Let $f:\mathbb R\to\mathbb R$ be a (bounded, if necessary) Lipschitz continuous function. Are we able to show that $\partial f^{-1}\left(\left\{0\right\}\right)$ has Lebesgue measure $0$. If not, are there mild conditions under which the claim holds true?
I don't have much to contribute, since I struggle to find a good starting point.
EDIT: The question seems to be related to the notion of Hausdorff measures and maybe Sard's theorem. Since I've never heard about Hausdorff measures before reading the Wikipedia article, I hope there is a solution to this problem which doesn't need this concept.
Best Answer
Let $C\subset {\mathbb R}$ be a fat Cantor set, i.e. a subset of ${\mathbb R}$ which has positive Lebesgue measure and is homeomorphic to the standard Cantor set (i.e. is nonempty, compact, perfect and has empty interior). Let $f(x)=d(x,C)$ be the distance function to $C$. As you know, $f$ is 1-Lipschitz. At the same time, $C=f^{-1}(0)= \partial C$ (since $C$ has empty interior). With a bit more work one can replace $f$ with a function $g$ which is infinitely differentiable on ${\mathbb R}$ and still have $C=g^{-1}(0)$.
As for Sard's theorem, it is about images not preimages, so is unrelated to your question.
Related: The Boundary of a Lipschitz domain has Lebesgue measure zero?