Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets

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

Assuming countable choice, $E$ is Lebesgue measurable if and only if for every real $\epsilon >0$ there is a closed $F\subseteq E$ with ${\lambda }_n^{\ast }(E\setminus F)<\epsilon$ (Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of ${\mathbb{R}}^n$, condition 3).

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra and ${\lambda }_n$ 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, 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 Lebesgue measurable subset of ${\mathbb{R}}^n$ has finite measure, and every compact subset of ${\mathbb{R}}^n$ is Lebesgue measurable of finite measure (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).

A subset $K\subseteq {\mathbb{R}}^n$ is compact if and only if $K$ is closed in ${\mathbb{R}}^n$ and bounded (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, claim 2), and a compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded, Bounded subset, diameter, distance from a point to a set, and distance between two sets in 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).

Closed balls are closed, for every $x\in X$ and every $r>0$ (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, claim 4), where $\bar{B}(x,r):=\{y\in X:d(x,y)\le r\}$ (Open ball, closed ball and sphere in a metric space).

Let $(E_n{)}_{n\in \mathbb{N}}$ be an increasing sequence of measurable sets for a measure $\mu$; then $\mu \left({\bigcup }_{n\in \mathbb{N}}{E}_n\right)={\sup }_{n\in \mathbb{N}}\mu (E_n)$ (Continuity from below for measures).

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).

If $A,B\in \mathcal{A}$ and $A\subseteq B$, then $\mu (A)\le \mu (B)$ (Measures are monotone).

Every complete ordered field $F$ is Archimedean: for every $x\in F$ there is a natural number $n\ge 1$ with $x<n\cdot 1_F$ (Every complete ordered field is Archimedean).

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