Real Analysis Folland 1.22a.)

Question

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.

Answer 1

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 $$\mu^*(C \setminus F) \le \mu^*(C \setminus N) + \mu^*(N \setminus F) \le \mu^*(C \setminus N) + \mu^*(N) = 0 + \mu(N) = 0.$$

This should work, let me know what you think!