# Why is the Hausdorff measure of this set zero?

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 $$\mathbb{R}^N$$.

• 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? Jan 1, 2021 at 17:40
• Does $\bar$ mean the closure, that is, the set you are interested in is contained in the boundary of $\Omega$? Jan 1, 2021 at 17:41
• @NateEldredge for example, $N=2$, $\Omega=(0,1)^2$, the same function (not depending on the second coordinate). Jan 1, 2021 at 17:44
• @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. Jan 1, 2021 at 18:29
• @bogdan I guess you mean the differential of $\phi$ at any zero x to be invertibile, not just nonzero Jan 1, 2021 at 19:46

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 $$\Omega$$. $$\Box$$
• In fact one may take $\Omega$ to be the union of the family of balls, skipping the extension & restriction steps Jan 2, 2021 at 0:07
• 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. Jan 2, 2021 at 13:44