Condition for $\mu^*$-measurable set 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. $\mu^*(E)=\inf\Big\{\sum_{i=1}^\infty \mu(A_i):E\subset \bigcup_{i=1}^\infty A_i,\;\ A_i\in\mathcal A\;\ \forall i\in\Bbb N\Big\}$ $\Big)$
Lets suppose that $\forall E\subset X$ and $\forall\epsilon>0\;\ \exists\ A\in\mathcal A$ s.t. $\mu^*(E\mathbin\triangle A)<\epsilon$, so I'm trying to prove that:
$$E\in\mathcal A^*$$
Proof: (What I've got)
So, since $\mathcal A^*=\{E\subset X:\mu^*(B)=\mu^*(B\setminus E)+\mu^*(B\cap E)\ \forall B\subset X\}$ is a $\sigma-$algebra and if $\mu^*(F)=0\Rightarrow F\in\mathcal A^*$, it's sufficient to prove that:
$$\mu^*(E)<\epsilon,\;\ \forall \epsilon>0$$
Thereby, let $E\subset X$ and $\epsilon>0$, so there$\;\exists\ A\in\mathcal A$ s.t. $\mu^*(E\mathbin\triangle A)<\frac{\epsilon}{2}$, but
$$E\setminus A\subset E\mathbin\triangle A\\
A\setminus E\subset E\mathbin\triangle A$$
so, since an outer measure is monotone,
$$\mu^*(E\setminus A)<\frac{\epsilon}{2}\\
\mu^*(A\setminus E)<\frac{\epsilon}{2}$$
and
$$\mu^*(E)=\mu^*(E\setminus A\ \cup\ E\cap A)\le \mu^*(E\setminus A)+\mu^*(E\cap A)<\frac{\epsilon}{2} + \mu^*(E\cap A)$$
so I need to show that $\mu^*(E\cap A)<\frac{\epsilon}{2}$, but got stuck here since I have tried some set rewriting and the only "usefull" thing I got so far is that $\mu^*(E\cap A)\le \mu^*(A)$ which doesn`t tell me much. And $\mathcal A-$covers of $E$ don't think could help or I can't see how. Any ideas would be appreciated.
 A: It is obvious you went wrong here:

$$\mu^*(E)<\epsilon,\;\ \forall \epsilon>0$$

$E$ is a general set. But here suddenly you are trying to prove that its outer measure is $0$. This statement is obviously false in general. If every set had measure $0$, then there would be no point to measure theory.
The reason you can't show that $\mu^*(E\cap A)<\frac{\epsilon}{2}$ is because it just isn't true for most sets $E$.

Edit:
First show that for $A \in \mathcal A$, $\mu^*(B)=\mu^*(B\setminus A)+\mu^*(B\cap A)$ from the definition of $\mu^*$. Then use the fact that $\mu^*(E\triangle A)<\epsilon$ to show the same for $E$.
A: We only need to prove that 
$$
μ^∗(B)\geqslant μ^∗(B\setminus E)+μ^∗(B∩E)\tag1
$$
for 
$$
μ^∗(B)\leqslant μ^∗(B\setminus E)+μ^∗(B∩E)
$$
follows from subadditivity of outer measure. 
Since $A$ is measurable
$$
μ^∗(B)=μ^∗(B\setminus A)+μ^∗(B∩A)\tag2
$$
Since $B=(B\setminus A)\cup (B∩A)$
$$
B\setminus E=(B\setminus A)\cup (B∩A)\setminus E=((B\setminus A)-E)\cup ((B∩A)\setminus E)\subset (B\setminus A)\cup (A\setminus E)
$$
So
$$
μ^∗(B\setminus E)\leqslant μ^∗(B\setminus A)+μ^∗(A\setminus E)\tag3
$$
Moreover 
$$
B∩E=((B\setminus A)\cup (B∩A))\cap E=((B\setminus A)\cap E)\cup((B∩A)\cap E)\subset (E\setminus A)\cup(B∩A)
$$
So
$$
μ^∗(B\cap E)\leqslant μ^∗(E\setminus A)+μ^∗(A\cap B)\tag4
$$
Since $\mu^*(E\triangle A)<\frac{\epsilon}{2}$, as well as $A\setminus E\subset E\triangle A$ and $E\setminus A\subset E\triangle A$, we have
$$
\mu^*(A\setminus E)\leqslant \mu^*(E\triangle A)<\frac{\epsilon}{2}\quad\text{and}\quad\mu^*(E\setminus A)\leqslant \mu^*(E\triangle A)<\frac{\epsilon}{2}\tag5
$$
Add $(3)$ and $(4)$, we have
\begin{align}
μ^∗(B\setminus E)+μ^∗(B\cap E)&\leqslant μ^∗(B\setminus A)+μ^∗(A\setminus E)+μ^∗(E\setminus A)+μ^∗(A\cap B)
\\
&=μ^∗(B)+μ^∗(A\setminus E)+μ^∗(E\setminus A)\tag{by (2)}
\\
&<μ^∗(B)+\epsilon\tag{by (5)}
\end{align}
Since $\epsilon$ is arbitrary small, $(1)$ is proved.
A: Ok so continuing with Paul Sinclair's idea:
Since we know that $\forall\;\epsilon>0\;\ \exists\ A\in\mathcal A\;\ \mu^*(E\triangle\ A)<\epsilon,\;\ $ let $\epsilon>0$ and take $A\in\mathcal A$ s.t. $\mu^*(E\triangle A)< \frac{\epsilon}{2}$ and we get that:
$$\mu^*(E\setminus A)\le \mu^*(E\triangle A)< \frac{\epsilon}{2}\\
\mu^*(A\setminus E)\le \mu^*(E\triangle A)< \frac{\epsilon}{2}$$
Then, since $A\in\mathcal A\subset\mathcal A^*$, we get that: 
$$\mu^*(B)=\mu^*(B\setminus A)+\mu^*(B\cap A)\;\ \forall B\subset X$$
So, since $(B\setminus E), (B\cap E)\subset X$, then:
$$\mu^*(B\setminus E)=\mu^*((B\setminus E)\setminus A)+\mu^*((B\setminus E)\cap A)\;\ \forall B\subset X\\
\mu^*(B\cap E)=\mu^*((B\cap E)\setminus A)+\mu^*((B\cap E)\cap A)\;\ \forall B\subset X$$
where ($\ \forall B\subset X\;$) is clear that:
$$(B\setminus E)\setminus A=B\cap E^c\cap A^c\subset B\cap A^c=B\setminus A\\
(B\setminus E)\cap A=B\cap E^c\cap A\subset A\cap E^c=A\setminus E\\
(B\cap E)\setminus A=B\cap E\cap A^c\subset E\cap A^c=E\setminus A\\
(B\cap E)\cap A=B\cap E\cap A\subset B\cap A$$
thus
$$\mu^*(B\setminus E)+\mu^*(B\cap E)\le\mu^*(B\setminus A)+\mu^*(A\setminus E)+\mu^*(E\setminus A)+\mu^*(B\cap A)\;\ \forall B\subset X\\
\Rightarrow \mu^*(B\setminus E)+\mu^*(B\cap E)<\mu^*(B\setminus A)+\frac{\epsilon}{2}+\frac{\epsilon}{2}+\mu^*(B\cap A)=\mu^*(B)+\epsilon\;\ \forall B\subset X\;\ \forall\epsilon>0\\
 \\
\Rightarrow\ \mu^*(B\setminus E)+\mu^*(B\cap E)<\mu^*(B)+\epsilon\;\ \forall B\subset X\;\ \forall\epsilon>0$$
so
$$\mu^*(B)\ge \mu^*(B\setminus E)+\mu^*(B\cap E)\;\ \forall B\subset X$$
thus $E\in\mathcal A^*$
