A C^1 diffeomorphism maps Lebesgue null sets to Lebesgue null sets

Given: Countable Choice, open sets $U,V\subseteq {\mathbb{R}}^n$, a $C^1$ diffeomorphism $T:U\to V$, and a Lebesgue null set $N\subseteq U$.

Lipschitz self-maps of Euclidean space send closed-cube-cover null sets to closed-cube-cover null sets (A Lipschitz map ${\mathbb{R}}^m\to {\mathbb{R}}^m$ sends null sets to null sets).

Under Countable Choice, a subset of ${\mathbb{R}}^n$ is Lebesgue null exactly when it is null in this closed-cube-cover sense (A subset of ${\mathbb{R}}^m$ has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers, The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).

Lebesgue measure is countably subadditive (Finite and countable subadditivity of measures).

The rational closed cubes of positive side length lying in $U$ form a countable family and their interiors cover $U$ ($\mathbb{Q}$ is countably infinite). For each such cube $Q$, continuity of $DT$ bounds its derivative on the compact convex cube, so the mean-value inequality makes $T{\mathrm{\mid }}_Q$ Lipschitz. The coordinatewise clamp $r_Q:{\mathbb{R}}^n\to Q$ is $1$-Lipschitz; hence $S_Q=T\circ r_Q$ is a global Lipschitz extension agreeing with $T$ on $Q$. These formulas define the extensions from the cubes without a countable selection.