Lebesgue measure is sigma-finite, and every metrically bounded subset of ${\mathbb{R}}^n$ has finite outer measure

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, and Lebesgue measure ${\lambda }_n$ on $\mathcal{L}({\mathbb{R}}^n)$.

Assuming countable choice, $\mathcal{L}({\mathbb{R}}^n)$ is a sigma-algebra, ${\lambda }_n$ is a complete measure on it, and ${\lambda }_n(B)=\mathrm{vol}(B)$ for every half-open box $B$ (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).

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).

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

For a nonempty box $\mathrm{vol}(B):={\prod }_{i<n}(b_i-a_i)$ when every $a_i$ and every $b_i$ is real, and $(u,v{]}^n:=B(\mathbf{u},\mathbf{v})$ (Half-open boxes in ${\mathbb{R}}^n$ and their volume, Integer powers $a^m$).

$\mu$ is sigma-finite if there is a sequence $(E_n{)}_{n\in \mathbb{N}}$ in $\mathcal{A}$ such that $X={\bigcup }_n{E}_n$ and $\mu (E_n)<+\mathrm{\infty }$ for every $n$ (Finite, sigma-finite, and semifinite measures).

$A$ is bounded if $A=\mathrm{\varnothing }$ or there are $x_0\in X$ and a real $r>0$ with $A\subseteq B(x_0,r)$, where $B(x_0,r):=\{y:d(x_0,y)<r\}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

For every $x\in {\mathbb{R}}^n$, $\parallel x{\parallel }_{\mathrm{\infty }}\le \parallel x{\parallel }_2$, and $\parallel x-y{\parallel }_2=d_2(x,y)$, $\parallel x-y{\parallel }_{\mathrm{\infty }}=d_{\mathrm{\infty }}(x,y)$, where $d_{\mathrm{\infty }}(x,y)=\max \{\mathrm{\mid }{x}_k-y_k\mathrm{\mid }:k<n\}$ (The finite and reverse triangle inequalities for a norm; and for $n\ge 1$ every norm $N$ on ${\mathbb{R}}^n$ satisfies $N(x)\le C\parallel x{\parallel }_1$ and is Lipschitz, hence continuous, for $d_2$, claim 3; Each $\parallel \cdot {\parallel }_p$ is a norm on ${\mathbb{R}}^n$, and the induced metrics are exactly $d_1$, $d_2$ and $d_{\mathrm{\infty }}$ of the published metric-spaces page, claim 3; The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$; ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it).

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; 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); and a compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded).

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).