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

Given: A natural number $n\ge 1$ and the Axiom of Countable Choice.

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra on ${\mathbb{R}}^n$ and every elementary set is Lebesgue measurable (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).

$\sigma ({\mathcal{H}}_n)=\sigma ({\mathcal{E}}_n)=\mathcal{B}({\mathbb{R}}^n)$, where ${\mathcal{H}}_n$ is the family of half-open boxes and ${\mathcal{E}}_n$ the family of elementary sets (The sigma-algebra generated by the half-open boxes of ${\mathbb{R}}^n$ is the Borel sigma-algebra).

At $m=1$ every half-open box is elementary (Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$).

${\sigma }_X(\mathcal{E})$ is the unique smallest sigma-algebra on $X$ containing $\mathcal{E}$ (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal, The sigma-algebra generated by a family of sets).

The Borel sigma-algebra of $X$ is the sigma-algebra generated by its open sets (The Borel sigma-algebra of a topological space); a sigma-algebra on $X$ is an algebra of subsets closed under countable unions (Sigma-algebras).

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 (${\mathrm{A}\mathrm{C}}_{\omega }$)).