Let $E \subseteq \Bbb R^d$. Show that the following are equivalent. 1. $E$ is Lebesgue measurable. 2. $E$ is a $G_\delta$ set with a null set removed.
Let $E$ be Lebesgue measurable. By outer regularity and the axiom of choice we can find a sequence of $(O_n)_n$ such that for all $n$ we have $E \subseteq O_n$ and $m(O_n) - m(E) < \varepsilon.$ By the excision property we have $$m(O_n \setminus E) < \varepsilon.$$
Let $O= \bigcap_{n} O_n$. Then $E \subseteq O$ from where we get that $$m(O\setminus E) \le m(O_n\setminus E) < \varepsilon \implies m(O \setminus E)=0.$$
This was the proof of $1 \implies 2$. I would like to know what is the usage of the axiom of choice here? It seems to justify the fact thath we can find the open sets $O_n$ for $n \in \Bbb N$ such that $E \subseteq O_n$, but I don't quite follow how the conclusion is made.
You know that the sets $M_n=\{O$ open$:m(O∖E)<1/n\}$ are not empty and you have to choose a sequence $(O_n)_{n\in\mathbb N}$ with $O_n\in M_n$ for every n. This is a countable version of the axiom of choice.