Approximating a set in outer measure Given $\mu$ is a measure on a field $\mathcal{F}$ of subsets of $\Omega$, $\mu^\star$ is the corresponding outer measure. Then, for any $E\subset \Omega$, $\mu^\star(E)=\text{inf}\{\sum \mu A_i:E\subset\bigcup A_i, A_i\in\mathcal{F}\}$.
I have to show that for any $E\subset\Omega$ with finite outer measure and for any given $\epsilon>0$, there is a $F\in \mathcal{F}$, such that $\mu^\star(E\Delta F) < \epsilon$
So far I have that, $E\subset \bigcup A_i$ such that $\mu^\star(E) \le \sum \mu A_i < \mu^\star(E) + \frac{\epsilon}{2}$. Then there is a $N\in\mathbb{N}$, such that $\sum_{i>N}\mu A_i < \frac{\epsilon}{2}$. Let $F=\sum_{i=1}^{N}\mu A_i$. Now, $E\setminus F\subset \bigcup_{i>N}A_i$ and so $\mu^\star(E\setminus F)<\frac{\epsilon}{2}$.
On the other hand, $F\setminus E\subset \bigcup A_i\setminus E$. But since I have an outer measure, I cannot obtain any suitable inequality.
Is my approach correct? Can someone point me in the right direction?! Any help regarding this is appreciated.
 A: Disclaimer: There are surely shorther proofs of what I do below, but here goes:
Your claim only holds if $E$ is $\mu^{\ast}$-measurable. To see
this, assume that your claim holds for $E$. Then for every $n\in\mathbb{N}$,
there is $F_{n}\in\mathcal{F}$ (which is in particular $\mu^{\ast}$-measurable
since $\mu$ is assumed to be a (pre)measure(?!) on $\mathcal{F}$) with
$$
\mu^{\ast}\left(E\Delta F_{n}\right)<\frac{1}{2^{n}}.
$$
Then, for $k\in\mathbb{N}$, we have
\begin{eqnarray*}
E\Delta\left(\bigcup_{n=k}^{\infty}F_{n}\right) & = & \left(E\setminus\bigcup_{n=k}^{\infty}F_{n}\right)\cup\left[\left(\bigcup_{n=k}^{\infty}F_{n}\right)\setminus E\right]\\
 & = & \left(E\cap\bigcap_{n=k}^{\infty}F_{n}^{c}\right)\cup\bigcup_{n=k}^{\infty}\left(F_{n}\setminus E\right)\\
 & = & \bigcap_{n=k}^{\infty}\left(E\cap F_{n}^{c}\right)\cup\bigcup_{n=k}^{\infty}\left(F_{n}\setminus E\right)\\
 & \subset & \bigcup_{n=k}^{\infty}\left(E\setminus F_{n}\right)\cup\left(F_{n}\setminus E\right)\\
 & = & \bigcup_{n=k}^{\infty}\left(E\Delta F_{n}\right)
\end{eqnarray*}
and thus
$$
\mu^{\ast}\left(E\Delta\bigcup_{n=k}^{\infty}F_{n}\right)\leq\sum_{n=k}^{\infty}\left(E\Delta F_{n}\right)\leq\sum_{n=k}^{\infty}2^{-n}=2^{1-k}.
$$
Let us set $E_{k}:=\bigcup_{n=k}^{\infty}F_{n}$ and note $E_{k+1}\subset E_{k}$
for all $n$. Thus, we have $F:=\bigcap_{k=1}^{\infty}E_{k}=\bigcap_{k=k_{0}}^{\infty}E_{k}$
for all $k_{0}\in\mathbb{N}$. Note that $F$ is $\mu^{\ast}$-measurable.
But
\begin{eqnarray*}
E\Delta F & = & E\Delta\bigcap_{k=k_{0}}^{\infty}E_{k}\\
 & = & \left(E\setminus\bigcap_{k=k_{0}}^{\infty}E_{k}\right)\cup\left[\left(\bigcap_{k=k_{0}}^{\infty}E_{k}\right)\setminus E\right]\\
 & = & \left(E\cap\bigcup_{k=k_{0}}^{\infty}E_{k}^{c}\right)\cup\bigcap_{k=k_{0}}^{\infty}\left(E_{k}\setminus E\right)\\
 & \subset & \bigcup_{k=k_{0}}^{\infty}\left[\left(E\setminus E_{k}\right)\cup\left(E_{k}\setminus E\right)\right]\\
 & = & \bigcup_{k=k_{0}}^{\infty}\left(E\Delta E_{k}\right)
\end{eqnarray*}
and hence
$$
\mu^{\ast}\left(E\Delta F\right)\leq\sum_{k=k_{0}}^{\infty}\mu^{\ast}\left(E\Delta E_{k}\right)\leq\sum_{k=k_{0}}^{\infty}2^{1-k}=2^{2-k_{0}}\xrightarrow[k_{0}\to\infty]{}0,
$$
i.e. $\mu^{\ast}\left(E\Delta F\right)=0$.
This entails
$$
\mu^{\ast}\left(E\setminus F\right)=\mu^{\ast}\left(F\setminus E\right)=0,
$$
since $E\setminus F,F\setminus E\subset E\Delta F$. In particular,
these sets are measurable (since by the Caratheodory construction, $\mu^\ast$ is a complete measure on the class of measurable sets). But then
$$
E\cap F=F\setminus\left(F\setminus E\right)
$$
is measurable, so that also
$$
E=\left(E\cap F\right)\cup\left(E\setminus F\right)
$$
is measurable.
Thus, your claim will fail for any nonmeasurable $E$, since if it is true, we just saw that $E$ has to be measurable.
