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 $\mathbb{R}^N$.
 A: 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 $\Omega$.
$\Box$
