Background information:
Exercise 18 – Let $\mathcal{A}\subset P(X)$ 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 in $\mathcal{A}_\sigma$. Let $\mu_{0}$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced 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 if and only if there exists $B\in A_{\sigma\delta}$ 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
a.) By the definition of outermeasure 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 $A = \bigcup_{j=1}^{\infty} A_{j}$ as above. Then $A \in \mathcal{A}_{\sigma}$ and $E \subset A$. For each $j$ we can construct a sequence $\{ B_{j}^{k} \} _{k=1}^{\infty}\subset A_\sigma$ such that $A_j\subset \bigcup_{j,k=1}^{\infty}B_{j}^{k}$. It follows that since $$\mu^{*}(A_j) = \inf\left\{ \sum_{j=1}^{\infty} \mu_{0} (B_{j}^{k} ) : B_{j}^{k} \in \mathcal{A}, A_j \subset \bigcup_{j,k=1}^{\infty} B_{j}^{k} \right \}$$ We have that, $$\mu^{*}(A_{j}) \leq \mu_{0}(A_{j}) + \epsilon 2^{-j}, \forall j \ \ \text{and} \ \ \epsilon>0$$ Thus: $$\mu^{*}(A) \leq \sum_{j=1}^{\infty} \mu^{*} ( A_{j} ) \leq \sum_{j=1}^{\infty} ( \mu_{0}(A_{j}) + \epsilon 2^{-j}) = \mu^{*}(E) + \epsilon$$ Since $\epsilon$ is arbitrary we are done.
b.) Suppose $E$ is $\mu^{*}$-measurable. From part (a), we know that $\forall n\in\mathbb{N}$ there exists $B_n\in A_\sigma$ with $E\subset B_n$ and $\mu^{*}(B_n) – \mu^{*}(E) \leq 1/n$. Let, $$B = \bigcap_{n = 1}^{\infty}B_n\in A_\sigma$$ since E is $\mu^{*}$-measurable, we have $\mu^{*}(B_n) = \mu^{*}(B_n\cap E) + \mu^{*}(B_n\cap E^{c})$ hence, $$\mu^{*}(B\cap E^{c})\leq \mu^{*}(B_n\cap E^{c})= \mu^{*}(B_n) – \mu^{*}(E)\leq 1/n$$ for every $n\in\mathbb{N}$. Hence we have $\mu^{*}(B\setminus E) = 0$\ To show the converse, let's suppose $B\in A_{\sigma\delta}$ with $E\subset B$ and $\mu^{*}(B\setminus E) = 0$. From part (a), we know that $\forall n\in\mathbb{N}$ there exists $A_n\in A_\sigma$ with $(B\setminus E)\subset A_n$ and $\mu^{*}(A_n) – \mu^{*}(B\setminus E)\leq 1/n$. But, since $\mu^{*}(B\setminus E) = 0$ then, $\mu^{*}(A_n)\leq 1/n$. Let, $$A = \bigcap_{n=1}^{\infty}A_n$$ then A is $\mu^{*}$-measurable (since $A\in A_{\sigma\delta}$ and the set of all $\mu^{*}$-measurable sets is a $\sigma$-algebra) such that $(B\setminus E)\subset A$ and $\mu^{*}(A) = 0$.\ By Carathedors's theorem we know that the restriction of $\mu^{*}$ to $\mu^{*}$-measurable sets is a complete measure. From this, we know that $(B\setminus E)$ is $\mu^{*}$-measurable. Also, since $B\in A_{\sigma\delta}$ then $B$ is also $\mu^{*}$-measurable and we can express $E$ as $$E = (B^{c}\cup (B\cap E^{c}))^{c}$$ Thus $E$ is $\mu^{*}$-measurable.
c.) Let $\mu_0$ be $\sigma$-finite, then let $$X = \bigcup_{1}^{\infty}X_i$$ where $X_i\in M$ and $\mu(X_i) < \infty$ Now, suppose $E$ is $\mu^{*}$-measurable and $\mu^{*}(E) = \infty$, set $$E_n = (E\cap \bigcup_{1}^{n}X_i)$$ then $\mu^{*}(E_n) < \infty$ and $E = \bigcup_{1}^{\infty}E_n$. Let $\epsilon > 0$, from part (a) $\forall n\in\mathbb{N} \exists C_n\in A_\sigma$ such that $E_n\subset C_n$ and $$\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) – \mu^{*}(E_n) \leq \epsilon/2^{-n}$$ Let $$C_{\epsilon} = \bigcup_{1}^{\infty}C_n$$ The $\epsilon$ in $C_\epsilon$ is there to stress the dependence upon $\epsilon$ of $C_n$'s and hence of $C$. Notice that, $C_\epsilon\in A_\delta$ being the countable union of unions of elements of $\mathcal{A}$ Intuitively, $C_\epsilon$ is a decent approximation pf $E$: $$\mu^{*}(C_\epsilon\cap E^{c}) = \mu^{*}\left(\bigcup_{1}^{\infty}(C_n\cap E^{c})\right) = \mu^{*}\left(\bigcup_{1}^{\infty}(C_n\cap E_n^{c})\right) \leq \sum_{1}^{\infty}\mu^{*}(C_n\setminus E_n) \leq \epsilon$$ Let's go back to our intuition: $C_{\epsilon}$ is an $\epsilon$-good approximation of $E$, therefore we would like to send $\epsilon$ to $0$ to get the best possible approximation. Since we want to end up in $\mathcal{A}_{\sigma \delta}$ what we need to do is to "discretize" the sets $C_{\epsilon}$ considering instead the sets $C_{\frac{1}{n}}$.Then clearly $$C := \bigcap_{n = 1}^{\infty}C_{\frac{1}{n}} \in \mathcal{A}_{\sigma \delta}$$ and satisfies $\mu^*(C \setminus E) = 0$, proving the claim. To show this last equality notice that $$\mu^*(C \setminus E) \le \mu^*(C_{\frac{1}{n}} \setminus E) \le \frac{1}{n} \to 0.$$
Exercise 19 – Let $\mu^*$ be an outer measure on $X$ induced from a finite premeasure $\mu_0$. If $E\subset X$, define the inner measure of $E$ to be $\mu_*(E) = \mu_0(X) – \mu^*(E^c)$. Then $E$ is $\mu^*$-measurable iff $\mu^*(E) = \mu_*(E)$ (Use Exercise 18).
Since $E\supset X$ then $E$ is $\mu^*$-measurable says $$\mu^*(X) = \mu^*(E) + \mu^*(E\cap X^c)$$ then $$\mu^*(E) = \mu^*(X) – \mu^*(E\cap X^c)$$ As you can see we have that $\mu^*(E)$ which is the outer measure of $E$ is equal to the inner measure of $E$. But I am not sure if this what we need for the $\Rightarrow$ part of the proof since the inner measure of $E$ is defined in a different way.
Best Answer
We begin with a small lemma that in fact highlights part of what has already been proved in Exercise 18 item b.).
Proof From Exercise 18 item a.) we know that
For any $E\subset X$ and $\epsilon > 0$ there exists $A\in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(E)\leq \mu^*(A) \leq \mu^*(E) + \epsilon$.
So, for each $n\in\mathbb{N}$, $n>0$, let $A_n\in \mathcal{A}_\sigma$ 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 $B\in \mathcal{A}_{\sigma\delta}$ 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.
First note that since $\mu_0$ is a finite premeasure, we have that for all $A\subset X$, $\mu^*(A)<+\infty$.
($\Rightarrow$) Suppose $E\subset X$. Note that $E^c=X\setminus E \subset X$. So,since $E$ is $\mu^*$-measurable, we have $$\mu^*(X) = \mu^*(X\cap E) + \mu^*(X\cap E^c)=\mu^*(E) + \mu^*(E^c)$$ then, since $\mu^*(E^c)<+\infty$ and $\mu^*(X)=\mu_0(X)$, we have $$\mu^*(E) = \mu^*(X) - \mu^*(E^c)=\mu_0(X)- \mu^*(E^c)=\mu_*(E)$$
($\Leftarrow$) If $\mu^*(E) =\mu_*(E)$ then we have, since $\mu^*(X)=\mu_0(X)$, $$\mu^*(E) =\mu_*(E)=\mu_0(X)- \mu^*(E^c) = \mu^*(X) - \mu^*(E^c)$$ So we can conclude that $$\mu^*(X) = \mu^*(E)+ \mu^*(E^c) \tag{1}$$ Now we apply our lemma to $E$ and $E^c$. Let $B, D \in \mathcal{A}_{\sigma\delta}$ such that $E\subset B$ and $\mu^*(B) = \mu^*(E)$ and $E^c\subset D$ and $\mu^*(D) = \mu^*(E^c)$. From $(1)$, we have $$\mu^*(X) = \mu^*(B)+ \mu^*(D) \tag{2}$$ On the other hand, since $D \in \mathcal{A}_{\sigma\delta}$, we have that $D$ is $\mu^*$-measurable, so $$ \mu^*(X) = \mu^*(D)+ \mu^*(D^c) \tag{3}$$ From $(2)$ and $(3)$, we get $$ \mu^*(D)+ \mu^*(D^c)= \mu^*(B)+ \mu^*(D) $$ Since $\mu^*(D) <\infty$, we have $$\mu^*(D^c) = \mu^*(B) \tag{4} $$ Note that since $E^c\subset D$ we have that $D^c\subset E$. So we actually have $$D^c\subset E \subset B \tag{5}$$ and, since $D$ are $\mu^*$-measurable, $D^c$ is also $\mu^*$-measurable. So $$\mu^*(B)=\mu^*(D^c) + \mu^*(B\setminus D^c) \tag {6}$$ Since $\mu^*(D^c)<+\infty$ (and $\mu^*(B)<+\infty$), we have from $(4)$ and $(6)$ that $$\mu^*(B\setminus D^c)=0$$ But from $(5)$ we have $B \setminus E\subset B\setminus D^c$, so $$\mu^*(B\setminus E)=0$$ By exercise 18 item b.), $E$ is $\mu^*$-measurable.