# Real Analysis, problem 1.4.22 Outer Measures

Exercise 22 - Let $$(X,M,\mu)$$ be a measure space, $$\mu^*$$ the outer measure induced by $$\mu$$ according to (1.12), $$M^*$$ the $$\sigma$$-algebra of $$\mu^*$$-measurable sets, and $$\overline{\mu} = \mu^*|M^*$$.

a.) If $$\mu$$ is $$\sigma$$-finite, then $$\overline{\mu}$$ is the completion of $$\mu$$ (Use exercise 18 found here)

b.) In general, $$\overline{\mu}$$ is the saturation of the completion of $$\mu$$

This exercises is from Follad's Real Analysis (Section 1.4. Outer Measures).

Attempted proof a.) - Let $$F\subset N$$, where $$N$$ is a measurable null set, i.e., $$N\in M$$ and $$\mu(N) = 0$$. I will prove that $$F\in M^*$$. Using 18b and 18c, I believe it suffices to show that there is a $$B\in \mathcal{A}_{\sigma \delta}$$ such that $$F\subset B$$ and $$\mu^*(B\setminus F) = 0$$.

Since $$M\subset M^*$$, $$N\in M^*$$ and hence there exists a $$C\in\mathcal{A}_{\sigma \delta}$$ such that $$N\subset C$$ and $$\mu^*(C\setminus N) = 0$$. Thus $$\mu^*(C\setminus F) \leq \mu^*(C\setminus N) + \mu^*(N\setminus F) \leq \mu^*(C\setminus N) + \mu^*(N) = 0$$

I am not sure if this is completely right. Also any suggestions on part b would be greatly appreciated. I just don't understand the statement "$$\overline{\mu}$$ is the saturation of the completion of $$\mu$$".

• for (a) could you show $\overline{\mathcal M}=\{E\cup F:E\in \mathcal M, \ F\subset N\text{ for some null set}N\}=\mathcal M^*$? (look at theorem 1.9 Folland). then use theorem 1.14.
– mac
Commented May 21, 2016 at 13:34
• @mac Didn't think of that but I believe my approach is correct as well, what do you think? Commented May 21, 2016 at 13:39
• It's enought to show that $\overline{\mathcal M}=\mathcal M^*$ Since by proposition 1.13and theorem 1.14 $\overline{\mu}$ is an extension of $\mu$. If you want to show two meaures are equal then you should show that their $\sigma$-algebras are equal. However in your approach if you want to show $F\in \mathcal M^*$ it is obvious that $\mu^*(F)\leq \mu^*(N)=\mu(N)=0$ and you don't need exercise 18 in this case. indeed you need 18 to show $\overline{\mathcal M}=\mathcal M^*$
– mac
Commented May 21, 2016 at 16:25
• 1. For item (a), your proof proves only that if $F\subset N$, where $N$ is a measurable null set, i.e., $N\in M$ and $\mu(N) = 0$, then $F$ is $\mu^*$-measurable. It is an important result, but it is not what item (a) is asking to be proved. 2. The definition of the saturation of a measure can be found in Folland's Exercise 1.3.16. Commented May 22, 2016 at 15:34
• I have posted a detailed answer. Please, let me know if you have any question regarding my answer. Commented May 22, 2016 at 15:59

Initial Remark: This is a long exercise and it depends on some previous exercises in Folland. I made this answer as self-contained as reasonable (I explictly used only the results Exercise 18). The definition of saturation can be found Exercise 1.3.16.

Notation: Let $$(X,M,\mu)$$ be a measure space, $$\mu^*$$ the outer measure induced by $$\mu$$ according to (1.12), $$M^*$$ the $$\sigma$$-algebra of $$\mu^*$$-measurable sets, and $$\overline{\mu} = \mu^*|M^*$$. Let $$\widehat{M}$$ be the $$\mu$$-completion of $$M$$ and $$\widehat{\mu}$$ the completion of $$\mu$$.

We begin with three lemmas.

Lemma 1: If $$E \in M^*$$ and $$\mu^*(E)< \infty$$, then $$E\in \widehat{M}$$.

Proof

Let $$E\subset X$$ be any $$\mu^*$$-measurable set such that $$\mu^*(E)< \infty$$.

From Exercise 18-b, there exists $$B\in M_{\sigma\delta} \subset M$$ with $$E\subset B$$ and $$\mu^{*}(B\setminus E)) = 0$$. So
$$\mu^*(B)\leq \mu^*(E)+ \mu^{*}(B\setminus E))= \mu^*(E)$$ Since $$E\subset B$$, we have $$\mu^*(B)= \mu^*(E)$$.

Now, we have that $$B\setminus E$$ is $$\mu^*$$-measurable set and $$\mu^*(B\setminus E)\leq \mu^*(B)= \mu^*(E) < \infty$$. From Exercise 18-b, again, there exists $$C\in M$$ with $$B\setminus E\subset C$$ and $$\mu^{*}(C\setminus (B\setminus E)) = 0$$

Define $$D=C\cap B$$. Then, since $$B\setminus E\subset C$$, we have $$B\setminus E\subset D$$ and so we have $$B\setminus D \subset E$$ and $$E\setminus (B \setminus D)= E\cap D= D\setminus (B\setminus E) = (C\setminus (B\setminus E))\cap B \subset C\setminus (B\setminus E)$$ So, since $$\mu^{*}(C\setminus (B\setminus E)) = 0$$, we have $$\mu^{*}(E\setminus (B\setminus D)) = 0$$.

So we have that $$D\in M$$, $$B\setminus D\subset E$$ and $$\mu^{*}(E \setminus (B\setminus D)) = 0$$.

So $$E =(B \setminus D) \cup (E\setminus (B\setminus D))$$ where $$B-D \in M$$ and $$\mu^{*}(E\setminus (B\setminus D)) = 0$$. So $$E\in \widehat{M}$$ (where $$\widehat{M}$$ is the $$\mu$$-completion of $$M$$).

Lemma 2: For any $$E\subset X$$ there exists $$B\in M$$ such that $$E\subset B$$ and $$\mu^*(B) = \mu^*(E)$$.

Proof From Exercise 18-a we know that

For any $$E\subset X$$ and $$\epsilon > 0$$ there exists $$A\in M$$ with $$E\subset A$$ and $$\mu^*(E)\leq \mu^*(A) \leq \mu^*(E) + \epsilon$$.

So, for each $$n\in\mathbb{N}$$, $$n>0$$, let $$A_n\in M$$ with $$E\subset A_n$$ and $$\mu^*(E)\leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$$.

Then let $$B=\bigcap_{n=1}^\infty A_n$$. Then we have $$B\in M$$ and $$E\subset B$$. Moreover, for all $$n\in\mathbb{N}$$, $$n>0$$, $$B\subset A_n$$ and $$\mu^*(E)\leq \mu^*(B) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$$ So, $$\mu^*(B)=\mu^*(E)$$.

Remark: $$E$$ don't need to be measurable. And as a consequence of item Exercise 18 item b.) $$\mu^*(B\setminus E)$$ may not be zero.

Lemma 3: If $$A \in \widehat{M}$$, then $$A \in M^*$$ and $$\mu^*(A)=\widehat{\mu}(A)$$.

Proof:

Let $$A\in \widehat{M}$$. So $$A =B\cup C$$ , where $$B\in M$$ and $$C\subset N$$ such that $$N \in M$$ and $$\mu(N)=0$$. In particular, we have $$0 \leq \mu^*(C)\leq \mu^*(N)=\mu(N)=0$$. So $$\mu^*(C)=0$$ and it follows that $$C$$ is $$\mu^*$$-measurable. So we have $$A =G\cup H$$, $$B\in M \subset M^*$$ and $$C\in M^*$$. So $$A\in M^*$$. Moreover $$\mu*(B) \leq \mu^*(A) \leq \mu*(B) + \mu*(C) = \mu*(B)$$ So $$\mu^*(A)= \mu*(B)= \mu(B)=\widehat{\mu}(A)$$

Exercise 22 - Let $$(X,M,\mu)$$ be a measure space, $$\mu^*$$ the outer measure induced by $$\mu$$ according to (1.12), $$M^*$$ the $$\sigma$$-algebra of $$\mu^*$$-measurable sets, and $$\overline{\mu} = \mu^*|M^*$$.

a.) If $$\mu$$ is $$\sigma$$-finite, then $$\overline{\mu}$$ is the completion of $$\mu$$ (Use exercise 18 found [here][1])

b.) In general, $$\overline{\mu}$$ is the saturation of the completion of $$\mu$$

Proof:

a.) Suppose $$\mu$$ is $$\sigma$$-finite. Then there is $$\{X_i\}_{i\in \mathbb{N}}$$ family of disjoint sets, such that, for all $$i$$, $$\mu(X_i)<+\infty$$ and $$X=\bigcup_{i=1}^\infty X_i$$.

Let $$E\subset X$$ be any $$\mu^*$$-measurable set. Then, we define, for all $$i$$, $$E_i=E\cap X_i$$, we have that $$\{E_i\}_{i\in \mathbb{N}}$$ is a family of disjoint sets, such that, for all $$i$$, $$\mu(X_i)<+\infty$$ and $$E=\bigcup_{i=1}^\infty E_i$$.

From the lemma 1, for all $$i$$, $$E_i\in \widehat{M}$$ (where $$\widehat{M}$$ is the $$\mu$$-completion of $$M$$).

Since $$\widehat{M}$$ is a $$\sigma$$-algebra, $$E = E=\bigcup_{i=1}^\infty E_i \in \widehat{M}$$.

b.) Let $$\widetilde{\widehat{\mu}}$$ be the saturation of the completion of $$\mu$$. We know that $$\widetilde{\widehat{\mu}}$$ is defined on $$\widetilde{\widehat{M}}= \{E : E \textrm{ is {\widehat{\mu}}-locally measurable} \}$$ On the other hand, $$\overline{\mu}$$ is defined on $$M^*$$. We want to prove that $$\widetilde{\widehat{\mu}}=\overline{\mu}$$.

We begin by proving that $$\widetilde{\widehat{M}}=M^*$$, it means, we begin by proving that $$E$$ is $${\widehat{\mu}}$$-locally measurable if and only if $$E$$ is $$\mu^*$$-measurable.

($$\Rightarrow$$) Suppose $$E$$ is $${\widehat{\mu}}$$-locally measurable. For any $$B \in M$$ such that $$\mu(B)<+\infty$$, we have $$B\in \widehat{M}$$ and $$\widehat{\mu}(B)=\mu(B)<\infty$$

So $$B\in \widehat{M}$$ and $$\widehat{\mu}(B)<\infty$$. Since $$E$$ is $${\widehat{\mu}}$$-locally measurable, we have that $$E\cap B \in \widehat{M}$$. So, by lemma 3, $$E\cap B \in M^*$$. And we also have $$E^c\cap B= B \setminus (E\cap B) \in M^*$$.

So we have proved that $$B \in M$$ and $$\mu(B)<+\infty$$, then $$E\cap B \in M^*$$ and $$E^c\cap B \in M^*$$

Now let us check the $$\mu*$$-measurability of $$E$$. Let $$A\subset X$$. If $$\mu^*(A)=\infty$$, we trivially have $$\mu^*(A) \geq \mu*(E\cap A)+\mu*(E^c\cap A)$$ If $$\mu^*(A)<\infty$$ then using lemma 2, we have $$B\in M$$ such that $$A\subset B$$ and $$\mu(B)=\mu^*(B)=\mu^*(A)<\infty$$. So we have that $$E\cap B \in M^*$$ and $$E^c\cap B \in M^*$$ and $$\mu^*(A)= \mu^*(B) = \mu^*(E\cap B ) + \mu^*(E^c\cap B) \geq \mu^*(E\cap A ) + \mu^*(E^c\cap A)$$

So E is $$\mu^*$$-measurable.

($$\Leftarrow$$) Suppose E is $$\mu^*$$-measurable. For any $$A\in \widehat{M}$$ such that $$\widehat{\mu}(A)<+\infty$$. By lemma 3, $$A\in M^*$$ and $$\mu*(A)= \widehat{\mu}(A)<+\infty$$. So $$E\cap A\in M^*$$ and $$\mu*(E\cap A)\leq \mu^*(A)<+\infty$$. So by lemma 1, $$E \cap A \in \widehat{M}$$. So, $$E$$ is $${\widehat{\mu}}$$-locally measurable.

So we have proved $$\widetilde{\widehat{M}}=M^*$$.

Now we must prove that $$\widetilde{\widehat{\mu}}$$ coincides to $$\overline{\mu}$$.

Since the extension of $$\mu$$ to $$\widehat{M}$$ (the completion of $$\mu$$) is unique, we have for all $$E \in \widehat{M}$$, $$\widetilde{\widehat{\mu}}(E)= \overline{\mu}(E)= \widehat{\mu}(E)$$

If $$E\in \widetilde{\widehat{M}} \setminus \widehat{M}$$, then by definition of $$\widetilde{\widehat{\mu}}$$, we have $$\widetilde{\widehat{\mu}}(E)=+\infty$$. Note $$E\in \widetilde{\widehat{M}} \setminus \widehat{M}= M^*\setminus \widehat{M}$$ and, by lemma 1, we have $$\mu^*(E)=+\infty$$, that is $$\overline{\mu}(E)=\mu^*(E)=+\infty$$. So, $$\widetilde{\widehat{\mu}}(E)=\overline{\mu}(E)$$.

So we have proved that $$\widetilde{\widehat{\mu}}=\overline{\mu}$$.

• Are the lemmas you present results from exercise 18? Commented May 23, 2016 at 9:53
• @Wolfy , Lemma 1 and lemma 2 are consequence of exercise 18 (items a and b). Of course, we can prove lemma 1 and lemma 2 without using exercise 18. In this case the proofs would be a little bit longer (since we would have to redo, at least partially, the work done prove exercise 18a and 18b). Lemma 3 does not use exercise 18. Commented May 23, 2016 at 13:14
• Ok, I see (not really). You seem pretty good at this stuff, are you possibly interested in tutoring via skype for ? Commented May 23, 2016 at 13:15
• @Wolfy , Note that lemma 2 is an important result, which is useful to prove several other results in Measure Theory. The set $B$ whose existence is stated in lemma 2 is called a "mensurable cover" of $E$. Note that lemma 1 and lemma 3 are kind of reciprocal to each other. Together, they describe the relationship between $M^*$ and $\widehat{M}$. Commented May 23, 2016 at 13:31
• @Wolfy Yes, I may be interested in tutoring via skype. We can try. Commented May 23, 2016 at 13:42