8.6.E: Problems on Integrability and Convergence Theorems
- Page ID
- 25028
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Fill in the missing details in the proofs of this section.
(i) Show that if \(f: S \rightarrow E^{*}\) is bounded and \(m\)-measurable on \(A,\) with \(m A<\infty,\) then \(f\) is \(m\) -integrable on \(A(\text { Theorem } 2)\) and
\[
\int_{A} f=c \cdot m A ,
\]
where inf \(f[A] \leq c \leq \sup f[A]\).
(ii) Prove that if \(f\) also has the Darboux property on \(A,\) then
\[
\left(\exists x_{0} \in A\right) \quad c=f\left(x_{0}\right) .
\]
[Hint: Take \(g=1 \text { in Theorem } 3 .]\)
(iii) What results if \(A=[a, b]\) and \(m=\) Lebesgue measure?
Prove Theorem 4 assuming that the \(f_{n}\) are measurable on \(A\) and that
\[
(\exists k) \quad \int_{A} f_{k}>-\infty
\]
instead of \(f_{n} \geq 0\).
[Hint: As \(\left\{f_{n}\right\} \uparrow\), show that
\[
(\forall n \geq k) \quad \int_{A} f_{n}>-\infty .
\]
If
\[
(\exists n) \quad \int_{A} f_{n}=\infty ,
\]
then
\[
\int_{A} f=\lim \int_{A} f_{n}=\infty .
\]
Otherwise,
\[
(\forall n \geq k) \quad\left|\int_{A} f_{n}\right|<\infty ;
\]
so \(f_{n}\) is integrable. (Why?) By Corollary 1 in §5, assume \(\left|f_{n}\right|<\infty .\) (Why?) Apply Theorem 4 to \(h_{n}=f_{n}-f_{k}(n \geq k),\) considering two cases:
\[
\left.\int_{A} h<\infty \text { and } \int_{A} h=\infty .\right]
\]
Show that if \(f_{n} \nearrow f\) (pointwise) on \(A \in \mathcal{M},\) there are \(\mathcal{M}\)-measurable maps \(F_{n} \geq f_{n}\) and \(F \geq f\) on \(A,\) with \(F_{n} \nearrow F\) (pointwise) on \(A,\) such that
\[
\int_{A} F=\overline{\int}_{A} f \text { and } \int_{A} F_{n}=\overline{\int}_{A} f_{n} .
\]
[Hint: By Lemma 2 of §5, fix measurable maps \(h \geq f\) and \(h_{n} \geq f_{n}\) with the same integrals. Let
\[
F_{n}=\inf _{k \geq n}\left(h \wedge h_{k}\right), \quad n=1,2, \ldots ,
\]
and \(F=\sup _{n} F_{n} \leq h .\) (Why?) Proceed.]
For \(A \in \mathcal{M}\) and any (even nonmeasurable) functions \(f, f_{n}: S \rightarrow E^{*},\) prove the following.
(i) If \(f_{n} \nearrow f(\text { a.e. })\) on \(A,\) then
\[
\overline{\int}_{A} f_{n} \nearrow \overline{\int}_{A} f ,
\]
provided
\[
(\exists n) \quad \overline{\int}_{A} f_{n}>-\infty .
\]
(ii) If \(f_{n} \searrow f(\text { a.e. })\) on \(A,\) then
\[
\underline{\int}_{A} f_{n} \searrow \underline{\int}_{A} f ,
\]
provided
\[
(\exists n) \quad \underline{\int}_{A} f_{n}<\infty .
\]
[Hint: Replace \(f, f_{n}\) by \(F, F_{n}\) as in Problem \(4 .\) Then apply Problem 3 to \(F_{n} ;\) thus obtain (i). For (ii), use (i) and Theorem \(1\left(\mathrm{e}^{\prime}\right)\) in §5. (All is orthodox; why?)]
Show by examples that
(i) the conditions
\[
\overline{\int}_{A} f_{n}>-\infty \text { and } \underline{\int}_{A} f_{n}<\infty
\]
in Problem 5 are essential; and
(ii) Problem \(5(\mathrm{i})\) fails for lower integrals. What about \(5(\mathrm{ii}) ?\)
[Hints: (i) Let \(A=(0,1) \subset E^{1}, m=\) Lebesgue measure, \(f_{n}=-\infty\) on \(\left(0, \frac{1}{n}\right), f_{n}=1\) elsewhere.
(ii) Let \(\mathcal{M}=\left\{E^{1}, \emptyset\right\}, m E^{1}=1, m \emptyset=0, f_{n}=1\) on \((-n, n), f_{n}=0\) elsewhere. If \(f=1\) on \(A=E^{1},\) then \(f_{n} \rightarrow f,\) but not
\[
\underline{\int}_{A} f_{n} \rightarrow \underline{\int}_{A} f .
\]
Explain!]
Given \(f_{n}: S \rightarrow E^{*}\) and \(A \in \mathcal{M},\) let
\[
g_{n}=\inf _{k \geq n} f_{k} \text { and } h_{n}=\sup _{k \geq n} f_{k} \quad(n=1,2, \ldots) .
\]
Prove that
(i) \(\overline{\int}_{A} \underline{\lim} f_{n} \leq \underline{\lim} \underline{\int}_{A} f_{n}\) provided \((\exists n) \overline{\int}_{A} g_{n}>-\infty ;\) and
(ii) \(\underline{\int}_{A} \overline{\lim } f_{n} \leq \overline{\lim } \underline{\int}_{A} f_{n} \text{provided}(\exists n) \underline{\int}_{A} h_{n}<\infty\).
[Hint: Apply Problem 5 to \(\left.g_{n} \text { and } h_{n} .\right]\)
(iii) Give examples for which
\[
\overline{\int}_{A} \underline{\lim} f_{n} \neq \overline{\lim}_{A} \overline{\int}_{A} f_{n} \text { and } \underline{\int}_{A} \overline{\lim } f_{n} \neq \underline{\lim } \underline{\int}_{A} f_{n} .
\]
(See Note 2).
Let \(f_{n} \geq 0\) on \(A \in \mathcal{M}\) and \(f_{n} \rightarrow f(\text { a.e. })\) on \(A .\) Let \(A \supseteq X, X \in \mathcal{M} .\)
Prove the following.
(i) If
\[
\overline{\int}_{A} f_{n} \rightarrow \overline{\int}_{A} f<\infty ,
\]
then
\[
\overline{\int}_{X} f_{n} \rightarrow \overline{\int}_{X} f .
\]
(ii) This fails for sign-changing \(f_{n}\).
[Hints: If (i) fails, then
\[
\underline{\lim} _{X} \overline{\int}_{X} f_{n}<\overline{\int}_{X} f \text { or } \underline{\lim}_{X} \overline{\int}_{X} f_{n}>\overline{\int}_{X} f .
\]
Find a subsequence of
\[
\left\{\overline{\int}_{X} f_{n}\right\} \text { or }\left\{\overline{\int}_{A-X} f_{n}\right\}
\]
contradicting Lemma 2.
(ii) Let \(m=\) Lebesgue measure; \(A=(0,1), X=\left(0, \frac{1}{2}\right)\),
\[
f_{n}=\left\{\begin{array}{ll}{n} & {\text { on }\left(0, \frac{1}{2 n}\right],} \\ {-n} & {\text { on }\left(1-\frac{1}{2 n}, 1\right[ .}\end{array}\right.
\]
\(\Rightarrow 9\). (i) Show that if \(f\) and \(g\) are \(m\)-measurable and nonnegative on \(A,\) then
\[
(\forall a, b \geq 0) \quad \int_{A}(a f+b g)=a \int_{A} f+b \int_{A} g .
\]
(ii) If, in addition, \(\int_{A} f<\infty\) or \(\int_{A} g<\infty,\) this formula holds for any \(a, b \in E^{1} .\)
[Hint: Proceed as in Theorem 1.]
\(\Rightarrow 10\). If
\[
f=\sum_{n=1}^{\infty} f_{n} ,
\]
with all \(f_{n}\) measurable and nonnegative on \(A,\) then
\[
\int_{A} f=\sum_{n=1}^{\infty} \int_{A} f_{n} .
\]
[Hint: Apply Theorem 4 to the maps
\[
g_{n}=\sum_{k=1}^{n} f_{k} \nearrow f .
\]
Use Problem 9.]
If
\[
q=\sum_{n=1}^{\infty} \int_{A}\left|f_{n}\right|<\infty
\]
and the \(f_{n}\) are \(m\)-measurable on \(A,\) then
\[
\sum_{n=1}^{\infty}\left|f_{n}\right|<\infty(a . e .) \text { on } A
\]
and \(f=\sum_{n=1}^{\infty} f_{n}\) is \(m\)-integrable on \(A,\) with
\[
\int_{A} f=\sum_{n=1}^{\infty} \int_{A} f_{n} .
\]
[Hint: Let \(g=\sum_{n=1}^{\infty}\left|f_{n}\right| .\) By Problem 10,
\[
\int_{A} g=\sum_{n=1}^{\infty} \int_{A}\left|f_{n}\right|=q<\infty ;
\]
so \(g<\infty(a . e .)\) on \(A .\) (Why?) Apply Theorem 5 and Note 1 to the maps
\[
g_{n}=\sum_{k=1}^{n} f_{k} ;
\]
note that \(\left.\left|g_{n}\right| \leq g .\right]\)
(Convergence in measure; see Problem 11(ii) of §3).
(i) Prove Riesz' theorem: If \(f_{n} \rightarrow f\) in measure on \(A \subseteq S\), there is a subsequence \(\left\{f_{n_{k}}\right\}\) such that \(f_{n_{k}} \rightarrow f\) (almost uniformly), hence (a.e.), on \(A\).
[Outline: Taking
\[
\sigma_{k}=\delta_{k}=2^{-k} ,
\]
pick, step by step, naturals
\[
n_{1}<n_{2}<\cdots<n_{k}<\cdots
\]
and sets \(D_{k} \in \mathcal{M}\) such that \((\forall k)\)
\[
m D_{k}<2^{-k}
\]
and
\[
\rho^{\prime}\left(f_{n_{k}}, f\right)<2^{-k}
\]
on \(A-D_{k} .\) (Explain!) Let
\[
E_{n}=\bigcup_{k=n}^{\infty} D_{k} ,
\]
\(m E_{n}<2^{1-n} .(\) Why?) Show that
\[
(\forall n)(\forall k>n) \quad \rho^{\prime}\left(f_{n_{k}}, f\right)<2^{1-n}
\]
on \(\left.A-E_{n} . \text { Use Problem } 11 \text { in } §3 .\right]\)
(ii) For maps \(f_{n}: S \rightarrow E\) and \(g: S \rightarrow E^{1}\) deduce that if
\[
f_{n} \rightarrow f
\]
in measure on \(A\) and
\[
(\forall n) \quad\left|f_{n}\right| \leq g(\text { a.e. }) \text { on } A ,
\]
then
\[
|f| \leq g(\text { a.e. }) \text { on } A .
\]
\(\left[\text { Hint: } f_{n_{k}} \rightarrow f(a . e .) \text { on } A .\right]\)
Continuing Problem \(12(\text { ii }),\) let
\[
f_{n} \rightarrow f
\]
in measure on \(A \in \mathcal{M}\left(f_{n}: S \rightarrow E\right)\) and
\[
(\forall n) \quad\left|f_{n}\right| \leq g(\mathrm{a.e.}) \text { on } A ,
\]
with
\[
\overline{\int_{A}} g<\infty .
\]
Prove that
\[
\lim _{n \rightarrow \infty} \overline{\int}_{A}\left|f_{n}-f\right|=0 .
\]
Does
\[
\overline{\int}_{A} f_{n} \rightarrow \overline{\int}_{A} f ?
\]
[Outline: From Corollary 1 of §5, infer that \(g=0\) on \(A-C,\) where
\[
C=\bigcup_{k=1}^{\infty} C_{k}(\text {disjoint}) ,
\]
\(m C_{k}<\infty .\) (We may assume \(g \mathcal{M}\)-measurable on \(A .\) Why?) Also,
\[
\infty>\int_{A} g=\int_{A-C} g+\int_{C} g=0+\sum_{k=1}^{\infty} \int_{C_{k}} g ;
\]
so the series converges. Hence
\[
(\forall \varepsilon>0)(\exists p) \quad \int_{A} g-\varepsilon<\sum_{k=1}^{p} \int_{C_{k}} g=\int_{H} g ,
\]
where
\[
H=\bigcup_{k=1}^{p} C_{k} \in \mathcal{M}
\]
and \(m H<\infty .\) As \(\left|f_{n}-f\right| \leq 2 g(\text { a.e. }),\) we get
\[
\text { (1) } \underline{\int}_{A}\left|f_{n}-f\right| \leq \overline{\int}_{A}\left|f_{n}-f\right| \leq \overline{\int}_{H}\left|f_{n}-f\right|+\int_{A-H} 2 g<\overline{\int_{H}}\left|f_{n}-f\right|+2 \varepsilon .
\]
(Explain!)
As \(m H<\infty,\) we can fix \(\sigma>0\) with
\[
\sigma \cdot m H<\varepsilon .
\]
Also, by Theorem \(6,\) fix \(\delta\) such that
\[
2 \int_{X} g<\varepsilon
\]
whenever \(A \supseteq X, X \in \mathcal{M}\) and \(m X<\delta\).
As \(f_{n} \rightarrow f\) in measure on \(H,\) we find \(\mathcal{M}\)-sets \(D_{n} \subseteq H\) such that
\[
\left(\forall n>n_{0}\right) \quad m D_{n}<\delta
\]
and
\[
\left|f_{n}-f\right|<\sigma \text { on } A_{n}=H-D_{n} .
\]
(We may use the standard metric, as \(|f|\) and \(\left|f_{n}\right|<\infty\) a.e. Why?) Thus from \((1),\) we get
\[
\begin{aligned} \overline{\int}_{A}\left|f_{n}-f\right| & \leq \overline{\int_{H}}\left|f_{n}-f\right|+2 \varepsilon \\ &=\overline{\int}_{A_{n}}\left|f_{n}-f\right|+\overline{\int}_{D_{n}}\left|f_{n}-f\right|+2 \varepsilon \\ &<\overline{\int}_{A_{n}}\left|f_{n}-f\right|+3 \varepsilon \\ & \leq \sigma \cdot m H+3 \varepsilon<4 \varepsilon \end{aligned}
\]
for \(n>n_{0} .\) (Explain!) Hence
\[
\lim \overline{\int_{A}}\left|f_{n}-f\right|=0 .
\]
See also Problem 7 in §5 and Note 1 of §6 (for measurable functions) as regards
\[
\left.\lim \overline{\int_{A}} f_{n} \cdot\right]
\]
Do Problem 12 in §3 (Lebesgue-Egorov theorems) for \(T=E,\) assuming
\[
(\forall n) \quad\left|f_{n}\right| \leq g(a . e .) \text { on } A ,
\]
with
\[
\int_{A} g<\infty
\]
(instead of \(m A<\infty)\).
[Hint: With \(H_{i}(k)\) as before, it suffices that
\[
\lim _{i \rightarrow \infty} m\left(A-H_{i}(k)\right)=0 .
\]
(Why?) Verify that
\[
(\forall n) \quad \rho^{\prime}\left(f_{n}, f\right)=\left|f_{n}-f\right| \leq 2 g(a . e .) \text { on } A ,
\]
and
\[
(\forall i, k) \quad A-H_{i}(k) \subseteq A\left(2 g \geq \frac{1}{k}\right) \cup Q(m Q=0) .
\]
Infer that
\[
(\forall i, k) \quad m\left(A-H_{i}(k)\right)<\infty.
\]
Now, as \((\forall k) H_{i}(k) \searrow \emptyset\) (why?), right continuity applies.]