Let $\mu:\mathcal A\to[0,\infty]$ be a measure, where $\mathcal A$ is an algebra and let $\mu^*:\mathcal P(X)\to[0,\infty]$ be the outer measure generated by $\mu$. $\Big($i.e. $\mu^*(E)=\inf\Big\{\sum_{i=1}^\infty \mu(A_i):E\subset \bigcup_{i=1}^\infty A_i,\;\ A_i\in\mathcal A\;\ \forall i\in\Bbb N\Big\}$ $\Big)$
Let $E\in\mathcal A^*$ with $\mu^*(E)<\infty$, so it follows that: $$\forall \epsilon>0\;\ \exists\ A_{\epsilon}\in \mathcal A\;\text{such that}\;\ \mu^*(E\triangle A_{\epsilon})<\epsilon $$
$\big($where $\mathcal A^*=\big\{E\subset X: \mu^*(B)=\mu^*(B\cap E)+\mu^*(B\cap E^C)\;\ \forall B\subset X\big\}$$\big)$
So I started:
Let $\epsilon>0$, so since I have to find or construct some set $A_{\epsilon}\in\mathcal A$ such that $\mu^*(E\triangle A_{\epsilon})<\epsilon$ I managed to get that: (looking for $A_{\epsilon}$)
$$\mu^*(E\triangle A_{\epsilon})=\mu^*(E\setminus A_{\epsilon}\ \cup\ A_{\epsilon}\setminus E)\le \mu^*(E\setminus A_{\epsilon})+ \mu^*(A_{\epsilon}\setminus E)\le \mu^*(E)+ \mu^*(A_{\epsilon}\setminus E)=\mu^*(E)+ \mu^*(A_{\epsilon})-\mu^*(E)=\mu^*(A_{\epsilon})$$
$$\Rightarrow\;\ \mu^*(E\triangle A_{\epsilon})\le \mu^*(A_{\epsilon})$$ And got stuck here since I think I have to construct $A_{\epsilon}\in \mathcal A$ in such a way that $\mu(A_{\epsilon})<\epsilon\;$(since $\mu^*(A_{\epsilon})=\mu(A_{\epsilon})$ by hypothesis), but can't figure out how. Any idea would be appreciated.
So, since $\epsilon>0$ we can get $(A_i)_{i\in\Bbb N}$ an $\mathcal A-$cover of $E$ such that:
$$\sum_{i=1}^\infty \mu(A_i)<\mu^*(E)+\frac{\epsilon}{2}$$
and it's clear that $\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)$, so we get that:
$$\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)<\mu^*(E)+\frac{\epsilon}{2}$$
thus $$0\le \sum_{i=1}^\infty \mu(A_i)-\mu^*(E)<\frac{\epsilon}{2}$$
So, $\exists\;N\in\Bbb N$ such that $\sum_{i=N+1}^\infty\mu(A_i)<\frac{\epsilon}{2}$, so let $A=\bigcup_{i=1}^N A_i\in\mathcal A\Rightarrow\ E\setminus A \subset \bigcup_{i=N+1}^\infty A_i$
$$\Rightarrow \mu^*(E\setminus A)\le \sum_{i=N+1}^\infty \mu(A_i)<\frac{\epsilon}{2}$$
Then, $$\mu^*(A\setminus E)=\mu^*(\bigcup_{i=1}^N A_i\setminus E)\le\mu^*(\bigcup_{i=1}^\infty A_i\setminus E)=\mu^*(\bigcup_{i=1}^\infty A_i)-\mu^*(E)\le \sum_{i=1}^\infty \mu(A_i)-\mu^*(E)<\frac{\epsilon}{2}$$
thus $$\mu^*(E\triangle A)\le\mu^*(E\setminus A)+\mu^*(A\setminus E)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
What do you think?