How does one show that for any $E \subset \mathbb{R}^n$, there is a $G \in G_\delta$ such that $E \subset G$ and $m_* (E) = m_*(G)$?
$m_*$ stands for outer measure and $G_\delta $ set is re sets that are the countable intersection of open sets.
I was thinking since we don't know anything about $E$, we could look at its $int(E)$ which is open and maybe construct something along the lines of $E \subset \bigcap_{r} int(E) \cup B_{r}(0)$.
I am mostly having trouble coming up the reasons for the existence of $G$.
EDIT: When I wrote this answer I mistakenly thought the question was about outer measure on $\Bbb R$, not $\Bbb R^n$. Things are not quite so trivial in $\Bbb R^n$. There will be a version for $\Bbb R^n$ appearing here when I have time. Thanks to Nate Eldrege for pointing out the problem.
This is obvious. Say $m_*(E)=\alpha$. By definition $\alpha$ is the infimum of the measure of $V$ such that $V$ is an open set with $E\subset V$. Hence there exist open sets $V_n$ with $E\subset V_n$ and $m(V_n)\to\alpha$.
Let $G=\bigcap V_n$. Then $G\subset V_n$ for every $n$, so $m_*(G)\le m_*(V_n)$ for every $n$, hence $m_*(G)\le\lim m_*(V_n)= \alpha$. On the other hand, $E\subset G$ so $\alpha=m_*(E)\le m_*(G)$.