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# 7.6.E: Problems on Measures and Outer Measures

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## Exercise $$\PageIndex{1}$$

Show that formulas (1) and (2) are equivalent.
[Hints: (i) Assume (1) and let $$X \subseteq A, Y \subseteq-A .$$
As $$X$$ in (1) is arbitrary, we may replace it by $$X \cup Y.$$ Simplifying, obtain (2) on noting that $$X \cap A=X, X \cap-A=\emptyset, Y \cap A=\emptyset,$$ and $$Y \cap-A=Y$$.
(ii) Assume (2). Take any $$X$$ and substitute $$X \cap A$$ and $$X-A$$ for $$X$$ and $$Y$$ in (2).]

## Exercise $$\PageIndex{2}$$

Given an outer measure space $$\left(S, \mathcal{M}^{*}, m^{*}\right)$$ and $$A \subseteq S,$$ set
$A \cap \mathcal{M}^{*}=\left\{A \cap X | X \in \mathcal{M}^{*}\right\}$ (SYMBOL!)
(all sets of the form $$A \cap X$$ with $$X \in \mathcal{M}^{*}$$).
Prove that $$A \cap \mathcal{M}^{*}$$ is a $$\sigma$$-field in $$A,$$ and $$m^{*}$$ is $$\sigma$$-additive on it. (SYMBOL!)
[Hint: Use Lemma 4, with $$X_{k}=A \cap A_{k} \in A \cap \mathcal{M}^{*}$$.] (SYMBOL!)

## Exercise $$\PageIndex{3}$$

Prove Lemmas 1 and 2, using formula (1).

## Exercise $$\PageIndex{3'}$$

Prove Corollary 1.

## Exercise $$\PageIndex{4}$$

Verify Examples (b),(c), and (d). Why is $$m$$ an outer measure as well?
[Hint: Use Corollary 2 in §5.]

## Exercise $$\PageIndex{5}$$

Fill in all details (induction, etc.) in the proofs of this section.

## Exercise $$\PageIndex{6}$$

Verify that $$m^{*}$$ is an outer measure and describe $$\mathcal{M}^{*}$$ under each of the following conditions.
(a) $$m^{*} A=1$$ if $$\emptyset \subset A \subseteq S; m^{*} \emptyset=0$$.
(b) $$m^{*} A=1$$ if $$\emptyset \subset A \subset S; m^{*} S=2; m^{*} \emptyset=0$$.
(c) $$m^{*} A=0$$ if $$A \subseteq S$$ is countable; $$m^{*} A=1$$ otherwise ($$S$$ is uncountable).
(d) $$S=N$$ (naturals); $$m^{*} A=1$$ if $$A$$ is infinite; $$m^{*} A=\frac{n}{n+1}$$ if $$A$$ has $$n$$ elements.

## Exercise $$\PageIndex{7}$$

Prove the following.
(i) An outer measure $$m^{*}$$ is $$\mathcal{M}^{*}$$-regular (Definition 5 in §5) iff
$(\forall A \subseteq S)\left(\exists B \in \mathcal{M}^{*}\right) \quad A \subseteq B \text { and } m^{*} A=m B.$
$$B$$ is called a measurable cover of $$A$$.
[Hint: If
$m^{*} A=\inf \left\{m X | A \subseteq X \in \mathcal{M}^{*}\right\},$
then
$(\forall n)\left(\exists X_{n} \in \mathcal{M}^{*}\right) \quad A \subseteq X_{n} \text { and } m X_{n} \leq m^{*} A+\frac{1}{n}.$
Set $$B=\bigcap_{n=1}^{\infty} X_{n}$$.]
(ii) If $$m^{*}$$ is as in Definition 3 of §5, with $$\mathcal{C} \subseteq \mathcal{M}^{*},$$ then $$m^{*}$$ is $$\mathcal{M}^{*}$$-regular.

## Exercise $$\PageIndex{8}$$

Show that if $$m^{*}$$ is $$\mathcal{M}^{*}$$-regular (Problem 7), it is left continuous.
[Hints: Let $$\left\{A_{n}\right\} \uparrow;$$ let $$B_{n}$$ be a measurable cover of $$A_{n};$$ set
$C_{n}=\bigcap_{k=n}^{\infty} B_{k}.$
Verify that $$\left\{C_{n}\right\} \uparrow, B_{n} \supseteq C_{n} \supseteq A_{n},$$ and $$m C_{n}=m^{*} A_{n}$$.
By the left continuity of $$m$$ (Theorem 2 in §4),
$\lim m^{*} A_{n}=\lim m C_{n}=m \bigcup_{n=1}^{\infty} C_{n} \geq m^{*} \bigcup_{n=1}^{\infty} A_{n}.$
Prove the reverse inequality as well.]

## Exercise $$\PageIndex{9}$$

Continuing Problems 6-8, verify the following.
(i) In 6(a), with $$S=N, m^{*}$$ is $$\mathcal{M}^{*}$$-regular, but not right continuous.
Hint: Take $$A_{n}=\{x \in N | x \geq n\}$$.
(ii) In 6(b), with $$S=N, m^{*}$$ is neither $$\mathcal{M}^{*}$$-regular nor left continuous.
(iii) In 6(d), $$m^{*}$$ is not $$\mathcal{M}^{*}$$-regular; yet it is left continuous. (Thus Problem 8 is not a necessary condition.)

## Exercise $$\PageIndex{10}$$

In Problem 2, let $$n^{*}$$ be the restriction of $$m^{*}$$ to $$2^{A}.$$ Prove the following.
(a) $$n^{*}$$ is an outer measure in $$A$$.
(b) $$A \cap \mathcal{M}^{*} \subseteq \mathcal{N}^{*}=\left\{n^{*} \text {-measurable sets}\right\}$$. (SYMBOL!)
(c) $$A \cap \mathcal{M}^{*}=\mathcal{N}^{*}$$ if $$A \in \mathcal{M}^{*},$$ or if $$m^{*}$$ is $$\mathcal{M}^{*}$$-regular (see Problem 7) and finite. (SYMBOL!)
(d) $$n^{*}$$ is $$\mathcal{N}^{*}$$-regular if $$m^{*}$$ is $$\mathcal{M}^{*}$$-regular.

## Exercise $$\PageIndex{11}$$

Show that if $$m^{*}$$ is $$\mathcal{M}^{*}$$-regular and finite, then $$A \subseteq S$$ is $$m^{*}$$-measurable iff
$m S=m^{*} A+m^{*}(-A).$
[Hint: Assume the latter. By Problem 7,
$(\forall X \subseteq S)\left(\exists B \in \mathcal{M}^{*}, B \supseteq X\right) \quad m^{*} X=m B;$
so
$m^{*} A=m^{*}(A \cap B)+m^{*}(A-B).$
Similarly for $$-A.$$ Deduce that
$m^{*}(A \cap B)+m^{*}(A-B)+m^{*}(B-A)+m^{*}(-A-B)=m S=m B+m(-B);$
hence
$m^{*} X=m B \geq m^{*}(B \cap A)+m^{*}(B-A) \geq m^{*}(X \cap A)+m^{*}(X-A),$
so $$A \in \mathcal{M}^{*}$$.]

## Exercise $$\PageIndex{12}$$

Using Problem 15 in §5, prove that if $$m^{*}$$ has the CP then each open set $$G \subseteq S$$ is in $$\mathcal{M}^{*}$$.
[Outline: Show that
$(\forall X \subseteq G)(\forall Y \subseteq-G) \quad m^{*}(X \cup Y) \geq m^{*} X+m^{*} Y,$
assuming $$m^{*} X<\infty.$$ (Why?) Set
$D_{0}=\{x \in X | \rho(x,-G) \geq 1\}$
and
$D_{k}=\left\{x \in X | \frac{1}{k+1} \leq \rho(x,-G)<\frac{1}{k}\right\}, \quad k \geq 1.$
Prove that
$X=\bigcup_{k=0}^{\infty} D_{k}$
and
$\rho\left(D_{k}, D_{k+2}\right)>0;$
so by Problem 15 in §5,
$\sum_{n=0}^{\infty} m^{*} D_{2 n}=m^{*} \bigcup_{n=0}^{\infty} D_{2 n} \leq m^{*} \bigcup_{n=0}^{\infty} D_{n}=m^{*} X<\infty.$
Similarly,
$\sum_{n=0}^{\infty} m^{*} D_{2 n+1} \leq m^{*} X<\infty.$
Hence
$\sum_{n=0}^{\infty} m^{*} D_{n}<\infty;$
so
$\lim _{n \rightarrow \infty} \sum_{k=n}^{\infty} m^{*} D_{k}=0.$
(Why?) Thus
$(\forall \varepsilon>0)(\exists n) \sum_{k=n}^{\infty} m^{*} D_{k}<\varepsilon.$
Also,
$X=\bigcup_{k=0}^{\infty} D_{k}=\bigcup_{k=0}^{n-1} D_{k} \cup \bigcup_{k=n}^{\infty} D_{k};$
so
$m^{*} X \leq m^{*} \bigcup_{k=0}^{n-1} D_{k}+\sum_{k=n}^{\infty} m^{*} D_{k}<m^{*} \bigcup_{k=0}^{n-1} D_{k}+\varepsilon.$
Adding $$m^{*} Y$$ on both sides, get
$m^{*} X+m^{*} Y \leq m^{*} \bigcup_{k=0}^{n-1} D_{k}+m^{*} Y+\varepsilon.$
Moreover,
$\rho\left(\bigcup_{k=0}^{n-1} D_{k}, Y\right)>0,$
for $$Y \subseteq-G$$ and
$\rho\left(D_{k},-G\right) \geq \frac{1}{k+1}.$
Hence by the CP,
$m^{*} Y+\sum_{k=0}^{n-1} m^{*} D_{k}=m^{*}\left(Y \cup \bigcup_{k=0}^{n-1} D_{k}\right)<m^{*}(Y \cup X).$
(Why?) Combining with (iii), obtain
$m^{*} X+m^{*} Y \leq m^{*}(X \cup Y)+\varepsilon.$
Now let $$\varepsilon \rightarrow 0$$.]

## Exercise $$\PageIndex{13}$$

$$\Rightarrow$$ Show that if $$m : \mathcal{M} \rightarrow E^{*}$$ is a measure, there is $$P \in \mathcal{M},$$ with
$m P=\max \{m X | X \in \mathcal{M}\}.$
[Hint: Let
$k=\sup \{m X | X \in \mathcal{M}\}$
in $$E^{*}.$$ As $$k \geq 0,$$ there is a sequence $$r_{n} \nearrow k, r_{n}<k.$$ (If $$k=\infty,$$ set $$r_{n}=n;$$ if $$\left.k<\infty, r_{n}=k-\frac{1}{n}.\right)$$ By lub properties,
$(\forall n)\left(\exists X_{n} \in \mathcal{M}\right) \quad r_{n}<m X_{n} \leq k,$
with $$\left\{X_{n}\right\} \uparrow$$ (Problem 9 in §3). Set
$P=\bigcup_{n=1}^{\infty} X_{n}.$
Show that
$m P=\lim _{n \rightarrow \infty} m X_{n}=k.]$

## Exercise $$\PageIndex{14}$$

$$\Rightarrow^{*}$$ Given a measure $$m : \mathcal{M} \rightarrow E^{*},$$ let
$\overline{\mathcal{M}}=\{\text {all sets of the form } X \cup Z \text { where } X \in \mathcal{M} \text { and } Z \text { is } m \text{-null}\}.$
Prove that $$\overline{\mathcal{M}}$$ is a $$\sigma$$-ring $$\supseteq \mathcal{M}$$.
[Hint: To prove that
$(\forall A, B \in \overline{\mathcal{M}}) \quad A-B \in \overline{\mathcal{M}},$
suppose first $$A \in \mathcal{M}$$ and $$B$$ is "null," i.e., $$B \subseteq U \in \mathcal{M}, m U=0$$.
Show that
$A-B=X \cup Z,$
with $$X=A-U \in \mathcal{M}$$ and $$Z=A \cap U-B m$$-null ($$Z$$ is shaded in Figure 31). Next, if $$A, B \in \overline{\mathcal{M}},$$ let $$A=X \cup Z$$ $$B=X^{\prime} \cup Z^{\prime},$$ where $$X, X^{\prime} \in \mathcal{M}$$ and $$Z, Z^{\prime}$$ are $$m$$-null. Hence
\begin{aligned} A-B &=(X \cup Z)-B \\ &=(X-B) \cup(Z-B) \\ &=(X-B) \cup Z^{\prime \prime}, \end{aligned}
where
$Z^{\prime \prime}=Z-B$
is $$m$$-null. Also, $$B=X^{\prime} \cup Z^{\prime}$$ implies
$X-B=\left(X-X^{\prime}\right)-Z^{\prime} \in \overline{\mathcal{M}},$
by the first part of the proof.
Deduce that
$A-B=(X-B) \cup Z^{\prime \prime} \in \overline{\mathcal{M}}$
(after checking closure under unions).]

## Exercise $$\PageIndex{15}$$

$$\Rightarrow^{*}$$ Continuing Problem 14, define $$\overline{m} : \overline{\mathcal{M}} \rightarrow E^{*}$$ by setting $$\overline{m} A=m X$$ whenever $$A=X \cup Z,$$ with $$X \in \mathcal{M}$$ and $$Z$$ $$m$$-null. (Show that $$\overline{m} A$$ does not depend on the particular representation of $$A$$ as $$X \cup Z$$.)
Prove the following.
(i) $$\overline{m}$$ is a complete measure (called the completion of $$m$$), with $$\overline{m}=m$$ on $$\mathcal{M}.$$
(ii) $$\overline{m}$$ is the least complete extension of $$m;$$ that is, if $$n : \mathcal{N} \rightarrow E^{*}$$ is another complete measure, with $$\mathcal{M} \subseteq \mathcal{N}$$ and $$n=m$$ on $$\mathcal{M},$$ then $$\overline{\mathcal{M}} \subseteq \mathcal{N}$$ and $$n=\overline{m}$$ on $$\overline{\mathcal{M}}.$$
(iii) $$m=\overline{m}$$ iff $$m$$ is complete.

## Exercise $$\PageIndex{16*}$$

Show that if $$m : \mathcal{M}^{*} \rightarrow E^{*}$$ is induced by an $$\mathcal{M}^{*}$$-regular outer measure $$\mu^{*},$$ then $$m$$ equals its Lebesgue extension $$m^{\prime}$$ and completion $$\overline{m}$$ (see Problem 15).
[Hint: By Definition 3 in §5, $$m$$ induces an outer measure $$m^{*}.$$ By Theorem 3 in §5,
$m^{*} A=\inf \left\{m X | A \subseteq X \in \mathcal{M}^{*}\right\}=\mu^{*} A$
(for $$\mu^{*}$$ is $$\mathcal{M}^{*}$$-regular).
As $$m^{*}=\mu^{*},$$ we get $$m^{\prime}=m.$$ Also, $$m=\overline{m},$$ by Problem 15(iii).]

## Exercise $$\PageIndex{17*}$$

Prove that if a measure $$\mu : \mathcal{M} \rightarrow E^{*}$$ is $$\sigma$$-finite (Definition 4 in §5), with $$S \in \mathcal{M},$$ then its Lebesgue extension $$m : \mathcal{M}^{*} \rightarrow E^{*}$$ equals its completion $$\overline{\mu}$$ (see Problem 15).
[Outline: It suffices to prove $$\mathcal{M}^{*} \subseteq \overline{\mathcal{M}}.$$ (Why?)
To start with, let $$A \in \mathcal{M}^{*}, m A<\infty.$$ By Problem 12 in §5,
$(\exists B \in \mathcal{M}) \quad A \subseteq B \text { and } m^{*} A=m A=m B<\infty;$
so
$m(B-A)=m B-m A=0.$
Also,
$(\exists H \in \mathcal{M}) \quad B-A \subseteq H \text { and } \mu H=m(B-A)=0.$
Thus $$B-A$$ is $$\mu$$-null; so $$B-A \in \overline{\mathcal{M}}.$$ (Why?) Deduce that
$A=B-(B-A) \in \overline{\mathcal{M}}.$
Thus $$\overline{\mathcal{M}}$$ contains any $$A \in \mathcal{M}^{*}$$ with $$m A<\infty.$$ Use the $$\sigma$$-finiteness of $$\mu$$ to show
$\left.\left(\forall x \in \mathcal{M}^{*}\right)\left(\exists\left\{A_{n}\right\} \subseteq \mathcal{M}^{*}\right) \quad m A_{n}<\infty \text { and } X=\bigcup_{n} A_{n} \in \overline{\mathcal{M}}.\right]$