[Math] Let $E$ have finite outer measure. Prove that there exist a $G_{ \delta}$ set $G$ with $E \subseteq G$

Question

Let $E$ have finite outer measure. Prove that there exist a $G_{\delta} \ni G$ with $E \subseteq G$ and $m^*(E) = m^*(G)$

Proof:

Since $m^*(E) < \infty$, then $\forall n \in \mathbb{N}(\exists\mathcal{O}_n \supset E )$, $\mathcal{O}_n$ open such that:
$$ m^*(E) + \frac{1}{n} > m^*(\mathcal{O}_n).$$
Define $G = \bigcap_{k\in \mathbb{N}}\mathcal{O}_n$, thus $G \in G_{\delta}$ and $E \subset G \Rightarrow m^*(E) \leq m^*(G)$.On the other hand, for a fixed $n \in \mathbb{N}(G \subset \mathcal{O}_n)$, thus:
$$m^*(G) \leq m^*(\mathcal{O}_n) < m^*(E) + \frac{1}{n}.$$
Since it works for any $n > 0$, it works when $n \rightarrow \infty$ and thus $$m^*(G) < m^*(E),$$
as required.

If this proof correct? If the proof correct can someone explain the following argument.

Since $m^*(E) < \infty$, then $\forall n \in \mathbb{N}(\exists\mathcal{O}_n \supset E )$ and $\mathcal{O}_n$ is open.

Answer 1

A correct and rigorous proof of this statement you can find it in Royden's Real Analysis in chapter 2 theorem 11. The argument you are asking is just almost the proper definition of the Lebesgue outer measure, remember that outer measure if defined as the inf{of possible outer coverings by open sets}, then all follows from the definition of inf, the monotonicity of Lebesgue outer measure and the excision property of measurable sets.