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

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, Lebesgue outer measure ${\lambda }_n^{\ast }$, and the family $\mathcal{L}({\mathbb{R}}^n)$ of sets Carathéodory measurable for it.

A set $E\subseteq {\mathbb{R}}^n$ is Lebesgue measurable when it is Carathéodory measurable for ${\lambda }_n^{\ast }$, the family of these is $\mathcal{L}({\mathbb{R}}^n)$, and ${\lambda }_n:={\lambda }_n^{\ast }\text{\hspace{0.17em}⁣}{\upharpoonright }_{\mathcal{L}({\mathbb{R}}^n)}$ (Lebesgue measurable sets, the family $\mathcal{L}({\mathbb{R}}^n)$, and the restricted set function ${\lambda }_n$).

Assuming countable choice, ${\lambda }_n^{\ast }$ is an outer measure on ${\mathbb{R}}^n$, and ${\lambda }_n^{\ast }(A)={\mu }_0(A)$ for every elementary set $A$ (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume).

${\lambda }_n^{\ast }$ is the outer set function induced by the premeasure ${\mu }_0$ on the algebra ${\mathcal{E}}_n$ of elementary sets (Lebesgue outer measure on ${\mathbb{R}}^n$).

Elementary volume ${\mu }_0$ is a sigma-finite premeasure on ${\mathcal{E}}_n$ (Elementary volume is a sigma-finite premeasure on the algebra of elementary sets); its value on a half-open box is the box volume (The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition, Half-open boxes in ${\mathbb{R}}^n$ and their volume), and ${\mathbb{R}}^n$ is elementary with infinite elementary volume (Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$).

For an outer measure ${\mu }^{\ast }$ on $X$, the Carathéodory measurable subsets form a sigma-algebra, and the restriction of the outer measure to it is a complete measure (Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure).

Assume the Axiom of Countable Choice. Every member of the source algebra is Carathéodory measurable for the induced outer measure (Assuming countable choice, every source-algebra set is measurable for the induced outer measure).

Assume the Axiom of Countable Choice. If ${\mu }_0$ is a premeasure on an algebra ${\mathcal{A}}_0$ of subsets of $X$ and ${\mu }^{\ast }$ is its induced outer set function, then ${\mathcal{A}}_0\subseteq {\mathcal{M}}_{{\mu }^{\ast }}$ and ${\mu }^{\ast }{\mathrm{\mid }}_{{\mathcal{A}}_0}={\mu }_0$ (Assuming countable choice, a premeasure extends through its induced outer measure).

Every set of outer measure zero, and every subset of it, is Carathéodory measurable and has outer measure zero (Every outer-null set is Carathéodory measurable).

A measure space $(X,\mathcal{A},\mu )$ is complete if every subset of every measurable $\mu$-null set is measurable (Complete measure spaces); and a measure on $(X,\mathcal{A})$ is a function $\mu :\mathcal{A}\to [0,+\mathrm{\infty }]$ with $\mu (\mathrm{\varnothing })=0$ that is countably additive on pairwise disjoint sequences (Measures on 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 (${\mathrm{A}\mathrm{C}}_{\omega }$)).