Universal real measurability transfers to finite-dimensional Euclidean spaces

Let $D\subseteq [0,1)$ consist of the reals whose canonical binary code has no eventually-$1$ subsequence in any residue class modulo $n$; its complement is the finite union of Borel null sets. Split the digits of a code in $D$ by residue modulo $n$. This is a Borel bijection $T:D\to [0,1{)}^n$ with Borel inverse given by interleaving the canonical coordinate codes. A length-$nk$ cylinder maps to the corresponding product of $n$ dyadic intervals, each of length $2^{-k}$; both sides have measure $2^{-nk}$. The monotone-class extension and F2–F3 make $T$ measure preserving on all Borel sets and send Borel null sets both ways. F3 assumes countable choice, supplied here exactly by internal DC through F4–F5; the digit map itself makes no selections.

For arbitrary $E\subseteq [0,1{)}^n$, F1 makes $T^{-1}(E)$ measurable. Completion gives Borel $B_0$ and Borel null $N_0$ with $T^{-1}(E)\mathrm{△}{B}_0\subseteq N_0$. Put $B=B_0\cap D$ and $N=N_0\cap D$, so both lie in the domain of $T$. Bimeasurability and step 1.1 give $E\mathrm{△}T(B)\subseteq T(N)$, with Borel measurable $T(B)$ and null $T(N)$; hence $E$ is measurable.

Cover ${\mathbb{R}}^n$ by the explicitly indexed disjoint half-open cubes $k+[0,1{)}^n$, $k\in {\mathbb{Z}}^n$. For each $k$, the set $E_k=(E\cap (k+[0,1{)}^n))-k$ belongs to $M$ and is measurable by step 2.1; F6 makes its translate $E_k+k$ measurable. The defining closure of the Lebesgue sigma-algebra under the displayed countable union now gives $E={\bigcup }_{k\in {\mathbb{Z}}^n}(E_k+k)$ measurable, with no selection of representatives. When $n=1$, there is one residue class, $D=[0,1)$ for canonical non-eventually-$1$ codes, and $T$ is the identity under that code, so step 2.1 is exactly F1. The case $n=0$ is outside the stated positive range.