Integration: More Exercises
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last updates: 18 March 2011
Integration: More Exercises
 If $\mu $ is a complex measure on a σalgebra $\mathcal{M}$,
and if $E\in \mathcal{M}$, define
$\lambda \left(E\right)=\mathrm{sup}\sum \left\mu \right({E}_{i}\left)\right,$
the supremum being taken over all finite partitions
$\left\{{E}_{i}\right\}$ of $E$.
Does it follow that $\lambda =\left\mu \right$?

Prove that the example given at the end of [Ru, Sec. 6.10] has the stated properties.
 Prove that the vector space $M\left(X\right)$ of
all complex regular Borel measures on a locally compact Hausdorff space $X$
is a Banch space if
$\Vert \mu \Vert =\left\mu \right\left(X\right)$. Hint:
Compare [Ru, Ch. 5, Exercise 8]. [That the difference of any two members of
$M\left(X\right)$ is in
$M\left(X\right)$ was used in the first paragraph of the proof
of [Ru, Theorem 6.19]; supply a proof of this fact.]
 Suppose $1\le p\le \infty $ and
$q$ is the exponent conjugate to $p$. Suppose $\mu $
is a positive σfinite measure and $g$ is a measurable function such that
$fg\in {L}^{1}\left(\mu \right)$ for every
$f\in {L}^{p}\left(\mu \right)$.
Prove that then
$g\in {L}^{q}\left(\mu \right)$.
 Suppose $X$ consists of two points $a$ and $b$;
define $\mu \left(\right\{a\left\}\right)=1$,
$\mu \left(\right\{b\left\}\right)=\mu \left(X\right)=\infty $, and
$\mu \left(\varnothing \right)=0$. Is it true, for this
$\mu $ that
${L}^{\infty}\left(\mu \right)$
is the dual space of
${L}^{1}\left(\mu \right)$?
 Suppose $1<p<\infty $ and
$q$ is the exponent conjugate to $p$. Prove that
${L}^{q}\left(\mu \right)$ is the dual space
of
${L}^{p}\left(\mu \right)$
even if $\mu $ is not σfinite.
 Suppose $\mu $ is a complex Borel measure on
$[0,2\pi )$ (or on the unit circle
$T$), and define the Fourier coefficients of $\mu $ by
$\hat{\mu}\left(n\right)=\int {e}^{i\pi t}d\mu \left(t\right)\phantom{\rule{2em}{0ex}}(n=0,\pm 1,\pm 2,\dots )$.
 
Assume that
$\hat{\mu}\left(n\right)\to 0$
as $n\to +\infty $
and prove that then
$\hat{\mu}\left(n\right)\to 0$
as $n\to \infty $.
Hint: The assumption also holds with
$f\phantom{\rule{0.1em}{0ex}}d\mu $ in
place of $d\mu ;$ if $f$ is any trigonometric polynomial,
hence if $f$ is continuous, hence if $f$ is any bounded Borel function,
hence if $d\mu $ is replaced by
$d\left\mu \right$.
 In the terminology of Exercise 7, find all $\mu $ such that
$\hat{\mu}$ is periodic, with period $k$.
[This means that
$\hat{\mu}(n+k)=\hat{\mu}\left(n\right)$ for all integers $n$; of course, $k$ is also assumed to be
an integer.]
 Suppose that $\left\{{g}_{n}\right\}$ is a sequence
of positive continuous functions on $I=[0,1]$, that $\mu $
is a positive Borel measure on $I$, and that
 (a) $\underset{n\to \infty}{\mathrm{lim}}}\phantom{\rule{0.2em}{0ex}}{g}_{n}\left(x\right)=0,\phantom{\rule{2em}{0ex}}\mathrm{a.e.}\left[\mathrm{m}\right]$,
 (b) ${\int}_{I}}\phantom{\rule{0.2em}{0ex}}{g}_{n}\phantom{\rule{0.2em}{0ex}}dm=1$,
for all $n$,
 (c)$\underset{n\to \infty}{\mathrm{lim}}{\int}_{I}}\phantom{\rule{0.2em}{0ex}}f{g}_{n}\phantom{\rule{0.2em}{0ex}}dm={\int}_{I}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}d\mu $,
for every $f\in C\left(I\right)$.
Does it follow that $\mu \perp m$?

Let $(X,\mathcal{M},\mu )$
be a positive measure space. Call a set $\Phi \subseteq {L}^{1}\left(\mu \right)$
uniformly integrable if to each $\epsilon <0$
corresponds a $\delta >0$ such that
${\int}_{E}}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}d\mu <\epsilon $,
 
whenever $f\in \Phi $
and $\mu \left(E\right)<\delta $.
 (a)
Prove that every finite subset of
${L}^{1}\left(\mu \right)$
is uniformly integrable.
 (b)
Prove the following convergence theorem of Vitali:
If
 (i)
$\mu \left(X\right)<\infty $,
 (ii)
$\left\{{f}_{n}\right\}$ is uniformly integrable,
 (iii)
${f}_{n}\to f\left(x\right)$ a.e.
as $n\to \infty $,
 (iv)
$\leftf\right(x\left)\right<\infty $
a.e.,
then
$f\in {L}^{1}\left(\mu \right)$ and
$\underset{n\to \infty}{\mathrm{lim}}{\int}_{X}}\phantom{\rule{0.2em}{0ex}}f{f}_{n}\phantom{\rule{0.2em}{0ex}}d\mu =0$.
 
Suggestion: Use Egoroff's theorem.
 (c)
Show that (b) fails if $\mu $ is Lebesgue measure on $(\infty ,\infty )$, even if
$\left\{{\Vert {f}_{n}\Vert}_{1}\right\}$ is assumed to be bounded. Hypothesis (i) can therefore not be omitted in (b).
 (d)
Show that hypothesis (iv) is redundant in (b) for some $\mu $ ( for instance, for Lebesgue
measure on a bounded interval), but that there are finite measures for which the omission of (iv)
would make (b) false.
 (e) Show that Vitali's theorem implies Lebesgue's dominated convergence theorem, for
finite measure spaces. Construct an example in which Vitali's theorem applies although the
hypotheses of Lebesgue's theorem do not hold.
 (f) Construct a sequence $\left\{{f}_{n}\right\}$,
say on $[0,1]$, so that
${f}_{n}\left(x\right)\to 0$,
but
$\left\{{f}_{n}\right\}$ is not uniformly
integrable (with respect to Lebesgue measure).
 (g) However, the following converse of Vitali's theorem is true: If
$\mu \left(X\right)<\infty $,
${f}_{n}\in {L}^{1}\left(\mu \right)$ and
$\underset{n\to \infty}{\mathrm{lim}}{\int}_{E}}\phantom{\rule{0.2em}{0ex}}{f}_{n}\phantom{\rule{0.2em}{0ex}}d\mu $ exists for every $E\in \mathcal{M}$,
then
$\left\{{f}_{n}\right\}$ is uniformly
integrable.
Prove this by completing the following outline.
Define $\rho (A,B)=\int {\chi}_{A}{\chi}_{B}\phantom{\rule{0.2em}{0ex}}d\mu $. Then $(\mathcal{M},\rho )$ is a complete metric space
(modulo sets of measure $0$), and $E\to {\int}_{E}\phantom{\rule{0.2em}{0ex}}{f}_{n}\phantom{\rule{0.2em}{0ex}}d\mu $ is continuous for each $n$. If $\epsilon >0$,
there exist ${E}_{0},\delta ,N$
(Exercise 13, Chapt. 5) so that
${\int}_{X}}\phantom{\rule{0.2em}{0ex}}(f{f}_{n})\phantom{\rule{0.2em}{0ex}}d\mu <\epsilon $,
if
$\rho (E,{E}_{0})<\delta $,
$n>N$.
 (*) 
If $\mu <\delta $, (*) holds with
$B={E}_{0}A$
and
$C={E}_{0}\cup A$
in place of $E$. Thus (*) holds with $A$ in place of
$E$ and $2\epsilon $ in place of $\epsilon $.
Now apply (a) to $\{{f}_{1},{f}_{2},\dots {f}_{n}\}$:
There exists $\delta \prime >0$ such that
${\int}_{X}}\phantom{\rule{0.2em}{0ex}}{f}_{n}\phantom{\rule{0.2em}{0ex}}d\mu <3\epsilon $,
if
$\mu \left(A\right)<\delta \prime $,
$n=1,2,3,\dots $.
 
 Suppose that $\mu $ is a positive measure on $X$,
$\mu \left(X\right)\le \infty $,
${f}_{n}\in {L}^{1}\left(\mu \right)$ for $n=1,2,3,\dots $, ${f}_{n}\to f\left(x\right)$ a.e., and there exists $p>1$ and $C<\infty $ such that
${\int}_{X}}\phantom{\rule{0.2em}{0ex}}{\left{f}_{n}\right}^{p}\phantom{\rule{0.2em}{0ex}}d\mu <C$ for all $n$. Prove that
$\underset{n\to \infty}{\mathrm{lim}}{\int}_{X}}\phantom{\rule{0.2em}{0ex}}f{f}_{n}\phantom{\rule{0.2em}{0ex}}d\mu =0$.
 
Hint: $\left\{{f}_{n}\right\}$ is uniformly integrable.
 Let $\mathcal{M}$ be the collection of all sets $E$ in the
unit interval $[0,1]$ such that either
$E$ or its complement is at most countable. Let $\mu $
be the counting measure on this σalgebra $\mathcal{M}$. If
$g\left(x\right)=x$ for
$0\le x\le 1$, show that $g$
is not $\mathcal{M}$measurable, although the mapping
$f\mapsto \sum xf\left(x\right)=\int fg\phantom{\rule{0.1em}{0ex}}d\mu $
 
makes sense for every $f\in {L}^{1}\left(\mu \right)$ and defines a bounded linear functional on
${L}^{1}\left(\mu \right)$.
Thus ${\left({L}^{\infty}\right)}^{*}\ne {L}^{1}$ in this situation.
 Let ${L}^{\infty}={L}^{\infty}\left(m\right)$,
where $m$ is Lebesgue measure on $I=[0,1]$. Show that there is
a bounded linear functional $\Lambda \ne 0$ on
${L}^{\infty}$ that is $0$ on
$C\left(I\right)$, and that therefore there is no
$g\in {L}^{1}\left(m\right)$
that satisfies $\Lambda f={\displaystyle {\int}_{I}}\phantom{\rule{0.2em}{0ex}}fg\phantom{\rule{0.2em}{0ex}}dm$ for every $f\in {L}^{\infty}$.
Thus ${\left({L}^{\infty}\right)}^{*}\ne {L}^{1}$.
Notes and References
These exercises are taken from [Ru, Chapt. 6] for a course in "Measure Theory" at the Masters level at University of Melbourne.
References
[Ru]
W. Rudin,
Real and complex analysis, Third edition, McGrawHill, 1987.
MR0924157.
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