This comes from an exercise from Real Analysis by Folland.
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
proof 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 $\{ B_{j}^{k} \} _{k=1}^{\infty}\subset \mathcal{A}$ 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 $$\mu^{*}(C_n\setminus E_n) = \mu^{*}(C_n) - \mu^{*}(E_n) \leq \epsilon/2^{n}$$ $$\mu^{*}(E_n) = \mu^{*}(E\cap \bigcup_{1}^{n}X_i) = \mu^{*}(E) \cap \mu^{*}(\bigcup_{1}^{n}X_i) = \infty$$ hence $\mu^{*}(C_n\setminus E_n) = 0$
I am not sure where to go from here, any hints or suggestions is greatly appreciated