Set containing another set but having same measure

A set of real numbers is said to be a $G_{\delta}$ set provided it is the intersection of a countable collection of open sets. Show that for any bounded set $E$, there is a $G_{\delta}$ set $G$ for which $$E\subseteq G\text{ and }m^*(G)=m^*(E)$$

I want to define $G_n$ to be the the union of open balls of size $1/n$ around every point in $E$, i.e. $G_n=\bigcup_{x\in E}B_{1/n}(x)$. Then take $G=\bigcap_{n\in\mathbb{Z}^+}G_n$

Clearly $E\subseteq G$. Also, $G_n$ is an open set, since it's a union of open balls. Then $G$ is an intersection of a countable collection of open sets.

I need to prove $m^*(G)=m^*(E)$. Since $E\subseteq G$, I know $m^*(G)\geq m^*(E)$. How can I prove they're equal?

As written, this is not the case. Consider the rationals in $[0,1]$. This is measure $0$ but each $G_n$ will have at least measure $1$.
Try directly using the definition of outer measure to get a countable collection of open sets that approximate the measure of $E$.
• I see. Choose $G_n$ to be the open intervals containing $E$ with measure in $[m(E), m(E)+1/n)$. That should do it. – PJ Miller Jun 29 '13 at 3:00