Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of ${\mathbb{R}}^n$ is the infimum of the measures of the open sets containing it

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, and a subset $E\subseteq {\mathbb{R}}^n$.

${\lambda }_n^{\ast }(E):=\inf \{{\sum }_{k=0}^{\mathrm{\infty }}{\mu }_0(A_k):A_k\in {\mathcal{E}}_n$ for every $k$ and $E\subseteq {\bigcup }_k{A}_k\}$ (Lebesgue outer measure on ${\mathbb{R}}^n$, Elementary sets: the finite unions of half-open boxes in ${\mathbb{R}}^n$).

Assuming countable choice, ${\lambda }_n^{\ast }$ is an outer measure on ${\mathbb{R}}^n$, hence monotone and countably subadditive, 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, Outer measures).

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra and ${\lambda }_n$ is a complete measure on it (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; in particular every open set is (Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable).

$A^{+\delta }$ is an elementary set determined by $A$ and $\delta$ alone, it contains $A$, and every point of $A$ is an interior point of $A^{+\delta }$; and for every real $\epsilon >0$ there is $m\in \mathbb{N}$ with ${\mu }_0\left(A^{+1\mathrm{/}(m+1)}\right)\le {\mu }_0(A)+\epsilon$ (Every elementary set is squeezed in volume between a compact subset and an elementary set whose interior contains it, claims 1 and 2).

Elementary volume is finitely additive on pairwise disjoint elementary sets, monotone, and finitely subadditive (Elementary volume is finitely additive, monotone and finitely subadditive on the elementary algebra, The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition).

The interior $\mathrm{int}(A)$ is open, and an arbitrary union of open sets is open (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, 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, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, claim 2).

The nonnegative extended sum of a sequence in $[0,+\mathrm{\infty }]$ is ${\sum }_{k=0}^{\mathrm{\infty }}{a}_k:={\sup }_{n\in \mathbb{N}}{s}_n$, the supremum of its nondecreasing partial sums (Series in the nonnegative extended real line).

Every nonempty subset $S\subseteq \mathbb{N}$ has a least element (The well-ordering principle).

If $\mathrm{\mid }r\mathrm{\mid }<1$ then ${\sum }_{k=0}^{\mathrm{\infty }}{r}^k=1\mathrm{/}(1-r)$; in particular ${\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k}=2$ (For $\mathrm{\mid }r\mathrm{\mid }<1$, ${\sum }_{k\ge 0}{r}^k=1\mathrm{/}(1-r)$, and for $\mathrm{\mid }r\mathrm{\mid }\ge 1$ the series diverges).

For sequences of reals, ${\sum }_{k<n}(a_k+b_k)={\sum }_{k<n}{a}_k+{\sum }_{k<n}{b}_k$, and if $a_k\le b_k$ whenever $0\le k<n$ then ${\sum }_{k<n}{a}_k\le {\sum }_{k<n}{b}_k$ (Laws of finite sums and finite products, claims 1 and 4; Finite sums and finite products, by recursion).

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