A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in ${\mathbb{R}}^n$

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, an index $i_0<n$ and a real $c$.

Every set $R$ with $R^{\circ }\subseteq R\subseteq \bar{R}$ is Lebesgue measurable with ${\lambda }_n(R)={\prod }_{i<n}(b_i-a_i)$, and it gives measure $0$ to all of them whenever $a_i=b_i$ for some $i<n$ (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, $\mathcal{L}({\mathbb{R}}^n)$ 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 (Assuming countable choice, every Borel subset of ${\mathbb{R}}^n$ is Lebesgue measurable).

For a measure $\mu$ and measurable $(E_k{)}_{k\in \mathbb{N}}$, $\mu \left({\bigcup }_{k\in \mathbb{N}}{E}_k\right)\le {\sum }_{k=0}^{\mathrm{\infty }}\mu (E_k)$ (Finite and countable subadditivity of measures).

$[a,b]:=\{x\in {\mathbb{R}}^m:a_j\le x_j\le b_j(j<m)\}$ (Axis-parallel rectangles in ${\mathbb{R}}^m$ and their volume).

A measurable set $N\in \mathcal{A}$ is $\mu$-null if $\mu (N)=0$ (Measure-null sets and almost-everywhere statements relative to a measure); a sigma-algebra is closed under countable unions (Sigma-algebras).

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).

$B(a,b):=\{x\in {\mathbb{R}}^n:a_i<x_i\le b_i\text{ for every }i<n\}$ (Half-open boxes in ${\mathbb{R}}^n$ and their volume).