For a Lebesgue measurable set and every positive $\epsilon$ there is an open superset whose difference from it has outer measure below $\epsilon$

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, a Lebesgue measurable set $E$, and a real $\epsilon >0$.

Assuming countable choice, ${\lambda }_n^{\ast }(E)=\inf \{{\lambda }_n(U):U\subseteq {\mathbb{R}}^n$ open and $E\subseteq U\}$ for every subset $E$ (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).

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra, ${\lambda }_n$ is a complete measure on it, and ${\lambda }_n$ is the restriction of ${\lambda }_n^{\ast }$ (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, ${\lambda }_n^{\ast }$ is an outer measure on ${\mathbb{R}}^n$, hence monotone and countably subadditive (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Outer measures).

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).

Every bounded subset $E\subseteq {\mathbb{R}}^n$ has ${\lambda }_n^{\ast }(E)<+\mathrm{\infty }$ (Lebesgue measure is sigma-finite, and every metrically bounded subset of ${\mathbb{R}}^n$ has finite outer measure).

Every set $R$ with $R^{\circ }\subseteq R\subseteq \bar{R}$ is Lebesgue measurable with ${\lambda }_n(R)={\prod }_{i<n}(b_i-a_i)$ (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included), and $(u,v{]}^n:=B(\mathbf{u},\mathbf{v})$ (Half-open boxes in ${\mathbb{R}}^n$ and their volume).

Let $\mu$ be a measure and let $A\subseteq B$ be measurable with $\mu (A)<+\mathrm{\infty }$; then $\mu (B)=\mu (A)+\mu (B\setminus A)$ (Measure of a set difference when the smaller set has finite measure).

An arbitrary union of open sets is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, claim 2; 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).

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).

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).

For sequences of reals, ${\sum }_{k<n}\lambda a_k=\lambda {\sum }_{k<n}{a}_k$, 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 2 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 (${\mathrm{A}\mathrm{C}}_{\omega }$)).