infty = \|a\|_\infty\) for every \(a \in \ell^\infty\), while \(\varphi(a) = \psi(a) = \lim_{k \to \infty} a_k\) whenever that limit exists.

Take \(\varphi \in V^{\prime}\) with \(\|\varphi\| = 1\) and \(\varphi(g) = \|g\|\), where \(g\) is the center of \(B = \{f \in V : \|f - g\| < r\}\) and \(r > 0\) its radius.

Such a \(\varphi\) is supplied by 6.72, since \(0 \notin B\) forces \(\|g\| \ge r > 0\) and in particular \(g \ne 0\). For \(f \in B\),

\begin{equation*} \begin{aligned} \operatorname{Re} \varphi(f) &= \operatorname{Re} \varphi(g) + \operatorname{Re} \varphi(f - g) \\ &= \| g \| + \operatorname{Re} \varphi(f - g) \\ &\geq \| g \| - | \varphi(f - g) | \\ &\geq \| g \| - \| \varphi \| \, \| f - g \| \\ &= \| g \| - \| f - g \| \\ &> \| g \| - r \\ &\geq 0, \end{aligned} \end{equation*}

using \(\varphi(g) = \|g\| \in \mathbb{R}\), then \(\operatorname{Re} z \geq -|z|\), then the definition of \(\|\varphi\|\), then \(\|f - g\| < r\), and finally \(\|g\| \geq r\).

For each \(n \in \mathbb{Z}^+\) the dual space \(V^{\prime}\) of an infinite-dimensional \(V\) contains a linearly independent list of length \(n\), which is impossible for a space spanned by fewer than \(n\) elements; hence \(V^{\prime}\) is infinite-dimensional.

First, \(V\) contains linearly independent \(f_1, \ldots, f_n\): starting from any \(f_1 \neq 0\), if \(f_1, \ldots, f_k\) is linearly independent then \(\operatorname{span}\{f_1, \ldots, f_k\} \neq V\) (else \(V\) would be finite-dimensional by 6.54), and any \(f_{k+1}\) outside that span extends the list, since a vanishing combination must have \(\alpha_{k+1} = 0\) and then all \(\alpha_j = 0\).

Let \(W = \operatorname{span}\{f_1, \ldots, f_n\}\) carry the norm inherited from \(V\), and let \(\psi_j : W \to \mathbb{F}\) be the \(j^{\text{th}}\) coordinate functional for the basis \(f_1, \ldots, f_n\) of \(W\), so \(\psi_j(f_k) = 1\) if \(j = k\) and \(0\) otherwise. Each \(\psi_j\) is bounded because \(W\) is finite-dimensional (Exercise 7 in Section 6D), so 6.69 extends it to some \(\varphi_j \in V^{\prime}\) with \(\varphi_j(f_k) = \psi_j(f_k)\).

If \(\alpha_1 \varphi_1 + \cdots + \alpha_n \varphi_n = 0\), evaluating at \(f_k\) gives \(\alpha_k = 0\) for each \(k\). Thus \(\varphi_1, \ldots, \varphi_n\) is linearly independent in \(V^{\prime}\).

Replace the maximal element that Zorn’s Lemma supplies in the proof of 6.69 by an explicitly defined recursion: extend one dimension at a time along a fixed enumeration of a countable dense set, then extend by continuity. Every step is given by a formula (a least index, a supremum, a limit), so no choices are made.

Fix an enumeration \(g_1, g_2, \ldots\) of a countable dense subset of \(V\), indexed by all of \(\mathbb{Z}^+\) (repeat elements if it is finite; if it is empty then \(V = \{0\}\) and 6.69 is trivial). Fixing one such enumeration is a single existential instantiation, not an application of the Axiom of Choice. Let \(\psi\) be a bounded linear functional on a subspace \(U\) of \(V\).

Case (i): \(\mathbb{F} = \mathbb{R}\). The Extension Lemma (6.63) becomes choice-free once \(c\) is specified: for a bounded linear functional \(\rho\) on a subspace \(W\) and \(h \in V \setminus W\), put

\begin{equation*} c = \sup_{f \in W} \bigl( -\| \rho \| \, \| f + h \| - \rho(f) \bigr) , \end{equation*}

a real number, since \(f = 0\) bounds it below by \(-\|\rho\| \, \|h\|\) and 6.66 bounds it above by \(\|\rho\| \, \|h\|\); with this \(c\), the computation in the proof of 6.63 makes \(\rho_h(f + \alpha h) = \rho(f) + \alpha c\) a norm-preserving extension of \(\rho\) to \(W + \mathbb{R}h\).

Recursively set \(U_0 = U\), \(\varphi_0 = \psi\), and given \((U_n, \varphi_n)\): if every \(g_k\) lies in \(U_n\), keep \((U_{n+1}, \varphi_{n+1}) = (U_n, \varphi_n)\); otherwise let \(m_n\) be the least index with \(g_{m_n} \notin U_n\), put \(U_{n+1} = U_n + \mathbb{R} g_{m_n}\), and let \(\varphi_{n+1}\) be the explicit extension above with \(h = g_{m_n}\). Each ingredient is determined by \((U_n, \varphi_n)\) and the enumeration, so the recursion theorem produces the sequence, and \(\|\varphi_n\| = \|\psi\|\) throughout.

Every \(g_k\) lies in some \(U_n\): otherwise \(m_n\) is defined for all \(n\) and strictly increases (as \(g_{m_n} \in U_{n+1}\)), so \(m_n \ge m_0 + n \to \infty\). Hence \(W = \bigcup_{n} U_n\), a subspace because the \(U_n\) increase, is dense in \(V\), and \(\varphi(f) = \varphi_n(f)\) for any \(n\) with \(f \in U_n\) is a well-defined linear functional on \(W\) extending \(\psi\) with \(|\varphi(f)| \le \|\psi\| \, \|f\|\).

For \(f \in V\) let \(k_n(f)\) be the least \(k\) with \(\|g_k - f\| < \frac{1}{n}\), which exists by density. Then

\begin{equation*} \bigl| \varphi(g_{k_n(f)}) - \varphi(g_{k_m(f)}) \bigr| \leq \| \psi \| \Bigl( \frac{1}{n} + \frac{1}{m} \Bigr) , \end{equation*}

so \(\tilde{\varphi}(f) = \lim_{n \to \infty} \varphi(g_{k_n(f)})\) exists by completeness of \(\mathbb{R}\) (itself choice-free, the limit being \(\sup_n \inf_{j \ge n}\) of the sequence). The same estimate shows \(\varphi(h_j) \to \tilde{\varphi}(f)\) for every sequence \(h_j \to f\) in \(W\); applying this along the explicit sequences \(h_j = g_{k_j(f)}\) and \(h_j^{\prime} = g_{k_j(f^{\prime})}\) gives

\begin{equation*} \tilde{\varphi}(f + \alpha f^{\prime}) = \lim_{j \to \infty} \bigl( \varphi(h_j) + \alpha \varphi(h_j^{\prime}) \bigr) = \tilde{\varphi}(f) + \alpha \tilde{\varphi}(f^{\prime}) . \end{equation*}

Taking \(h_j = f\) for \(f \in W\) shows \(\tilde{\varphi}|_W = \varphi\), so \(\tilde{\varphi}\) extends \(\psi\); and \(|\varphi(g_{k_n(f)})| \le \|\psi\| \, \|g_{k_n(f)}\|\) with \(g_{k_n(f)} \to f\) gives \(\|\tilde{\varphi}\| \le \|\psi\|\), hence \(\|\tilde{\varphi}\| = \|\psi\|\).

Case (ii): \(\mathbb{F} = \mathbb{C}\). The reduction in the proof of 6.69 uses no choice: set \(\psi_1 = \operatorname{Re} \psi\), apply case (i) to the real normed vector space \(V\) (same norm, hence the same countable dense set, so separable) to get \(\varphi_1\) extending \(\psi_1\) with \(\|\varphi_1\| = \|\psi_1\|\), and put \(\varphi(f) = \varphi_1(f) - i \varphi_1(if)\); the verifications in 6.69 that \(\varphi\) is \(\mathbb{C}\)-linear, extends \(\psi\), and has \(\|\varphi\| \le \|\psi\|\) are computations.

Take a countable dense \(\{\varphi_1, \varphi_2, \ldots\}\) in \(V^{\prime}\) and vectors \(f_n \in V\) with \(\|f_n\| \le 1\) and \(|\varphi_n(f_n)| > \frac{1}{2} \|\varphi_n\|\) (possible by the definition of \(\|\varphi_n\|\) as a supremum when \(\varphi_n \ne 0\); put \(f_n = 0\) otherwise); then the rational span of \(\{f_n\}\) is a countable dense subset of \(V\).

Let \(W = \operatorname{span}\{f_n : n \in \mathbb{Z}^+\}\) and let \(D\) consist of the finite combinations \(\sum_{n \in \Omega} \alpha_n f_n\) with all \(\alpha_n \in \mathbb{Q}\) (in \(\mathbb{Q} + i\mathbb{Q}\) when \(\mathbb{F} = \mathbb{C}\)), a countable set. Then \(W \subseteq \overline{D}\), because

\begin{equation*} \Bigl\| \sum_{n \in \Omega} (\alpha_n - \beta_n) f_n \Bigr\| \leq \sum_{n \in \Omega} | \alpha_n - \beta_n | \, \| f_n \| \end{equation*}

is arbitrarily small for suitable rational \(\beta_n\); so it suffices to show \(\overline{W} = V\).

If not, take \(h \in V \setminus \overline{W}\). Since \(W\) is a subspace, 6.73 gives \(\varphi \in V^{\prime}\) with \(\varphi|_W = 0\) and \(\varphi(h) \ne 0\), and dividing by \(\|\varphi\| \ne 0\) we may take \(\|\varphi\| = 1\). By density choose \(n\) with \(\|\varphi - \varphi_n\| < \frac{1}{4}\), so \(\|\varphi_n\| \ge \|\varphi\| - \|\varphi - \varphi_n\| > \frac{3}{4}\); in particular \(\varphi_n \ne 0\), so \(f_n\) is as displayed above. But \(f_n \in W\) forces \(\varphi(f_n) = 0\), whence

\begin{equation*} \tfrac{1}{2} \| \varphi_n \| < | \varphi_n(f_n) - \varphi(f_n) | \leq \| \varphi_n - \varphi \| \, \| f_n \| < \tfrac{1}{4} , \end{equation*}

contradicting \(\|\varphi_n\| > \frac{3}{4}\). Hence \(\overline{D} \supseteq \overline{W} = V\) and \(V\) is separable.

The evaluation functionals \(\varphi_t(f) = f(t)\), for \(t \in [0,1]\), form an uncountable family in \(C([0,1])^{\prime}\) that is pairwise at distance \(2\), leaving no room for a countable dense subset.

Each \(\varphi_t\) is linear with \(|\varphi_t(f)| = |f(t)| \le \|f\|\), so \(\varphi_t \in C([0,1])^{\prime}\) with \(\|\varphi_t\| = 1\), the constant function \(1\) attaining the supremum. For \(s \ne t\) put \(\delta = |s - t| > 0\) and

\begin{equation*} f(x) = \max \Bigl\{ -1, \ 1 - \frac{2 | x - s |}{\delta} \Bigr\} , \end{equation*}

continuous with \(\|f\| \le 1\), \(f(s) = 1\), and \(f(t) = -1\); hence

\begin{equation*} 2 = | f(s) - f(t) | \leq \| \varphi_s - \varphi_t \| \leq \| \varphi_s \| + \| \varphi_t \| = 2 . \end{equation*}

Suppose \(\{\mu_1, \mu_2, \ldots\}\) were dense in \(C([0,1])^{\prime}\), and let \(\kappa(t)\) be the least \(j\) with \(\|\mu_j - \varphi_t\| < 1\) (such \(j\) exists by density). If \(\kappa(s) = \kappa(t) = j\) then

\begin{equation*} \| \varphi_s - \varphi_t \| \leq \| \varphi_s - \mu_j \| + \| \mu_j - \varphi_t \| < 2 , \end{equation*}

so \(s = t\) by the previous paragraph. Thus \(\kappa\) is a one-to-one map from the uncountable set \([0,1]\) into \(\mathbb{Z}^+\), which is impossible.

Each \(\Phi f\) is linear on \(V^{\prime}\) by the definition of the vector space operations on \(V^{\prime}\), and the definition of \(\|\varphi\|\) gives

\begin{equation*} | (\Phi f)(\varphi) | = | \varphi(f) | \leq \| \varphi \| \, \| f \| \quad \text{for all } \varphi \in V^{\prime} , \end{equation*}

so \(\Phi f \in V^{\prime\prime}\) with \(\|\Phi f\| \le \|f\|\).

For the reverse inequality, both sides vanish when \(f = 0\); and when \(f \ne 0\), 6.72 supplies \(\varphi \in V^{\prime}\) with \(\|\varphi\| = 1\) and \(\varphi(f) = \|f\|\), whence

\begin{equation*} \| \Phi f \| \geq | (\Phi f)(\varphi) | = | \varphi(f) | = \| f \| . \end{equation*}

Thus \(\|\Phi f\| = \|f\|\) for every \(f \in V\).

Take \(U = \{f \in V : \rho(f) > 0\}\), where \(\rho : V \to \mathbb{R}\) is a discontinuous \(\mathbb{R}\)-linear functional.

Convexity (8.25) involves only real scalars, so regard \(V\) as a real normed vector space; it is still infinite-dimensional, since a finite real spanning list would span \(V\) over \(\mathbb{C}\) as well. Hence 6.62 supplies a discontinuous linear functional \(\rho : V \to \mathbb{R}\), which is unbounded by 6.48 and in particular is not identically \(0\).

\(\operatorname{null} \rho\) is dense in \(V\) (this is (d) \(\Rightarrow\) (a) of 6.52 in contrapositive form): unboundedness gives \(h_j\) with \(\|h_j\| \leq 1\) and \(|\rho(h_j)| > j\), obtained by normalizing any \(f_j\) with \(|\rho(f_j)| > j \|f_j\|\), and then for \(g \in V\) the vectors \(k_j = g - \frac{\rho(g)}{\rho(h_j)} h_j\) satisfy \(\rho(k_j) = 0\) and

\begin{equation*} \| k_j - g \| = \frac{| \rho(g) |}{| \rho(h_j) |} \| h_j \| \leq \frac{| \rho(g) |}{j} \longrightarrow 0 . \end{equation*}

Both \(U\) and \(V \setminus U = \{f : \rho(f) \le 0\}\) are convex: for \(t \in [0,1]\),

\begin{equation*} \rho\bigl( (1-t) f + t g \bigr) = (1-t) \rho(f) + t \rho(g) , \end{equation*}

which is \(> 0\) when \(\rho(f), \rho(g) > 0\) (both terms are \(\ge 0\) and one of \(t\), \(1-t\) is nonzero) and is \(\le 0\) when \(\rho(f), \rho(g) \le 0\).

Density of \(V \setminus U\) is immediate from \(\operatorname{null} \rho \subseteq V \setminus U\). For \(U\), fix \(h = h_0 / \rho(h_0)\) with \(\rho(h_0) \ne 0\), so \(\rho(h) = 1\) and \(h \ne 0\); given \(g \in V\) and \(\varepsilon > 0\), pick \(k \in \operatorname{null} \rho\) with \(\|k - g\| < \varepsilon/2\) and set \(\delta = \frac{\varepsilon}{2\|h\| + 2} > 0\) and \(u = k + \delta h\). Then \(\rho(u) = \delta > 0\), so \(u \in U\), while

\begin{equation*} \| u - g \| \leq \| k - g \| + \delta \| h \| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon . \end{equation*}

Hence \(\overline{U} = V = \overline{V \setminus U}\).

Both directions are 6.7, which says \(\overline{U}\) consists of the \(g \in V\) with \(B(g,\varepsilon) \cap U \ne \emptyset\) for every \(\varepsilon > 0\).

(i) Suppose \(\overline{U} = V\) and let \(G\) be a nonempty open subset of \(V\). Pick \(f \in G\) and, by 6.4, an \(r > 0\) with \(B(f,r) \subseteq G\). Then \(f \in \overline{U}\) gives \(B(f,r) \cap U \ne \emptyset\), and any of its elements lies in \(G \cap U\).

(ii) Conversely, let \(f \in V\) and \(\varepsilon > 0\). The ball \(B(f,\varepsilon)\) is open (6.5) and contains \(f\), so by hypothesis it meets \(U\); as \(\varepsilon\) was arbitrary, 6.7 gives \(f \in \overline{U}\). Hence \(\overline{U} = V\).

Both directions run through Exercise 1: \(V \setminus U\) is dense exactly when every nonempty open subset of \(V\) meets \(V \setminus U\), that is, when no nonempty open set is contained in \(U\).

(i) Suppose \(\operatorname{int} U = \emptyset\) and some nonempty open \(G\) satisfies \(G \subseteq U\). Any \(f \in G\) then has \(B(f,r) \subseteq G \subseteq U\) for some \(r > 0\) by 6.4, so \(f \in \operatorname{int} U\) by 6.74, a contradiction. Hence every nonempty open set meets \(V \setminus U\), and Exercise 1 gives density.

(ii) Conversely, suppose \(V \setminus U\) is dense and \(f \in \operatorname{int} U\), so \(B(f,r) \subseteq U\) for some \(r > 0\). The ball \(B(f,r)\) is nonempty and open (6.5), so by Exercise 1 it contains an element of \(V \setminus U\), a contradiction. Hence \(\operatorname{int} U = \emptyset\).

False: in \(V = \mathbb{R}\) take \(U = \mathbb{Q}\) and \(W = \mathbb{R} \setminus \mathbb{Q}\). Each is dense in \(\mathbb{R}\) and is the complement of the other, so Exercise 2 gives \(\operatorname{int} U = \operatorname{int} W = \emptyset\), whereas

\begin{equation*} \operatorname{int}(U \cup W) = \operatorname{int} \mathbb{R} = \mathbb{R} \neq \emptyset = (\operatorname{int} U) \cup (\operatorname{int} W) . \end{equation*}

Only the inclusion \(\subseteq\) holds in general: if \(f \in \operatorname{int} U\) then \(B(f,r) \subseteq U \subseteq U \cup W\) for some \(r > 0\), so \(f \in \operatorname{int}(U \cup W)\) by 6.74, and symmetrically with \(W\) in place of \(U\).

True; by 6.74, \(f \in \operatorname{int} E\) means \(f \in E\) and \(B(f,r) \subseteq E\) for some \(r > 0\).

(i) If \(f \in (\operatorname{int} U) \cap (\operatorname{int} W)\), choose \(r_1, r_2 > 0\) with \(B(f,r_1) \subseteq U\) and \(B(f,r_2) \subseteq W\); then \(f \in U \cap W\) and \(B(f, \min\{r_1, r_2\}) \subseteq U \cap W\), so \(f \in \operatorname{int}(U \cap W)\).

(ii) If \(f \in \operatorname{int}(U \cap W)\), choose \(r > 0\) with \(B(f,r) \subseteq U \cap W\); since \(U \cap W\) is contained in each of \(U\) and \(W\), this gives \(f \in \operatorname{int} U\) and \(f \in \operatorname{int} W\).

(a) \(X\) is a closed subset of the complete metric space \(\mathbb{R}\), hence complete.

Closedness: since \(X \subseteq [0,1]\) and every \(x \in (0,1)\) not of the form \(1/k\) satisfies \(\frac{1}{k+1} < x < \frac{1}{k}\) for exactly one \(k \in \mathbb{Z}^+\),

\begin{equation*} \mathbb{R} \setminus X = (-\infty, 0) \cup (1, \infty) \cup \bigcup_{k=1}^{\infty} \left( \tfrac{1}{k+1}, \tfrac{1}{k} \right) , \end{equation*}

a union of open sets, so \(X\) is closed by 6.6. A Cauchy sequence in \(X\) is Cauchy in \(\mathbb{R}\), hence converges to some \(x \in \mathbb{R}\), and \(x \in X\) by 6.9(e); as \(d\) is the restriction of the metric of \(\mathbb{R}\), this is convergence in \((X,d)\).

(b) Baire’s Theorem [6.76(a)] concerns empty interior in the complete space at hand, here \(X\) rather than \(\mathbb{R}\), and each singleton \(\{\frac{1}{k}\}\) has nonempty interior in \(X\).

Indeed, with \(r = \frac{1}{k(k+1)}\) the only element of \(X\) within \(r\) of \(\frac{1}{k}\) is \(\frac{1}{k}\) itself: \(|\frac{1}{k} - 0| = \frac{1}{k} > r\), while \(|\frac{1}{k} - \frac{1}{j}| \ge \frac{1}{k} - \frac{1}{k+1} = r\) for \(j > k\) and \(|\frac{1}{k} - \frac{1}{j}| \ge \frac{1}{k-1} - \frac{1}{k} > r\) for \(j < k\). So \(B(\frac{1}{k}, r) = \{\frac{1}{k}\}\), which by 6.74 gives \(\operatorname{int}\{\frac{1}{k}\} = \{\frac{1}{k}\}\) in \(X\). (Only \(\{0\}\) has empty interior in \(X\), since every \(B(0,r)\) contains some \(\frac{1}{k} \ne 0\).) Thus the displayed decomposition is not a union of closed sets with empty interior in \(X\), and 6.76(a) is not contradicted.

Take \(\mathbb{Q}\) with \(d(x,y) = |x-y|\) and an enumeration \(r_1, r_2, \ldots\) of its elements, so that

\begin{equation*} \mathbb{Q} = \bigcup_{j=1}^{\infty} \{ r_j \} . \end{equation*}

Each \(\{r_j\}\) is closed in \(\mathbb{Q}\) by 6.6, since \(B(x, |x - r_j|) \subseteq \mathbb{Q} \setminus \{r_j\}\) for every rational \(x \ne r_j\). Each has empty interior in \(\mathbb{Q}\): given \(r > 0\), density of \(\mathbb{Q}\) in \(\mathbb{R}\) puts a rational \(q \ne r_j\) in \((r_j, r_j + r)\), so \(B(r_j, r) \not\subseteq \{r_j\}\) and 6.74 applies.

(a) \(f\) is continuous exactly at the irrational numbers. Write \(n(a)\) for the least positive integer with \(a = m/n(a)\) for some \(m \in \mathbb{Z}\), so \(f(a) = 1/n(a) > 0\) for rational \(a\) and \(f = 0\) on the irrationals.

Discontinuity at rational \(a\): take \(\varepsilon = f(a) > 0\). Every interval \((a - \delta, a + \delta)\) contains an irrational \(x\), by density of \(\mathbb{R} \setminus \mathbb{Q}\), and then \(|f(x) - f(a)| = f(a) = \varepsilon\), so no \(\delta\) works in 6.10.

Continuity at irrational \(a\): here \(f(a) = 0\). Given \(\varepsilon > 0\), choose \(N \in \mathbb{Z}^+\) with \(\frac{1}{N} < \varepsilon\) and let

\begin{equation*} S = \left\{ x \in (a-1, a+1) : f(x) \geq \varepsilon \right\}. \end{equation*}

Each \(x \in S\) has \(f(x) > 0\), hence is rational with \(\frac{1}{n(x)} \ge \varepsilon > \frac{1}{N}\), so \(n(x) < N\); and for each fixed \(n \le N\) the condition \(\frac{m}{n} \in (a-1, a+1)\) confines \(m\) to an interval of length \(2n \le 2N\), so at most \(2N+1\) integers. Hence \(S\) is finite, and \(a \notin S\) because \(a\) is irrational. Put

\begin{equation*} \delta = \min \left( \{1\} \cup \{ |a - s| : s \in S \} \right) > 0 . \end{equation*}

If \(|x - a| < \delta\) then \(x \in (a-1, a+1) \setminus S\), so \(|f(x) - f(a)| = f(x) < \varepsilon\).

(b) Suppose \(G_1, G_2, \ldots\) are open with \(\bigcap_{k} G_k = \mathbb{Q}\). Each \(G_k\) contains \(\mathbb{Q}\), hence is dense. Enumerating \(\mathbb{Q} = \{r_1, r_2, \ldots\}\), each \(H_j = \mathbb{R} \setminus \{r_j\}\) is open, and dense because every nonempty open subset of \(\mathbb{R}\) is infinite (Exercise 1). But then \(G_1, H_1, G_2, H_2, \ldots\) are countably many dense open subsets of the complete metric space \(\mathbb{R}\) with

\begin{equation*} \Bigl( \bigcap_{k=1}^{\infty} G_k \Bigr) \cap \Bigl( \bigcap_{j=1}^{\infty} H_j \Bigr) = \mathbb{Q} \cap (\mathbb{R} \setminus \mathbb{Q}) = \emptyset , \end{equation*}

contradicting Baire’s Theorem [6.76(b)].

(c) The set of continuity points of any \(f : \mathbb{R} \to \mathbb{R}\) is a countable intersection of open sets, so by (b) it is never \(\mathbb{Q}\).

For \(a \in \mathbb{R}\) and \(\delta > 0\) let \(\omega(a, \delta) = \sup\{|f(x) - f(y)| : x, y \in (a-\delta, a+\delta)\} \in [0,\infty]\) and \(\omega(a) = \inf_{\delta > 0} \omega(a,\delta)\). Then \(f\) is continuous at \(a\) if and only if \(\omega(a) = 0\): continuity gives \(\delta\) with \(|f(x) - f(a)| < \varepsilon/3\) on \((a-\delta, a+\delta)\), so \(\omega(a) \le \omega(a,\delta) \le 2\varepsilon/3 < \varepsilon\) for every \(\varepsilon > 0\); conversely \(\omega(a,\delta) < \varepsilon\) gives \(|f(x) - f(a)| \le \omega(a,\delta) < \varepsilon\) for \(|x - a| < \delta\), taking \(y = a\) in the supremum.

Each \(G_n = \{a : \omega(a) < \frac{1}{n}\}\) is open: choosing \(\delta\) with \(\omega(a,\delta) < \frac{1}{n}\), any \(b\) with \(|b - a| < \delta/2\) has \((b - \frac{\delta}{2}, b + \frac{\delta}{2}) \subseteq (a - \delta, a + \delta)\) and hence \(\omega(b) \le \omega(b, \delta/2) \le \omega(a,\delta) < \frac{1}{n}\), so \(B(a, \delta/2) \subseteq G_n\) (6.4). The set of continuity points is therefore \(\bigcap_{n=1}^{\infty} G_n\), which by (b) cannot equal \(\mathbb{Q}\).

By Exercise 1 in Section 6E, density of \(\bigcap_{k=1}^{\infty} G_k\) amounts to: every nonempty open \(U \subseteq X\) meets it. So fix such a \(U\), pick \(f \in U\), and use the fact proved just before 6.76 to get \(r > 0\) with

\begin{equation*} B(f, r) \subseteq Y := \overline{B(f, r)} \subseteq U . \end{equation*}

Then \(Y\) is nonempty and closed in \(X\), so \((Y, d)\) is a complete metric space by 6.16(b).

Set \(H_k = G_k \cap Y\), open in \(Y\) because \(G_k\) is open in \(X\). Each \(H_k\) is dense in \(Y\): given \(y \in Y\) and \(\varepsilon > 0\), the point \(y\) lies in the closure of \(B(f,r)\), so there is \(x \in B(f, r)\) with \(d(x, y) < \varepsilon/2\); the set \(B(f, r) \cap B(x, \varepsilon/2)\) is open and nonempty, so density of \(G_k\) supplies a point \(z\) of it lying in \(G_k\), and then \(z \in G_k \cap B(f,r) \subseteq H_k\) with

\begin{equation*} d(z, y) \le d(z, x) + d(x, y) < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

Baire’s Theorem [6.76(b)], applied to the dense open subsets \(H_1, H_2, \ldots\) of the complete metric space \(Y\), produces

\begin{equation*} g \in \bigcap_{k=1}^{\infty} H_k = Y \cap \bigcap_{k=1}^{\infty} G_k \subseteq U \cap \bigcap_{k=1}^{\infty} G_k , \end{equation*}

as required.

Suppose \(V\) is an infinite-dimensional Banach space with a countable basis in the sense of 6.54. The basis cannot be finite (else \(V\) would be finite-dimensional), so relabel it as \(e_1, e_2, \ldots\) and put \(V_n = \operatorname{span}\{e_1, \ldots, e_n\}\).

(i) \(V = \bigcup_{n=1}^{\infty} V_n\), because every \(f \in V\) is a finite linear combination \(\sum_{j \in \Omega} \alpha_j e_j\), so \(f \in V_n\) for \(n = \max \Omega\).

(ii) Each \(V_n\) is closed in \(V\) by Exercise 9 in Section 6D.

(iii) Each \(V_n\) has empty interior. Otherwise \(B(g, \rho) \subseteq V_n\) for some \(g \in V_n\) and \(\rho > 0\); then for \(f \in V\) with \(f \ne 0\),

\begin{equation*} \Big\| \Big( g + \tfrac{\rho}{2 \|f\|} f \Big) - g \Big\| = \frac{\rho}{2} < \rho , \end{equation*}

so \(g + \frac{\rho}{2\|f\|} f \in V_n\), and subtracting \(g \in V_n\) and rescaling gives \(f \in V_n\). Thus \(V = V_n\) is finite-dimensional, a contradiction.

Because \(V\) is a Banach space, \((V, d)\) with \(d(f,g) = \|f - g\|\) is a complete metric space, so (i)–(iii) exhibit \(V\) as a countable union of closed sets with empty interior, contradicting Baire’s Theorem [6.76(a)].

Take \(V = \ell^1\) with \(\|a\|_1 = \sum_{k=1}^{\infty} |a_k|\), take \(W = \ell^1\) with \(\|a\|_{\infty} = \sup_{k} |a_k|\), let \(T : V \to W\) be the identity map, and let \(G = \{a \in \ell^1 : \|a\|_1 < 1\}\) be the open unit ball of \(V\).

Then \(V\) is a Banach space (second bullet of Example 6.38) and \(W\) is a normed vector space (second bullet of Example 6.39). The map \(T\) is linear and onto \(W\), being the identity on the common underlying vector space, and \(|a_k| \le \sum_{j} |a_j|\) for each \(k\) gives

\begin{equation*} \|Ta\|_{\infty} \le \|a\|_1 , \end{equation*}

so \(T\) is bounded with \(\|T\| \le 1\).

But \(T(G) = G\) contains no \(\|\cdot\|_{\infty}\)-ball about \(0\): given \(\varepsilon > 0\), choose \(n > 2/\varepsilon\) and let \(a\) have \(n\) entries equal to \(\varepsilon/2\) followed by zeros. Then

\begin{equation*} \|a\|_{\infty} = \frac{\varepsilon}{2} < \varepsilon , \qquad \|a\|_1 = n \cdot \frac{\varepsilon}{2} > \frac{2}{\varepsilon} \cdot \frac{\varepsilon}{2} = 1 , \end{equation*}

so \(a \notin T(G)\). Hence \(T(G)\) is not an open subset of \(W\).

Take \(W = \ell^1\) with \(\|f\|_1 = \sum_{k=1}^{\infty} |f_k|\), a Banach space by the second bullet of Example 6.38 and infinite-dimensional (the coordinate vectors \(e_1, e_2, \ldots\) form an infinite linearly independent family, so no finite list spans \(\ell^1\)). By 6.62 there is a discontinuous linear functional \(\varphi : W \to \mathbf{F}\), which by 6.48 means: no \(c \in (0,\infty)\) satisfies \(|\varphi(f)| \le c\|f\|_1\) for all \(f \in W\). Now take

\begin{equation*} V = \ell^1 \text{ with } \|f\|_V = \|f\|_1 + |\varphi(f)| \end{equation*}

(a norm, by linearity of \(\varphi\) and because \(\|f\|_V = 0\) forces \(\|f\|_1 = 0\)), let \(T : V \to W\) be the identity map, and let \(G = \{f \in V : \|f\|_V < 1\}\).

Then \(T\) is linear and onto \(W\), and bounded with \(\|T\| \le 1\) since \(\|Tf\|_1 \le \|f\|_V\).

Suppose \(T(G) = \{f \in \ell^1 : \|f\|_1 + |\varphi(f)| < 1\}\) were open in \(W\). Since \(0 \in T(G)\), there would be \(\varepsilon > 0\) with \(\|f\|_1 < \varepsilon \implies |\varphi(f)| < 1\). Applying this to \(g = \frac{\varepsilon}{2\|f\|_1} f\) for \(f \ne 0\), which has \(\|g\|_1 = \varepsilon/2 < \varepsilon\), and using homogeneity of \(\varphi\) gives

\begin{equation*} |\varphi(f)| \le \frac{2}{\varepsilon} \|f\|_1 \quad \text{for all } f \in W \end{equation*}

(the case \(f = 0\) being trivial), contradicting the unboundedness of \(\varphi\). Hence \(T(G)\) is not open in \(W\).

Write \(R = T(V)\), a subspace of \(W\) by linearity of \(T\).

(i) Suppose \(\|f\| \le c \|Tf\|\) for all \(f \in V\), where \(c \in (0, \infty)\). Then \(Tf = 0\) forces \(f = 0\), so \(T\) is injective. For closedness of \(R\), let \(g_k = Tf_k \to g\) in \(W\). The sequence \(g_1, g_2, \ldots\) is Cauchy (6.13), and

\begin{equation*} \|f_j - f_k\| \le c \|T(f_j - f_k)\| = c \|g_j - g_k\| \end{equation*}

makes \(f_1, f_2, \ldots\) Cauchy in the Banach space \(V\), say \(f_k \to f\). Since \(T\) is bounded it is continuous (6.48), so \(g = \lim_k Tf_k = Tf \in R\); thus \(R\) is closed by 6.9(e).

(ii) Conversely, suppose \(T\) is injective with \(R\) closed. Being a closed subset of the Banach space \(W\), \(R\) is complete [6.16(b)], hence itself a Banach space under the inherited norm. Viewed as a map \(V \to R\), \(T\) is a bounded linear bijection between Banach spaces, so the Bounded Inverse Theorem (6.83) gives \(b \in [0, \infty)\) with \(\|T^{-1} g\| \le b \|g\|\) for all \(g \in R\). Taking \(g = Tf\) yields

\begin{equation*} \|f\| \le b \|Tf\| \le c \|Tf\| , \qquad c := \max\{b, 1\} \in (0, \infty) , \end{equation*}

so \(T\) is bounded below.

Take the example of Exercise 10: \(V = \ell^1\) with \(\|a\|_1 = \sum_{k=1}^{\infty}|a_k|\), a Banach space by the second bullet of Example 6.38; \(W = \ell^1\) with \(\|a\|_{\infty} = \sup_k |a_k|\), a normed vector space by the second bullet of Example 6.39; and \(T : V \to W\) the identity map, which is linear, one-to-one, onto, and bounded with \(\|T\| \le 1\) since \(\|Ta\|_{\infty} \le \|a\|_1\).

Its inverse \(T^{-1} : W \to V\) is again the identity, and it is unbounded: for \(a^{(n)} = (1, \ldots, 1, 0, 0, \ldots)\) with \(n\) ones,

\begin{equation*} \|a^{(n)}\|_{\infty} = 1 , \qquad \|T^{-1} a^{(n)}\|_1 = n , \end{equation*}

so \(\|T^{-1} g\|_1 \le c\|g\|_{\infty}\) for all \(g \in W\) would force \(n \le c\) for every \(n\).

Take the construction of Exercise 11: \(W = \ell^1\) with \(\|\cdot\|_1\), a Banach space by the second bullet of Example 6.38 and infinite-dimensional, so 6.62 and 6.48 supply an unbounded linear functional \(\varphi : W \to \mathbf{F}\); \(V = \ell^1\) with the norm \(\|f\|_V = \|f\|_1 + |\varphi(f)|\); and \(T : V \to W\) the identity map, which is linear, one-to-one, onto, and bounded with \(\|T\| \le 1\) since \(\|Tf\|_1 \le \|f\|_V\).

Its inverse \(T^{-1} : W \to V\) is again the identity, and it is unbounded: were \(\|T^{-1}g\|_V \le c\|g\|_1\) for all \(g \in W\), then

\begin{equation*} |\varphi(f)| \le \|f\|_1 + |\varphi(f)| = \|f\|_V \le c \|f\|_1 \end{equation*}

for all \(f \in \ell^1\), making \(\varphi\) bounded.

Give \(V \times W\) the coordinatewise operations and \(\|(f,g)\| = \max\{\|f\|, \|g\|\}\).

Norm. Here \(\|(f,g)\| = 0\) forces \(\|f\| = \|g\| = 0\), and \(\|\lambda(f,g)\| = |\lambda| \max\{\|f\|,\|g\|\}\) because multiplication by \(|\lambda| \ge 0\) commutes with maxima. For the triangle inequality,

\begin{equation*} \|f_1 + f_2\| \le \|f_1\| + \|f_2\| \le \|(f_1,g_1)\| + \|(f_2,g_2)\| , \end{equation*}

and the same bound holds for \(\|g_1 + g_2\|\), hence for their maximum.

Coordinatewise convergence. Because \((f_k,g_k) - (f,g) = (f_k - f, g_k - g)\),

\begin{equation*} \|(f_k,g_k) - (f,g)\| = \max\{\|f_k - f\|, \|g_k - g\|\} , \end{equation*}

and \(a, b \le \max\{a,b\} \le a + b\) for \(a, b \ge 0\) shows this tends to \(0\) precisely when \(\|f_k - f\| \to 0\) and \(\|g_k - g\| \to 0\).

Completeness. If \((f_1,g_1), (f_2,g_2), \ldots\) is Cauchy in \(V \times W\), the same display makes \(f_1, f_2, \ldots\) Cauchy in \(V\) and \(g_1, g_2, \ldots\) Cauchy in \(W\); as \(V\) and \(W\) are Banach spaces, \(f_k \to f\) and \(g_k \to g\), so \((f_k, g_k) \to (f,g)\) by the previous paragraph.

Write \(V_\varphi\) for \(V\) with the norm \(\|\cdot\|_\varphi\), a Banach space by hypothesis, and let \(I : V_\varphi \to V\) be the identity map. It is a one-to-one linear map onto \(V\), bounded with \(\|I\| \le 1\) because

\begin{equation*} \|I f\| = \|f\| \le \|f\| + |\varphi(f)| = \|f\|_\varphi . \end{equation*}

Both \(V_\varphi\) and \(V\) are Banach spaces, so the Bounded Inverse Theorem (6.83) makes \(I^{-1}\) bounded; with \(c = \|I^{-1}\| < \infty\) this says \(\|f\|_\varphi \le c\|f\|\) for all \(f \in V\). Hence

\begin{equation*} |\varphi(f)| \le \|f\| + |\varphi(f)| = \|f\|_\varphi \le c \|f\| , \end{equation*}

so \(\varphi\) is bounded on \((V, \|\cdot\|)\) with \(\|\varphi\| \le c\), and therefore continuous by 6.48.

Remark. The converse is also true and easy: if \(\varphi\) is continuous on \(V\) with \(\|\varphi\| = c^{\prime}\), then \(\|f\| \le \|f\|_\varphi \le (1 + c^{\prime})\|f\|\), so the two norms are equivalent and \(V_\varphi\) is complete. Thus \(V_\varphi\) is a Banach space if and only if \(\varphi\) is continuous.

\(T\) is linear because limits in a normed vector space respect sums and scalar multiples (the triangle inequality and homogeneity of the norm): \(T_k(f + h) = T_k f + T_k h \to Tf + Th\) and \(T_k(\lambda f) = \lambda T_k f \to \lambda\, Tf\), while these same sequences converge to \(T(f+h)\) and \(T(\lambda f)\), and limits in a metric space are unique.

For boundedness, fix \(f \in V\). The convergent sequence \(T_1 f, T_2 f, \ldots\) is bounded: choosing \(n\) with \(\|T_k f - Tf\| \le 1\) for \(k \ge n\) gives

\begin{equation*} \sup_{k} \|T_k f\| \le \max\big(\{\|T_j f\| : j < n\} \cup \{\|Tf\| + 1\}\big) < \infty , \end{equation*}

a maximum over a finite set. So \(\{T_k : k \in \mathbf{Z}^+\}\) is a pointwise bounded family of bounded linear maps from the Banach space \(V\) to \(W\), and the Principle of Uniform Boundedness (6.86) gives \(M := \sup_k \|T_k\| < \infty\). Since \(\|T_k f\| \le M\|f\|\) for each \(k\) and \(\big|\,\|T_k f\| - \|Tf\|\,\big| \le \|T_k f - Tf\| \to 0\),

\begin{equation*} \|Tf\| = \lim_{k \to \infty} \|T_k f\| \le M \|f\| , \end{equation*}

so \(T\) is bounded with \(\|T\| \le \sup_k \|T_k\|\).

Run the Principle of Uniform Boundedness on \(V^{\prime}\) rather than on \(V\): by 6.71 and 6.47, \(V^{\prime} = \mathcal{B}(V, \mathbf{F})\) is a Banach space because \(\mathbf{F}\) is complete, whereas \(V\) is only normed.

For \(f \in V\) define \(\widehat{f} : V^{\prime} \to \mathbf{F}\) by \(\widehat{f}(\varphi) = \varphi(f)\), which is linear because the operations on \(V^{\prime}\) are pointwise. Then \(|\widehat{f}(\varphi)| = |\varphi(f)| \le \|\varphi\| \, \|f\|\) puts \(\widehat{f} \in (V^{\prime})^{\prime}\) with \(\|\widehat{f}\| \le \|f\|\); conversely \(\|\widehat{f}\| \ge \|f\|\), trivially for \(f = 0\) and otherwise by taking the \(\varphi \in V^{\prime}\) that 6.72 supplies with \(\|\varphi\| = 1\) and \(\varphi(f) = \|f\|\). Hence \(\|\widehat{f}\| = \|f\|\).

The hypothesis says exactly that the family \(\{\widehat{f} : f \in B\}\) of bounded linear maps from \(V^{\prime}\) to \(\mathbf{F}\) is pointwise bounded, since

\begin{equation*} \sup_{f \in B} |\widehat{f}(\varphi)| = \sup_{f \in B} |\varphi(f)| < \infty \end{equation*}

for each \(\varphi \in V^{\prime}\). So 6.86 gives \(\sup_{f \in B} \|\widehat{f}\| < \infty\), that is, \(\sup_{f \in B} \|f\| < \infty\).

Apply the Closed Graph Theorem (6.85): \(V\) and \(W\) are Banach spaces, and \(\operatorname{graph}(T) = \{(f, Tf) : f \in V\}\) is a subspace of \(V \times W\) by 6.68(a), so it suffices to show the graph is closed in the norm \(\|(f,g)\| = \max\{\|f\|,\|g\|\}\) of 6.84.

Suppose \((f_k, Tf_k) \to (f, g)\), which by the convergence criterion of 6.84 means \(f_k \to f\) in \(V\) and \(Tf_k \to g\) in \(W\). Fix \(\varphi \in W^{\prime}\). Being bounded, \(\varphi\) is continuous (6.48), so \(\varphi(Tf_k) \to \varphi(g)\); and \(\varphi \circ T \in V^{\prime}\) is likewise continuous, so \(\varphi(Tf_k) \to \varphi(Tf)\). Uniqueness of limits in \(\mathbf{F}\) gives

\begin{equation*} \varphi(g - Tf) = 0 \qquad \text{for every } \varphi \in W^{\prime} . \end{equation*}

Were \(g - Tf \ne 0\), then 6.72 would supply \(\varphi \in W^{\prime}\) with \(\varphi(g - Tf) = \|g - Tf\| > 0\). Hence \(g = Tf\), so \((f,g) \in \operatorname{graph}(T)\) and the graph is closed by 6.9(e); 6.85 now makes \(T\) bounded.

Everything follows from the fact that the infimum in 7.1 is attained. Indeed, put \(c = \|f\|_\infty < \infty\); for each \(n\) there is \(t_n < c + \frac1n\) with \(\mu(\{|f| > t_n\}) = 0\), whence \(\mu(\{|f| > c + \frac1n\}) = 0\), and countable subadditivity (2.58) applied to

\begin{equation*} \{|f| > c\} = \bigcup_{n=1}^{\infty} \{|f| > c + \tfrac1n\} \end{equation*}

gives \(\mu(\{|f| > \|f\|_\infty\}) = 0\); that is, \(|f| \le \|f\|_\infty\) almost everywhere.

Triangle inequality. For \(f, g \in L^\infty(\mu)\) the set

\begin{equation*} E = \{|f| > \|f\|_\infty\} \cup \{|g| > \|g\|_\infty\} \end{equation*}

has \(\mu(E) = 0\) by the above and subadditivity, and off \(E\) we have \(|f + g| \le |f| + |g| \le \|f\|_\infty + \|g\|_\infty\). So every \(t > \|f\|_\infty + \|g\|_\infty\) has \(\{|f+g| > t\} \subseteq E\) and hence measure \(0\); taking the infimum over such \(t\),

\begin{equation*} \|f + g\|_\infty \le \|f\|_\infty + \|g\|_\infty < \infty . \end{equation*}

Since \(f + g\) is \(\mathcal{S}\)-measurable (2.46, applied to real and imaginary parts when \(\mathbf{F} = \mathbf{C}\)), \(f + g \in L^\infty(\mu)\).

Homogeneity. For \(\alpha = 0\) both sides are \(0\). For \(\alpha \ne 0\) and every \(t > 0\) we have \(\{|\alpha f| > t\} = \{|f| > t/|\alpha|\}\), so writing \(A_f = \{t > 0 : \mu(\{|f| > t\}) = 0\}\) we get \(A_{\alpha f} = |\alpha| A_f\), and multiplication by \(|\alpha| > 0\) commutes with infima:

\begin{equation*} \|\alpha f\|_\infty = \inf A_{\alpha f} = |\alpha| \inf A_f = |\alpha| \, \|f\|_\infty \end{equation*}

(both sides being \(\infty\) when \(A_f = \varnothing\)). With \(\alpha f\) measurable (2.46), \(\alpha f \in L^\infty(\mu)\).

As \(0 \in L^\infty(\mu)\), these two closure properties make \(L^\infty(\mu)\) a subspace of \(\mathbf{F}^X\), hence a vector space.

By 7.6, \(p^{\prime} = \frac{p}{p-1}\), so \(\frac{p^{\prime}}{p} = p^{\prime} - 1 = \frac{1}{p-1}\).

(i) \(b = 0\). The equation reads \(0 = \frac{a^p}{p}\) and the condition reads \(a^p = 0\); both say \(a = 0\).

(ii) \(b > 0\). As in the proof of Young’s inequality (7.8), set

\begin{equation*} f(a) = \frac{a^p}{p} + \frac{b^{p^{\prime}}}{p^{\prime}} - ab , \end{equation*}

so the equation in question says \(f(a) = 0\). Here \(f^{\prime}(a) = a^{p-1} - b\), and since \(a \mapsto a^{p-1}\) is strictly increasing on \((0,\infty)\) this is negative for \(0 < a < a_0\) and positive for \(a > a_0\), where \(a_0 = b^{1/(p-1)} > 0\). Hence \(f(a) > f(a_0)\) for all \(a \ge 0\) with \(a \ne a_0\). Because \(a_0^{\,p} = b^{p/(p-1)} = b^{p^{\prime}}\) and \(a_0 b = b^{\frac{1}{p-1} + 1} = b^{p^{\prime}}\),

\begin{equation*} f(a_0) = b^{p^{\prime}} \Bigl( \frac1p + \frac{1}{p^{\prime}} \Bigr) - b^{p^{\prime}} = 0 , \end{equation*}

so \(f(a) = 0\) if and only if \(a = a_0\). Finally \(a = b^{1/(p-1)}\) gives \(a^p = b^{p/(p-1)} = b^{p^{\prime}}\), and conversely \(a^p = b^{p^{\prime}} > 0\) gives \(a = b^{p^{\prime}/p} = b^{1/(p-1)}\).

Apply Hölder’s inequality (7.9) with \(p = 5\) and \(p^{\prime} = 5/4\) (by 7.6) on \(X = \{1, \ldots, n\}\) with counting measure, where every function is measurable and integration is summation (3.6 and 3.7). Take \(f(k) = a_k\) and \(h(k) = 1\), so that

\begin{equation*} \|f h\|_1 = \sum_{k=1}^{n} a_k , \qquad \|h\|_{5/4} = n^{4/5} , \qquad \|f\|_5 = \Bigl( \sum_{k=1}^{n} a_k^{\,5} \Bigr)^{1/5} , \end{equation*}

all finite. Hölder’s inequality then gives

\begin{equation*} \sum_{k=1}^{n} a_k \le n^{4/5} \Bigl( \sum_{k=1}^{n} a_k^{\,5} \Bigr)^{1/5} , \end{equation*}

and raising both nonnegative sides to the fifth power yields the claim.

Method (2): \(t \mapsto t^5\) is convex on \([0,\infty)\) (its second derivative \(20t^3\) is nonnegative), so the finite form of Jensen’s inequality with equal weights \(\frac1n\) gives

\begin{equation*} \Bigl( \frac{a_1 + \cdots + a_n}{n} \Bigr)^{5} \le \frac{a_1^{\,5} + \cdots + a_n^{\,5}}{n} ; \end{equation*}

multiply by \(n^5\).

By 7.7 we have \(1^{\prime} = \infty\) and \(\infty^{\prime} = 1\), and \(|f h| = |h f|\), so the case \(p = \infty\) is the case \(p = 1\) with \(f\) and \(h\) interchanged. Thus it suffices to prove \(\|f h\|_1 \le \|f\|_1 \|h\|_\infty\). Two facts recur: \(\mu(E) = 0\) forces \(\int u \, d\mu = 0\) for any \(u \ge 0\) vanishing off \(E\), because \(u \le \infty \cdot \chi_E\) and 3.8 with 3.7 give \(\int u \, d\mu \le \infty \cdot \mu(E) = 0\); and \(\mu(\{|h| > \|h\|_\infty\}) = 0\) whenever \(\|h\|_\infty < \infty\), by Exercise 1 in this section.

(i) \(\|f\|_1 = 0\). With \(E_n = \{|f| > \frac1n\}\), the bound \(\frac1n \chi_{E_n} \le |f|\) and 3.8, 3.7 give \(\frac1n \mu(E_n) \le 0\), so \(\mu(E_n) = 0\); as \(\{f \ne 0\} = \bigcup_{n} E_n\), countable subadditivity (2.58) makes \(\mu(\{f \ne 0\}) = 0\). Since \(|f h|\) vanishes off that set, \(\|f h\|_1 = 0\).

(ii) \(\|h\|_\infty = 0\). Then \(\mu(\{|h| > 0\}) = 0\), and \(|f h|\) vanishes off that set, so \(\|f h\|_1 = 0\).

(iii) \(\|f\|_1 > 0\), \(\|h\|_\infty > 0\), and one of them is \(\infty\). Then the right side is \(\infty\).

(iv) \(0 < \|f\|_1 < \infty\) and \(0 < \|h\|_\infty < \infty\). Put \(E = \{|h| > \|h\|_\infty\}\), so \(\mu(E) = 0\). Pointwise

\begin{equation*} |f h| \le \|h\|_\infty |f| + \infty \cdot \chi_E \end{equation*}

(on \(E\) the right side is \(\infty\); off \(E\) we have \(|h| \le \|h\|_\infty\)), so 3.8, additivity 3.16, homogeneity 3.20, and 3.7 give

\begin{equation*} \|f h\|_1 \le \|h\|_\infty \|f\|_1 + \infty \cdot \mu(E) = \|f\|_1 \|h\|_\infty . \end{equation*}

Both \(\|f\|_p\) and \(\|h\|_{p^{\prime}}\) are finite, and three facts from Chapter 3 are used throughout.

(F1) If \(u \ge 0\) is measurable with \(\int u \, d\mu = 0\), then \(u = 0\) almost everywhere: with \(E_n = \{u > \frac1n\}\), the bound \(\frac1n \chi_{E_n} \le u\) and 3.8, 3.7 give \(\mu(E_n) = 0\), and \(\{u > 0\} = \bigcup_n E_n\) is null by 2.58.

(F2) Nonnegative measurable functions agreeing outside a null set \(E\) have equal integrals, since \(u \le v + \infty \cdot \chi_E\) with 3.8, 3.16, 3.7 gives \(\int u \, d\mu \le \int v \, d\mu\), and symmetrically.

(F3) Hence \(\|f\|_p = 0\) if and only if \(f = 0\) almost everywhere.

(i) Suppose \(a |f|^p = b |h|^{p^{\prime}}\) almost everywhere, with \(a, b \ge 0\) not both \(0\). If \(b = 0\) then \(a > 0\), so \(f = 0\) almost everywhere and both sides of Hölder’s inequality vanish by (F2) and (F3); the case \(a = 0\) is symmetric. So let \(a, b > 0\) and \(c = b/a\), giving \(|f|^p = c |h|^{p^{\prime}}\) almost everywhere and hence \(\|f\|_p^{\,p} = c \|h\|_{p^{\prime}}^{\,p^{\prime}}\) by (F2). If \(\|h\|_{p^{\prime}} = 0\) both sides vanish again, so assume \(\|h\|_{p^{\prime}} > 0\); then \(\|f\|_p > 0\) too. Put \(F = |f| / \|f\|_p\) and \(H = |h| / \|h\|_{p^{\prime}}\), so that \(\int F^p \, d\mu = \int H^{p^{\prime}} \, d\mu = 1\) and, almost everywhere,

\begin{equation*} F^p = \frac{|f|^p}{\|f\|_p^{\,p}} = \frac{c \, |h|^{p^{\prime}}}{c \, \|h\|_{p^{\prime}}^{\,p^{\prime}}} = H^{p^{\prime}} . \end{equation*}

By Exercise 2 in this section, Young’s inequality is therefore an equality almost everywhere, and integrating

\begin{equation*} F H = \frac{F^p}{p} + \frac{H^{p^{\prime}}}{p^{\prime}} \end{equation*}

via (F2), 3.16, and 3.20 gives \(\int F H \, d\mu = \frac1p + \frac{1}{p^{\prime}} = 1\). Since \(|f h| = \|f\|_p \|h\|_{p^{\prime}} F H\), homogeneity yields \(\|f h\|_1 = \|f\|_p \|h\|_{p^{\prime}}\).

(ii) Conversely, suppose \(\|f h\|_1 = \|f\|_p \|h\|_{p^{\prime}}\). If \(\|f\|_p = 0\) take \(a = 1, b = 0\), and if \(\|h\|_{p^{\prime}} = 0\) take \(a = 0, b = 1\); in each case (F3) makes both sides vanish almost everywhere. Otherwise both norms lie in \((0, \infty)\); with \(F, H\) as above, \(\int F^p \, d\mu = \int H^{p^{\prime}} \, d\mu = 1\) and homogeneity turns the assumed equality into \(\int F H \, d\mu = 1\). The function

\begin{equation*} g = \frac{F^p}{p} + \frac{H^{p^{\prime}}}{p^{\prime}} - F H \end{equation*}

is nonnegative by Young’s inequality (7.8), and \(g + F H = \frac{F^p}{p} + \frac{H^{p^{\prime}}}{p^{\prime}}\) with 3.16 and 3.20 gives \(\int g \, d\mu + \int F H \, d\mu = \frac1p + \frac{1}{p^{\prime}} = 1\); subtracting the finite quantity \(\int F H \, d\mu = 1\) leaves \(\int g \, d\mu = 0\), so \(g = 0\) almost everywhere by (F1). By Exercise 2 this forces \(F^p = H^{p^{\prime}}\) almost everywhere, that is,

\begin{equation*} a |f|^p = b |h|^{p^{\prime}} \quad \text{almost everywhere, where } a = \|h\|_{p^{\prime}}^{\,p^{\prime}} > 0 , \ b = \|f\|_p^{\,p} > 0 . \end{equation*}

Write \(M = \|h\|_\infty < \infty\) and \(N = \|f\|_1 < \infty\), and use facts (F1) and (F2) from Exercise 5 in this section.

Reduction. The set \(E_0 = \{|h| > M\}\) is null by Exercise 1 in this section, so replacing \(h\) by \(h \chi_{X \setminus E_0}\) changes neither \(\|h\|_\infty\) (the essential supremum is unaffected by modification on a null set; see the discussion following 7.1), nor \(\|f h\|_1\) (by (F2)), nor the measure of \(\{f \ne 0 \text{ and } |h| \ne M\}\). So assume \(|h| \le M\) everywhere.

Then \(u := (M - |h|) |f|\) is nonnegative and measurable with \(u + |f h| = M|f|\) pointwise, so additivity (3.16) and homogeneity (3.20) give

\begin{equation*} \int u \, d\mu + \|f h\|_1 = M \|f\|_1 = M N . \end{equation*}

Hölder’s inequality (7.9) for \(p = 1\), proved in Exercise 4 of this section, makes \(\|f h\|_1 \le M N < \infty\), so that term may be subtracted:

\begin{equation*} \int u \, d\mu = \|f\|_1 \|h\|_\infty - \|f h\|_1 . \end{equation*}

Hence \(\|f h\|_1 = \|f\|_1 \|h\|_\infty\) if and only if \(\int u \, d\mu = 0\), which by (F1) holds if and only if \(u = 0\) almost everywhere. Because \(|h| \le M\),

\begin{equation*} \{u > 0\} = \{f \ne 0 \text{ and } |h| < M\} = \{f \ne 0 \text{ and } |h| \ne M\} , \end{equation*}

so that condition says exactly that \(|h(x)| = \|h\|_\infty\) for almost every \(x \in X\) with \(f(x) \ne 0\).

Apply Hölder’s inequality (7.9) to \(|f|^r\) and \(|h|^r\) with exponent \(P = p/r\). Here \(1 < P < \infty\), since \(\frac1r = \frac1p + \frac1q > \frac1p\) forces \(r < p\), and \(Q := q/r\) is the dual exponent of \(P\) because

\begin{equation*} \frac1P + \frac1Q = r \Bigl( \frac1p + \frac1q \Bigr) = 1 , \end{equation*}

so \(Q = P^{\prime}\) by 7.6. As \(|f|^r\) and \(|h|^r\) are measurable by 6.20, 7.9 gives

\begin{equation*} \bigl\| \, |f|^r |h|^r \, \bigr\|_1 \le \bigl\| \, |f|^r \, \bigr\|_P \; \bigl\| \, |h|^r \, \bigr\|_Q . \end{equation*}

By 7.1 the left side is \(\int |f h|^r \, d\mu = \|f h\|_r^{\,r}\), while \(rP = p\) gives

\begin{equation*} \bigl\| \, |f|^r \, \bigr\|_P = \Bigl( \int |f|^{p} \, d\mu \Bigr)^{r/p} = \|f\|_p^{\,r} , \end{equation*}

and likewise \(\bigl\| \, |h|^r \, \bigr\|_Q = \|h\|_q^{\,r}\) from \(rQ = q\). Hence

\begin{equation*} \|f h\|_r^{\,r} \le \bigl( \|f\|_p \, \|h\|_q \bigr)^{r} \end{equation*}

(using the convention \(0 \cdot \infty = 0\) under which 7.9 is stated), and taking \(r\)-th roots, which preserves order on \([0,\infty]\), completes the proof.

Induct on \(n\), with Exercise 7 in this section as the two-factor step. The statement to prove is: if \(p_1, \ldots, p_n, r\) are positive numbers with \(\sum_{k=1}^{n} \frac{1}{p_k} = \frac1r\), then

\begin{equation*} \|f_1 f_2 \cdots f_n\|_r \le \|f_1\|_{p_1} \|f_2\|_{p_2} \cdots \|f_n\|_{p_n} . \end{equation*}

For \(n = 1\) we have \(r = p_1\) and the two sides agree. Assume the statement for \(n\), and let \(p_1, \ldots, p_{n+1}, r\) be positive with \(\sum_{k=1}^{n+1} \frac{1}{p_k} = \frac1r\). Define \(s\) by \(\frac1s = \sum_{k=1}^{n} \frac{1}{p_k}\), a positive finite number, so that \(\frac1s + \frac{1}{p_{n+1}} = \frac1r\). The induction hypothesis bounds \(\|f_1 \cdots f_n\|_s\), and Exercise 7 applied to \(f_1 \cdots f_n\) and \(f_{n+1}\) gives

\begin{equation*} \|f_1 \cdots f_n f_{n+1}\|_r \le \|f_1 \cdots f_n\|_s \, \|f_{n+1}\|_{p_{n+1}} \le \prod_{k=1}^{n+1} \|f_k\|_{p_k} , \end{equation*}

the second inequality because multiplication by a fixed element of \([0,\infty]\) preserves order under the convention \(0 \cdot \infty = 0\) used in 7.9. Taking \(r = 1\) yields the exercise.

With \(p = \infty\) and \(p^{\prime} = 1\) (7.6 and 7.7), the formula in 7.12 reads

\begin{equation*} \|f\|_{\infty} = \sup\Bigl\{ \int f h \, d\mu : h \in \mathcal{L}^{1}(\mu) \text{ and } \|h\|_1 \le 1 \Bigr\} . \end{equation*}

For any admissible \(h\), Hölder’s inequality (7.9) with \(p = \infty\), \(p^{\prime} = 1\) gives

\begin{equation*} \int f h \, d\mu \le \int |f h| \, d\mu = \|f h\|_1 \le \|f\|_{\infty} \|h\|_1 \le \|f\|_{\infty} , \end{equation*}

so the supremum is at most \(\|f\|_{\infty}\); as \(h = 0\) is admissible the supremum is also at least \(0\), settling the case \(\|f\|_\infty = 0\). Assume now \(\|f\|_{\infty} > 0\) and fix \(t \in (0, \|f\|_{\infty})\).

Then \(E = \{|f| > t\} \in \mathcal{S}\) has \(\mu(E) > 0\), since \(t\) lies below the infimum in 7.1. Writing \(X = \bigcup_{j=1}^{\infty} X_j\) with each \(\mu(X_j) < \infty\) by \(\sigma\)-finiteness, countable subadditivity (2.58) rules out \(\mu(E \cap X_j) = 0\) for every \(j\), so \(A := E \cap X_j\) satisfies \(0 < \mu(A) < \infty\) for some \(j\). Put

\begin{equation*} h = \frac{\overline{f}}{\mu(A) \, |f|} \, \chi_A , \end{equation*}

which is defined because \(|f| > t > 0\) on \(A\). Then \(|h| = \frac{1}{\mu(A)} \chi_A\), so \(\|h\|_1 = 1\) by 3.15 and \(h\) is admissible; moreover \(f h = \frac{|f|}{\mu(A)} \chi_A \ge \frac{t}{\mu(A)} \chi_A\) pointwise, so 3.8 and 3.15 give

\begin{equation*} \int f h \, d\mu \ge \frac{t \, \mu(A)}{\mu(A)} = t . \end{equation*}

Hence the supremum is at least \(t\) for every \(t < \|f\|_{\infty}\), and the two bounds give equality.

Here \(\mu\) is counting measure on \(\mathbf{Z}^{+}\), so by 7.2 and 7.4 we have \(\|a\|_p = (\sum_{k} |a_k|^p)^{1/p}\) for \(p < \infty\), \(\|a\|_{\infty} = \sup_k |a_k|\), and \(\ell^p = \{a : \|a\|_p < \infty\}\).

(a) Immediate from (b): if \(a \in \ell^p\) then \(\|a\|_q \le \|a\|_p < \infty\), so \(a \in \ell^q\).

(b) If \(\|a\|_p = \infty\) there is nothing to prove, and if \(\|a\|_p = 0\) then every \(a_k = 0\) and both sides vanish. So assume \(0 < \|a\|_p < \infty\) and put \(b = a / \|a\|_p\), which has \(\|b\|_p = 1\) by homogeneity (7.5(b)); thus \(\sum_{k=1}^{\infty} |b_k|^p = 1\), and in particular \(|b_k| \le 1\) for every \(k\).

(i) \(q = \infty\). Then \(\|b\|_{\infty} = \sup_k |b_k| \le 1\).

(ii) \(q < \infty\). Then \(0 \le |b_k| \le 1\) and \(q > p\) give \(|b_k|^q \le |b_k|^p\), so

\begin{equation*} \|b\|_q^{\,q} = \sum_{k=1}^{\infty} |b_k|^q \le \sum_{k=1}^{\infty} |b_k|^p = 1 . \end{equation*}

Either way \(\|b\|_q \le 1\), so homogeneity of the \(q\)-norm (7.5(b), or Exercise 1 in this section when \(q = \infty\)) gives

\begin{equation*} \|a\|_q = \bigl\| \, \|a\|_p \, b \, \bigr\|_q = \|a\|_p \, \|b\|_q \le \|a\|_p . \end{equation*}

Take \(a = (a_1, a_2, \ldots)\) with \(a_k = \frac1k\). Since \(\ell^1 \subseteq \ell^p\) for every \(p > 1\) by Exercise 10(a) in this section, it suffices to show \(a\) lies in every \(\ell^p\) with \(p > 1\) but not in \(\ell^1\).

Not in \(\ell^1\). The block \(\{2^m, \ldots, 2^{m+1}-1\}\) has \(2^m\) terms, each at least \(2^{-(m+1)}\), so it contributes at least \(\frac12\); summing the blocks \(m = 0, \ldots, M-1\),

\begin{equation*} \sum_{k=1}^{2^M - 1} \frac1k \ge \frac{M}{2} . \end{equation*}

The partial sums are unbounded, so \(\|a\|_1 = \infty\).

In \(\ell^p\) for \(p > 1\). For \(p = \infty\) this is clear, as \(\sup_k \frac1k = 1\). For \(1 < p < \infty\) the same block has \(2^m\) terms, each at most \(2^{-mp}\), so it contributes at most \((2^{1-p})^m\); since \(0 < 2^{1-p} < 1\), summing the geometric series bounds the partial sums of the nonnegative series:

\begin{equation*} \sum_{k=1}^{\infty} \frac{1}{k^p} \le \sum_{m=0}^{\infty} \bigl( 2^{1-p} \bigr)^{m} = \frac{1}{1 - 2^{1-p}} < \infty . \end{equation*}

Take \(f = \sum_{n=1}^{\infty} n \, \chi_{A_n}\) on \([0,1]\) (with \(f(0) = 0\)), where \(A_n = (2^{-n}, 2^{-n+1}]\) are disjoint Borel sets with \(\lambda(A_n) = 2^{-n}\); here \(f\) is Borel measurable because \(f^{-1}((t,\infty))\) is the union of those \(A_n\) with \(n > t\), together with \(\{0\}\) when \(t < 0\).

The containment. If \(g \in \mathcal{L}^{\infty}([0,1])\) and \(0 < p < \infty\), then \(N = \{|g| > \|g\|_{\infty}\}\) is null by Exercise 1 in this section, so \(|g|^p \le \|g\|_{\infty}^{\,p} \chi_{[0,1]} + \infty \cdot \chi_N\) pointwise and 3.8, 3.16, 3.15 give

\begin{equation*} \int |g|^p \, d\lambda \le \|g\|_{\infty}^{\,p} \lambda([0,1]) + \infty \cdot \lambda(N) = \|g\|_{\infty}^{\,p} < \infty . \end{equation*}

Thus \(\mathcal{L}^{\infty}([0,1]) \subseteq \bigcap_{p<\infty} \mathcal{L}^p([0,1])\), and it suffices to place \(f\) in the right side but not the left.

\(f \notin \mathcal{L}^{\infty}([0,1])\). Given \(t > 0\), any integer \(n > t\) gives \(\lambda(\{|f| > t\}) \ge \lambda(A_n) > 0\), so the set whose infimum defines \(\|f\|_{\infty}\) in 7.1 is empty and \(\|f\|_{\infty} = \infty\).

\(f \in \mathcal{L}^p([0,1])\) for \(0 < p < \infty\). Disjointness gives \(|f|^p = \sum_{n} n^p \chi_{A_n}\), whose increasing partial sums are simple with integrals \(\sum_{n \le N} n^p 2^{-n}\) by 3.15, so the Monotone Convergence Theorem (3.11) gives

\begin{equation*} \int |f|^p \, d\lambda = \sum_{n=1}^{\infty} \frac{n^p}{2^n} < \infty , \end{equation*}

the series converging because \(c_n = n^p 2^{-n}\) satisfies \(\frac{c_{n+1}}{c_n} = \frac12 (1 + \frac1n)^p \to \frac12\), hence \(c_{n+1} \le \frac34 c_n\) for large \(n\), which dominates the tail by a geometric series.

Take \(f = \sum_{n=1}^{\infty} \frac{2^n}{n^2} \, \chi_{A_n}\) on \([0,1]\), with \(A_n = (2^{-n}, 2^{-n+1}]\) as in Exercise 12 in this section: disjoint Borel sets with \(\lambda(A_n) = 2^{-n}\), and \(f(0) = 0\); as there, \(f\) is Borel measurable because \(f^{-1}((t,\infty))\) is a union of \(A_n\)’s together with \(\{0\}\) when \(t < 0\).

The containment \(\bigcup_{p>1} \mathcal{L}^p([0,1]) \subseteq \mathcal{L}^1([0,1])\) holds by 7.10 when \(1 < p < \infty\), since \(\lambda_{[0,1]}\) is a finite measure, and by the first paragraph of Exercise 12 when \(p = \infty\). So it suffices that \(f\) lie in \(\mathcal{L}^1([0,1])\) but in no \(\mathcal{L}^p([0,1])\) with \(p > 1\). Exactly as in Exercise 12 (disjointness, 3.15, and the Monotone Convergence Theorem 3.11), for \(0 < p < \infty\),

\begin{equation*} \int |f|^p \, d\lambda = \sum_{n=1}^{\infty} \Bigl( \frac{2^n}{n^2} \Bigr)^{p} \lambda(A_n) = \sum_{n=1}^{\infty} \frac{2^{n(p-1)}}{n^{2p}} . \end{equation*}

(i) \(p = 1\). The sum is \(\sum_{n} \frac{1}{n^2} < \infty\), since \(\frac{1}{n^2} \le \frac{1}{n-1} - \frac1n\) for \(n \ge 2\) bounds the partial sums by \(2\). Hence \(f \in \mathcal{L}^1([0,1])\).

(ii) \(1 < p < \infty\). Choose an integer \(m > 2p\); then \(e^t \ge t^m / m!\) with \(t = n(p-1)\ln 2\) gives

\begin{equation*} \frac{2^{n(p-1)}}{n^{2p}} \ge \frac{((p-1)\ln 2)^m}{m!} \, n^{m - 2p} \longrightarrow \infty , \end{equation*}

so the terms of the series do not tend to \(0\) and \(\int |f|^p \, d\lambda = \infty\).

(iii) \(p = \infty\). The same device with \(m = 3\) and \(t = n \ln 2\) gives \(2^n \ge (\ln 2)^3 n^3 / 6\), so \(2^n / n^2 \to \infty\) and every \(t > 0\) admits \(n\) with \(\lambda(\{|f| > t\}) \ge \lambda(A_n) > 0\); hence \(\|f\|_{\infty} = \infty\).

By symmetry assume \(p < q\), so \(p \in (0,\infty)\); put \(r = q\) if \(q < \infty\) and \(r = p+1\) if \(q = \infty\), and take

\begin{equation*} f = x^{-1/r} \chi_{(0,1)} , \qquad g = x^{-1/p} \chi_{(1,\infty)} . \end{equation*}

Each is continuous and nonnegative on the interval carrying it, so its Lebesgue integrals are the improper Riemann integrals

\begin{equation*} \int |f|^s \, d\lambda = \int_0^1 x^{-s/r} \, dx , \qquad \int |g|^s \, d\lambda = \int_1^{\infty} x^{-s/p} \, dx , \end{equation*}

the first finite exactly when \(s < r\), the second exactly when \(s > p\).

(i) \(f \in \mathcal{L}^p(\mathbb{R}) \setminus \mathcal{L}^q(\mathbb{R})\). Since \(p < r\), \(\|f\|_p < \infty\). If \(q < \infty\), then \(r = q\) and \(\int |f|^q \, d\lambda = \int_0^1 dx/x = \infty\). If \(q = \infty\), then for every \(t > 0\) the set \(\{|f| > t\}\) contains an interval \((0,\delta)\) of positive measure, so \(\|f\|_{\infty} = \infty\).

(ii) \(g \in \mathcal{L}^q(\mathbb{R}) \setminus \mathcal{L}^p(\mathbb{R})\). Here \(\int |g|^p \, d\lambda = \int_1^{\infty} dx/x = \infty\). If \(q < \infty\), then \(q > p\) gives \(\|g\|_q < \infty\); if \(q = \infty\), then \(|g| \le 1\).

Thus \(\mathcal{L}^p(\mathbb{R}) \not\subseteq \mathcal{L}^q(\mathbb{R})\) and \(\mathcal{L}^q(\mathbb{R}) \not\subseteq \mathcal{L}^p(\mathbb{R})\).

Take \(f(x) = 1 / \bigl(\sqrt{x}\,\lvert \ln x\rvert\bigr)\) for \(x \in \left(0,\tfrac12\right) \cup (2,\infty)\) and \(f(x) = 0\) otherwise; on each of those intervals \(f\) is continuous and nonnegative, so its Lebesgue integrals are the improper Riemann integrals below.

(i) \(p = 2\). The substitution \(u = \ln x\), \(du = dx/x\), gives

\begin{equation*} \|f\|_2^{\,2} = \int_0^{1/2} \frac{dx}{x(\ln x)^2} \ + \ \int_2^{\infty} \frac{dx}{x(\ln x)^2} = \int_{-\infty}^{-\ln 2} \frac{du}{u^2} + \int_{\ln 2}^{\infty} \frac{du}{u^2} = \frac{2}{\ln 2} , \end{equation*}

so \(f \in \mathcal{L}^2(\mathbb{R})\).

(ii) \(0 < p < 2\); put \(q = p/2 \in (0,1)\). On \((2,\infty)\) we have \(\lvert f\rvert^p = x^{-q} (\ln x)^{-p}\), and \((\ln x)^p \leq x^{(1-q)/2}\) for all \(x \geq M\), some \(M \geq 2\), whence

\begin{equation*} \lvert f(x)\rvert^p \geq x^{-(1+q)/2} \quad (x \geq M) , \qquad \tfrac{1+q}{2} < 1 , \end{equation*}

so \(\int \lvert f\rvert^p \, d\lambda = \infty\).

(iii) \(2 < p < \infty\); put \(q = p/2 > 1\) and fix \(r \in (1,q)\). On \(\left(0,\tfrac12\right)\) we have \(\lvert f\rvert^p = x^{-q}\lvert \ln x\rvert^{-p}\), and \(x^{q-r}\lvert \ln x\rvert^p \to 0\) as \(x \downarrow 0\) (a positive power of \(x\) beats any power of \(\lvert \ln x\rvert\)), so \(\lvert \ln x\rvert^p \leq x^{-(q-r)}\) on some \((0,\delta)\); there \(\lvert f(x)\rvert^p \geq x^{-r}\) with \(r > 1\), so \(\int \lvert f\rvert^p \, d\lambda = \infty\).

(iv) \(p = \infty\). Since \(\sqrt{x}\,\lvert \ln x\rvert \to 0\) as \(x \downarrow 0\), each set \(\{\lvert f\rvert > t\}\) contains an interval \((0,\delta_t)\) of positive measure, so the infimum in 7.1 is over the empty set and \(\|f\|_{\infty} = \infty\).

The two bounds \(\|f\|_p \leq \|f\|_{\infty}\mu(X)^{1/p}\) and \(\|f\|_p \geq t\,\mu(\{\lvert f\rvert > t\})^{1/p}\) give the limit, since both measure factors tend to \(1\). In detail: if \(\mu(X) = 0\), then every set is null, so \(\|f\|_p = 0 = \|f\|_{\infty}\) for all \(p\); assume henceforth \(0 < \mu(X) < \infty\).

(i) \(\limsup_{p \to \infty} \|f\|_p \leq \|f\|_{\infty}\). Only \(\|f\|_{\infty} < \infty\) needs proof. Then \(\lvert f\rvert \leq \|f\|_{\infty}\) almost everywhere, because

\begin{equation*} \bigl\{ \lvert f\rvert > \|f\|_{\infty}\bigr\} = \bigcup_{n=1}^{\infty} \bigl\{ \lvert f \rvert > \|f\|_{\infty} + \tfrac1n \bigr\} \end{equation*}

and each set on the right is null by the definition of the infimum in 7.1; countable subadditivity finishes it. Hence \(\int \lvert f\rvert^p \, d\mu \leq \|f\|_{\infty}^{\,p}\,\mu(X)\), so \(\|f\|_p \leq \|f\|_{\infty}\mu(X)^{1/p}\), and \(\mu(X)^{1/p} \to 1\) because \(0 < \mu(X) < \infty\).

(ii) \(\liminf_{p \to \infty}\|f\|_p \geq \|f\|_{\infty}\). Only \(\|f\|_{\infty} > 0\) needs proof. Fix \(t \in (0, \|f\|_{\infty})\); then \(t\) fails the defining condition in 7.1, so \(c := \mu(\{\lvert f\rvert > t\})\) satisfies \(0 < c \leq \mu(X) < \infty\), and

\begin{equation*} \int \lvert f\rvert^p \, d\mu \geq t^p c \qquad \text{for every } p \in (0,\infty) . \end{equation*}

Thus \(\|f\|_p \geq t\,c^{1/p} \to t\), giving \(\liminf_{p\to\infty}\|f\|_p \geq t\); now let \(t \uparrow \|f\|_{\infty}\).

Take \(g\) from the simple approximations supplied by 2.89, applied to \(\operatorname{Re} f\) and \(\operatorname{Im} f\). Precisely, 2.89(b) and (c) give simple \(\mathcal{S}\)-measurable \(u_k \to \operatorname{Re} f\) and \(v_k \to \operatorname{Im} f\) pointwise on \(X\) with \(\lvert u_k\rvert \leq \lvert \operatorname{Re} f\rvert\) and \(\lvert v_k\rvert \leq \lvert \operatorname{Im} f\rvert\); then \(g_k := u_k + iv_k\) is simple and \(\mathcal{S}\)-measurable, with

\begin{equation*} \lvert g_k\rvert \leq \lvert \operatorname{Re} f\rvert + \lvert \operatorname{Im} f\rvert \leq 2\lvert f\rvert \qquad\text{and}\qquad g_k \to f \ \text{ pointwise}, \end{equation*}

so \(\|g_k\|_p \leq 2\|f\|_p < \infty\) and \(g_k \in \mathcal{L}^p(\mu)\) for every \(p \in (0,\infty]\).

(i) \(0 < p < \infty\). Then \(\lvert f - g_k\rvert^p \leq 3^p \lvert f\rvert^p\), which is integrable because \(f \in \mathcal{L}^p(\mu)\), and \(\lvert f - g_k\rvert^p \to 0\) pointwise, so the Dominated Convergence Theorem 3.31 gives \(\|f - g_k\|_p \to 0\); take \(g = g_k\) for \(k\) large.

(ii) \(p = \infty\). Put \(M = \|f\|_{\infty} < \infty\), \(E = \{\lvert f\rvert \leq M\}\), and \(\tilde f = f\chi_E\); as in 7A.16(i), \(\mu(X \setminus E) = 0\), so \(\|f - \tilde f\|_{\infty} = 0\). Since \(\operatorname{Re}\tilde f\) and \(\operatorname{Im}\tilde f\) are bounded, 2.89(d) supplies simple \(\mathcal{S}\)-measurable \(a, b\) with \(\sup_X \lvert \operatorname{Re}\tilde f - a\rvert < \varepsilon/4\) and \(\sup_X \lvert \operatorname{Im}\tilde f - b\rvert < \varepsilon/4\). Then \(g = a + ib\) is simple with \(\lvert g\rvert \leq M + \varepsilon/2\), so \(g \in \mathcal{L}^{\infty}(\mu)\), and by the triangle inequality of 7A.1,

\begin{equation*} \|f - g\|_{\infty} \leq \|f - \tilde f\|_{\infty} + \|\tilde f - g\|_{\infty} \leq 0 + \frac{\varepsilon}{2} < \varepsilon . \end{equation*}

Approximate \(f\) first by a simple function using 7A.17, then replace each of its level sets by a finite union of intervals; with \(q = \min\{p,1\}\), the errors add through the \(q\)-triangle inequality

\begin{equation*} \Bigl\| \sum_{k=1}^m f_k \Bigr\|_p^{\,q} \leq \sum_{k=1}^m \|f_k\|_p^{\,q} . \end{equation*}

That inequality is Minkowski 7.14 with induction on \(m\) when \(p \geq 1\) (so \(q = 1\)); when \(0 < p < 1\) it is \(q = p\) and follows by integrating \(\bigl(\sum_k \lvert f_k\rvert\bigr)^p \leq \sum_k \lvert f_k\rvert^p\), which iterates \((a+b)^p \leq a^p + b^p\) for \(a, b \geq 0\) (divide by \((a+b)^p\) and use \(s^p \geq s\) for \(s \in [0,1]\)).

Interval approximation. If \(A \subseteq \mathbb{R}\) is Lebesgue measurable with \(\lvert A\rvert < \infty\) and \(\delta > 0\), there is a union \(E\) of finitely many disjoint bounded open intervals with \(\|\chi_A - \chi_E\|_p^{\,p} = \lvert A\setminus E\rvert + \lvert E \setminus A\rvert < \delta\). Indeed, 2.71(e) gives an open \(G \supseteq A\) with \(\lvert G\setminus A\rvert < \delta/2\), so \(\lvert G\rvert < \infty\); writing \(G = \bigcup_{j=1}^{\infty} J_j\) with the \(J_j\) disjoint open intervals (bounded, as \(\lvert G\rvert < \infty\)), choose \(N\) with \(\sum_{j>N}\lvert J_j\rvert < \delta/2\) and set \(E = J_1 \cup \cdots \cup J_N\); then

\begin{equation*} A \setminus E \subseteq \bigcup_{j > N} J_j \quad\text{and}\quad E \setminus A \subseteq G \setminus A . \end{equation*}

Now let \(\varepsilon > 0\). By 7A.17 there is a simple function \(h \in \mathcal{L}^p(\mathbb{R})\) with

\begin{equation*} \|f - h\|_p^{\,q} < \frac{\varepsilon^q}{2} . \end{equation*}

If \(h = 0\), take \(g = 0\). Otherwise write \(h = a_1\chi_{A_1} + \cdots + a_n \chi_{A_n}\) in the standard representation following 2.88, the \(a_k\) distinct and nonzero and the \(A_k\) disjoint and measurable; disjointness gives \(\|h\|_p^{\,p} = \sum_k \lvert a_k\rvert^p \lvert A_k\rvert < \infty\), so each \(\lvert A_k\rvert < \infty\). Choose \(\delta_k > 0\) with \(\lvert a_k\rvert^q \delta_k^{\,q/p} < \varepsilon^q/(2n)\) and then \(E_k\) as above with \(\|\chi_{A_k} - \chi_{E_k}\|_p^{\,p} < \delta_k\), and set

\begin{equation*} g = a_1 \chi_{E_1} + \cdots + a_n \chi_{E_n} . \end{equation*}

Expanding each \(\chi_{E_k}\) over its finitely many bounded intervals exhibits \(g\) as a step function; it is bounded with bounded support, so \(g \in \mathcal{L}^p(\mathbb{R})\). By the \(q\)-triangle inequality and the homogeneity 7.5(b),

\begin{equation*} \|h - g\|_p^{\,q} \leq \sum_{k=1}^n \lvert a_k\rvert^{q}\,\|\chi_{A_k} - \chi_{E_k}\|_p^{\,q} < \sum_{k=1}^n \lvert a_k \rvert^q \delta_k^{\,q/p} < \frac{\varepsilon^q}{2} . \end{equation*}

Therefore

\begin{equation*} \|f - g\|_p^{\,q} \leq \|f - h\|_p^{\,q} + \|h - g\|_p^{\,q} < \frac{\varepsilon^q}{2} + \frac{\varepsilon^q}{2} = \varepsilon^q , \end{equation*}

which gives \(\|f - g\|_p < \varepsilon\).

Round off the corners of the step function supplied by 7A.18: replace each \(\chi_I\) by a continuous trapezoid. Throughout, \(q = \min\{p,1\}\) and errors add through the \(q\)-triangle inequality \(\|\sum_k f_k\|_p^{\,q} \leq \sum_k \|f_k\|_p^{\,q}\) proved in 7A.18.

Let \(\varepsilon > 0\). By 7A.18 there is a step function \(h \in \mathcal{L}^p(\mathbb{R})\) with

\begin{equation*} \|f - h\|_p^{\,q} < \frac{\varepsilon^q}{2} . \end{equation*}

If \(h = 0\), take \(g = 0\). Otherwise write \(h = a_1\chi_{I_1} + \cdots + a_n\chi_{I_n}\) with the \(a_k\) nonzero and the \(I_k\) intervals, which may be taken disjoint: the finitely many endpoints cut \(\mathbb{R}\) into finitely many pieces on each of which \(h\) is constant, and we discard the pieces where \(h = 0\) and relabel. Disjointness then gives \(\|h\|_p^{\,p} = \sum_k \lvert a_k\rvert^p \lvert I_k\rvert < \infty\), so each \(I_k\) is bounded, say with endpoints \(b_k \leq c_k\).

Trapezoids. For a bounded interval \(I\) with endpoints \(b \leq c\) and for \(\delta > 0\), let \(\varphi = 0\) if \(c - b \leq 2\delta\), and otherwise let \(\varphi\) be the continuous piecewise-linear function that equals \(1\) on \([b+\delta, c-\delta]\), equals \(0\) off \((b,c)\), and is linear on \([b, b+\delta]\) and on \([c-\delta, c]\). In the first case \(\|\chi_I - \varphi\|_p^{\,p} = c - b \leq 2\delta\); in the second, \(\chi_I - \varphi\) is bounded by \(1\) and vanishes off \([b,b+\delta] \cup [c-\delta,c]\), a set of measure \(2\delta\). Either way \(\varphi\) is continuous, vanishes off \([b,c]\), and

\begin{equation*} \|\chi_I - \varphi\|_p^{\,p} \leq 2\delta . \end{equation*}

Choose \(\delta > 0\) small enough that

\begin{equation*} \lvert a_k\rvert^{q} (2\delta)^{q/p} < \frac{\varepsilon^q}{2n} \qquad \text{for } k = 1, \ldots, n, \end{equation*}

possible since \((2\delta)^{q/p} \to 0\) as \(\delta \downarrow 0\); let \(\varphi_k\) be the trapezoid for \(I_k\) and this \(\delta\), and set

\begin{equation*} g = a_1 \varphi_1 + \cdots + a_n \varphi_n . \end{equation*}

Then \(g\) is continuous and vanishes outside the bounded set \([b_1,c_1] \cup \cdots \cup [b_n,c_n]\), hence is bounded and lies in \(\mathcal{L}^p(\mathbb{R})\). By the \(q\)-triangle inequality and 7.5(b),

\begin{equation*} \|h - g\|_p^{\,q} \leq \sum_{k=1}^n \lvert a_k \rvert^{q}\,\|\chi_{I_k} - \varphi_k\|_p^{\,q} \leq \sum_{k=1}^n \lvert a_k\rvert^{q} (2\delta)^{q/p} < \frac{\varepsilon^q}{2} , \end{equation*}

so \(\|f - g\|_p^{\,q} \leq \|f-h\|_p^{\,q} + \|h-g\|_p^{\,q} < \varepsilon^q\).

Normalize and use strict convexity of \(t \mapsto t^p\). Two pointwise facts do the work.

Fact 1. For \(z, w \in \mathbf{F}\), \(\lvert z + w\rvert = \lvert z\rvert + \lvert w\rvert\) if and only if \(z\overline{w} \in [0,\infty)\): comparing

\begin{equation*} \lvert z + w\rvert^2 = \lvert z\rvert^2 + 2\operatorname{Re}(z\overline w) + \lvert w\rvert^2 , \quad (\lvert z\rvert + \lvert w\rvert)^2 = \lvert z\rvert^2 + 2\lvert z \overline w\rvert + \lvert w\rvert^2 , \end{equation*}

equality holds exactly when \(\operatorname{Re}(z \overline w) = \lvert z \overline w\rvert\), that is, when \(z \overline w \geq 0\).

Fact 2. For \(p > 1\), \(s \in (0,1)\), and \(\alpha, \beta \geq 0\) we have \(\bigl(s\alpha + (1-s)\beta\bigr)^p \leq s\alpha^p + (1-s)\beta^p\), with equality if and only if \(\alpha = \beta\); indeed \(t \mapsto t^p\) has second derivative \(p(p-1)t^{p-2} > 0\) on \((0,\infty)\), hence is strictly convex there, while for \(\alpha = 0 < \beta\) the left side is \((1-s)^p \beta^p < (1-s)\beta^p\).

(\(\Leftarrow\)) Suppose \(af = bg\) almost everywhere with \(a, b \geq 0\) not both \(0\). If \(b = 0\), then \(a > 0\), so \(f = 0\) almost everywhere and \(\|f+g\|_p = \|g\|_p = \|f\|_p + \|g\|_p\). If \(b > 0\), then \(g = cf\) almost everywhere with \(c = a/b \geq 0\), so 7.5(b) gives \(\|f+g\|_p = (1+c)\|f\|_p = \|f\|_p + \|g\|_p\).

(\(\Rightarrow\)) Suppose \(\|f+g\|_p = \|f\|_p + \|g\|_p\). If \(\|f\|_p = 0\), take \((a,b) = (1,0)\); if \(\|g\|_p = 0\), take \((a,b) = (0,1)\). Otherwise \(A := \|f\|_p\) and \(B := \|g\|_p\) lie in \((0,\infty)\); put \(u = f/A\), \(v = g/B\), and \(s = A/(A+B) \in (0,1)\), so that \(\|u\|_p = \|v\|_p = 1\) by 7.5(b) and \(f + g = (A+B)\bigl(su + (1-s)v\bigr)\). The hypothesis becomes \(\|su + (1-s)v\|_p = 1\), while pointwise

\begin{equation*} \bigl\lvert su + (1-s)v \bigr\rvert^p \leq \bigl(s\lvert u\rvert + (1-s)\lvert v\rvert\bigr)^p \leq s \lvert u\rvert^p + (1-s)\lvert v\rvert^p \end{equation*}

by the triangle inequality in \(\mathbf{F}\) and Fact 2. Integrating turns this chain into \(1 \leq 1 \leq 1\), so each of the two differences is a nonnegative function of integral \(0\) (all three terms are integrable, being dominated by the right-hand side) and hence vanishes almost everywhere by the second bullet point of 3.43. Thus for almost every \(x \in X\):

\begin{equation*} \text{(i)}\quad \bigl\lvert s u(x) + (1-s) v(x)\bigr\rvert = s\lvert u(x)\rvert + (1-s)\lvert v(x)\rvert , \end{equation*}

\begin{equation*} \text{(ii)}\quad \bigl(s \lvert u(x)\rvert + (1-s)\lvert v(x)\rvert\bigr)^p = s\lvert u(x)\rvert^p + (1-s)\lvert v(x)\rvert^p . \end{equation*}

where (i) uses the injectivity of \(t \mapsto t^p\) on \([0,\infty)\). At such an \(x\), Fact 2 turns (ii) into \(\lvert u(x)\rvert = \lvert v(x)\rvert\), and Fact 1 (with \(z = su(x)\), \(w = (1-s)v(x)\)) turns (i) into \(u(x)\overline{v(x)} \geq 0\). If \(u(x) = 0\), then \(v(x) = 0\); otherwise

\begin{equation*} u(x)\overline{v(x)} = \lvert u(x)\rvert\,\lvert v(x)\rvert = \lvert u(x)\rvert^2 = u(x)\overline{u(x)} , \end{equation*}

and dividing by \(u(x)\) gives \(v(x) = u(x)\). Hence \(u = v\) almost everywhere, that is, \(Bf = Ag\) almost everywhere; take \(a = B\) and \(b = A\).

The two sides differ by the integral of the nonnegative function

\begin{equation*} \psi = \lvert f\rvert + \lvert g\rvert - \lvert f + g\rvert , \end{equation*}

so the equality holds if and only if \(\psi = 0\) almost everywhere. Indeed \(\psi \geq 0\) by the triangle inequality in \(\mathbf{F}\), and \(\lvert f\rvert\), \(\lvert g\rvert\), \(\lvert f+g\rvert\) all have finite integral (as \(f, g \in \mathcal{L}^1(\mu)\)), so additivity of the integral gives

\begin{equation*} \int \psi \, d\mu = \|f\|_1 + \|g\|_1 - \|f+g\|_1 , \end{equation*}

and a nonnegative function has integral \(0\) exactly when it vanishes almost everywhere, by the second bullet point of 3.43.

Finally \(\psi(x) = 0\) says \(\lvert f(x) + g(x)\rvert = \lvert f(x)\rvert + \lvert g(x)\rvert\), which holds if and only if \(f(x)\overline{g(x)} \geq 0\) by the equality case of the triangle inequality in \(\mathbf{F}\) established as Fact 1 in 7A.20.

Apply Tonelli’s Theorem 5.28 to \(|f|^p\). That function is nonnegative and \(\mathcal{S} \otimes \mathcal{T}\)-measurable, and directly from the definition 5.7 of cross sections,

\begin{equation*} \bigl[\,|f|^p\,\bigr]_x = \bigl|[f]_x\bigr|^p \qquad\text{and}\qquad \bigl[\,|f|^p\,\bigr]^y = \bigl|[f]^y\bigr|^p . \end{equation*}

Both measure spaces are \(\sigma\)-finite, so 5.28 says that

\begin{equation*} h \colon X \to [0,\infty] , \qquad h(x) = \int_Y \bigl|[f]_x\bigr|^p \, d\nu \end{equation*}

is \(\mathcal{S}\)-measurable with \(\int_X h \, d\mu = \int_{X \times Y} |f|^p \, d(\mu \times \nu) = \|f\|_p^p < \infty\). Writing \(E = \{h = \infty\} \in \mathcal{S}\), the bound \(c\chi_E \leq h\) gives

\begin{equation*} c\,\mu(E) \leq \int_X h \, d\mu = \|f\|_p^p \qquad \text{for every } c > 0 , \end{equation*}

so \(\mu(E) = 0\). Thus for every \(x \notin E\) the cross section \([f]_x\) is \(\mathcal{T}\)-measurable (5.9) with \(\int_Y |[f]_x|^p \, d\nu < \infty\), that is, \([f]_x \in \mathcal{L}^p(\nu)\) for almost every \(x \in X\). Part (b) of 5.28 gives \([f]^y \in \mathcal{L}^p(\mu)\) for almost every \(y \in Y\) by the same argument.

Both parts reduce, via the density result 7A.19, to a continuous function vanishing off a bounded set; the reduction runs on the two invariance identities

\begin{equation*} \int_{-\infty}^{\infty} h(x-t) \, dx = \int_{-\infty}^{\infty} h , \qquad \int_{-\infty}^{\infty} h(tx) \, dx = \frac{1}{t}\int_{-\infty}^{\infty} h \end{equation*}

for measurable \(h \geq 0\), with \(t \in \mathbb{R}\) in the first and \(t > 0\) in the second. Both hold for \(h = \chi_E\) because \(|t + E| = |E|\) by 2.7 and \(|t^{-1}E| = t^{-1}|E|\) by the definition 2.2 of outer measure applied to dilated covers (translates and dilates of measurable sets are measurable, by 2.70 in the Lebesgue case), hence for simple \(h\) by 3.7, hence for all measurable \(h \geq 0\) by 2.89 and the Monotone Convergence Theorem 3.11. Taking \(h = |f|^p\) gives \(f_t \in \mathcal{L}^p(\mathbb{R})\) with

\begin{equation*} \|f_t\|_p = \|f\|_p \quad \text{in (a)} , \qquad \|f_t\|_p = t^{-1/p}\|f\|_p \quad \text{in (b)} . \end{equation*}

Also, for each \(\varepsilon > 0\), 7A.19 supplies a continuous \(g\) vanishing outside some \([-M,M]\) with \(\|f - g\|_p < \varepsilon\); such a \(g\) is uniformly continuous on \(\mathbb{R}\), being uniformly continuous on the compact set \([-M-2, M+2]\) and zero on a neighbourhood of the complement. (Check!)

(a) Put \(\varphi(t) = \|f - f_t\|_p\). Minkowski 7.14 and translation invariance give \(\varphi(t) \leq \|f\|_p + \|f_t\|_p = 2\|f\|_p\), so \(\varphi\) is bounded. Since \((f_a)_b = f_{a+b}\), we have \(f_t - f_s = (f - f_{s-t})_t\), whence \(\|f_t - f_s\|_p = \varphi(s-t)\), and 7.14 applied twice gives

\begin{equation*} \lvert \varphi(t) - \varphi(s)\rvert \leq \|f_t - f_s\|_p = \varphi(s-t) . \end{equation*}

So uniform continuity follows once \(\varphi(u) \to 0\) as \(u \to 0\). Given \(\varepsilon > 0\), take \(g\) as above with \(\|f-g\|_p < \varepsilon/3\); then \(\|f_u - g_u\|_p = \|f-g\|_p\), so 7.14 gives \(\varphi(u) \leq \tfrac{2\varepsilon}{3} + \|g - g_u\|_p\). For \(\lvert u\rvert \leq 1\) the function \(g - g_u\) vanishes outside \([-M-1, M+1]\), so

\begin{equation*} \|g - g_u\|_p^{\,p} \leq (2M+2) \Bigl( \sup_{x \in \mathbb{R}} \lvert g(x) - g(x-u)\rvert \Bigr)^{p} , \end{equation*}

and the supremum tends to \(0\) as \(u \to 0\) by uniform continuity of \(g\). Hence \(\varphi(u) < \varepsilon\) for all small \(\lvert u\rvert\).

(b) Given \(\varepsilon > 0\), take \(g\) as above with \(\|f - g\|_p < \varepsilon\). For \(t \in [\tfrac12, 2]\), dilation invariance gives \(\|f_t - g_t\|_p = t^{-1/p}\|f-g\|_p \leq 2^{1/p}\varepsilon \leq 2\varepsilon\) (here \(p \geq 1\)), so by 7.14

\begin{equation*} \|f - f_t\|_p \leq \|f-g\|_p + \|g - g_t\|_p + \|g_t - f_t\|_p < 3\varepsilon + \|g - g_t\|_p . \end{equation*}

For such \(t\), if \(\lvert x\rvert > 2M\) then \(\lvert x \rvert > M\) and \(\lvert tx\rvert \geq \lvert x\rvert/2 > M\), so \(g - g_t\) vanishes outside \([-2M, 2M]\) and

\begin{equation*} \|g - g_t\|_p^{\,p} \leq 4M \Bigl( \sup_{\lvert x\rvert \leq 2M} \lvert g(x) - g(tx)\rvert \Bigr)^{p} . \end{equation*}

Since \(\lvert tx - x\rvert \leq 2M\lvert t - 1\rvert\) for \(\lvert x\rvert \leq 2M\), uniform continuity of \(g\) sends that supremum to \(0\) as \(t \to 1\). Hence \(\limsup_{t \to 1}\|f - f_t\|_p \leq 3\varepsilon\) for every \(\varepsilon > 0\).

Apply the Lebesgue Differentiation Theorem 4.10 to the countably many localized functions

\begin{equation*} h_{r,n} = |f - r|^p \, \chi_{(-n,n)} , \qquad r \in D, \ n \in \mathbb{Z}^{+}, \end{equation*}

where \(D\) is a countable dense subset of \(\mathbf{F}\) (take \(\mathbb{Q}\) or \(\mathbb{Q} + i\mathbb{Q}\)), and then let \(r \to f(b)\). Convexity of \(s \mapsto s^p\) on \([0,\infty)\) (here \(p \geq 1\)) gives

\begin{equation*} |u + v|^p \leq \bigl(|u| + |v|\bigr)^p \leq 2^{p-1}\bigl(|u|^p + |v|^p\bigr) \qquad (u, v \in \mathbf{F}) , \end{equation*}

so \(\int h_{r,n} \leq 2^{p-1}\bigl(\|f\|_p^p + 2n|r|^p\bigr) < \infty\) and \(h_{r,n} \in \mathcal{L}^1(\mathbb{R})\). Hence 4.10 supplies null sets \(N_{r,n}\) off which the averages of \(|h_{r,n} - h_{r,n}(b)|\) over \((b-t,b+t)\) tend to \(0\); put \(N = \bigcup_{r,n} N_{r,n}\), which is null by 2.8.

Fix \(b \notin N\) and \(r \in D\), and choose \(n > |b| + 1\). For \(0 < t < 1\) we have \((b-t,b+t) \subseteq (-n,n)\), so \(h_{r,n} = |f - r|^p\) there and \(h_{r,n}(b) = |f(b) - r|^p\), whence

\begin{equation*} \left| \frac{1}{2t}\int_{b-t}^{b+t} |f - r|^p - |f(b) - r|^p \right| \leq \frac{1}{2t} \int_{b-t}^{b+t} \bigl| h_{r,n} - h_{r,n}(b) \bigr| \longrightarrow 0 . \end{equation*}

Now let \(\varepsilon > 0\) and pick \(r \in D\) with \(|f(b) - r| < \varepsilon\). The displayed inequality with \(u = f(x) - r\) and \(v = r - f(b)\), integrated over \((b-t, b+t)\), gives

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)|^p \leq 2^{p-1}\left( \frac{1}{2t}\int_{b-t}^{b+t} |f - r|^p + |r - f(b)|^p \right) , \end{equation*}

so by the previous paragraph

\begin{equation*} \limsup_{t \downarrow 0} \ \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)|^p \leq 2^p |f(b) - r|^p < 2^p \varepsilon^p . \end{equation*}

As \(\varepsilon > 0\) was arbitrary and the averages are nonnegative, the limit is \(0\) for every \(b \notin N\).

The triangle inequality fails. Take \(a = (1,0,\ldots,0)\) and \(b = (0,1,0,\ldots,0)\), available because \(n > 1\). Then \(\|a\| = \|b\| = 1\), while \(1/p > 1\) gives

\begin{equation*} \|a + b\| = (1^p + 1^p)^{1/p} = 2^{1/p} > 2 = \|a\| + \|b\| . \end{equation*}

(a) Yes: the sequences with finitely many nonzero terms, all of them in \(\mathbb{D}\), where \(\mathbb{D} = \mathbb{Q}\) or \(\mathbb{Q} + i\mathbb{Q}\) is a countable dense subset of \(\mathbb{F}\). Call this set \(D\); it is countable, being \(\bigcup_{n=1}^{\infty} D_n\) with \(D_n\) (vanishing after coordinate \(n\)) in bijection with \(\mathbb{D}^n\), and \(D \subseteq \ell^p\). Given \(a \in \ell^p\) and \(\varepsilon > 0\), convergence of \(\sum_k |a_k|^p\) gives \(n\) with \(\sum_{k > n}|a_k|^p < \varepsilon^p/2\); choosing \(b_k \in \mathbb{D}\) with \(|a_k - b_k|^p < \varepsilon^p/(2n)\) for \(k \leq n\) and setting \(b = (b_1, \ldots, b_n, 0, 0, \ldots) \in D\) gives

\begin{equation*} \|a - b\|_p^p = \sum_{k=1}^n |a_k - b_k|^p + \sum_{k > n} |a_k|^p < \varepsilon^p . \end{equation*}

(b) No: the set \(S\) of sequences with entries in \(\{0,1\}\) lies in \(\ell^\infty\), is uncountable (it is in bijection with the subsets of \(\mathbb{Z}^+\)), and any two distinct members are at distance exactly \(1\). If \(C \subseteq \ell^\infty\) has closure \(\ell^\infty\), choose for each \(a \in S\) some \(c(a) \in C\) with \(\|a - c(a)\|_\infty < \tfrac12\). This map is injective, since \(c(a) = c(b)\) would give

\begin{equation*} \|a - b\|_\infty \leq \|a - c(a)\|_\infty + \|c(b) - b\|_\infty < 1 , \end{equation*}

forcing \(a = b\). So \(C\) is uncountable.

(a) Yes: with \(\mathbb{D} = \mathbb{Q}\) or \(\mathbb{Q} + i\mathbb{Q}\) (countable and dense in \(\mathbb{F}\)), take

\begin{equation*} D = \Bigl\{ \textstyle\sum_{j=1}^{m} c_j \chi_{(a_j, b_j)} : m \in \mathbb{Z}^+ ,\ c_j \in \mathbb{D} ,\ a_j < b_j \text{ rational} \Bigr\} . \end{equation*}

For each \(m\) these sums form the image of the countable set \((\mathbb{D}\times\mathbb{Q}\times\mathbb{Q})^m\), so \(D\) is countable, and each of its elements is bounded with bounded support, so \(D \subseteq \mathcal{L}^p(\mathbb{R})\).

Let \(f \in \mathcal{L}^p(\mathbb{R})\) and \(\varepsilon > 0\). By 7A.18 there is a step function \(h\) with \(\|f - h\|_p < \varepsilon/2\), and as in 7A.19 we may write \(h = \sum_{j=1}^m a_j \chi_{I_j}\) with the \(I_j\) disjoint bounded intervals. Given \(\delta \in (0,1)\), enlarge each \(I_j\) to an interval \(J_j\) with rational endpoints within \(\delta\) of those of \(I_j\), so that \(\lambda(J_j \setminus I_j) < 2\delta\) and hence \(\|\chi_{I_j} - \chi_{J_j}\|_p \leq (2\delta)^{1/p}\), and pick \(q_j \in \mathbb{D}\) with \(|a_j - q_j| < \delta\). Then \(u = \sum_{j=1}^m q_j \chi_{J_j} \in D\), and Minkowski 7.14 gives

\begin{equation*} \|h - u\|_p \leq (2\delta)^{1/p} \sum_{j=1}^m |a_j| \ + \ \delta \sum_{j=1}^m \lambda(J_j)^{1/p} , \end{equation*}

which is less than \(\varepsilon/2\) once \(\delta\) is small (the \(\lambda(J_j) \leq \lambda(I_j) + 2\) stay bounded). Hence \(\|f - u\|_p < \varepsilon\), so the closure of \(D\) is \(L^p(\mathbb{R})\).

(b) No: for \(t \in (0,1)\) put \(g_t = \chi_{(0,t)} \in L^\infty(\mathbb{R})\). If \(0 < s < t < 1\), then \(g_t - g_s = \chi_{[s,t)}\), whose every level set \(\{|\cdot| > r\}\) with \(r \in (0,1)\) has measure \(t - s > 0\), so \(\|g_t - g_s\|_\infty = 1\) by 7.1. Thus \(\{g_t\}\) is an uncountable family at pairwise distance \(1\). If \(C \subseteq L^\infty(\mathbb{R})\) had closure \(L^\infty(\mathbb{R})\), choosing \(c_t \in C\) with \(\|g_t - c_t\|_\infty < \tfrac12\) would give an injection \(t \mapsto c_t\) of \((0,1)\) into \(C\), since \(c_s = c_t\) forces

\begin{equation*} \|g_s - g_t\|_\infty \leq \|g_s - c_s\|_\infty + \|c_t - g_t\|_\infty < 1 , \end{equation*}

hence \(s = t\). So \(C\) is uncountable.

If \(f \notin \mathcal{L}^p(\mu)\), a single \(h \in \mathcal{L}^{p^{\prime}}(\mu)\) built from a \(\sigma\)-finite exhaustion violates the hypothesis. Fix \(X_1 \subseteq X_2 \subseteq \cdots\) in \(\mathcal{S}\) with \(\mu(X_n) < \infty\) and \(\bigcup_n X_n = X\), and recall \(\tfrac1p + \tfrac1{p^{\prime}} = 1\) (7.6). We use once and for all that \(\int u\chi_N \, d\mu = 0\) for measurable \(u \geq 0\) and \(\mu(N) = 0\), so that \(\int u \, d\mu = \int u \chi_{X \setminus N}\, d\mu\) (Check!); this converts a nonnegative function that is finite almost everywhere into an \(\mathbb{F}\)-valued one with the same norm.

(i) \(p = 1\), so \(p^{\prime} = \infty\). The constant function \(1\) lies in \(\mathcal{L}^{\infty}(\mu)\), so \(f = f \cdot 1 \in \mathcal{L}^1(\mu)\) by hypothesis.

(ii) \(p = \infty\), so \(p^{\prime} = 1\). Suppose \(\|f\|_{\infty} = \infty\). By 7.1 each \(A_k = \{|f| > 4^k\}\) has \(\mu(A_k) > 0\), so continuity from below supplies \(n_k\) with \(B_k = A_k \cap X_{n_k}\) satisfying \(0 < \mu(B_k) < \infty\). Put \(h = \sum_{k=1}^{\infty} 2^{-k}\mu(B_k)^{-1}\chi_{B_k}\); the Monotone Convergence Theorem 3.11 applied to the partial sums gives \(\int h \, d\mu = 1\), so \(h < \infty\) almost everywhere and \(\hat h = h\chi_{X \setminus N}\), with \(N = \{h = \infty\}\), lies in \(\mathcal{L}^1(\mu)\). But for every \(k\),

\begin{equation*} \int |f \hat h| \, d\mu = \int |f| h \, d\mu \geq \frac{2^{-k}}{\mu(B_k)} \int_{B_k} |f| \, d\mu \geq 2^{-k} 4^k = 2^k , \end{equation*}

contradicting \(f\hat h \in \mathcal{L}^1(\mu)\). Hence \(f \in \mathcal{L}^{\infty}(\mu)\).

(iii) \(1 < p < \infty\). Put \(E_n = \{x \in X_n : |f(x)| \leq n\}\) and \(g_n = |f|\chi_{E_n}\). The \(E_n\) increase with union \(X\), so 3.11 gives \(\|g_n\|_p^p \to \int |f|^p \, d\mu\), while \(g_n \leq n \chi_{X_n}\) gives \(\|g_n\|_p^p \leq n^p \mu(X_n) < \infty\). Suppose \(\int |f|^p \, d\mu = \infty\). Choose \(n_1 < n_2 < \cdots\) with \(c_k := \|g_{n_k}\|_p \geq 4^k\), write \(G_k = g_{n_k}\), and set \(h_k = G_k^{\,p-1}/c_k^{\,p-1}\). Since \((p-1)p^{\prime} = p\),

\begin{equation*} \|h_k\|_{p^{\prime}}^{p^{\prime}} = \frac{1}{c_k^{\,p}} \int G_k^{\,p} \, d\mu = 1 \qquad\text{and}\qquad \int G_k h_k \, d\mu = \frac{c_k^{\,p}}{c_k^{\,p-1}} = c_k \geq 4^k . \end{equation*}

Put \(h = \sum_{k=1}^{\infty} 2^{-k} h_k\). Minkowski 7.14 bounds the \(p^{\prime}\)-norm of each partial sum by \(\sum_k 2^{-k} = 1\), so 3.11 gives \(\|h\|_{p^{\prime}} \leq 1\); hence \(h < \infty\) almost everywhere and \(\hat h = h \chi_{X \setminus N}\) lies in \(\mathcal{L}^{p^{\prime}}(\mu)\). But \(|f| \geq G_k\) and \(\hat h \geq 2^{-k}h_k \chi_{X \setminus N}\) give

\begin{equation*} \int |f \hat h| \, d\mu \geq 2^{-k} \int G_k h_k \, d\mu \geq 2^{-k} 4^k = 2^k \end{equation*}

for every \(k\), a contradiction. Hence \(\int |f|^p \, d\mu < \infty\).

(a) Everything follows from the strict convexity inequality

\begin{equation*} \left| \frac{a+b}{2} \right|^p \leq \left( \frac{|a| + |b|}{2}\right)^{p} \leq \frac{|a|^p + |b|^p}{2} \qquad (a, b \in \mathbb{F}) , \end{equation*}

in which equality forces \(a = b\). The first step is the triangle inequality in \(\mathbb{F}\) with \(t \mapsto t^p\) increasing; the second is convexity of \(t \mapsto t^p\) on \([0,\infty)\), strict because \(\varphi^{\prime}(u) = pu^{p-1}\) is strictly increasing there (this is where \(p > 1\) enters). So overall equality forces \(|a| = |b| =: r\) from the second step and \(|a + b| = 2r\) from the first; if \(r > 0\), then expanding \(|a+b|^2 = 2r^2 + 2\operatorname{Re}(a\overline b) = 4r^2\) gives \(\operatorname{Re}(a\overline b) = r^2 = |a \overline b|\), so \(a\overline b = r^2\) and \(a = r^2 b/|b|^2 = b\); and if \(r = 0\), then \(a = b = 0\).

Now let \(c = \|f\|_p = \|g\|_p = \bigl\|\tfrac{f+g}{2}\bigr\|_p < \infty\) and put

\begin{equation*} u = \frac{|f|^p + |g|^p}{2} - \left| \frac{f+g}{2}\right|^{p} \geq 0 . \end{equation*}

Additivity 3.16 gives \(\int u \, d\mu + c^p = c^p\), and \(c^p < \infty\) permits the subtraction, so \(\int u \, d\mu = 0\). By the second bullet point of 3.43, \(u = 0\) almost everywhere, and by the equality case above this says \(f = g\) almost everywhere, that is, \(f = g\) in \(L^p(\mu)\).

(b) Counting measure on \(\{1,2\}\), with \(f = (1,0)\) and \(g = (0,1)\): here \(\|f\|_1 = \|g\|_1 = 1\) and \(\tfrac{f+g}{2} = (\tfrac12, \tfrac12)\) also has \(1\)-norm \(1\), yet \(f \neq g\).

(c) The same space with \(f = (1,1)\) and \(g = (1,0)\): here \(\|f\|_\infty = \|g\|_\infty = 1\) and \(\tfrac{f+g}{2} = (1,\tfrac12)\) has \(\infty\)-norm \(1\), yet \(f \neq g\).

Integrate the pointwise inequality \(|f + g|^p \leq |f|^p + |g|^p\). For that, it suffices that

\begin{equation*} (a+b)^p \leq a^p + b^p \qquad (a, b \geq 0) , \end{equation*}

since \(|f+g| \leq |f| + |g|\) and \(t \mapsto t^p\) is increasing. To prove it, assume \(a + b > 0\) and put \(s = a/(a+b)\), \(t = b/(a+b)\); then \(s, t \in [0,1]\) with \(s + t = 1\), and \(u^p \geq u\) for \(u \in [0,1]\) (as \(u^{p-1} \geq 1\) there, \(p < 1\)), so

\begin{equation*} \frac{a^p + b^p}{(a+b)^p} = s^p + t^p \geq s + t = 1 . \end{equation*}

Integrating the pointwise bound, 3.8 and additivity 3.16 give

\begin{equation*} \|f+g\|_p^p = \int |f+g|^p \, d\mu \leq \int |f|^p \, d\mu + \int |g|^p \, d\mu = \|f\|_p^p + \|g\|_p^p , \end{equation*}

an inequality in \([0,\infty]\).

Everything rests on two facts: \(Z(\mu)\) is a subspace of \(\mathcal{L}^p(\mu)\), and \(\|\cdot\|_p\) is constant on its cosets. Then \(L^p(\mu)\) is the quotient \(\mathcal{L}^p(\mu)/Z(\mu)\) with the operations of 7.16.

(i) \(\mathcal{L}^p(\mu)\) is a vector space: this is 7.5 for \(p < \infty\), and for \(p = \infty\) it follows from the triangle inequality and homogeneity of \(\|\cdot\|_{\infty}\) proved in 7A.1. Its subset \(Z(\mu)\) of functions vanishing almost everywhere is a subspace, the union of two null sets being null, and \(Z(\mu) \subseteq \mathcal{L}^p(\mu)\) because such functions have \(\|\cdot\|_p = 0\).

(ii) \(\widetilde f = \widetilde F\) if and only if \(f - F \in Z(\mu)\), that is, \(f = F\) almost everywhere: one direction uses that \(Z(\mu)\) is closed under addition and negation, the other that \(f \in \widetilde f\). (Check!) Consequently, if \(\widetilde f = \widetilde F\) and \(\widetilde g = \widetilde G\), then

\begin{equation*} (f+g) - (F+G) = (f-F) + (g-G) \in Z(\mu) , \qquad \alpha f - \alpha F = \alpha(f - F) \in Z(\mu) , \end{equation*}

so the operations of 7.16 do not depend on the representatives. Each vector space axiom for \(L^p(\mu)\) then transfers from the corresponding axiom in \(\mathcal{L}^p(\mu)\) through \(f \mapsto \widetilde f\), which by definition carries sums to sums and scalar multiples to scalar multiples; the additive identity is \(\widetilde 0 = Z(\mu)\) and \(-\widetilde f = (-f)^{\sim}\). (Check!)

(iii) \(\|\cdot\|_p\) is well defined on \(L^p(\mu)\). Suppose \(f = F\) off a set \(N\) with \(\mu(N) = 0\). If \(p < \infty\), then \(|f|^p\chi_{X \setminus N} = |F|^p \chi_{X\setminus N}\) and a null set contributes nothing to an integral of a nonnegative function, so \(\int |f|^p \, d\mu = \int |F|^p \, d\mu\). If \(p = \infty\), then \(\{|f| > t\}\) and \(\{|F| > t\}\) differ only inside \(N\), so one is null exactly when the other is, and the two infima in 7.1 agree. Since \(f \in \mathcal{L}^p(\mu)\), the common value \(\|\widetilde f\|_p = \|f\|_p\) lies in \([0,\infty)\), as 6.33 requires.

(iv) The three conditions of 6.33. Homogeneity is 7.5(b) for \(p < \infty\) and 7A.1 for \(p = \infty\), read on representatives; the triangle inequality is Minkowski 7.14 likewise. For positive definiteness, \(\|\widetilde f\|_p = 0\) forces \(f = 0\) almost everywhere: for \(p < \infty\) apply the second bullet point of 3.43 to \(|f|^p\); for \(p = \infty\) each set \(\{|f| > \tfrac1n\}\) is null by 7.1, and \(\{f \neq 0\}\) is their union. Hence \(\widetilde f = \widetilde 0\), and the converse is (iii) with \(F = 0\).

Take \(f\) to be the pointwise limit of \(f_1, f_2, \ldots\) off the null set where the essential suprema fail to bound. As in 7A.16(i), \(\mu(\{|g| > \|g\|_{\infty}\}) = 0\) for every \(\mathcal{S}\)-measurable \(g\), so

\begin{equation*} E = \bigcup_{j,k} \bigl\{ |f_j - f_k| > \|f_j - f_k\|_{\infty} \bigr\} \ \cup \ \bigcup_k \bigl\{ |f_k| > \|f_k\|_{\infty} \bigr\} \end{equation*}

is a countable union of null sets, hence \(E \in \mathcal{S}\) with \(\mu(E) = 0\). On \(X \setminus E\) we have \(|f_j - f_k| \leq \|f_j - f_k\|_{\infty}\) for all \(j, k\), so the Cauchy hypothesis makes \(f_1, f_2, \ldots\) uniformly Cauchy there, hence pointwise convergent by completeness of \(\mathbf{F}\). Define \(f = \lim_{k \to \infty} f_k \chi_{X \setminus E}\), which is \(\mathcal{S}\)-measurable by 2.48.

Given \(\varepsilon > 0\), choose \(n\) with \(\|f_j - f_k\|_{\infty} < \varepsilon\) for all \(j, k \geq n\). For \(k \geq n\) and \(x \in X \setminus E\),

\begin{equation*} |f_k(x) - f(x)| = \lim_{j \to \infty} |f_k(x) - f_j(x)| \leq \varepsilon , \end{equation*}

so \(\{|f_k - f| > t\} \subseteq E\) is null for every \(t > \varepsilon\) and thus \(\|f_k - f\|_{\infty} \leq \varepsilon\); hence \(\|f_k - f\|_{\infty} \to 0\). Taking \(\varepsilon = 1\) with its \(n\), we get \(|f| \leq |f_n| + 1 \leq \|f_n\|_{\infty} + 1\) on \(X \setminus E\), so \(\|f\|_{\infty} \leq \|f_n\|_{\infty} + 1 < \infty\) and \(f \in \mathcal{L}^{\infty}(\mu)\).

The telescoping proof of 7.20 runs unchanged with \(\|\cdot\|_p^{\,p}\) in place of the norm, because 7B.6 gives \(\|u+v\|_p^p \leq \|u\|_p^p + \|v\|_p^p\) and hence, by induction, \(\|\sum_{k=1}^m u_k\|_p^p \leq \sum_{k=1}^m \|u_k\|_p^p\).

It suffices to find a convergent subsequence: if \(\|f_{k_m} - f\|_p \to 0\), then given \(\varepsilon > 0\) and \(n\) with \(\|f_j - f_k\|_p^p < \varepsilon/2\) for \(j,k \geq n\), choosing \(k_m \geq n\) with \(\|f_{k_m} - f\|_p^p < \varepsilon/2\) gives \(\|f_k - f\|_p^p < \varepsilon\) for all \(k \geq n\). So pick \(n_1 < n_2 < \cdots\) with \(\|f_j - f_k\|_p^p < 2^{-m}\) for \(j, k \geq n_m\), pass to the subsequence \((f_{n_m})\) without relabelling, and set \(f_0 = 0\); then

\begin{equation*} \sum_{k=1}^{\infty} \|f_k - f_{k-1}\|_p^p < \infty . \end{equation*}

Let \(g = \sum_{k=1}^{\infty} |f_k - f_{k-1}|\) with partial sums \(g_m\). Finite subadditivity gives \(\int g_m^{\,p} \, d\mu \leq \sum_{k=1}^m \|f_k - f_{k-1}\|_p^p\), and \(g_m^{\,p} \uparrow g^p\) pointwise, so the Monotone Convergence Theorem 3.11 gives

\begin{equation*} \int g^p \, d\mu \leq \sum_{k=1}^{\infty} \|f_k - f_{k-1}\|_p^p < \infty . \end{equation*}

Hence \(A = \{g < \infty\}\) satisfies \(\mu(X \setminus A) = 0\), since \(c\chi_{X \setminus A} \leq g^p\) for every \(c > 0\). On \(A\) the series \(\sum_k (f_k - f_{k-1})\) converges absolutely with \(m\)-th partial sum \(f_m\), so \(f := \lim_{m \to \infty} f_m \chi_A\) exists pointwise on \(X\) and is \(\mathcal{S}\)-measurable by 2.48; and \(|f| \leq g\) on \(A\) gives \(\|f\|_p^p \leq \int g^p \, d\mu < \infty\), so \(f \in \mathcal{L}^p(\mu)\).

Finally, let \(\varepsilon > 0\) and take \(n\) with \(\|f_j - f_k\|_p^p < \varepsilon\) for \(j,k \geq n\). For \(k \geq n\) the functions \(|f_k - f_j|^p\) converge almost everywhere to \(|f_k - f|^p\), so Fatou’s Lemma (3A.17) gives

\begin{equation*} \|f_k - f\|_p^p \leq \liminf_{j \to \infty} \|f_k - f_j\|_p^p \leq \varepsilon . \end{equation*}

Choose \(k_1 < k_2 < \cdots\) with \(\int |f_{k_m} - f|^p \, d\mu < 4^{-m}\), possible because \(\|f_k - f\|_p^p \to 0\). Markov’s inequality 4.1, applied to \(h_m = |f_{k_m} - f|^p \in \mathcal{L}^1(\mu)\) with constant \(2^{-m}\), gives

\begin{equation*} \mu(E_m) \leq 2^m \|h_m\|_1 < 2^{-m} , \qquad E_m = \{h_m \geq 2^{-m}\} . \end{equation*}

Put \(F = \bigcap_{M=1}^{\infty} \bigcup_{m \geq M} E_m\). Monotonicity 2.57 and countable subadditivity 2.58 give \(\mu(F) \leq \sum_{m \geq M} 2^{-m} = 2^{-M+1}\) for every \(M\), so \(\mu(F) = 0\).

If \(x \notin F\), then \(x \notin E_m\) for all \(m\) beyond some \(M\), so \(|f_{k_m}(x) - f(x)| < 2^{-m/p} \to 0\). Hence \(f_{k_m} \to f\) pointwise on \(X \setminus F\), that is, almost everywhere.

Take \(a = (2^{-1}, 2^{-2}, \ldots)\), which lies in the set \(A\) in question: every term is nonzero, and \(\|a\|_p^p = \sum_k 2^{-kp} < \infty\) for \(p < \infty\) while \(\|a\|_{\infty} = \tfrac12\). Given \(\varepsilon > 0\), pick \(n\) with \(2^{-n} < \varepsilon\) and let \(b\) agree with \(a\) except that \(b_n = 0\). Then \(b \in \ell^p\), \(b \notin A\), and \(a - b = 2^{-n}e_n\), so

\begin{equation*} \|a - b\|_p = 2^{-n}\|e_n\|_p = 2^{-n} < \varepsilon \end{equation*}

for every \(p \in [1,\infty]\). So no ball centred at \(a\) lies in \(A\); as \(a \in A\), the set \(A\) is not open.

Take the typewriter sequence with growing heights. For \(n \in \mathbf{Z}^+\) and \(j \in \{0,1,\ldots,2^n-1\}\) put \(I_{n,j} = [j2^{-n}, (j+1)2^{-n}]\), and index the pairs by

\begin{equation*} k = 2^n - 1 + j , \qquad f_k = n \, \chi_{I_{n,j}} , \end{equation*}

which is a bijection onto \(\mathbf{Z}^+\) because the blocks \(\{2^n-1, \ldots, 2^{n+1}-2\}\) partition \(\mathbf{Z}^+\); write \(n(k)\) for the resulting \(n\). Then

\begin{equation*} \|f_k\|_1 = n(k)\,\lambda\bigl(I_{n(k),j(k)}\bigr) = \frac{n(k)}{2^{n(k)}} , \end{equation*}

so each \(f_k \in \mathcal{L}^1([0,1])\); and \(k \geq 2^{n(k)} - 1\) forces \(n(k) \to \infty\), whence \(\|f_k\|_1 \to 0\) because \(n2^{-n} \to 0\).

Fix \(x \in [0,1]\) and \(n \in \mathbf{Z}^+\). The intervals \(I_{n,0}, \ldots, I_{n,2^n-1}\) cover \([0,1]\), so \(x \in I_{n,j}\) for some \(j\), and the corresponding \(k = 2^n - 1 + j\) gives \(f_k(x) = n\). Hence \(\sup_k f_k(x) = \infty\).

By 7.23 some subsequence satisfies \(f_{k_m} \to f\) off a set \(E\) with \(\mu(E) = 0\), while by hypothesis \(f_k \to g\) off a set \(F\) with \(\mu(F) = 0\). For \(x \notin E \cup F\), a null set by 2.58, the full sequence \(f_k(x)\) converges to \(g(x)\), hence so does the subsequence; since limits in \(\mathbf{F}\) are unique and \(f_{k_m}(x) \to f(x)\), we get \(f(x) = g(x)\). Thus \(f = g\) almost everywhere.

(a) Take \(X = \{1\}\), \(\mathcal{S} = \{\emptyset, X\}\), and \(\mu(X) = \infty\). Then \(\int |f| \, d\mu\) is \(0\) or \(\infty\) according as \(f(1) = 0\) or not, so

\begin{equation*} L^1(\mu) = \{0\} \qquad\text{and hence}\qquad \bigl(L^1(\mu)\bigr)^{\prime} = \{0\} , \end{equation*}

whereas \(\|f\|_{\infty} = |f(1)|\) by 7.1 and \(\emptyset\) is the only null set, so \(L^{\infty}(\mu)\) is one-dimensional. For \(h\) with \(h(1) = 1\) we get \(\|\varphi_h\| = 0 \neq 1 = \|h\|_{\infty}\), and \(\varphi_h = \varphi_0\) with \(h \neq 0\), so \(h \mapsto \varphi_h\) is not one-to-one either. (This \(\mu\) is not semifinite: \(X\) has infinite measure but no subset of positive finite measure.)

(b) For every measure, Holder’s inequality 7.9 with \(p = 1\), \(p^{\prime} = \infty\) gives \(\|fh\|_1 \leq \|f\|_1\|h\|_{\infty}\), so \(\varphi_h\) is a bounded linear functional on \(L^1(\mu)\) with \(\|\varphi_h\| \leq \|h\|_{\infty}\), and \(h \mapsto \varphi_h\) is linear; only the reverse inequality needs \(\sigma\)-finiteness.

Assume \(\|h\|_{\infty} > 0\) and fix \(t \in (0, \|h\|_{\infty})\). Then \(E = \{|h| > t\}\) has \(\mu(E) > 0\), since \(\mu(E) = 0\) would give \(\|h\|_{\infty} \leq t\) by 7.1. Writing \(X = \bigcup_n X_n\) with \(\mu(X_n) < \infty\), countable subadditivity 2.58 supplies \(n\) with \(A = E \cap X_n\) satisfying \(0 < \mu(A) \leq \mu(X_n) < \infty\). Put

\begin{equation*} f = \frac{1}{\mu(A)} \cdot \frac{\overline{h}}{|h|} \, \chi_A , \end{equation*}

well defined because \(|h| > t > 0\) on \(A\). Then \(|f| = \chi_A/\mu(A)\), so \(\|f\|_1 = 1\), while \(fh = |h|/\mu(A)\) on \(A\) and \(0\) elsewhere gives

\begin{equation*} \varphi_h(f) = \frac{1}{\mu(A)}\int_A |h| \, d\mu \geq t , \end{equation*}

whence \(\|\varphi_h\| \geq t\). Letting \(t \uparrow \|h\|_{\infty}\) gives \(\|\varphi_h\| = \|h\|_{\infty}\). One-to-oneness follows: \(\varphi_{h_1} = \varphi_{h_2}\) gives \(\|h_1 - h_2\|_{\infty} = \|\varphi_{h_1 - h_2}\| = 0\).

(a) \(c_0\) is a closed subspace of \(\ell^\infty\), which is a Banach space by 7.24 (it is \(L^\infty(\mu)\) for \(\mu\) counting measure on \(\mathbb{Z}^+\), where no nonempty set is null, so the essential supremum is the supremum). It is a subspace because \(\lim_{k \to \infty}(a_k + \alpha a_k^{\prime}) = 0\) whenever \(a, a^{\prime} \in c_0\). For closedness, suppose \(a^{(n)} \in c_0\) and \(\|a^{(n)} - a\|_\infty \to 0\); given \(\varepsilon > 0\), choose \(n\) with \(\|a^{(n)} - a\|_\infty < \varepsilon/2\) and then \(K\) with \(|a_k^{(n)}| < \varepsilon/2\) for \(k > K\), so that

\begin{equation*} |a_k| \le |a_k - a^{(n)}_k| + |a^{(n)}_k| < \varepsilon \qquad \text{for } k > K , \end{equation*}

giving \(a \in c_0\). A Cauchy sequence in \(c_0\) is Cauchy in \(\ell^\infty\), hence converges there, and the limit lies in the closed set \(c_0\); thus \(c_0\) is a Banach space.

(b) The identification is \(b \mapsto \varphi_b\) from \(\ell^1\) onto \((c_0)^{\prime}\), where

\begin{equation*} \varphi_b(a) = \sum_{k=1}^\infty a_k b_k \qquad (a \in c_0) . \end{equation*}

This map is linear, and \(\sum_{k=1}^\infty |a_k b_k| \le \|a\|_\infty \|b\|_1\) shows the series converges absolutely and \(\|\varphi_b\| \le \|b\|_1\). For the reverse, fix \(n\) and let \(a^{(n)}\) have \(k^{\text{th}}\) term \(\overline{b_k}/|b_k|\) when \(k \le n\) and \(b_k \ne 0\), and \(0\) otherwise; then \(a^{(n)} \in c_0\) (finitely many nonzero terms) with \(\|a^{(n)}\|_\infty \le 1\), and

\begin{equation*} \sum_{k=1}^n |b_k| = \varphi_b\bigl(a^{(n)}\bigr) \le \|\varphi_b\| \, \|a^{(n)}\|_\infty \le \|\varphi_b\| . \end{equation*}

Let \(n \to \infty\): \(\|\varphi_b\| = \|b\|_1\), so \(b \mapsto \varphi_b\) is a linear isometry, in particular one-to-one.

For surjectivity, let \(\varphi \in (c_0)^{\prime}\) and set \(b_k = \varphi(e_k)\), where \(e_k\) is the \(k^{\text{th}}\) standard basis sequence. Since \(a^{(n)}\) above is a linear combination of \(e_1, \ldots, e_n\), the same display with \(\varphi\) in place of \(\varphi_b\) gives \(\|b\|_1 \le \|\varphi\| < \infty\), so \(b \in \ell^1\). Now \(\varphi\) and \(\varphi_b\) agree at each \(e_k\), hence on the finitely supported sequences, which are dense in \(c_0\) because \(\|a - (a_1, \ldots, a_n, 0, 0, \ldots)\|_\infty = \sup_{k > n}|a_k| \to 0\). Both are continuous by 6.48, so \(\varphi = \varphi_b\).

(a) Put \(r = p/2 \in (0,1]\), \(\alpha = |w+z|^2\), \(\beta = |w-z|^2\); the parallelogram equality gives \(\alpha + \beta = 2(|w|^2 + |z|^2)\). Concavity of \(t \mapsto t^r\) on \([0,\infty)\) for \(r \le 1\) (second derivative \(r(r-1)t^{r-2} \le 0\)), applied at \(\alpha, \beta\) with equal weights, together with the subadditivity \((\sigma+\tau)^r \le \sigma^r + \tau^r\) for \(\sigma, \tau \ge 0\), yields

\begin{equation*} \frac{|w+z|^p + |w-z|^p}{2} = \frac{\alpha^r + \beta^r}{2} \le \Bigl(\frac{\alpha+\beta}{2}\Bigr)^{r} = \bigl(|w|^2 + |z|^2\bigr)^{r} \le |w|^p + |z|^p , \end{equation*}

which is the left-hand inequality. (Subadditivity: with \(t = \sigma/(\sigma+\tau) \in [0,1]\) one has \(t^r \ge t\) and \((1-t)^r \ge 1-t\), so \(t^r + (1-t)^r \ge 1\); multiply by \((\sigma+\tau)^r\).) Applying this with \(w+z, w-z\) in place of \(w, z\), and using \((w+z)+(w-z) = 2w\), \((w+z)-(w-z) = 2z\), gives

\begin{equation*} 2^{p-1}\bigl(|w|^p + |z|^p\bigr) = \frac{|2w|^p + |2z|^p}{2} \le |w+z|^p + |w-z|^p ; \end{equation*}

divide by \(2^{p-1}\) for the right-hand inequality.

(b) Apply (a) at each \(x\) with \(w = f(x)\) and \(z = g(x)\):

\begin{equation*} |f(x) + g(x)|^p + |f(x) - g(x)|^p \le 2\bigl(|f(x)|^p + |g(x)|^p\bigr) \end{equation*}

and

\begin{equation*} 2^{p-1}\bigl(|f(x)|^p + |g(x)|^p\bigr) \le |f(x) + g(x)|^p + |f(x) - g(x)|^p . \end{equation*}

Each of \(|f+g|^p, |f-g|^p, |f|^p, |g|^p\) is nonnegative, measurable, and has finite integral, since \(L^p(\mu)\) is a vector space by 7.18 and \(p < \infty\). Integrating the two displays, using that integration is order preserving (3.8) and additive on nonnegative measurable functions (3.16), gives

\begin{equation*} \|f+g\|_p^p + \|f-g\|_p^p \le 2\bigl(\|f\|_p^p + \|g\|_p^p\bigr) \end{equation*}

and \(2^{p-1}(\|f\|_p^p + \|g\|_p^p) \le \|f+g\|_p^p + \|f-g\|_p^p\). Divide the first by \(2\) and the second by \(2^{p-1}\).

(a) Put \(r = p/2 \ge 1\), \(\alpha = |w+z|^2\), \(\beta = |w-z|^2\); the parallelogram equality gives \(\alpha + \beta = 2(|w|^2 + |z|^2)\). Convexity of \(t \mapsto t^r\) on \([0,\infty)\) for \(r \ge 1\) (second derivative \(r(r-1)t^{r-2} \ge 0\)), applied at \(\alpha, \beta\) with equal weights, together with the superadditivity \(\sigma^r + \tau^r \le (\sigma+\tau)^r\) for \(\sigma, \tau \ge 0\), yields

\begin{equation*} \frac{|w+z|^p + |w-z|^p}{2} = \frac{\alpha^r + \beta^r}{2} \ge \Bigl(\frac{\alpha+\beta}{2}\Bigr)^{r} = \bigl(|w|^2 + |z|^2\bigr)^{r} \ge |w|^p + |z|^p , \end{equation*}

which is the right-hand inequality. (Superadditivity: with \(t = \sigma/(\sigma+\tau) \in [0,1]\) one has \(t^r \le t\) and \((1-t)^r \le 1-t\), so \(t^r + (1-t)^r \le 1\); multiply by \((\sigma+\tau)^r\).) Applying this with \(w+z, w-z\) in place of \(w, z\), and using \((w+z)+(w-z) = 2w\), \((w+z)-(w-z) = 2z\), gives

\begin{equation*} |w+z|^p + |w-z|^p \le \frac{|2w|^p + |2z|^p}{2} = 2^{p-1}\bigl(|w|^p + |z|^p\bigr) ; \end{equation*}

divide by \(2^{p-1}\) for the left-hand inequality.

(b) Apply (a) at each \(x\) with \(w = f(x)\) and \(z = g(x)\):

\begin{equation*} |f(x)+g(x)|^p + |f(x)-g(x)|^p \le 2^{p-1}\bigl(|f(x)|^p + |g(x)|^p\bigr) \end{equation*}

and

\begin{equation*} 2\bigl(|f(x)|^p + |g(x)|^p\bigr) \le |f(x)+g(x)|^p + |f(x)-g(x)|^p . \end{equation*}

Each of \(|f+g|^p, |f-g|^p, |f|^p, |g|^p\) is nonnegative, measurable, and has finite integral, since \(L^p(\mu)\) is a vector space by 7.18 and \(p < \infty\). Integrating the two displays, using that integration is order preserving (3.8) and additive on nonnegative measurable functions (3.16), gives

\begin{equation*} \|f+g\|_p^p + \|f-g\|_p^p \le 2^{p-1}\bigl(\|f\|_p^p + \|g\|_p^p\bigr) \end{equation*}

and \(2(\|f\|_p^p + \|g\|_p^p) \le \|f+g\|_p^p + \|f-g\|_p^p\). Divide the first by \(2^{p-1}\) and the second by \(2\).

Apply the Closed Graph Theorem 6.85 to \(Tf = hf\), which by hypothesis maps \(L^p(\mu)\) into \(L^q(\mu)\) and is linear because \(h(f_1 + \alpha f_2) = hf_1 + \alpha hf_2\) pointwise; both spaces are Banach by 7.24, so it remains only to show \(\operatorname{graph}(T)\), a subspace by linearity, is closed.

Suppose \((f_k, hf_k) \to (f, g)\), that is (by 6.84) \(\|f_k - f\|_p \to 0\) and \(\|hf_k - g\|_q \to 0\). By 7.23 some subsequence has \(f_{k_m} \to f\) pointwise almost everywhere, hence

\begin{equation*} h(x) f_{k_m}(x) \longrightarrow h(x) f(x) \quad \text{for a.e.\ } x . \end{equation*}

Since \(\|hf_{k_m} - g\|_q \to 0\) as well, 7.23 gives a further subsequence with \(hf_{k_{m_j}} \to g\) almost everywhere. Off the union of the two exceptional null sets, uniqueness of limits in \(\mathbf{F}\) forces \(g = hf\); thus \(g = Tf\) in \(L^q(\mu)\) and the graph is closed.

Hence \(T\) is bounded by 6.85, so there is \(c < \infty\) with \(\|hf\|_q \le c\|f\|_p\) for all \(f \in L^p(\mu)\), and \(T\) is continuous by 6.48.

Let \(p^{\prime} = p/(p-1)\), so \(1 < p^{\prime} < \infty\) and \((p^{\prime})^{\prime} = p\). By 7.26 the map \(\Lambda_p b = \varphi_b\), where \(\varphi_b(a) = \sum_{k=1}^\infty a_k b_k\), is a norm-preserving linear bijection from \(\ell^{p^{\prime}}\) onto \((\ell^p)^{\prime}\); since \(1 \le p^{\prime} < \infty\), 7.26 applies equally with \(p^{\prime}\) in place of \(p\), giving a norm-preserving linear bijection

\begin{equation*} \Lambda_{p^{\prime}} : \ell^{p} \to (\ell^{p^{\prime}})^{\prime}, \qquad (\Lambda_{p^{\prime}} a)(b) = \sum_{k=1}^\infty b_k a_k . \end{equation*}

The canonical isometry \(\Phi : \ell^p \to (\ell^p)^{\prime\prime}\), \((\Phi a)(\varphi) = \varphi(a)\), is norm preserving by Exercise 20 in Section 6D, so only surjectivity is at issue.

Given \(\psi \in (\ell^p)^{\prime\prime}\), the composition \(\psi \circ \Lambda_p\) is linear and bounded, since \(|\psi(\Lambda_p b)| \le \|\psi\| \|\Lambda_p b\| = \|\psi\| \|b\|_{p^{\prime}}\); hence \(\psi \circ \Lambda_p \in (\ell^{p^{\prime}})^{\prime}\). Surjectivity of \(\Lambda_{p^{\prime}}\) gives \(a \in \ell^p\) with \(\Lambda_{p^{\prime}} a = \psi \circ \Lambda_p\), and evaluating at \(b \in \ell^{p^{\prime}}\),

\begin{equation*} (\Phi a)(\varphi_b) = \varphi_b(a) = \sum_{k=1}^\infty b_k a_k = (\Lambda_{p^{\prime}} a)(b) = \psi(\varphi_b) . \end{equation*}

Every \(\varphi \in (\ell^p)^{\prime}\) is some \(\varphi_b\) because \(\Lambda_p\) is onto, so \(\Phi a = \psi\).

The element \(\psi = \tau \circ \Lambda^{-1}\) of \((\ell^1)^{\prime\prime}\) constructed below is not in the range of the canonical isometry. Here \(\Lambda b = \varphi_b\), with \(\varphi_b(a) = \sum_{k=1}^\infty a_k b_k\), is the norm-preserving linear bijection of \(\ell^\infty\) onto \((\ell^1)^{\prime}\) supplied by 7.26 with \(p = 1\), so \(\Lambda^{-1}\) is linear with \(\|\Lambda^{-1}\varphi\|_\infty = \|\varphi\|\).

Let \(c_0\) be as in Exercise 15 and \(\mathbf{1} = (1,1,1,\ldots)\). Then \(c_0\) is closed in \(\ell^\infty\) by Exercise 15(a) and \(\mathbf{1} \notin c_0\), so 6.73 (with \(V = \ell^\infty\), \(U = c_0\)) gives \(\tau \in (\ell^\infty)^{\prime}\) with \(\tau|_{c_0} = 0\) and \(\tau(\mathbf{1}) \ne 0\). Then \(\psi = \tau \circ \Lambda^{-1}\) is linear with \(|\psi(\varphi)| \le \|\tau\| \|\varphi\|\), so \(\psi \in (\ell^1)^{\prime\prime}\) and \(\psi(\varphi_b) = \tau(b)\) for all \(b \in \ell^\infty\).

If \(\Phi a = \psi\) for some \(a \in \ell^1\), where \((\Phi a)(\varphi) = \varphi(a)\), then for every \(b \in \ell^\infty\)

\begin{equation*} \sum_{k=1}^\infty a_k b_k = \varphi_b(a) = (\Phi a)(\varphi_b) = \psi(\varphi_b) = \tau(b) . \end{equation*}

Taking \(b = e_j\), the \(j^{\text{th}}\) standard basis sequence, which lies in \(c_0\), gives \(a_j = \tau(e_j) = 0\) for every \(j\); so \(a = 0\) and hence \(\tau \equiv 0\), contradicting \(\tau(\mathbf{1}) \ne 0\). Thus \(\Phi\) is not surjective.

The claim is that \(\bigl(\Lambda_\infty^{-1} \circ \Gamma\bigr)(\Phi a) = a\) for every \(a \in c_0\), where the three identifications are as follows. By Exercise 15(b), \(\Lambda_1 b = \psi_b\) with \(\psi_b(a) = \sum_{k=1}^\infty a_k b_k\) is a norm-preserving linear bijection of \(\ell^1\) onto \((c_0)^{\prime}\). By 7.26 with \(p = 1\), \(\Lambda_\infty d = \varphi_d\) with \(\varphi_d(b) = \sum_{k=1}^\infty b_k d_k\) is a norm-preserving linear bijection of \(\ell^\infty\) onto \((\ell^1)^{\prime}\). Finally

\begin{equation*} \Gamma : (c_0)^{\prime\prime} \to (\ell^1)^{\prime}, \qquad \Gamma\theta = \theta \circ \Lambda_1 \end{equation*}

is linear, is surjective (\(\Gamma(\sigma \circ \Lambda_1^{-1}) = \sigma\)), and is norm preserving because \(\Lambda_1\) carries the closed unit ball of \(\ell^1\) onto that of \((c_0)^{\prime}\):

\begin{equation*} \|\Gamma\theta\| = \sup_{\|b\|_1 \le 1} |\theta(\Lambda_1 b)| = \sup_{\|\psi\| \le 1} |\theta(\psi)| = \|\theta\| . \end{equation*}

Now fix \(a \in c_0 \subseteq \ell^\infty\) and let \(\Phi\) be the canonical isometry, \((\Phi a)(\psi) = \psi(a)\). For every \(b \in \ell^1\),

\begin{equation*} \bigl(\Gamma(\Phi a)\bigr)(b) = (\Phi a)(\psi_b) = \psi_b(a) = \sum_{k=1}^\infty a_k b_k = \varphi_a(b) = (\Lambda_\infty a)(b) . \end{equation*}

Hence \(\Gamma(\Phi a) = \Lambda_\infty a\), so \(\Phi\) becomes \(a \mapsto a\) from \(c_0\) into \(\ell^\infty\): the inclusion map.

The norm is squeezed between two constant multiples of the product norm \(|(f,g)|_{\infty} = \max\{\|f\|, \|g\|\}\) of 6.84:

\begin{equation*} |(f,g)|_{\infty} \le \|(f,g)\| \le 2^{1/p} \, |(f,g)|_{\infty} , \end{equation*}

call this \((\ast)\); it follows from \(\max\{a,b\}^p \le a^p + b^p \le 2\max\{a,b\}^p\) and the monotonicity of \(t \mapsto t^{1/p}\). Here \(V \times W\) carries the componentwise operations of Exercise 10 in Section 6B.

Positivity holds because \(\|f\|^p + \|g\|^p = 0\) forces \(\|f\| = \|g\| = 0\), and homogeneity because

\begin{equation*} \|\alpha(f,g)\| = \bigl(|\alpha|^{p}\|f\|^{p} + |\alpha|^{p}\|g\|^{p}\bigr)^{1/p} = |\alpha| \, \|(f,g)\| , \end{equation*}

by homogeneity of the norms on \(V\) and \(W\).

For the triangle inequality, let \(\nu\) be counting measure on \(X = \{1,2\}\) with \(\sigma\)-algebra \(2^{X}\); integration against \(\nu\) is summation (Example 3.6), so every \(h : X \to \mathbb{R}\) lies in \(L^p(\nu)\) with \(\|h\|_p = (|h(1)|^p + |h(2)|^p)^{1/p}\). Given \((f_1,g_1), (f_2,g_2)\), define \(u, v : X \to \mathbb{R}\) by

\begin{equation*} u(1) = \|f_1\|, \quad u(2) = \|g_1\|, \qquad v(1) = \|f_2\|, \quad v(2) = \|g_2\| . \end{equation*}

Then \(\|f_1+f_2\| \le (u+v)(1)\) and \(\|g_1+g_2\| \le (u+v)(2)\) by the triangle inequalities in \(V\) and \(W\), and \(t \mapsto t^p\), \(t \mapsto t^{1/p}\) are increasing, so

\begin{equation*} \begin{aligned} \|(f_1,g_1) + (f_2,g_2)\| &= \bigl(\|f_1+f_2\|^{p} + \|g_1+g_2\|^{p}\bigr)^{1/p} \\ &\le \Bigl(\bigl((u+v)(1)\bigr)^{p} + \bigl((u+v)(2)\bigr)^{p}\Bigr)^{1/p} \\ &= \|u+v\|_{p} \\ &\le \|u\|_{p} + \|v\|_{p} \\ &= \|(f_1,g_1)\| + \|(f_2,g_2)\|, \end{aligned} \end{equation*}

the second inequality being Minkowski’s inequality 7.14 for \(\nu\).

For completeness, let \((f_k, g_k)\) be Cauchy in \(V \times W\). The left half of \((\ast)\) applied to the differences makes \((f_k)\) Cauchy in \(V\) and \((g_k)\) Cauchy in \(W\), so \(f_k \to f \in V\) and \(g_k \to g \in W\); the right half of \((\ast)\) then gives

\begin{equation*} \|(f_k, g_k) - (f,g)\| \le 2^{1/p} \max\{\|f_k - f\|, \|g_k - g\|\} \longrightarrow 0 . \end{equation*}

All four conditions of 8.1 hold. Writing \(\|h\|_\infty = \sup_{\mathbb{R}} |h|\), finite since the elements of \(V\) are bounded, the estimate

\begin{equation*} \left| \frac{f(r_k)\,\overline{g(r_k)}}{2^k} \right| \le \frac{\|f\|_\infty \|g\|_\infty}{2^k} \end{equation*}

together with \(\sum_k 2^{-k} = 1\) makes every series below absolutely convergent, which legitimizes splitting sums and pulling out scalars.

Positivity. \(\langle f, f \rangle = \sum_{k=1}^\infty |f(r_k)|^2 / 2^k \in [0,\infty)\).

Definiteness. If \(\langle f, f \rangle = 0\), every term of that nonnegative series vanishes, so \(f = 0\) on \(\mathbb{Q}\); since \(\mathbb{Q}\) is dense and \(f\) is continuous, \(f = 0\). (Continuity is used only here.)

Linearity in the first slot. For \(f, g, h \in V\) and \(\alpha \in \mathbf{F}\),

\begin{equation*} \begin{aligned} \langle f + g, h \rangle &= \sum_{k=1}^{\infty} \frac{f(r_k)\overline{h(r_k)}}{2^k} + \sum_{k=1}^{\infty} \frac{g(r_k)\overline{h(r_k)}}{2^k} = \langle f, h \rangle + \langle g, h \rangle , \\ \langle \alpha f, h \rangle &= \alpha \sum_{k=1}^{\infty} \frac{f(r_k)\overline{h(r_k)}}{2^k} = \alpha \langle f, h \rangle . \end{aligned} \end{equation*}

Conjugate symmetry. Conjugation is continuous and additive, so it passes through the partial sums, and \(2^k\) is real:

\begin{equation*} \overline{\langle g, f \rangle} = \sum_{k=1}^{\infty} \overline{\left(\frac{g(r_k)\overline{f(r_k)}}{2^k}\right)} = \sum_{k=1}^{\infty} \frac{f(r_k)\overline{g(r_k)}}{2^k} = \langle f, g \rangle . \end{equation*}

Write \(A = \|f\|^2 = \int |f|^2 d\mu\), \(B = \|g\|^2 = \int |g|^2 d\mu\), and \(C = \langle f, g \rangle = \int f\overline{g} \, d\mu\) (8.2, 8.5); then \(A, B < \infty\), and \(f\overline{g}, g\overline{f} \in L^1(\mu)\) with \(\int g \overline{f} \, d\mu = \overline{C}\) by Cauchy–Schwarz 8.14 and conjugate symmetry. Both integrations below use only linearity of the integral, so Tonelli is not needed.

Fix \(x\); expanding \(|w|^2 = w\overline{w}\) with \(w = f(x)g(y) - g(x)f(y)\) gives, for each \(y\),

\begin{equation*} \begin{aligned} |f(x)g(y) - g(x)f(y)|^2 &= |f(x)|^2 |g(y)|^2 + |g(x)|^2 |f(y)|^2 \\ &\qquad - f(x)\overline{g(x)}\, g(y)\overline{f(y)} \quad - \overline{f(x)}\,g(x)\, f(y)\overline{g(y)} , \end{aligned} \end{equation*}

a linear combination in \(y\), with finite scalar coefficients, of \(|g|^2, |f|^2, g\overline{f}, f\overline{g} \in L^1(\mu)\). Hence by linearity

\begin{equation*} \int |f(x)g(y) - g(x)f(y)|^2 \, d\mu(y) = |f(x)|^2 B + |g(x)|^2 A \quad - f(x)\overline{g(x)}\,\overline{C} - \overline{f(x)}g(x)\, C . \end{equation*}

This is in turn a linear combination in \(x\) of the same four \(L^1(\mu)\) functions, so integrating again by linearity,

\begin{equation*} \begin{aligned} \int \int |f(x)g(y) - g(x)f(y)|^2 \, d\mu(y) \, d\mu(x) &= AB + BA - \overline{C}\, C - C\, \overline{C} \\ &= 2AB - 2|C|^2 . \end{aligned} \end{equation*}

Halving gives \(AB - |C|^2 = \|f\|^2\|g\|^2 - |\langle f, g \rangle|^2\).

The hypothesis says exactly that \(\operatorname{Re}\langle f, g \rangle = 0\), since by 8.4, linearity in the first slot, 8.3(b), and conjugate symmetry,

\begin{equation*} \begin{aligned} \|f+g\|^2 &= \langle f + g, f + g \rangle \\ &= \langle f, f \rangle + \langle g, f \rangle + \langle f, g \rangle + \langle g, g \rangle \\ &= \|f\|^2 + \|g\|^2 + \langle f, g \rangle + \overline{\langle f, g \rangle} \\ &= \|f\|^2 + \|g\|^2 + 2 \operatorname{Re} \langle f, g \rangle, \end{aligned} \end{equation*}

(a) If \(\mathbf{F} = \mathbb{R}\) then \(\langle f, g \rangle\) is real, so \(\langle f, g \rangle = \operatorname{Re}\langle f, g \rangle = 0\): orthogonal by 8.7.

(b) Take \(V = \mathbb{C}\) with \(\langle w, z\rangle = w\overline{z}\) (8.2 with \(n = 1\)), \(f = 1\), \(g = i\). Then \(\|f\|^2 = \|g\|^2 = 1\) and \(\|f+g\|^2 = |1+i|^2 = 2\), so the displayed equation holds, while \(\langle f, g \rangle = \overline{i} = -i \ne 0\).

Take

\begin{equation*} a = \left(\tfrac{1}{2},\, 3,\, \tfrac{3}{2}\right), \qquad b = \left(\tfrac{9}{2},\, 1,\, -\tfrac{7}{2}\right) . \end{equation*}

These come from the orthogonal decomposition 8.10 applied to \(u = (5,4,-2)\) and \(v = (1,6,3) \ne 0\) in \(\mathbb{R}^3\) with its standard inner product (8.2): \(a = \frac{\langle u, v\rangle}{\|v\|^2} v\) and \(b = u - a\), where \(\langle u, v \rangle = 5 + 24 - 6 = 23\) and \(\|v\|^2 = 1 + 36 + 9 = 46\), so the scalar is \(\tfrac{23}{46} = \tfrac12\). Then \(\langle b, v \rangle = \tfrac92 + 6 - \tfrac{21}{2} = 0\) and \(a + b = (5,4,-2)\). (Check!)

Apply Cauchy–Schwarz 8.11 in \(\mathbb{R}^4\) with its standard inner product (8.2) to the nonzero vectors

\begin{equation*} u = \bigl(\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}\bigr), \qquad v = \Bigl(\tfrac{1}{\sqrt{a}}, \tfrac{1}{\sqrt{b}}, \tfrac{1}{\sqrt{c}}, \tfrac{1}{\sqrt{d}}\Bigr) , \end{equation*}

which are defined because \(a, b, c, d > 0\). Since \(\langle u, v \rangle = 4\), \(\|u\|^2 = a+b+c+d\), and \(\|v\|^2 = \frac1a + \frac1b + \frac1c + \frac1d\), squaring \(|\langle u,v\rangle| \le \|u\| \|v\|\) gives

\begin{equation*} 16 \le (a + b + c + d)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right) . \end{equation*}

By 8.11 equality holds if and only if one of \(u, v\) is a scalar multiple of the other; as both are nonzero this means \(v = \lambda u\) with \(\lambda \ne 0\), and comparing coordinates gives \(\lambda = \frac1a = \frac1b = \frac1c = \frac1d\), that is, \(a = b = c = d\). Conversely \(a = b = c = d\) gives \((4a)(4/a) = 16\).

Apply Cauchy–Schwarz 8.11 in \(\mathbb{R}^n\) (8.2) to \(u = (x_1, \ldots, x_n)\) and \(v = (1, \ldots, 1)\), where \(x_1, \ldots, x_n\) is the list (so \(n \ge 2\)). Since \(\langle u, v\rangle = x_1 + \cdots + x_n\), \(\|u\|^2 = x_1^2 + \cdots + x_n^2\), and \(\|v\|^2 = n\), squaring \(|\langle u, v\rangle| \le \|u\| \|v\|\) and dividing by \(n^2\) gives

\begin{equation*} \left(\frac{x_1 + \cdots + x_n}{n}\right)^{2} \le \frac{x_1^2 + \cdots + x_n^2}{n} . \end{equation*}

Equality would force, by 8.11 and \(v \ne 0\), that \(u = \lambda v\), that is \(x_1 = \cdots = x_n\), contradicting the presence of two distinct entries. Hence the inequality is strict.

Write \(s = \|f\|\) and \(t = \|g\|\), both in \([0,1]\) by hypothesis, so the square roots are defined. Cauchy–Schwarz 8.11 gives \(|\langle f, g \rangle| \le st\), so it suffices to prove \(\sqrt{1-s^2}\sqrt{1-t^2} \le 1 - st\). Both sides are nonnegative (\(1 - st \ge 0\) since \(s, t \le 1\)), so this is equivalent to the squared form, and

\begin{equation*} (1 - s^2)(1 - t^2) \le (1 - st)^2 \iff 2st \le s^2 + t^2 \iff 0 \le (s-t)^2 . \end{equation*}

Hence

\begin{equation*} \sqrt{1 - \|f\|^2}\,\sqrt{1 - \|g\|^2} \le 1 - st \le 1 - |\langle f, g \rangle| . \end{equation*}

Compare two evaluations of \(\|a-b\|^2\). Algebraically, by additivity in each slot and symmetry of the standard (real) inner product on \(\mathbb{R}^2\),

\begin{equation*} \begin{aligned} \|a-b\|^2 &= \langle a-b,\, a-b\rangle\\ &= \langle a,a\rangle - \langle a,b\rangle - \langle b,a\rangle + \langle b,b\rangle\\ &= \|a\|^2 + \|b\|^2 - 2\langle a,b\rangle. \end{aligned} \end{equation*}

Geometrically, when \(O\), \(A = a\), \(B = b\) are not collinear the triangle \(OAB\) has sides \(\|a\|, \|b\|, \|a-b\|\) with angle \(\theta\) at \(O\), so the law of cosines gives

\begin{equation*} \|a-b\|^2 = \|a\|^2 + \|b\|^2 - 2\|a\|\,\|b\|\cos\theta . \end{equation*}

Cancelling \(\|a\|^2 + \|b\|^2\) and dividing by \(-2\) gives \(\langle a,b\rangle = \|a\|\|b\|\cos\theta\).

In the collinear case write \(b = ta\) with \(t \ne 0\), and use \(\|b\| = |t| \|a\|\) (8.6).

(i) \(t > 0\): \(\theta = 0\) and \(\langle a,b\rangle = t\|a\|^2 = \|a\|\,\|b\| = \|a\|\|b\|\cos\theta\).

(ii) \(t < 0\): \(\theta = \pi\) and \(\langle a,b\rangle = t\|a\|^2 = -\|a\|\,\|b\| = \|a\|\|b\|\cos\theta\).

Because \(\arccos\), the inverse of the bijection \(\cos : [0,\pi] \to [-1,1]\), has domain \([-1,1]\). So the definition makes sense only if

\begin{equation*} \frac{\langle a,b\rangle}{\|a\|\,\|b\|} \in [-1,1] , \end{equation*}

which, since \(\|a\| \|b\| > 0\) for nonzero \(a, b\), says exactly that \(|\langle a,b\rangle| \le \|a\| \|b\|\) — the Cauchy–Schwarz inequality 8.11 in \(\mathbb{R}^n\). Granted it, each pair of nonzero vectors gets a unique angle in \([0,\pi]\).

(a) Both conditions say \(\|f\|^2 = \|g\|^2\). Indeed, symmetry of the real inner product (8.1), additivity, and 8.3(b),(c) give

\begin{equation*} \begin{aligned} \langle f+g,\, f-g\rangle &= \langle f,f\rangle - \langle f,g\rangle \quad + \langle g,f\rangle - \langle g,g\rangle\\ &= \|f\|^2 - \langle f,g\rangle + \langle f,g\rangle - \|g\|^2\\ &= \|f\|^2 - \|g\|^2 . \end{aligned} \end{equation*}

By 8.7, orthogonality of \(f+g\) and \(f-g\) means this vanishes, that is \(\|f\|^2 = \|g\|^2\), equivalently \(\|f\| = \|g\|\) since norms are nonnegative.

(b) Translate the parallelogram so its vertices are \(0, f, f+g, g\), where \(f, g \in \mathbb{R}^2\) are the nonzero non-parallel spanning vectors (orthogonality in \(\mathbb{R}^2\) is perpendicularity, by the discussion after 8.8). Its diagonals are the arrows \(f+g\) and \(g-f\), and by 8.3(c) with \(\alpha = -1\),

\begin{equation*} \langle f+g,\, g-f\rangle = -\langle f+g,\, f-g\rangle , \end{equation*}

so the diagonals are perpendicular exactly when \(\langle f+g, f-g\rangle = 0\), which by (a) means \(\|f\| = \|g\|\). The four side lengths are \(\|f\|, \|g\|, \|-f\| = \|f\|, \|-g\| = \|g\|\) by 8.6, so they are all equal — the parallelogram is a rhombus — under the same condition \(\|f\| = \|g\|\).

Everything follows from the identity

\begin{equation*} \|sf+tg\|^2 - \|tf+sg\|^2 = (s^2 - t^2)\bigl(\|f\|^2 - \|g\|^2\bigr) \qquad (s, t \in \mathbb{R}) . \end{equation*}

To prove it, expand by additivity in each slot, homogeneity in the first, and 8.3(c) in the second (the real scalars \(s, t\) are their own conjugates):

\begin{equation*} \begin{aligned} \|sf+tg\|^2 &= s^2\langle f,f\rangle + st\langle f,g\rangle \quad + ts\langle g,f\rangle + t^2\langle g,g\rangle\\ &= s^2\|f\|^2 + t^2\|g\|^2 + st\big(\langle f,g\rangle + \overline{\langle f,g\rangle}\big)\\ &= s^2\|f\|^2 + t^2\|g\|^2 + 2st\operatorname{Re}\langle f,g\rangle, \end{aligned} \end{equation*}

using \(\langle g,f\rangle = \overline{\langle f,g\rangle}\); interchanging \(s\) and \(t\) leaves the cross term \(2st\operatorname{Re}\langle f,g\rangle\) fixed, so subtracting gives the identity.

If \(\|f\| = \|g\|\), the right side vanishes for all \(s, t\), so \(\|sf+tg\| = \|tf+sg\|\). Conversely, taking \(s = 1\), \(t = 0\) in the hypothesis gives \(\|f\| = \|g\|\) at once.

\(\|f-g\| = 0\), so \(f = g\) by the definiteness requirement in 8.1. Indeed \(\langle g,f\rangle = \overline{\langle f,g\rangle} = 1\) by conjugate symmetry, so expanding by additivity in each slot and 8.3(c) with \(\alpha = -1\),

\begin{equation*} \begin{aligned} \|f-g\|^2 &= \langle f-g,\, f-g\rangle\\ &= \langle f,f\rangle - \langle f,g\rangle - \langle g,f\rangle + \langle g,g\rangle\\ &= \|f\|^2 - \langle f,g\rangle - \langle g,f\rangle + \|g\|^2\\ &= 1 - 1 - 1 + 1\\ &= 0 . \end{aligned} \end{equation*}

Method (2): the hypotheses give \(|\langle f,g\rangle| = 1 = \|f\|\|g\|\), equality in Cauchy–Schwarz 8.11, so one of \(f, g\) is a scalar multiple of the other; both being nonzero, \(f = \alpha g\) for some scalar \(\alpha\), and then \(1 = \langle f,g\rangle = \alpha\|g\|^2 = \alpha\).

Expand both squared norms. Conjugate symmetry over \(\mathbb{R}\) reads \(\langle g,f\rangle = \langle f,g\rangle\), so by additivity in each slot and 8.3(b),

\begin{equation*} \begin{aligned} \|f+g\|^2 &= \langle f+g,\, f+g\rangle\\ &= \langle f,f\rangle + \langle f,g\rangle + \langle g,f\rangle + \langle g,g\rangle\\ &= \|f\|^2 + \|g\|^2 + 2\langle f,g\rangle . \end{aligned} \end{equation*}

Replacing \(g\) by \(-g\) (and using 8.3(c) with \(\alpha = -1\), so that \(\langle f,-g\rangle = -\langle f,g\rangle\), together with \(\|-g\| = \|g\|\) from 8.6) gives

\begin{equation*} \|f-g\|^2 = \|f\|^2 + \|g\|^2 - 2\langle f,g\rangle . \end{equation*}

Subtracting, \(\|f\|^2 + \|g\|^2\) cancels and \(\|f+g\|^2 - \|f-g\|^2 = 4\langle f,g\rangle\); divide by \(4\).

The four terms recover the real and imaginary parts of \(\langle f,g\rangle = x + yi\) separately. Throughout, \(\langle g,f\rangle = \overline{\langle f,g\rangle}\) and \(\langle f,\alpha g\rangle = \overline{\alpha}\langle f,g\rangle\) by 8.3(c). Expanding as in the proof of 8.15,

\begin{equation*} \begin{aligned} \|f+g\|^2 &= \langle f,f\rangle + \langle f,g\rangle + \langle g,f\rangle + \langle g,g\rangle \\ &= \|f\|^2 + \|g\|^2 + 2\operatorname{Re}\langle f,g\rangle,\\ \|f-g\|^2 &= \langle f,f\rangle - \langle f,g\rangle - \langle g,f\rangle + \langle g,g\rangle \\ &= \|f\|^2 + \|g\|^2 - 2\operatorname{Re}\langle f,g\rangle, \end{aligned} \end{equation*}

since \(\langle f,g\rangle + \langle g,f\rangle = 2\operatorname{Re}\langle f,g\rangle\); subtracting gives \(\|f+g\|^2 - \|f-g\|^2 = 4x\).

Applying the first identity with \(ig\) in place of \(g\), where \(\|ig\| = \|g\|\) by 8.6 and \(\langle f, ig\rangle = \overline{i}\langle f,g\rangle = -i(x+yi)\) has real part \(y\), gives \(\|f+ig\|^2 = \|f\|^2 + \|g\|^2 + 2y\); with \(-ig\) in place of \(g\) the real part is \(-y\), so \(\|f-ig\|^2 = \|f\|^2 + \|g\|^2 - 2y\). Subtracting, \(\|f+ig\|^2 - \|f-ig\|^2 = 4y\). Hence

\begin{equation*} \|f+g\|^2 - \|f-g\|^2 + \|f+ig\|^2\,i - \|f-ig\|^2\,i = 4x + 4yi = 4\langle f,g\rangle . \end{equation*}

Apply the parallelogram equality 8.20 to \(u = h-f\) and \(v = h-g\), for which \(u+v = 2\bigl(h - \tfrac12(f+g)\bigr)\) and \(u - v = g-f\):

\begin{equation*} 4\left\|h - \tfrac{1}{2}(f+g)\right\|^2 + \|f-g\|^2 = 2\|h-f\|^2 + 2\|h-g\|^2 , \end{equation*}

using \(\|2w\| = 2\|w\|\) and \(\|g-f\| = \|f-g\|\) (8.6). Solve for the first term and divide by \(4\).

Take the polarization formulas of 8A.13 and 8A.14 — the only candidates — and verify the axioms 8.1; here \(\|\cdot\|\) satisfies \(\|u+v\|^2 + \|u-v\|^2 = 2\|u\|^2 + 2\|v\|^2\) throughout.

(1) Real case. For \(\mathbf{F} = \mathbf{R}\) define

\begin{equation*} \langle f, g\rangle = \frac{\|f+g\|^2 - \|f-g\|^2}{4} \qquad \text{for } f, g \in V. \end{equation*}

(i) Taking \(g = f\) gives \(\langle f, f\rangle = \tfrac14\|2f\|^2 = \|f\|^2\), which is nonnegative and vanishes only for \(f = 0\), and gives \(\|f\| = \langle f,f\rangle^{1/2}\).

(ii) Symmetry: \(\|g-f\| = \|f-g\|\), and all values are real.

(iii) Additivity in the first slot. Applying the parallelogram equality to the pairs \(f_1 + g, f_2\) and \(f_1 - g, f_2\) gives

\begin{equation*} \begin{aligned} \|f_1 + f_2 + g\|^2 + \|f_1 - f_2 + g\|^2 &= 2\|f_1+g\|^2 + 2\|f_2\|^2, \\ \|f_1 + f_2 - g\|^2 + \|f_1 - f_2 - g\|^2 &= 2\|f_1-g\|^2 + 2\|f_2\|^2 . \end{aligned} \end{equation*}

Subtracting the second equation from the first and dividing by \(4\) gives

\begin{equation*} \langle f_1 + f_2, g\rangle + \langle f_1 - f_2, g\rangle = \frac{\|f_1+g\|^2 - \|f_1-g\|^2}{2} = 2\langle f_1, g\rangle. \qquad (*) \end{equation*}

Since \(\langle 0, g\rangle = 0\), taking \(f_2 = f_1\) in \((*)\) gives \(\langle 2f, g\rangle = 2\langle f, g\rangle\). Now \((*)\) with \(f_1 = \frac{u+v}{2}\), \(f_2 = \frac{u-v}{2}\), so that \(f_1 + f_2 = u\) and \(f_1 - f_2 = v\), gives

\begin{equation*} \langle u, g\rangle + \langle v, g\rangle = 2\left\langle \tfrac{u+v}{2}, g\right\rangle = \langle u+v, g\rangle \end{equation*}

by that doubling identity: additivity.

(iv) Homogeneity in the first slot. From the definition,

\begin{equation*} \langle -f, g\rangle = \frac{\|{-f}+g\|^2 - \|{-f}-g\|^2}{4} = \frac{\|f-g\|^2 - \|f+g\|^2}{4} = -\langle f, g\rangle . \end{equation*}

With additivity and induction this gives \(\langle nf, g\rangle = n\langle f,g\rangle\) for \(n \in \mathbf{Z}\), whence (applying it to \(\frac1m f\)) \(\langle tf, g\rangle = t\langle f,g\rangle\) for all rational \(t\). Both \(t \mapsto \langle tf, g\rangle\) and \(t \mapsto t\langle f, g\rangle\) are continuous on \(\mathbf{R}\), the first because \(\bigl|\|tf \pm g\| - \|sf \pm g\|\bigr| \le |t-s| \|f\|\), so agreeing on the dense set \(\mathbf{Q}\) they agree on \(\mathbf{R}\).

By (i)-(iv) the requirements of 8.1 hold, so this is an inner product with associated norm \(\|\cdot\|\).

(2) Complex case. Restricting scalars to \(\mathbf{R}\) leaves the norm and the parallelogram equality intact, so by (1)

\begin{equation*} \langle f, g\rangle_{\mathbf{R}} = \frac{\|f+g\|^2 - \|f-g\|^2}{4} \end{equation*}

is a real inner product on \(V\) as a real vector space: real valued, symmetric, additive and \(\mathbf{R}\)-homogeneous in each slot, with \(\langle f,f\rangle_{\mathbf{R}} = \|f\|^2\). Define

\begin{equation*} \langle f, g\rangle = \langle f, g\rangle_{\mathbf{R}} + i\,\langle f, ig\rangle_{\mathbf{R}} . \end{equation*}

Since \(\|iu\| = \|u\|\), we have \(\langle if, ig\rangle_{\mathbf{R}} = \langle f, g\rangle_{\mathbf{R}}\); call this \((\dagger)\). Replacing \(g\) by \(ig\) there and using \(i(ig) = -g\) gives \(\langle f, ig\rangle_{\mathbf{R}} = -\langle if, g\rangle_{\mathbf{R}}\), call it \((\ddagger)\).

(i) \(\langle f, if\rangle_{\mathbf{R}} = \tfrac14(|1+i|^2 - |1-i|^2)\|f\|^2 = 0\), so \(\langle f,f\rangle = \|f\|^2\): nonnegative, zero only at \(f = 0\), and \(\|f\| = \langle f,f\rangle^{1/2}\).

(ii) \(f \mapsto \langle f, g\rangle\) is additive and \(\mathbf{R}\)-homogeneous, since \(f \mapsto \langle f, g\rangle_{\mathbf{R}}\) and \(f \mapsto \langle f, ig\rangle_{\mathbf{R}}\) are, by (1).

(iii) By the definition, then \((\dagger)\), then \((\ddagger)\),

\begin{equation*} \langle if, g\rangle = \langle if, g\rangle_{\mathbf{R}} + i\langle if, ig\rangle_{\mathbf{R}} = -\langle f, ig\rangle_{\mathbf{R}} + i\langle f, g\rangle_{\mathbf{R}} = i\langle f, g\rangle . \end{equation*}

Hence for \(\alpha = s + it\) with \(s, t \in \mathbf{R}\),

\begin{equation*} \langle \alpha f, g\rangle = \langle sf, g\rangle + \langle t(if), g\rangle = s\langle f, g\rangle + t\,i\langle f, g\rangle = \alpha \langle f, g\rangle , \end{equation*}

so the first slot is \(\mathbf{C}\)-linear.

(iv) Conjugate symmetry, by symmetry of \(\langle \cdot,\cdot\rangle_{\mathbf{R}}\), then \((\ddagger)\), the two real terms being real:

\begin{equation*} \langle g, f\rangle = \langle f, g\rangle_{\mathbf{R}} + i\langle if, g\rangle_{\mathbf{R}} = \langle f, g\rangle_{\mathbf{R}} - i\langle f, ig\rangle_{\mathbf{R}} = \overline{\langle f, g\rangle} . \end{equation*}

(a) This is Cauchy–Schwarz 8.14 applied to \(g(x) = xf(x)\) and \(h(x) = 1/x\), for which \(gh = f\), \(g^2(x) = x^2 f(x)^2\), and

\begin{equation*} \|h\|_2^2 = \int_1^{\infty} \frac{1}{x^2} \, d\lambda(x) = \lim_{n \to \infty}\left(1 - \frac{1}{n}\right) = 1 \end{equation*}

by the Monotone Convergence Theorem 3.11 applied to \(x^{-2}\chi_{[1,n]}\), the integral over \([1,n]\) being the Riemann integral there (3.34). We may assume \(g \in L^2(\lambda)\), as otherwise the right side is \(\infty\). Since \(g, h \ge 0\), \(\langle g, h\rangle = \int_1^\infty f \, d\lambda \ge 0\), so

\begin{equation*} \int_1^{\infty} f \, d\lambda = \bigl|\langle g, h\rangle\bigr| \le \|g\|_2 \|h\|_2 = \left(\int_1^{\infty} x^2 f(x)^2 \, d\lambda(x)\right)^{1/2} ; \end{equation*}

square both sides.

(b) Equality holds exactly when \(\int_1^\infty f \, d\lambda = \infty\), or \(f(x) = c/x^2\) for almost every \(x\) and some \(c \in [0,\infty)\).

(i) If \(\int_1^\infty f \, d\lambda = \infty\), then (a) forces both sides to be \(\infty\). If \(f = c/x^2\) a.e., both sides equal \(c^2\), by the computation of \(\int_1^\infty x^{-2} d\lambda = 1\) above.

(ii) Conversely, suppose equality holds with \(\int_1^\infty f \, d\lambda < \infty\); then \(\|g\|_2^2 = (\int_1^\infty f d\lambda)^2 < \infty\), so \(g \in L^2(\lambda)\) and the equality reads \(|\langle g, h\rangle| = \|g\|_2 \|h\|_2\). By the equality condition in 8.11, and since \(h \ne 0\), we get \(g = \alpha h\) in \(L^2(\lambda)\), where

\begin{equation*} \alpha = \alpha\|h\|_2^2 = \langle \alpha h, h\rangle = \langle g, h\rangle = \int_1^{\infty} f \, d\lambda \ge 0 . \end{equation*}

Thus \(xf(x) = \alpha/x\), that is \(f(x) = \alpha/x^2\), for almost every \(x\).

(a) The integral converges absolutely, because \(0 \le (|f| - |g|)^2\) gives \(|f\overline{g}| = |f|\,|g| \le \frac{1}{2}(|f|^2 + |g|^2)\) pointwise, whence by 3.8

\begin{equation*} \int |f\overline{g}| \, d\mu \le \tfrac{1}{2}\bigl(\|f\|_2^2 + \|g\|_2^2\bigr) < \infty \end{equation*}

for \(f, g \in L^2(\mu)\); thus \(\langle f, g\rangle \in \mathbf{F}\) (6.21). Now check the four requirements of 8.1. Linearity in the first slot is homogeneity and additivity of the integral (3.20, 3.21), all three integrands being absolutely integrable by the display:

\begin{equation*} \begin{aligned} \langle \alpha f_1 + f_2, g\rangle &= \alpha \int f_1 \overline{g}\,d\mu + \int f_2\overline{g}\, d\mu \\ &= \alpha\langle f_1, g\rangle + \langle f_2, g\rangle . \end{aligned} \end{equation*}

Conjugate symmetry follows from \(\overline{f\overline{g}} = g\overline{f}\) and 6.24, and positivity from \(\langle f, f\rangle = \int |f|^2 d\mu = \|f\|_2^2 \ge 0\). For definiteness, suppose \(\int |f|^2 d\mu = 0\) and put \(E_n = \{|f|^2 > 1/n\}\); then \(\frac{1}{n}\chi_{E_n} \le |f|^2\) forces \(\mu(E_n) = 0\) by 3.8, so \(\{f \ne 0\} = \bigcup_n E_n\) is null by 2.58 and \(f\) is the \(0\) element of \(L^2(\mu)\). Hence \(\langle\cdot,\cdot\rangle\) is an inner product, with associated norm (8.4) equal to \(\|\cdot\|_2\); Hölder’s inequality was nowhere used.

(b) Apply Cauchy–Schwarz (8.11) to \(|f|, |g| \in L^2(\mu)\), which are real valued, so \(\bigl\langle |f|, |g|\bigr\rangle = \int |f|\,|g|\,d\mu\):

\begin{equation*} \|fg\|_1 = \int |f|\,|g| \, d\mu \le \bigl\||f|\bigr\|_2 \, \bigl\||g|\bigr\|_2 = \|f\|_2 \, \|g\|_2 , \end{equation*}

the first equality because \(|f||g| = |fg|\).

Each of the four requirements of 8.1 for \(V = V_1 \times \cdots \times V_m\) (a vector space over the common field \(\mathbf{F}\) under coordinatewise operations) reduces coordinatewise to the same requirement in \(V_k\). Write \(f = (f_1,\ldots,f_m)\), \(g = (g_1,\ldots,g_m)\), \(h = (h_1,\ldots,h_m)\).

Positivity and definiteness: \(\langle f, f\rangle = \sum_{k=1}^m \langle f_k, f_k\rangle = \sum_{k=1}^m \|f_k\|^2 \ge 0\), and this sum of nonnegative terms vanishes exactly when every \(\|f_k\| = 0\), which by definiteness in \(V_k\) says \(f_k = 0\) for each \(k\), that is, \(f = 0\).

Linearity in the first slot, from linearity of \(u \mapsto \langle u, g_k\rangle\) on each \(V_k\) and \(\alpha f + h = (\alpha f_1 + h_1, \ldots, \alpha f_m + h_m)\):

\begin{equation*} \langle \alpha f + h, g\rangle = \sum_{k=1}^{m} \bigl(\alpha \langle f_k, g_k\rangle + \langle h_k, g_k\rangle\bigr) = \alpha\langle f, g\rangle + \langle h, g\rangle . \end{equation*}

Conjugate symmetry, from conjugate symmetry in each \(V_k\) and additivity of conjugation (6.24):

\begin{equation*} \langle g, f\rangle = \sum_{k=1}^{m}\overline{\langle f_k, g_k\rangle} = \overline{\langle f, g\rangle} . \end{equation*}

Hence \(\langle\cdot,\cdot\rangle\) is an inner product on \(V\), with associated norm \(\|f\| = \bigl(\|f_1\|^2 + \cdots + \|f_m\|^2\bigr)^{1/2}\).

Fix \((f,g) \in V \times V\) and \(\varepsilon > 0\); the required \(\delta\) is

\begin{equation*} \delta = \min\left\{1, \ \frac{\varepsilon}{\|f\| + \|g\| + 1}\right\} > 0 . \end{equation*}

By Exercise 19 the norm on \(V \times V\) is \(\|(f_1,f_2)\| = (\|f_1\|^2 + \|f_2\|^2)^{1/2}\), so \(\|(u,v) - (f,g)\| < \delta\) forces both \(\|u - f\| < \delta\) and \(\|v - g\| < \delta\). Linearity in the first slot together with 8.3(b) gives \(\langle u,v\rangle - \langle f,g\rangle = \langle u-f, v\rangle + \langle f, v-g\rangle\), so Cauchy–Schwarz (8.11) and \(\|v\| \le \|g\| + \|v-g\| < \|g\| + 1\) (by 8.15 and \(\delta \le 1\)) yield

\begin{equation*} \bigl|\langle u, v\rangle - \langle f, g\rangle\bigr| \le \|u - f\| \, \|v\| + \|f\| \, \|v - g\| < \delta\bigl(\|f\| + \|g\| + 1\bigr) \le \varepsilon . \end{equation*}

Thus \((f,g) \mapsto \langle f,g\rangle\) is continuous at each point of \(V \times V\).

In both parts the answer is \(p = 2\): for \(p = 2\) the norm is the one associated with the standard inner product (8.2 and 8.5), and for \(p \ne 2\) two disjointly supported unit vectors violate the parallelogram equality 8.20,

\begin{equation*} \|u+v\|^2 + \|u-v\|^2 = 2\|u\|^2 + 2\|v\|^2 , \end{equation*}

which every norm coming from an inner product satisfies.

(a) Take \(u = (1,0,0,\ldots)\) and \(v = (0,1,0,\ldots)\) in \(\ell^p\), so \(\|u\|_p = \|v\|_p = 1\).

(i) \(p < \infty\): here \(\|u+v\|_p = \|u-v\|_p = 2^{1/p}\), so 8.20 reads \(2 \cdot 2^{2/p} = 4\), that is \(2^{1+2/p} = 2^2\); injectivity of \(t \mapsto 2^t\) forces \(p = 2\).

(ii) \(p = \infty\): here \(\|u+v\|_\infty = \|u-v\|_\infty = 1\), so the left side of 8.20 is \(2\) and the right side is \(4\).

(b) Take \(u = \chi_{(0,1)}\) and \(v = \chi_{(1,2)}\), which lie in \(L^p(\mathbf{R})\) for every \(p \in [1,\infty]\) and satisfy \(\|u\|_p = \|v\|_p = 1\). Since \(\lambda(\{1\}) = 0\),

\begin{equation*} u + v = \chi_{(0,2)} \quad\text{and}\quad |u - v| = \chi_{(0,2)} \qquad \text{almost everywhere,} \end{equation*}

so \(\|u+v\|_p = \|u-v\|_p = 2^{1/p}\) when \(p < \infty\) and \(= 1\) when \(p = \infty\). Cases (i) and (ii) of part (a) now apply verbatim.

The identity is the parallelogram equality 8.20 applied to the two sides emanating from the vertex opposite the side of length \(c\). Work in \(\mathbf{R}^2\) with the standard inner product, whose associated norm is Euclidean distance (8.5), and label the vertices \(P, Q, R\) so that

\begin{equation*} c = \lVert P - Q \rVert, \qquad b = \lVert P - R \rVert, \qquad a = \lVert Q - R \rVert . \end{equation*}

Setting \(f = P - R\) and \(g = Q - R\) gives \(\lVert f \rVert = b\), \(\lVert g \rVert = a\), and \(\lVert f - g \rVert = \lVert P - Q \rVert = c\); moreover the midpoint \(M = \frac{1}{2}(P+Q)\) satisfies \(M - R = \frac{1}{2}(f+g)\), so homogeneity of the norm (8.6) gives \(\lVert f + g \rVert = 2d\). Hence 8.20 reads

\begin{equation*} (2d)^2 + c^2 = 2b^2 + 2a^2 , \end{equation*}

and dividing by \(2\) gives \(a^2 + b^2 = \frac{1}{2}c^2 + 2d^2\).

Each of the two spaces in 8.23 carries a Cauchy sequence with no limit in it, so by 8.21 neither is a Hilbert space.

(i) \(\ell^1\) with \(\|a\| = \bigl(\sum_k |a_k|^2\bigr)^{1/2}\). Take \(a^{(n)} = (1, \frac12, \ldots, \frac1n, 0, 0, \ldots) \in \ell^1\). For \(m < n\),

\begin{equation*} \|a^{(n)} - a^{(m)}\|^2 = \sum_{k=m+1}^{n} \frac{1}{k^2} \le \sum_{k=m+1}^{\infty} \frac{1}{k^2} \to 0 \end{equation*}

as \(m \to \infty\), so the sequence is Cauchy. If \(b \in \ell^1\) had \(\|a^{(n)} - b\| \to 0\), then \(|a^{(n)}_j - b_j| \le \|a^{(n)} - b\|\) forces \(b_j = 1/j\) for every \(j\), contradicting \(\sum_j 1/j = \infty\).

(ii) \(C([0,1])\) with \(\|f\| = \bigl(\int_0^1 |f|^2\bigr)^{1/2}\). Take the continuous ramps

\begin{equation*} f_n(x) = \begin{cases} 0 & \text{if } 0 \le x \le \frac12, \\ n\left(x - \frac12\right) & \text{if } \frac12 \le x \le \frac12 + \frac1n, \\ 1 & \text{if } \frac12 + \frac1n \le x \le 1, \end{cases} \end{equation*}

for \(n \ge 3\). Since \(f_n = f_m\) off \(\bigl(\frac12, \frac12 + \frac1m\bigr)\) and \(|f_n - f_m| \le 1\), we get \(\|f_n - f_m\|^2 \le \frac1m\) for \(3 \le m \le n\), so the sequence is Cauchy. Suppose \(f \in C([0,1])\) and \(\|f_n - f\| \to 0\). Because \(f_n = 0\) on \([0,\frac12]\) and \(f_n = 1\) on \([a,1]\) once \(\frac12 + \frac1n < a\),

\begin{equation*} \int_0^{1/2} |f|^2 \le \|f_n - f\|^2 \to 0, \qquad \int_a^1 |1-f|^2 \le \|f_n - f\|^2 \to 0 , \end{equation*}

so both integrals vanish; continuity of the nonnegative integrands then gives \(f = 0\) on \([0,\frac12]\) and \(f = 1\) on \([a,1]\) for every \(a \in (\frac12, 1)\). Continuity of \(f\) at \(\frac12\) forces \(f(\frac12) = 1\) as well as \(f(\frac12) = 0\), a contradiction.

Disproof: the ramps

\begin{equation*} \varphi_n(x) = \begin{cases} 0 & \text{if } x \le 0, \\ nx & \text{if } 0 \le x \le \frac1n, \\ 1 & \text{if } x \ge \frac1n \end{cases} \end{equation*}

form a Cauchy sequence with no limit in the space \(V\) of bounded continuous functions on \(\mathbf{R}\) with \(\|f\|^2 = \sum_k 2^{-k}|f(r_k)|^2\), so \(\|\cdot\|\) is not complete and 8.21 rules out a Hilbert space. Each \(\varphi_n\) is continuous with \(0 \le \varphi_n \le 1\), hence lies in \(V\).

Cauchy: for \(m \le n\) the functions \(\varphi_n\) and \(\varphi_m\) agree outside \((0, \frac1m)\) and differ by at most \(1\), so with \(A_m = \{k : 0 < r_k < \frac1m\}\),

\begin{equation*} \|\varphi_n - \varphi_m\|^2 \le \sum_{k \in A_m} 2^{-k} \longrightarrow 0 \quad \text{as } m \to \infty , \end{equation*}

since the \(A_m\) decrease with \(\bigcap_m A_m = \emptyset\) and \(\sum_k 2^{-k} < \infty\).

No limit: if \(f \in V\) had \(\|\varphi_n - f\| \to 0\), then \(2^{-j/2}|\varphi_n(r_j) - f(r_j)| \le \|\varphi_n - f\|\) forces \(f(r_j) = \lim_n \varphi_n(r_j)\), which is \(0\) for \(r_j \le 0\) and \(1\) for \(r_j > 0\). Taking rationals \(-1/n \uparrow 0\) and \(1/n \downarrow 0\), continuity of \(f\) at \(0\) gives both \(f(0) = 0\) and \(f(0) = 1\).

First, \(V\) is a subspace of \(V_1 \times V_2 \times \cdots\): it is closed under scalar multiplication because \(\sum_k \|\alpha f_k\|^2 = |\alpha|^2 \sum_k \|f_k\|^2\), and under addition because 8.15 gives \(\|f_k + g_k\|^2 \le (\|f_k\| + \|g_k\|)^2 \le 2\|f_k\|^2 + 2\|g_k\|^2\).

Second, the defining series converges absolutely for \(f, g \in V\), by Cauchy–Schwarz (8.11) in \(V_k\) and \(st \le \frac12(s^2+t^2)\):

\begin{equation*} \sum_{k=1}^{\infty}\bigl|\langle f_k, g_k\rangle\bigr| \le \sum_{k=1}^{\infty} \|f_k\|\,\|g_k\| \le \tfrac12 \sum_{k=1}^{\infty}\bigl(\|f_k\|^2 + \|g_k\|^2\bigr) < \infty . \end{equation*}

The four requirements of 8.1 now hold termwise exactly as in Exercise 19 of Section 8A, absolutely convergent series being added and conjugated termwise; in particular \(\langle f, f\rangle = \sum_k \|f_k\|^2\), which vanishes only when every \(f_k = 0\), and the associated norm is \(\|f\| = \bigl(\sum_k \|f_k\|^2\bigr)^{1/2}\).

Completeness. Let \(f^{(1)}, f^{(2)}, \ldots\) be Cauchy in \(V\). Since \(\|f^{(n)}_k - f^{(m)}_k\| \le \|f^{(n)} - f^{(m)}\|\) for each \(k\), completeness of \(V_k\) supplies \(f_k \in V_k\) with \(f^{(n)}_k \to f_k\). Given \(\varepsilon > 0\), choose \(N\) with \(\|f^{(n)} - f^{(m)}\| \le \varepsilon\) for \(m, n \ge N\); for \(n \ge N\) and each \(K\), letting \(m \to \infty\) in the finite sum \(\sum_{k=1}^{K} \|f^{(n)}_k - f^{(m)}_k\|^2 \le \varepsilon^2\) (legitimate since \(\bigl|\|u\| - \|v\|\bigr| \le \|u-v\|\)) and then \(K \to \infty\) gives

\begin{equation*} \sum_{k=1}^{\infty} \|f^{(n)}_k - f_k\|^2 \le \varepsilon^2 \qquad \text{for all } n \ge N . \end{equation*}

Thus \(f^{(N)} - f \in V\), so \(f = f^{(N)} - (f^{(N)} - f) \in V\) and the display reads \(\|f^{(n)} - f\| \le \varepsilon\) for \(n \ge N\). Hence \(V\) is complete, so by 8.21 it is a Hilbert space.

Write \([\cdot,\cdot]\) for the proposed form and \(A = \langle f_1, f_2\rangle + \langle g_1, g_2\rangle\), \(B = \langle g_1, f_2\rangle - \langle f_1, g_2\rangle\), so \([f_1 + ig_1, f_2 + ig_2] = A + iB\) with \(A, B \in \mathbf{R}\); here \(\langle\cdot,\cdot\rangle\) is the inner product of \(V\), symmetric and \(\mathbf{R}\)-bilinear.

Positivity and definiteness: taking \(f_1 = f_2 = f\), \(g_1 = g_2 = g\) makes \(B = 0\) by symmetry, so

\begin{equation*} [f + ig, f + ig] = \|f\|^2 + \|g\|^2 , \end{equation*}

which is nonnegative and vanishes exactly when \(f = g = 0\).

Additivity in the first slot is additivity of each \(\langle\cdot, f_2\rangle\) and \(\langle\cdot, g_2\rangle\) in its first argument, since \((f_1 + ig_1) + (f_1’ + ig_1’) = (f_1 + f_1’) + i(g_1 + g_1’)\). Complex homogeneity: for \(\lambda = \alpha + i\beta\) we have \(\lambda(f_1 + ig_1) = (\alpha f_1 - \beta g_1) + i(\alpha g_1 + \beta f_1)\), so \(\mathbf{R}\)-bilinearity gives

\begin{equation*} \begin{aligned} \bigl[\lambda(f_1 + ig_1),\, f_2 + ig_2\bigr] &= \langle \alpha f_1 - \beta g_1, f_2\rangle + \langle \alpha g_1 + \beta f_1, g_2\rangle \\ &\qquad + i\bigl(\langle \alpha g_1 + \beta f_1, f_2\rangle - \langle \alpha f_1 - \beta g_1, g_2\rangle\bigr) \\ &= (\alpha A - \beta B) + i(\alpha B + \beta A) = \lambda (A + iB) . \end{aligned} \end{equation*}

Conjugate symmetry: swapping the two arguments fixes \(A\) and negates \(B\), by symmetry of \(\langle\cdot,\cdot\rangle\), so \([f_2 + ig_2, f_1 + ig_1] = A - iB = \overline{[f_1 + ig_1, f_2 + ig_2]}\). Hence \([\cdot,\cdot]\) is an inner product on \(V_{\mathbf{C}}\), with associated norm \(\|f + ig\|_{\mathbf{C}} = (\|f\|^2 + \|g\|^2)^{1/2}\).

Completeness: since \((f_1 + ig_1) - (f_2 + ig_2) = (f_1 - f_2) + i(g_1 - g_2)\), a Cauchy sequence \(f_n + ig_n\) has \(\|f_n - f_m\|, \|g_n - g_m\| \le \|(f_n + ig_n) - (f_m + ig_m)\|_{\mathbf{C}}\), so \(f_n \to f\) and \(g_n \to g\) for some \(f, g \in V\) by completeness of \(V\), and then

\begin{equation*} \|(f_n + ig_n) - (f + ig)\|_{\mathbf{C}}^2 = \|f_n - f\|^2 + \|g_n - g\|^2 \to 0 . \end{equation*}

By 8.21, \(V_{\mathbf{C}}\) is a complex Hilbert space.

For \(g, h \in B(f,r)\) and \(t \in [0,1]\), the identity \(\bigl((1-t)g + th\bigr) - f = (1-t)(g-f) + t(h-f)\) together with the triangle inequality and homogeneity of the norm gives

\begin{equation*} \bigl\|\bigl((1-t)g + th\bigr) - f\bigr\| \le (1-t)\|g - f\| + t\,\|h - f\| < (1-t)r + tr = r , \end{equation*}

the last inequality strict because \(\|g-f\| < r\), \(\|h-f\| < r\), and \(t\) cannot be both \(0\) and \(1\), so at least one of the two terms is bounded strictly. Hence \((1-t)g + th \in B(f,r)\), which is convexity (8.25).

(a) Let \(f, g \in U\), let \(t \in (0,1)\) (the values \(t = 0, 1\) being trivial), and set \(h = (1-t)f + tg\). Since \(\overline{B} = \{f : \|f\| \le 1\}\) (the norm is continuous, and \((1-\frac1n)f \to f\) from inside \(B\)), we have \(\|f\|, \|g\| \le 1\), so by 8.15

\begin{equation*} \|h\| \le (1-t)\|f\| + t\|g\| \le (1-t) + t = 1 . \end{equation*}

(i) \(\|h\| < 1\): then \(h \in B \subseteq U\).

(ii) \(\|h\| = 1\): then equality holds throughout, which forces \(\|f\| = \|g\| = 1\) (since \(1-t, t > 0\)). Put \(u = (1-t)f\) and \(v = tg\), so \(u, v \ne 0\) and \(\|u+v\| = 1 = \|u\| + \|v\|\). Equality in the triangle inequality (8.18) makes one of \(u, v\) a nonnegative multiple of the other, so \(v = cu\) with \(c > 0\); taking norms gives \(c = t/(1-t)\), whence \(tg = tf\) and \(g = f\), so \(h = f \in U\).

(b) Take \(V = \mathbf{R}^2\) with \(\|(x_1,x_2)\| = \max\{|x_1|, |x_2|\}\), a Banach space because convergence in this norm is coordinatewise convergence. Then \(B\) and \(\overline{B}\) are the open and closed unit squares, and

\begin{equation*} U = B \cup \{(1,1), (1,-1)\} \end{equation*}

satisfies \(B \subseteq U \subseteq \overline{B}\) but is not convex, since \(\frac12(1,1) + \frac12(1,-1) = (1,0) \notin U\).

If \(U\) is convex, take \(t = \frac12\) in 8.25. Conversely, suppose \(\frac{f+g}{2} \in U\) whenever \(f, g \in U\), fix \(f, g \in U\), and set

\begin{equation*} D = \bigl\{ t \in [0,1] : (1-t)f + tg \in U \bigr\} ; \end{equation*}

we show \(D = [0,1]\), which is convexity.

\(D\) is closed under averaging: for \(s, t \in D\) the hypothesis applied to \((1-s)f + sg\) and \((1-t)f + tg\) puts their average, namely \(\bigl(1 - \frac{s+t}{2}\bigr)f + \frac{s+t}{2}g\), in \(U\), so \(\frac{s+t}{2} \in D\). Since \(0, 1 \in D\), induction on \(n\) now gives \(k/2^n \in D\) for all \(0 \le k \le 2^n\): an even \(k\) is covered by the previous stage, and an odd \(k = 2j+1\) satisfies \(\frac{k}{2^{n+1}} = \frac12\bigl(\frac{j}{2^n} + \frac{j+1}{2^n}\bigr)\). (Check!)

\(D\) is closed in \([0,1]\): the map \(\gamma(t) = (1-t)f + tg\) satisfies \(\|\gamma(t) - \gamma(s)\| = |t-s|\,\|g-f\|\), hence is continuous, and \(U\) is closed, so \(D = \gamma^{-1}(U)\) is closed. Being closed and containing the dense set of dyadic rationals, \(D = [0,1]\).

Given \(f, g \in \overline{U}\) and \(t \in [0,1]\), pick sequences \(f_k, g_k \in U\) with \(f_k \to f\) and \(g_k \to g\) (each ball \(B(f, 1/k)\) meets \(U\), else \(f\) would lie in the open set \(V \setminus \overline{U}\)). Convexity of \(U\) puts \(h_k = (1-t)f_k + tg_k\) in \(U\), and the triangle inequality with homogeneity gives

\begin{equation*} \bigl\|\bigl((1-t)f + tg\bigr) - h_k\bigr\| \le (1-t)\|f - f_k\| + t\|g - g_k\| \longrightarrow 0 . \end{equation*}

Hence \((1-t)f + tg \in \overline{U}\), so \(\overline{U}\) is convex (8.25).

Given \(f, g \in \operatorname{int} U\) with \(B(f,r) \subseteq U\) and \(B(g,s) \subseteq U\), and \(t \in [0,1]\), the ball \(B(h, \rho)\) around \(h = (1-t)f + tg\) with \(\rho = \min\{r,s\}\) lies in \(U\), so \(h \in \operatorname{int} U\). Indeed, for \(k \in B(h,\rho)\) put \(u = k - h\); then \(\|u\| < \rho\) places \(f + u\) in \(B(f,r) \subseteq U\) and \(g+u\) in \(B(g,s) \subseteq U\), so convexity of \(U\) gives

\begin{equation*} (1-t)(f+u) + t(g+u) = h + \bigl((1-t) + t\bigr)u = k \in U . \end{equation*}

Hence \(\operatorname{int} U\) is convex (vacuously so if it is empty).

Perturb \(g\) toward \(h\) inside \(U\). For \(h \in U\) and \(t \in (0,1]\), convexity (8.25) puts \((1-t)g + th = g + t(h-g)\) in \(U\), so minimality of \(\|g\|\) and expansion of the norm give

\begin{equation*} \|g\|^2 \le \|g + t(h-g)\|^2 = \|g\|^2 + 2t\operatorname{Re}\langle g, h-g\rangle + t^2\|h-g\|^2 , \end{equation*}

the cross terms combining because \(\langle h-g, g\rangle = \overline{\langle g, h-g\rangle}\) and \(z + \bar z = 2\operatorname{Re} z\). Subtracting \(\|g\|^2\) and dividing by \(2t > 0\) gives \(0 \le \operatorname{Re}\langle g, h-g\rangle + \frac{t}{2}\|h-g\|^2\) for every \(t \in (0,1]\); letting \(t \to 0^+\) yields \(\operatorname{Re}\langle g, h-g\rangle \ge 0\). Since \(\operatorname{Re}\langle g, h-g\rangle = \operatorname{Re}\langle g,h\rangle - \|g\|^2\) by additivity in the second slot, the inequality \(\operatorname{Re}\langle g,h\rangle \ge \|g\|^2\) follows.

Let \(C\) be closed and convex and let \(I\) be the intersection of all closed half-spaces containing \(C\). Then \(C \subseteq I\) trivially, so only \(I \subseteq C\) needs proof: given \(f \notin C\), we exhibit a closed half-space containing \(C\) but not \(f\).

(i) \(C = \varnothing\): taking \(h = 0\) and \(c = 1\) makes \(\{g : \operatorname{Re}\langle g, 0\rangle \ge 1\} = \varnothing\) itself a closed half-space containing \(C\), so \(I = \varnothing = C\).

(ii) \(C \ne \varnothing\): by 8.28 (\(V\) a Hilbert space, \(C\) nonempty closed convex) there is \(g \in C\) with \(\|f - g\| = \operatorname{distance}(f, C)\), and \(u = f - g \ne 0\). For \(k \in C\) and \(t \in (0,1]\), convexity puts \(g + t(k-g)\) in \(C\), so minimality gives

\begin{equation*} \|u\|^2 \le \|u - t(k-g)\|^2 = \|u\|^2 - 2t\operatorname{Re}\langle u, k-g\rangle + t^2\|k-g\|^2 , \end{equation*}

whence \(\operatorname{Re}\langle u, k-g\rangle \le \frac{t}{2}\|k-g\|^2\) for all such \(t\), and letting \(t \to 0^+\) gives the variational inequality \(\operatorname{Re}\langle u, k-g\rangle \le 0\) for every \(k \in C\). Now set

\begin{equation*} H = \bigl\{k \in V : \operatorname{Re}\langle k, -u\rangle \ge -\operatorname{Re}\langle g, u\rangle \bigr\} , \end{equation*}

a closed half-space. It contains \(C\), since for \(k \in C\) the variational inequality and \(\operatorname{Re}\langle k, u\rangle = \operatorname{Re}\langle u, k\rangle\) give \(\operatorname{Re}\langle k, u\rangle \le \operatorname{Re}\langle g, u\rangle\). It omits \(f\), since \(f = g + u\) gives

\begin{equation*} \operatorname{Re}\langle f, -u\rangle = -\operatorname{Re}\langle g, u\rangle - \|u\|^2 < -\operatorname{Re}\langle g, u\rangle . \end{equation*}

Hence \(f \notin I\), so \(I \subseteq C\) and \(C = I\).

Take \(a = 0\) and \(U = \{u_k : k \in \mathbf{Z}^+\}\), where \(u_k = \bigl(1 + \frac1k\bigr)e_k\) and \(e_k \in \ell^2\) is the \(k\)-th standard basis sequence.

\(U\) is closed: distinct \(u_j, u_k\) are supported on disjoint coordinates, so

\begin{equation*} \|u_j - u_k\|^2 = \Bigl(1 + \tfrac1j\Bigr)^2 + \Bigl(1 + \tfrac1k\Bigr)^2 > 2 , \end{equation*}

and a convergent sequence in \(U\), being Cauchy, must therefore be eventually constant, with its limit that constant value in \(U\).

The distance is not attained: \(\operatorname{distance}(0, U) = \inf_k \bigl(1 + \frac1k\bigr) = 1\), while \(\|0 - u_k\| = 1 + \frac1k > 1\) for every \(k\).

Take \(U = \{(x,y) : x \ge 1\}\) and \(z = (0,0)\); every \(w = (1,y)\) with \(|y| \le 1\) is then a closest point.

\(U\) is closed as the inverse image of \([1,\infty)\) under the \(\|\cdot\|_\infty\)-continuous map \((x,y) \mapsto x\), and convex because \((1-t)x + tx’ \ge 1\) whenever \(x, x’ \ge 1\) and \(t \in [0,1]\). For \(w = (x,y) \in U\) we have \(\|z - w\|_\infty = \max\{|x|,|y|\} \ge x \ge 1\), with \(\|z - (1,0)\|_\infty = 1\), so \(\operatorname{distance}(z,U) = 1\). Finally

\begin{equation*} \|z - (1,y)\|_\infty = \max\{1, |y|\} = 1 \qquad \text{for all } |y| \le 1 , \end{equation*}

giving infinitely many minimizers.

Both directions run off the expansion

\begin{equation*} \|f + \alpha g\|^2 = \|f\|^2 + 2\operatorname{Re}\bigl(\overline{\alpha}\langle f, g\rangle\bigr) + |\alpha|^2\|g\|^2 , \end{equation*}

which holds for all \(\alpha \in \mathbf{F}\) by additivity in each slot, \(\langle f, \alpha g\rangle = \overline{\alpha}\langle f,g\rangle\), and \(\langle \alpha g, f\rangle = \overline{\overline{\alpha}\langle f,g\rangle}\).

If \(\langle f, g\rangle = 0\), the middle term vanishes and \(\|f + \alpha g\|^2 = \|f\|^2 + |\alpha|^2\|g\|^2 \ge \|f\|^2\).

Conversely, suppose \(\|f\| \le \|f + \alpha g\|\) for all \(\alpha\). If \(g = 0\) then \(\langle f, g\rangle = 0\); otherwise take \(\alpha = -\langle f,g\rangle / \|g\|^2\), for which \(\overline{\alpha}\langle f,g\rangle = -|\langle f,g\rangle|^2/\|g\|^2\) is real and \(|\alpha|^2\|g\|^2 = |\langle f,g\rangle|^2/\|g\|^2\), so the expansion becomes

\begin{equation*} \|f\|^2 \le \|f + \alpha g\|^2 = \|f\|^2 - \frac{|\langle f, g\rangle|^2}{\|g\|^2} . \end{equation*}

As \(\|g\|^2 > 0\), this forces \(\langle f, g\rangle = 0\).

Since \(f - P_U f\) is orthogonal to every element of \(U\) (8.37(a)) and \(P_U f \in U\), the Pythagorean Theorem (8.9) applied to \(f = (f - P_U f) + P_U f\) gives

\begin{equation*} \|f\|^2 = \|f - P_U f\|^2 + \|P_U f\|^2 \ge \|P_U f\|^2 . \end{equation*}

Equality holds exactly when \(\|f - P_U f\| = 0\), that is, when \(f = P_U f\), which happens if and only if \(f \in U\): one direction because \(P_U f \in U\), the other because \(f \in U\) makes \(f\) itself the unique closest point of \(U\) to \(f\) (8.28).

Take \(U = \operatorname{range} P\).

\(U\) is a closed subspace: \(P^2 = P\) gives \(\operatorname{range} P = \{f : Pf = f\} = \operatorname{null}(I - P)\), and \(\|Pf\| \le \|f\|\) makes \(P\), hence \(I - P\), continuous, so this null space is the inverse image of the closed set \(\{0\}\).

Every \(h \in \operatorname{null} P\) is orthogonal to every \(g \in U\): linearity and \(Pg = g\) give \(P(g + \alpha h) = g\) for all \(\alpha \in \mathbf{F}\), so the hypothesis applied to \(g + \alpha h\) yields

\begin{equation*} \|g\| = \|P(g + \alpha h)\| \le \|g + \alpha h\| \qquad \text{for all } \alpha \in \mathbf{F} , \end{equation*}

whence \(\langle g, h\rangle = 0\) by Exercise 14 in this section.

Now let \(f \in V\). Then \(Pf \in U\), and \(P(f - Pf) = Pf - P^2 f = 0\) puts \(f - Pf\) in \(\operatorname{null} P\), so \(\langle f - Pf, g\rangle = 0\) for every \(g \in U\) (conjugating the previous paragraph). Hence \(Pf = P_U f\) by 8.37(b).

Take

\begin{equation*} T = \widetilde{S} \circ P_{\overline{U}} : V \to W , \end{equation*}

where \(\overline{U}\) is the closure of \(U\), a closed subspace of \(V\) (Exercise 12 in Section 6C), and \(\widetilde{S} : \overline{U} \to W\) is the extension of \(S\) with \(\widetilde{S}|_U = S\) and \(\|\widetilde{S}\| = \|S\|\) supplied by Exercise 14 in Section 6C (\(W\) is a Banach space, so the limits defining it exist).

\(T\) is linear as a composition of linear maps, \(P_{\overline{U}}\) being linear by 8.37(c). For \(f \in U\) we have \(P_{\overline{U}} f = f\), so \(Tf = \widetilde{S} f = Sf\) and \(T|_U = S\). Finally 8.37(d) gives \(\|P_{\overline{U}} f\| \le \|f\|\), so

\begin{equation*} \|Tf\| \le \|\widetilde{S}\| \, \|P_{\overline{U}} f\| \le \|S\| \, \|f\| , \end{equation*}

whence \(\|T\| \le \|S\|\); and \(\|T\| \ge \|S\|\) because \(T\) extends \(S\). Thus \(\|T\| = \|S\|\).

Each direction is one application of a numbered result. If \(\overline{U} = \overline{W}\), then 8.40(d) gives

\begin{equation*} U^{\perp} = \bigl(\overline{U}\bigr)^{\perp} = \bigl(\overline{W}\bigr)^{\perp} = W^{\perp} . \end{equation*}

If \(U^{\perp} = W^{\perp}\), then 8.41 (applicable since \(U\) and \(W\) are subspaces of the Hilbert space \(V\)) gives

\begin{equation*} \overline{U} = \bigl(U^{\perp}\bigr)^{\perp} = \bigl(W^{\perp}\bigr)^{\perp} = \overline{W} . \end{equation*}

Both directions are the statement \(W \subseteq U^{\perp}\) read through 8.45(a), which gives \(\operatorname{null} P_U = U^{\perp}\) and \(\operatorname{range} P_W = W\).

If \(\langle f, g\rangle = 0\) for all \(f \in U\) and \(g \in W\), then \(W \subseteq U^{\perp}\) by 8.38, so for every \(h\) in the Hilbert space, \(P_W h \in W \subseteq \operatorname{null} P_U\) and hence \(P_U P_W h = 0\).

Conversely, if \(P_U P_W = 0\) and \(g \in W\), then \(P_W g = g\) (as \(W\) is a closed subspace containing \(g\)), so

\begin{equation*} P_U g = P_U P_W g = 0 , \end{equation*}

putting \(g\) in \(\operatorname{null} P_U = U^{\perp}\), that is, \(\langle f, g\rangle = 0\) for every \(f \in U\).

Write \(W = \{g \in L^2(\mathbf{R}) : g = 0 \text{ almost everywhere on } [0,\infty)\}\); the whole example follows from the splitting \(h = h\chi_{[0,\infty)} + h\chi_{(-\infty,0)}\) into a member of \(U\) and a member of \(W\).

\(U\) is a closed subspace: it is closed under linear combinations (a union of two null sets is null), and if \(f_k \in U\) with \(f_k \to f\) in \(L^2\), then

\begin{equation*} \int_{(-\infty,0)} |f|^2 \, d\lambda = \int_{(-\infty,0)} |f - f_k|^2 \, d\lambda \le \|f - f_k\|^2 \longrightarrow 0 , \end{equation*}

forcing \(f = 0\) almost everywhere on \((-\infty, 0)\), that is, \(f \in U\).

\(U^{\perp} = W\). If \(g \in W\) and \(f \in U\), then \(f\overline{g} = 0\) almost everywhere (each factor vanishes on one half-line), so \(\langle f, g\rangle = 0\) and \(W \subseteq U^{\perp}\). Conversely, if \(g \in U^{\perp}\), then \(f = g\chi_{[0,\infty)}\) lies in \(U\) (and in \(L^2\), since \(|f| \le |g|\)), so

\begin{equation*} 0 = \langle f, g\rangle = \int_{[0,\infty)} |g|^2 \, d\lambda , \end{equation*}

giving \(g = 0\) almost everywhere on \([0,\infty)\), that is, \(g \in W\).

The projections. For \(h \in L^2(\mathbf{R})\), the element \(h\chi_{[0,\infty)}\) lies in \(U\) and \(h - h\chi_{[0,\infty)} = h\chi_{(-\infty,0)}\) lies in \(W = U^{\perp}\), hence is orthogonal to every element of \(U\); so \(P_U h = h\chi_{[0,\infty)}\) by 8.37(b). Symmetrically \(h\chi_{(-\infty,0)} \in U^{\perp}\) (a closed subspace by 8.40(a)) and \(h - h\chi_{(-\infty,0)} = h\chi_{[0,\infty)} \in U\) is orthogonal to every element of \(U^{\perp}\), so 8.37(b) applied to \(U^{\perp}\) gives \(P_{U^{\perp}} h = h\chi_{(-\infty,0)}\). Consequently

\begin{equation*} P_{U^{\perp}} h = h\bigl(1 - \chi_{[0,\infty)}\bigr) = h - P_U h = (I - P_U)h , \end{equation*}

confirming 8.45(c) here.

Take \(B = \overline{V}\), the closure of \(V\) in a Banach space \(C\) that contains the normed space \(V\) isometrically (Exercise 13 in Section 6C). Then \(B\) is a subspace of \(C\) (Exercise 12 in Section 6C) and is closed, hence complete; so \(B\) is a Banach space in which \(V\) is dense.

Extend the inner product of \(V\) to \(B\) by continuity. Given \(f, g \in B\), choose \(f_k \to f\) and \(g_k \to g\) with \(f_k, g_k \in V\); convergent sequences are bounded, say by \(M\), so the Cauchy-Schwarz inequality (8.11) in \(V\) gives

\begin{equation*} \begin{aligned} \big| \langle f_k, g_k \rangle_V - \langle f_j, g_j \rangle_V \big| &= \big| \langle f_k - f_j, g_k \rangle_V + \langle f_j, g_k - g_j \rangle_V \big| \\ &\leq M \big( \|f_k - f_j\| + \|g_k - g_j\| \big) . \end{aligned} \end{equation*}

The right side tends to \(0\), so \(\langle f_k, g_k \rangle_V\) converges; the same estimate with \(f_k^{\prime}, g_k^{\prime}\) in place of \(f_j, g_j\) shows the limit is independent of the approximating sequences. Hence

\begin{equation*} \langle f, g \rangle = \lim_{k \to \infty} \langle f_k, g_k \rangle_V \qquad (f, g \in B) \end{equation*}

is well defined, and constant sequences give \(\langle f, g \rangle = \langle f, g \rangle_V\) for \(f, g \in V\).

Linearity in the first slot and conjugate symmetry pass to the limit because addition, scalar multiplication, and conjugation are continuous (Check!). For positivity, continuity of the norm gives

\begin{equation*} \langle f, f \rangle = \lim_{k \to \infty} \|f_k\|^2 = \|f\|^2 , \end{equation*}

so \(\langle f, f \rangle \geq 0\) with equality only when \(f = 0\), and the norm induced by \(\langle \cdot, \cdot \rangle\) is the norm of \(B\). Being complete in that norm, \(B\) is a Hilbert space (8.21) containing \(V\) as a subspace with its original inner product.

Take \(h \neq 0\) spanning the one-dimensional subspace \(\operatorname{range} T\), and take \(g\) to be the Riesz vector of the functional \(\varphi\) below.

For each \(f \in V\) the equation \(T f = \varphi(f) h\) determines the scalar \(\varphi(f)\) uniquely, because \(h \neq 0\). That uniqueness converts the linearity of \(T\) into linearity of \(\varphi\): applying it to

\begin{equation*} \varphi(f_1 + \alpha f_2) h = T(f_1 + \alpha f_2) = \bigl( \varphi(f_1) + \alpha \varphi(f_2) \bigr) h \end{equation*}

gives \(\varphi(f_1 + \alpha f_2) = \varphi(f_1) + \alpha \varphi(f_2)\). Furthermore

\begin{equation*} |\varphi(f)| \, \|h\| = \|T f\| \le \|T\| \, \|f\| , \end{equation*}

so \(\varphi\) is bounded, with \(\|\varphi\| \le \|T\| / \|h\|\). Because \(V\) is a Hilbert space, the Riesz Representation Theorem (8.47) supplies \(g \in V\) with \(\varphi(f) = \langle f, g \rangle\) for all \(f \in V\). Hence

\begin{equation*} T f = \varphi(f) h = \langle f , g \rangle h \qquad \text{for all } f \in V . \end{equation*}

(a) Take \(V = \ell^1\), a Banach space by 7.24, and

\begin{equation*} \varphi(a) = \sum_{k=1}^{\infty} \Bigl( 1 - \frac{1}{k} \Bigr) a_k \qquad \text{for } a = (a_1, a_2, \dots) \in \ell^1 , \end{equation*}

the series converging absolutely since \(|(1 - 1/k) a_k| \le |a_k|\).

Here \(\|\varphi\| = 1\): the same bound gives \(|\varphi(a)| \le \|a\|_1\), while \(\varphi(e_k) = 1 - 1/k\) and \(\|e_k\|_1 = 1\) for the \(k\)th standard basis vector \(e_k\), so \(\|\varphi\| \ge 1 - 1/k\) for every \(k\). If \(a \neq 0\), choose \(m\) with \(a_m \neq 0\); then

\begin{equation*} |\varphi(a)| \le \|a\|_1 - \sum_{k=1}^{\infty} \frac{|a_k|}{k} \le \|a\|_1 - \frac{|a_m|}{m} < \|a\|_1 = \|\varphi\| \, \|a\|_1 , \end{equation*}

all quantities being finite because \(a \in \ell^1\).

(b) On a Hilbert space every bounded linear functional attains its norm, so the strict inequality always fails somewhere. Indeed, let \(V \neq \{0\}\) be a Hilbert space and \(\varphi\) a bounded linear functional on \(V\); the Riesz Representation Theorem (8.47) gives \(h \in V\) with \(\varphi(f) = \langle f, h \rangle\) for all \(f\) and \(\|\varphi\| = \|h\|\).

(i) If \(h \neq 0\), then \(f = h\) is nonzero and

\begin{equation*} |\varphi(h)| = \|h\|^2 = \|\varphi\| \, \|h\| . \end{equation*}

(ii) If \(h = 0\), then \(\varphi = 0\) and \(\|\varphi\| = 0\), so \(|\varphi(f)| = 0 = \|\varphi\| \, \|f\|\) for any nonzero \(f \in V\).

(For \(V = \{0\}\) the condition in (a) is vacuous.)

(a) The hypothesis forces equality in Cauchy-Schwarz for the two Riesz vectors. By 8.47 there are \(g, h \in V\) with \(\varphi = \langle \cdot , g \rangle\), \(\psi = \langle \cdot , h \rangle\), \(\|\varphi\| = \|g\|\), and \(\|\psi\| = \|h\|\); since \(\varphi + \psi = \langle \cdot , g + h \rangle\), the uniqueness in 8.47 gives \(\|\varphi + \psi\| = \|g + h\|\). So \(\|g + h\| = \|g\| + \|h\|\), and squaring both sides yields

\begin{equation*} \operatorname{Re} \langle g , h \rangle = \|g\| \, \|h\| . \end{equation*}

As \(\operatorname{Re} \langle g, h \rangle \le |\langle g , h \rangle| \le \|g\| \, \|h\|\) by the Cauchy–Schwarz inequality (8.11), both inequalities are equalities, and the equality case of 8.11 makes one of \(g, h\) a scalar multiple of the other. If \(g = \alpha h\), then

\begin{equation*} \varphi(f) = \langle f , \alpha h \rangle = \overline{\alpha} \, \psi(f) \qquad \text{for all } f \in V , \end{equation*}

so \(\varphi = \overline{\alpha} \psi\); symmetrically, \(h = \alpha g\) gives \(\psi = \overline{\alpha} \varphi\).

(b) Take \(V = \mathbf{F}^2\) with \(\|(a_1, a_2)\| = \max \{ |a_1| , |a_2| \}\), a Banach space because convergence in this norm is coordinatewise convergence in the complete field \(\mathbf{F}\), and take

\begin{equation*} \varphi(a_1, a_2) = a_1 , \qquad \psi(a_1, a_2) = a_2 . \end{equation*}

Then \(\|\varphi\| = \|\psi\| = 1\), attained at \((1,0)\) and \((0,1)\), while \(|a_1 + a_2| \le 2\|(a_1,a_2)\|\) with equality at \((1,1)\) gives

\begin{equation*} \|\varphi + \psi\| = 2 = \|\varphi\| + \|\psi\| . \end{equation*}

Yet \(\varphi = c \psi\) fails at \((1,0)\) and \(\psi = c \varphi\) fails at \((0,1)\).