Assuming the Axiom of Choice, a Vitali set is not Lebesgue measurable

AC supplies countable choice: for a sequence of nonempty sets $(X_n)$, apply AC to $\{X_n:n\in \mathbb{N}\}$ and put $f(n)=g(X_n)$ for its choice function $g$. Thus [L7] and the countable-choice hypotheses in [L2] and [L3] apply. Finite additivity follows from [L5] by padding with empty sets; monotonicity follows by writing a measurable $B\supseteq A$ as $A\bigsqcup (B\setminus A)$.

Take a repetition-free enumeration of $\mathbb{Q}$ from [L4] and retain its terms in $[-1,1]$ in their original order, obtaining $(q_k{)}_{k\in \mathbb{N}}$. There are infinitely many retained terms, since $1\mathrm{/}(j+1)$ are distinct members for $j\in \mathbb{N}$; taking successive least retained indices requires no choice. The translates $V+q_k$ are pairwise disjoint: an equality $v_1+q_i=v_2+q_j$ gives $v_1-v_2\in \mathbb{Q}$, so [L1] gives $v_1=v_2$ and then $i=j$. They cover $[0,1]$, since each $x\in [0,1]$ has a representative $v\in V$ with $x-v\in \mathbb{Q}\cap [-1,1]$, and all lie inside $[-1,2]$.

Suppose, for contradiction, that $V$ is Lebesgue measurable with $\lambda (V)=0$. Then every translate $V+q_k$ is measurable with measure $0$ by [L2], and countable additivity applied to the pairwise disjoint family of step 1.2 gives $1=\lambda ([0,1])\le \lambda \text{\hspace{0.17em}⁣}({\bigcup }_{k\in \mathbb{N}}(V+q_k))={\sum }_{k=0}^{\mathrm{\infty }}\lambda (V+q_k)=0$, contradicting [L3].

Suppose instead that $V$ is Lebesgue measurable with $\lambda (V)>0$. Since $V\subseteq [0,1]$, step 1.1 and [L3] give $0<\lambda (V)\le 1$. Apply [L6] to the real number $3\mathrm{/}\lambda (V)$ to choose $m\ge 1$ with $3<m\lambda (V)$. The first $m$ translates from step 1.2 are disjoint and lie in $[-1,2]$, so finite additivity and translation invariance give $3\ge {\sum }_{k<m}\lambda (V+q_k)=m\lambda (V)$, a contradiction.

Steps 2.1 and 2.2 rule out both possible values of the measure of a measurable Vitali set, so $V$ is not Lebesgue measurable.