A locally finite translation-invariant Borel measure on ${\mathbb{R}}^d$ is a multiple of Lebesgue measure

Fix a coordinate $i$ and either face $F$ of $[0,1{]}^d$ in that coordinate. For each integer $n\ge 1$, translate the lower face by $j\mathrm{/}(n+1)e_i$, or the upper face by $-j\mathrm{/}(n+1)e_i$, for $j=1,\dots ,n$. These $n$ translates are pairwise disjoint compact subsets of $[0,1{]}^d$, each has measure $m(F)$ by translation invariance, and $m([0,1{]}^d)<\mathrm{\infty }$ by local finiteness. Hence $nm(F)\le m([0,1{]}^d)$ for every $n$, which forces $m(F)=0$. The boundary of $[0,1{]}^d$ is a finite union of such faces.

The symmetric difference of $Q$ and $(0,1{]}^d$ is contained in that boundary, so [F1] and step 1.1 give $m(Q)=m((0,1{]}^d)=c$.

If $c=0$, the integer translates $Q+k$, $k\in {\mathbb{Z}}^d$, are a countable disjoint partition of ${\mathbb{R}}^d$. Translation invariance gives $m(Q+k)=0$ for every $k$, and countable additivity gives $m({\mathbb{R}}^d)=0$. Monotonicity then gives $m(B)=0$ for every Borel $B$.

If $c>0$, $m\mathrm{/}c$ is a translation-invariant Borel measure and, by step 2.1, $(m\mathrm{/}c)((0,1{]}^d)=1$. Applying [F2] gives $m\mathrm{/}c={\lambda }_d$, hence $m(B)=c{\lambda }_d(B)$ for every Borel $B$.

The cases $c=0$ and $c>0$ exhaust the nonnegative value $m(Q)$, proving the statement. The proof uses Countable Choice only through [F2]; no stronger choice principle is used.