A subset of ${\mathbb{R}}^n$ with open supersets of arbitrarily small excess is Lebesgue measurable

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, and a set $E\subseteq {\mathbb{R}}^n$ admitting open supersets of arbitrarily small outer excess.

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra, ${\lambda }_n$ is a complete measure on it, and every $S\subseteq {\mathbb{R}}^n$ with ${\lambda }_n^{\ast }(S)=0$ is Lebesgue measurable with ${\lambda }_n(S)=0$ (Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra containing every elementary set and ${\lambda }_n$ is a complete measure extending elementary volume).

Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable (Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable).

Assuming countable choice, ${\lambda }_n^{\ast }$ is an outer measure on ${\mathbb{R}}^n$, hence monotone (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Outer measures).

$A$ is a $G_{\delta }$ set of $X$ when there is a sequence $(V_n{)}_{n\in \mathbb{N}}$ of open subsets of $X$ with $A={\bigcap }_{n\in \mathbb{N}}{V}_n$ ($G_{\delta }$ and $F_{\sigma }$ subsets of a topological space, agreeing with the real-line notion).

The product topology on ${\mathbb{R}}^n$ is the metric topology of $d_2$ (A subset of ${\mathbb{R}}^n$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology, claim 1; The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

For every real $\epsilon >0$ there is a natural number $k\ge 1$ with $1\mathrm{/}k<\epsilon$ (For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$).

The Axiom of Countable Choice says that for every family $(X_n{)}_{n\in \mathbb{N}}$ of nonempty sets indexed by $\mathbb{N}$ there is a function $f$ with domain $\mathbb{N}$ such that $f(n)\in X_n$ for every $n\in \mathbb{N}$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).