A $C^1$ map sends a compact set of content zero to a set of content zero

If $A=\mathrm{\varnothing }$ then $\psi [A]=\mathrm{\varnothing }$, which is covered by the single degenerate cube ${\prod }_{j<m}[0,0]$ of volume $0$, so it has content zero by [F1] and the assertion holds. For the rest of the proof assume $A\mathrm{\ne }\mathrm{\varnothing }$.

Since $A$ has content zero, [F2] makes $A$ null.

A $C^1$ map is continuous, since by [F5] and [L6] each component is totally differentiable and hence continuous at every point of $W$. So $\psi [A]$ is a compact subset of ${\mathbb{R}}^m$ by [L8].

Every point $c\in A$ lies in the open set $W$, so some closed cube $Q$ centred at $c$ with positive edge is contained in $W$, and the interior of $Q$ contains $c$. Those interiors form an open cover of the compact $A$, so by [F4] and [L9] finitely many of them cover $A$: there are closed cubes $Q_1,\dots ,Q_N\subseteq W$ with $N\ge 1$ whose union contains $A$.

Fix $i$ with $1\le i\le N$. The $m^2$ functions ${\mathrm{\partial }}_j{\psi }_k$ are continuous on $W$ by [F5], and $Q_i$ is a nonempty compact subset of $W$ by [L9], so [L7] bounds each of them on $Q_i$: there is $C\ge 0$ with $\mathrm{\mid }{\mathrm{\partial }}_j{\psi }_k(c)\mathrm{\mid }\le C$ for all $c\in Q_i$ and all $j,k<m$. Put $M:=m^{3\mathrm{/}2}C$. For $c\in Q_i$ and $v\in {\mathbb{R}}^m$, [L5] and [L6] give $D\psi (c)v=J\psi (c)v$, whose $k$th coordinate is ${\sum }_{j<m}{\mathrm{\partial }}_j{\psi }_k(c)v_j$, of absolute value at most $mC\parallel v{\parallel }_2$ because $\mathrm{\mid }{v}_j\mathrm{\mid }\le \parallel v{\parallel }_2$ by [F6]; hence $\parallel D\psi (c)v{\parallel }_2\le \sqrt{m}mC\parallel v{\parallel }_2=M\parallel v{\parallel }_2$, again by [F6].

Let $x,y\in Q_i$. A cube is convex, so $\gamma (t):=x+t(y-x)$ lies in $Q_i\subseteq W$ for $0\le t\le 1$. The map $\gamma$ is differentiable with ${\gamma }^{\mathrm{\prime }}(t)=y-x$, and $\psi$ is totally differentiable on $W$ by [F5] and [L6], so [L4] and [L5] make $t\mapsto \psi (\gamma (t))$ differentiable on $[0,1]$ with derivative $D\psi (\gamma (t))(y-x)$, of norm at most $M\parallel y-x{\parallel }_2$ by step 3.1. Hence [L3] on $[0,1]$ gives $\parallel \psi (y)-\psi (x){\parallel }_2\le M\parallel y-x{\parallel }_2$, so the restriction $\psi {\mathrm{\mid }}_{Q_i}$ is Lipschitz with constant $M$ in the sense of [F3].

Write $Q_i={\prod }_{j<m}[{\alpha }_j,{\beta }_j]$ and let ${\rho }_i:{\mathbb{R}}^m\to Q_i$ be the coordinatewise clamp, ${\rho }_i(v{)}_j=\min \{\max \{v_j,{\alpha }_j\},{\beta }_j\}$. Each scalar clamp satisfies $\mathrm{\mid }\min \{\max \{s,\alpha \},\beta \}-\min \{\max \{s^{\mathrm{\prime }},\alpha \},\beta \}\mathrm{\mid }\le \mathrm{\mid }s-s^{\mathrm{\prime }}\mathrm{\mid }$, so $\parallel {\rho }_i(v)-{\rho }_i(v^{\mathrm{\prime }}){\parallel }_2\le \parallel v-v^{\mathrm{\prime }}{\parallel }_2$ by [F6] and ${\rho }_i$ is Lipschitz with constant $1$; therefore $T_i:=\psi \circ {\rho }_i$ is defined on all of ${\mathbb{R}}^m$, agrees with $\psi$ on $Q_i$ since ${\rho }_i$ fixes $Q_i$ pointwise, and is Lipschitz with constant $M$ by step 4.1 and [F3].

For each $i$, the set $A\cap Q_i$ is a subset of the null set $A$ of step 1.2, hence null by [L1]; so [L2] applied to the Lipschitz map $T_i$ of step 5.1 makes $T_i[A\cap Q_i]$ null, and that set is $\psi [A\cap Q_i]$ because $T_i$ agrees with $\psi$ on $Q_i$.

By step 2.1 the union of the $Q_i$ contains $A$, so $\psi [A]={\bigcup }_{i=1}^N\psi [A\cap Q_i]$. Let $\epsilon >0$. By step 6.1 and [F1] each of the $N$ sets admits a sequence of closed cubes covering it with volume sum at most $\epsilon \mathrm{/}N$; concatenating those $N$ sequences gives one sequence of closed cubes covering $\psi [A]$ with volume sum at most $\epsilon$, so $\psi [A]$ is null by [F1]. The index set is finite, so only finitely many covers are named and no choice principle is used.

The set $\psi [A]$ is compact by step 1.3 and null by step 7.1, so [L10] gives that it has content zero.