Let $\mu$ be a measure on a semi-ring $A$ $\subseteq$ $P(\Omega)$ and let $\mu^*$ be the outer measure generated by $(A, \mu)$. Prove that for any $E$ $\subseteq$ $\Omega$ $\exists$ $B$ $\in$ $\sigma(A)$ such that $E$ $\subseteq$ $B$ and $\mu^*(E)$ $=$ $\mu(B).$
So far I know that $\sigma(A)$ $\subseteq$ $M$ where M is the sigma-field containing all $\mu^*$ measurable subsets of $A$. This means: $\mu^*(E) = \mu^*(E $ $\cap$ $B)$ $+$ $\mu^*(E $ $\cap$ $B^c)$.
I let B be a cover of E such that:
$\mu^*(E)$ $\leq$ $\sum_{n=1}^\infty \mu(B_n)$ $\leq$ $\mu^*(E)$ $+$ $1/m$ $\forall$ $m$ $\in$ $N$
How can I use this to get the result that $\mu^*(E)$ $=$ $\mu(B)$?
Let $E \subset \Omega$. For $n \in \mathbb{N}$ let $B_n = \bigcup_{m=1}^\infty B_m^n$ with $B_m^n \in A$ for each $m \in \mathbb{N}$ and $$ \mu^*(E) \leq \sum_{m = 1}^\infty \mu(B_m^n) + \frac{1}{n}.$$ Since the outer measure is monotonic the set $B:= \bigcap_{n = 1}^\infty B_n \in \sigma(A)$ satisfies for each $n \in \mathbb{N}$ $$\mu^*(E) \leq \mu^*(B) = \mu(B) \leq \sum_{m = 1}^\infty \mu(B_m^n) \leq \mu^*(E) + \frac{1}{n}. $$ As $n$ was arbitrary this yiels $\mu(B) = \mu^*(E)$.