Every at most countable subset of ${\mathbb{R}}^n$ is Lebesgue null; in particular ${\lambda }_1(\mathbb{Q})=0$

Given: A natural number $n\ge 1$, the Axiom of Countable Choice, and an at most countable set $E\subseteq {\mathbb{R}}^n$.

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 this 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).

$A$ is at most countable if it is finite or countably infinite (Finite, countably infinite, countable, uncountable); a nonempty $A$ is at most countable if and only if there is a surjection $s:\mathbb{N}\to A$ (A nonempty set is at most countable iff it is a surjective image of $\mathbb{N}$).

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).

$\mathbb{Q}\approx \mathbb{N}$: the rationals are countably infinite ($\mathbb{Q}$ is countably infinite), and ${\mathbb{Q}}_{\mathbb{R}}$ denotes the image of $\mathbb{Q}$ in $\mathbb{R}$ under the canonical order-preserving field embedding (The rationals embed densely in the reals).

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).

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