18. Let $\mathcal{A}\subset\mathcal{P}(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$, and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets inf $\mathcal{A}_\sigma$. Let $\mu_0$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induces outer measure.
a. For any $E\subset X$ and $\epsilon>0$ there exists $A\in\mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(A)\leq\mu^*(E)+\epsilon$.
b. If $\mu^*(E)<\infty$, then $E$ is $\mu^*$-measurable iff there exists $B\in\mathcal{A}_{\sigma\delta}$ with $E\subset B$ and $\mu^*(B\setminus E)=0$.
c. If $\mu_0$ is $\sigma$-finite, the restriction $\mu^*(E)<\infty$ in (b) is superfluous.
Now there is a slight twist to part b).
Let $\mu^*(E) < \infty$, then I want to show that $E$ is $\mu^*$-measurable iff there exists $\mathcal{A}_{\sigma \delta}$ and $C \in \mathcal{A}_{\delta \sigma }$ with $ C \subset E \subset B$ such that $\mu^*(B / C)=0$.
I've proven part b). But for the forward direction, I'm not sure how to construct such $C \in \mathcal{A}_{\delta \sigma }$.
I thought about using the fact that we have $\mathcal{A}_{\delta \sigma } \subset M(\mathcal{A}) \subset M^*$, where $ M(\mathcal{A})$ is the $\sigma$-algebra generated by $\mathcal{A}$, and $M^*$(a $\sigma$-algebra) is the set of $\mu^* $-measurable sets.
But that didn't me get anywhere.
Thanks in advance!
Best Answer
You have already proven that:
Now, if $\mu^*(E)<\infty$ and $E$ is $\mu^*$-measurable, take $B\in\mathcal{A}_{\sigma\delta}$ with $E\subset B$ and $\mu^*(B\setminus E)=0$.
Since $\mu^*(E)<\infty$, there is $B_1\in\mathcal{A}_{\sigma}$ with $E\subset B_1$ and $\mu^*(E) \leqslant \mu^*(B_1) \leqslant \mu^*(E)+1$.
Consider $B_1 \setminus E$. It is easy to see that $\mu^*(B_1 \setminus E)<\infty$ and $B_1 \setminus E$ is $\mu^*$-measurable. So applying item b again we have, that there is $D\in\mathcal{A}_{\sigma\delta}$ with $B_1 \setminus E\subset D$ and $\mu^*(D\setminus (B_1 \setminus E))=0$. Note that $$ D\setminus (B_1 \setminus E)=D \cap (B_1 \cap E^c)^c= (D\cap B_1^{\phantom{1}c})\cup (D\cap E) $$ So, from $\mu^*(D\setminus (B_1 \setminus E))=0$, we can conclude that $\mu^*(D\cap E)=0$.
Take $C = E \setminus D$. Since $B_1 \setminus E\subset D$, we have that $C= E \setminus D= B_1 \setminus D$. It is immediate that $C \in \mathcal{A}_{\delta \sigma }$ and $ C \subset E$. So, we have $ C \subset E \subset B$ and
$$B\setminus C= (B\setminus E)\cup (E\setminus C)=(B\setminus E)\cup (E\setminus (E \setminus D))= (B\setminus E) \cup (E\cap D) $$ So $$ \mu^*(B\setminus C)\leqslant \mu^*(B\setminus E)+ \mu^*(E\cap D)=0+0=0$$