A translation-invariant measure on the Borel sets of ${\mathbb{R}}^n$ giving the unit cube measure one is the restriction of Lebesgue measure

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, and a translation-invariant measure $\mu$ on $\mathcal{B}({\mathbb{R}}^n)$ with $\mu ((0,1{]}^n)=1$.

If $\mu$ is a measure on the Borel sets of ${\mathbb{R}}^n$ that is translation invariant and gives the unit cube measure $1$, then $\mu (Q)=2^{-kn}$ for every dyadic cube $Q$ of generation $k$ (A translation-invariant Borel measure giving the unit cube measure one gives each generation-$k$ dyadic cube measure $2^{-kn}$, Dyadic cubes of generation $k$ in ${\mathbb{R}}^n$).

Every open $U\subseteq {\mathbb{R}}^n$ is the union of an at most countable family of pairwise disjoint dyadic cubes (Every open subset of ${\mathbb{R}}^n$ is the union of a countable pairwise disjoint family of dyadic cubes).

$\mathrm{vol}(Q_{k,m})=2^{-kn}$ (For each generation, the dyadic cubes of that generation are pairwise disjoint and cover ${\mathbb{R}}^n$).

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) and ${\lambda }_n$ is a measure on $\mathcal{L}({\mathbb{R}}^n)$ with ${\lambda }_n(B)=\mathrm{vol}(B)$ for every half-open box (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).

${\lambda }_n^{\ast }(E+h)={\lambda }_n^{\ast }(E)$ for every subset $E$, $E$ is Lebesgue measurable if and only if $E+h$ is, and ${\lambda }_n(E+h)={\lambda }_n(E)$ for measurable $E$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a subset of ${\mathbb{R}}^n$).

Let $\mathcal{P}$ be a pi-system on $X$ generating $\mathcal{A}$, and let $\mu ,\nu$ be measures on $(X,\mathcal{A})$ that agree on $\mathcal{P}$; suppose there is an increasing sequence $(P_n)$ in $\mathcal{P}$ with $X={\bigcup }_n{P}_n$ and $\mu (P_n)=\nu (P_n)<+\mathrm{\infty }$ for every $n$; then $\mu =\nu$ on $\mathcal{A}$ (Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system).

A pi-system on $X$ is a nonempty family $\mathcal{P}\subseteq \mathcal{P}(X)$ closed under binary intersections (Pi-systems).

The Borel sigma-algebra of $X$ is the sigma-algebra generated by its open sets (The Borel sigma-algebra of a topological space), and ${\sigma }_X(\mathcal{E})$ is the unique smallest sigma-algebra on $X$ containing $\mathcal{E}$ (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal); a sigma-algebra is closed under complements and countable unions (Sigma-algebras).

A subset $U$ is open in $(X,d)$ when every $x\in U$ has a ball $B(x,r)\subseteq U$ (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, Open ball, closed ball and sphere in a metric space); a finite intersection 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 3); and $d_2(x,y):=\sqrt{{\sum }_{k<n}(x_k-y_k{)}^2}$ is a metric on ${\mathbb{R}}^n$ (${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it).

A measure is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras), and an at most countable family may be presented as a sequence (Finite, countably infinite, countable, uncountable).

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).