Real Analysis Folland 1.22a.)

Refer to exercise 18 here: 1.18 Let $(X,M,\mu)$ be a measure space, $\mu^{*}$ the outer measure induced by $\mu$ according to (1.12)(I will define this in the attempted proof), $M^{*}$ 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)

proof: By the definition of outer measure we know that $$\mu^{*}(E) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} ( A_{j} ) : A_{j} \in \mathcal{A}, E \subset \bigcup_{j=1}^{\infty} A_{j} \right \}$$ Let $B_n = E\cap (A_n\setminus \cup_{1}^{n-1}A_j)$ then the $B_n$'s are disjoint members of $\mathcal{A}$ whose union is $E$, so $$\mu_{0}(E) = \sum_{1}^{\infty}\mu_{0}(B_j) \leq \sum_{1}^{\infty}\mu_{0}(A_j)$$ It follows that $\mu_{0}(E) \leq \mu^{*}(E)$

I am struggling with this one, not sure if this is the correct method, any suggestions is greatly appreciated.

• Hi Morgan, it is not clear what you are trying to prove, did you forget to write the statement? Commented Aug 31, 2015 at 21:43
• oh whoops, I will edit it Commented Aug 31, 2015 at 21:53
• What is $\mu_0$? You suddenly produce a formula for it without introduction of this symbol (and the same is true of $\mathcal{A}$, but there the intent is clear). Commented Aug 31, 2015 at 22:27
• @PaulSinclair: I checked on Folland's book, for this problem $\mu_0$ is the measure $\mu$ in the statement. Commented Aug 31, 2015 at 22:28

Let $F \subset N$, where $N$ is a measurable null set, i.e. $N \in \mathcal{M}$ and $\mu(N) = 0$. We want to prove that $F \in \mathcal{M}^*$.
Using part $(b)$ and $(c)$ of the problem we solved yesterday, it is enough to show that there is $B \in \mathcal{A}_{\sigma \delta}$ such that $F \subset B$ and $\mu^*(B \setminus F) = 0$.
Since $\mathcal{M} \subset \mathcal{M}^*$, we have that $N \in \mathcal{M}^*$ and hence there is $C \in \mathcal{A}_{\sigma \delta}$ such that $N \subset C$ and $\mu^*(C \setminus N) = 0$. But this clearly implies that $$\mu^*(C \setminus F) \le \mu^*(C \setminus N) + \mu^*(N \setminus F) \le \mu^*(C \setminus N) + \mu^*(N) = 0 + \mu(N) = 0.$$
• Look's really good, so does $\mu^{*}(N) = 0$ as well as $\mu(N)$? Commented Aug 31, 2015 at 23:11
• Yes, $\mu^*$ restricted to $\mathcal{M}$ is $\mu$ so by assumption $\mu^*(N) = \mu(N) = 0$. (as you can see I have used this in very last two equalities.) If you meant to ask weather $\mu^*(F) = 0$, then the answer is yes again: once we know that it is $\mathcal{M}^*$ measurable then you have $\mu^*(F) \le \mu^*(N) = 0$. :) Commented Aug 31, 2015 at 23:24