Consider Lebesgue measure on $\mathbb{R}$ and let $f:\mathbb{R}\to\mathbb{R}$ be differentiable. Does $f$ necessarily preserve measure zero sets? Does $f$ necessarily preserve measurable sets?
Note that if $f$ is $C^1$ then $f$ preserves measure zero sets since $C^1$ functions are locally Lipschitz. Therefore $C^1$ functions also preserve measurable sets since a measurable set is the union of an $F_\sigma$ set and an measure zero set, and continuous functions preserve $F_\sigma$ sets. More generally if $f$ is absolutely continuous on each interval then $f$ preserves both measure zero sets and measurable sets.
However, I'm not sure about the differentiable case. I would guess that the answer to both questions is no. I'm interested in a counter example or proof in each case.
Best Answer
If $f$ preserves null-sets, it also preserves Lebesgue-measurability (but not necessarily Borel measurability), because by regularity, every Lebesgue measurable set $M$ can be written as
$$ M = N \cup \bigcup K_n $$
with $K_n$ compact and $N$ a null-set. By continuity, $f$ preserves compact sets.
Now a theorem in Rudin, Real and Complex Analysis (Lemma 7.25) shows in particular that every everywhere differentiable function maps null-sets to null-sets, so that your claim is true.