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?
Best Answer
For each integer $n\geq 1$, let $\{I_{n,j}\}$ be a countable collection of open intervals which covers $E$ such that $$\sum_{j=1}^{+\infty}m^*(I_{n,j})\leqslant m^*(E)+n^{-1}.$$ Such a collection exists because the outer measure of $E$ is finite, as $E$ is a bounded subset of $\Bbb R$.
Let $G:=\bigcap_{n\geqslant 1}\bigcup_{j\geq 1}I_{n,j}$. As $\bigcup_{j\geqslant 1}I_{n,j}$ is open for each $n$, $G$ is a countable intersection of open sets, and it contains $E$. For $n\geqslant 1$, the following inequalities hold $$m^*(G)\leqslant m^*\left(\bigcup_{j=1}^{+\infty}I_{n,j}\right) \leqslant\sum_{j=1}^{+\infty}m^*\left(I_{n,j}\right) \leqslant m^*(E)+n^{-1},$$ which concludes the proof.