Is the image of a null set under a differentiable map always null? Let $n$ be a positive integer.  $\;  $  Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere Frechet differentiable.

Let $S$ be a subset of $\mathbb{R}^n$ with Lebesgue measure zero.
Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?
(I know it would if $D(f)$ is continuous.)
 A: More generally, if $f:\ X \to Y$ is a locally Lipschitz function from one $\sigma$-compact metric space to another and $S \subseteq X$ has  $d$-dimensional Hausdorff measure 0, then $f(S)$ also has $d$-dimensional Hausdorff measure 0.  This is pretty much immediate from the definition of Hausdorff measures.  In your case, Lebesgue measure on ${\mathbb R}^n$ coincides with $n$-dimensional Hausdorff measure.
A: Yes. In fact, if $f$ has a Frechet derivative $Df$ on a measurable set $S\subseteq{R}^n$ then
$$
\begin{align}
\mu(f(S))\le\int_S\left\vert{\rm det} Df\right\vert\,d\mu&&(1)
\end{align}
$$
where $\mu$ is the Lebesgue measure on $\mathbb{R}^n$. In particular, if $\mu(S)=0$ then the right hand side is zero regardless of the value of $Df$. Actually, (1) is an immediate consequence of the change of variables formula for multidimensional integration, which can be shown to apply in the generality required here. Note that a general form of Sard's theorem also follows from (1), and says that the image of the set of points where $Df$ is singular has zero measure.
The Vitali covering lemma gives a quick proof that $f(S)$ has zero Lebesgue measure. If $f$ is Frechet differentiable at a point $x$ then, for any $K > \Vert Df\Vert$ we have $f(B_r(x))\subseteq B_{Kr}(f(x))$ for all small enough $r > 0$. So, $\mu(f(B_r(x))\le K^n\mu(B_r(x))$. Now, if $\Vert Df\Vert < K$ on a measurable set $S$ then you can choose open $U\supseteq S$ with $\mu(U)$ as close to $\mu(S)$ as you wish. For each $x\in S$ choose $r_x\in(0,1)$ such that $B_{r_x}(x)\subseteq U$ and $f(B_{r}(x))\subseteq B_{Kr}(f(x))$ for all $r\le 5r_x$. The Vitali covering lemma implies that there is a countable subset $S^\prime\subseteq S$ such that $\{B_{r_x}(x)\colon x\in S^\prime\}$ are disjoint and
$$
\bigcup_{x\in S^\prime}B_{5r_x}(x)\supseteq\bigcup_{x\in S}B_{r_x}(x)\supseteq S.
$$
So,
$$
\begin{align}
\mu(f(S))&\le\sum_{x\in S^\prime}\mu(f(B_{5r_x}(x)))\\
&\le 5^nK^n\sum_{x\in S^{\prime}}\mu(B_{r_x}(x))\\
&=5^nK^n\mu\left(\bigcup_{x\in S^\prime}B_{r_x}(S)\right)\\
&\le5^nK^n\mu(U).
\end{align}
$$
Letting $\mu(U)$ decrease to $\mu(S)$ gives $\mu(f(S))\le5^nK^n\mu(S)$. In particular, if $\mu(S)=0$ then $\mu(f(S))=0$. Even if the Frechet derivative just exists on $S$, but is not bounded, we can let $S_n$ be the subset of $S$ on which $\Vert Df\Vert\le n$. Then, using monotone convergence, $\mu(f(S))=\lim_{n\to\infty}\mu(f(S_n))=0$.
This gives a positive answer to the question asked, where $\mu(S)=0$. We can go further and use the Vitali covering theorem to prove (1). In fact, the full change of variables formula
$$
\int_S\vert{\rm det} Df\vert\,d\mu=\int_{f(S)}\#\left(f^{-1}(\{y\})\cap S\right)\,d\mu(y)
$$
can be shown to hold in the generality asked here. This can be proven with a similar argument as above involving the Vitali covering theorem, and also using the limits $\lim_{r\to0}\mu(f(B_r(x))/\mu(B_r(x))=\vert{\rm det}Df(x)\vert$ on $S$.
