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.

  • $\begingroup$ Hint: Every closed subset can be realized as zero level set of 1-Lipschitz function. $\endgroup$ – Moishe Kohan May 5 '19 at 16:50
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    $\begingroup$ @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. $\endgroup$ – 0xbadf00d May 5 '19 at 17:57
  • $\begingroup$ Take a Cantor set of positive linear measure... $\endgroup$ – Moishe Kohan May 5 '19 at 21:04
  • $\begingroup$ @MoisheKohan Could you elaborate on your approach? $\endgroup$ – 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 $C=g^{-1}(0)$.

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?

  • $\begingroup$ 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 $\endgroup$ – DanielWainfleet May 6 '19 at 14:04
  • $\begingroup$ @DanielWainfleet sure $\endgroup$ – Moishe Kohan May 6 '19 at 15:14
  • $\begingroup$ 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 $\mathbb R^d$. $\endgroup$ – 0xbadf00d May 8 '19 at 8:39
  • $\begingroup$ @0xbadf00d No, it is not a submanifold. You can even get infinitely differentiable function. $\endgroup$ – Moishe Kohan May 8 '19 at 11:41
  • $\begingroup$ Sorry, I've intended to assume that $0$ is a regular value of $f$. Asked for that here: math.stackexchange.com/q/3218322/47771. $\endgroup$ – 0xbadf00d May 8 '19 at 11:45

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