# Given a bounded set $E$, there is a $G_\delta$ set such that the outer measures are equal

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?

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You should denote $I_{n,j}$ the collection of intervals which covers $E$ and approach up to $2^{-n}$ the outer measure of $E$ (which is finite). Here you take $n$ as an index, but we see $n$ outside the sum, and we don't know what it is. But you saw the idea and you just have to take now $G:=\bigcap_{j\geq 1}\bigcup_{n\geq 1}I_{n,j}$. –  Davide Giraudo Sep 24 '12 at 7:10

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