Question: Show that for any bounded set $E \in \mathbb{R}$, there is a $G_\delta$ set $G$ for which $E \subseteq G$ and $m^*(E)=m^*(G)$.
Let $\{I_n\}$ be a countable collection of open intervals such that $E \subset \bigcup\limits_{n=1}^{\infty} I_n$. Observe the following: $m^*(E) \leq m^*(\bigcup\limits_{n=1}^{\infty} I_n) \leq \sum\limits_{n=1}^\infty l(I_n) < m^*(E)+\frac{1}{2^n}.$
Now let $G=\bigcap\bigcup I_n$. $G$ is a $G_\delta$ set. From this, we see that $E \subset G$. Thus for each $n \in \mathbb{N}$: $m^*(E) \leq m^*(G) \leq m^*(\bigcup\limits_{n=1}^{\infty} I_n) \leq \sum\limits_{n=1}^\infty l(I_n) < m^*(E)+\frac{1}{2^n}.$
So we take $n \rightarrow \infty$.
Is my proof correct?