Outer measure of outer measure

I have a homework question, and I'm totally confused... I would appreciate every help or a clue.

Let $S$ be an algebra of subsets of a set $X$, and let $\mu$ be a pre-measure on $S$.

We can construct an outer measure $\mu^*$, as shown here:

$$\mu^* \left({E}\right) = \inf \ \left\{{\sum_{n=0}^\infty \mu\left({A_n}\right) : A_n \in S, \ E \subseteq \bigcup_{n=0}^\infty A_n}\right\}.$$

Now, we can look at the $\sigma$-algebra of all $\mu^*$-measurable sets, let's denote it $M$.

The outer measure $\mu^*$ is then a measure on $M$.

Now, we can take this measure on $\mu^*|_M$, and again construct an outer measure:

$$\mu^{**} \left({E}\right) = \inf \ \left\{{\sum_{n=0}^\infty \mu^*\left({A_n}\right) : A_n \in M, \ E \subseteq \bigcup_{n=0}^\infty A_n}\right\}.$$

I have to prove that $\mu^*(E)=\mu^{**}(E)$ for every $E\subseteq X$.

Solution 1:

I think the following works but please check.

First, $S\subset M$ so $\mu^{**}(E)\leq\mu^{*}(E)$. On the other hand, given $\varepsilon>0$, there exist $A_{n}\in M$ such that $E\subset\bigcup A_{n}$ and $\sum\mu^{*}(A_{n})<\mu^{**}(E)+\varepsilon$. Then you can see why $\mu^{*}(E)\leq\mu^{**}(E)$.