# Does the boundary of a level set of a Lipschitz continuous function have Lebesgue measure $0$?

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.

• Hint: Every closed subset can be realized as zero level set of 1-Lipschitz function. – Moishe Kohan May 5 '19 at 16:50
• @MoisheKohan That's clear to me. If $(E,d)$ is a metric space and $A\subseteq E$, then $E\ni x\mapsto d(x,A)$ is $1$-Lipschitz continuous. And if $A$ is closed, its zero level set is precisely $A$. But I don't see how this helps here. – 0xbadf00d May 5 '19 at 17:57
• Take a Cantor set of positive linear measure... – Moishe Kohan May 5 '19 at 21:04
• @MoisheKohan Could you elaborate on your approach? – 0xbadf00d May 6 '19 at 4:29

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 $$C=g^{-1}(0)$$.
• Any set which is closed and nowhere dense and of positive measure will do . So if we want $f$ to be bounded , replace $C$ with $C\cup \Bbb Z.$................+1 – DanielWainfleet May 6 '19 at 14:04
• Does the situation change if $f$ is assumed to be continuously differentiable? In that case, $f^{-1}(\{0\})$ is a $(d-1)$-dimensional submanifold of $\mathbb R^d$. – 0xbadf00d May 8 '19 at 8:39
• Sorry, I've intended to assume that $0$ is a regular value of $f$. Asked for that here: math.stackexchange.com/q/3218322/47771. – 0xbadf00d May 8 '19 at 11:45