Let $\Omega\subseteq\mathbb{R}^N$ be an open and bounded set, and let $\phi:\Omega\to\mathbb{R}^N$ be a $C^1$ function with the property that $\phi^{-1}(0)\neq\emptyset$, and $\nabla\phi(x)\neq 0,\ \forall\ x\in \phi^{-1}(0)$.

How can we prove or disprove that:

$$\mathcal{H}^{N-1}\left (\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0)\right )=0$$

It is well-known that $\mathcal{L}^N(\phi^{-1}(0))=\mathcal{H}^N(\phi^{-1}(0))=0$ (Lebesgue measure of the zero level set is null).

I denote by $\mathcal{H}^{N-1}$ the $N-1$ dimensional Hausdorff measure on $\mathbb{R}^N$.

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    $\begingroup$ So what if I take $N=1$, $\Omega = (0,1)$, and $\phi(x) = \sin(1/x)$? Then $\overline{\phi^{-1}(0)} \setminus \phi^{-1}(0) = \{0\}$ whose 0-dimensional Hausdorff measure (counting measure) is not zero. I realize this is pretty trivial, but couldn't one make similar examples in higher dimensions? $\endgroup$ Jan 1, 2021 at 17:40
  • $\begingroup$ Does $\bar$ mean the closure, that is, the set you are interested in is contained in the boundary of $\Omega$? $\endgroup$ Jan 1, 2021 at 17:41
  • $\begingroup$ @NateEldredge for example, $N=2$, $\Omega=(0,1)^2$, the same function (not depending on the second coordinate). $\endgroup$ Jan 1, 2021 at 17:44
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    $\begingroup$ @Bogdan: How about $\Omega$ as in my previous example and $\phi(x) = x^3 \sin(1/x)$? Or $x^n \sin(1/x)$ if you want more derivatives. $\endgroup$ Jan 1, 2021 at 18:29
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    $\begingroup$ @bogdan I guess you mean the differential of $\phi$ at any zero x to be invertibile, not just nonzero $\endgroup$ Jan 1, 2021 at 19:46

1 Answer 1


The next result answers the question in the negative.

Theorem. There is $\phi:\mathbb{R}^n\supset\Omega\to\mathbb{R}^n$ of class $C^\infty$ such that $\phi$ is a local diffeomorphism in a neighborhood of $\phi^{-1}(0)$, but the Lebesgue measure of the following set is positive: $$ (*)\quad \mathcal{L}^n\left (\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0)\right )>0. $$

Proof. Let $\Omega\subset\mathbb{R}^n$ be a open set such that $\mathcal{L}^n(\partial\Omega)>0$. It is well known that such sets exist and in fact they can be homeomorphic to a ball.

Let $E=\{x_i\}_{i=1}^\infty\subset\Omega$ be a countable set such that $\partial\Omega\subset\overline{E}$. Let $r_i>0$ be such that $$ \overline{B}(x_i,r_i)\subset\Omega \quad \text{and} \quad \overline{B}(x_i,r_i)\cap\overline{B}(x_j,r_j)=\emptyset. $$ Define $$ \phi:\bigcup_{i=1}^\infty\overline{B}(x_i,r_i)\to B(0,1) $$ as a similarity in each ball and extend it to $\Omega$ as a $C^\infty$ map. Then $E\subset\phi^{-1}(0)$ and hence $$ \partial\Omega\subset \overline{E}\setminus\Omega\subset\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0). $$ proves ($*$). Clearly, $\phi$ is a local diffeomorphism in a neighborhood of $E\subset\phi^{-1}(0)$, but there might be points $x\in \phi^{-1}(0)\setminus E$ where the Jacobian $J_\phi=0$ equals zero. To avoid this problem we simply remove a small neighborhood of this set from $\Omega$. $\Box$

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    $\begingroup$ In fact one may take $\Omega$ to be the union of the family of balls, skipping the extension & restriction steps $\endgroup$ Jan 2, 2021 at 0:07
  • $\begingroup$ @PietroMajer You are right. I actually wanted to have a domain homeomorphic to a ball so the extension, but then I realized that I have to remove something. Perhaps I will modify my answer following your suggestion. $\endgroup$ Jan 2, 2021 at 12:57
  • $\begingroup$ I think one can do the extension as you do for any given $\Omega$ without introducing new zeros. Join the balls with a set of arcs so to form a tree (a Christmas tree indeed). Since this is contractible, we can map $\Omega$ to this tree by a smooth map $h$, that also map each ball in itself homeomorphically. We then compose $h$ with a map like you do, that maps each ball by a similarity to the unit ball, and each arc somewhere in the complement. $\endgroup$ Jan 2, 2021 at 13:44

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