$G_\delta$ where $E\setminus G_\delta$ has postive measure Let $E$ be a non-measurable set of finite outer measure. Show that there is a $G_\delta$ set containing $E$ such that $m^*(G)=m^*(E)$ and $m^*(G \setminus E)>0$.
Attempt
The equality is easy to show since there is an open set containing $E$ where $m^*(O\setminus E) < \varepsilon$. It is getting the other side of the inequality greater than 0 that I'm unsure on.
 A: I don't think you can deduce that there is an open set $O$ such that $m^{\ast}( O\setminus E)<\epsilon$. For the Lebesgue measure on $\mathbb{R}^d$ this can be taken to be the definition of a measurable set.
$\fbox{1}$ However, for every $\epsilon>0$ you can find an open set $O$ such that $O\supset E$ and $m^{\ast}(O)-m^{\ast}(E) <\epsilon$. 
This does not imply what you wrote, because it is not necessarily true that $$m^{\ast}(A\cup B)=m^{\ast}(A)+m^{\ast}( B)$$ for disjoint sets that are not measurable. 
The outer measure is sub-additive, which would imply $ m^{\ast}(O)-m^{\ast}(E)\leq  m^{\ast}(O\setminus E)  $, so you see the inequality is in the wrong way.
However, $\fbox{1}$ is good enough to give you the desired $G_{\delta}$ set. Indeed, for $\epsilon= \frac{1}{n}$ take $O_n$ such that $m^{\ast}(O)-m^{\ast}(E) <\frac{1}{n}$. Then, the set that does the job is $\mathcal{O}=\bigcap_{n\geq 1} O_n$.
For the other equality if $m^{\ast}(\mathcal{O}\setminus E)=0$ then $\mathcal{O}\setminus E$ is measurable. But, then $E=\mathcal{O}\setminus(\mathcal{O}\setminus E) $ a contradiction since all the sets involved are measurable. Therefore, $m^{\ast}(\mathcal{O}\setminus E)>0$.
