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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.

0xbadf00d
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1 Answers1

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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?

Moishe Kohan
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  • 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 06 '19 at 14:04
  • @DanielWainfleet sure – Moishe Kohan May 06 '19 at 15:14
  • 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 08 '19 at 08:39
  • @0xbadf00d No, it is not a submanifold. You can even get infinitely differentiable function. – Moishe Kohan May 08 '19 at 11:41
  • Sorry, I've intended to assume that $0$ is a regular value of $f$. Asked for that here: https://math.stackexchange.com/q/3218322/47771. – 0xbadf00d May 08 '19 at 11:45
  • @0xbadf00d then of course this holds by the implicit function theorem : even Lipschitz hypersurface has measure zero. – Moishe Kohan May 08 '19 at 14:11
  • It's clear to me that ${f=0}$ has measure zero by the implicit function theorem, but not how we can show this for the boundary. If you know how we can show it (under the assumption that $0$ is a regular value), then it would be great if you could provide an answer to the other question. – 0xbadf00d May 08 '19 at 14:39
  • The hypersurface ${f=0}$ (if $0$ is a regular value) has empty interior and equals its own boundary. – Moishe Kohan May 08 '19 at 16:47