Continuity of Hausdorff measure on level sets Let $\Omega\subset\mathbb{R}^2$ a open and bounded set with smooth boundary and $\phi:\Omega\to\mathbb{R}$ a smooth function such that:
$\bullet$ $\phi^{-1}(0)\neq\emptyset$;
$\bullet$ $\nabla\phi(x)\neq 0$ on a neighborhood $W\subset\Omega$ of the curve $\phi^{-1}(0)$.
WLOG we can assume that $W=\{x\in\mathbb{R}^2\ |\ |\phi(x)|<\varepsilon_0\}\subset\Omega$.
How can we prove that:
$$\lim\limits_{\varepsilon\to 0^+} \mathcal{H}^1\big (\{x\in\Omega\ |\ \phi(x)=\varepsilon\}\big )=\mathcal{H}^1\big (\{x\in\Omega\ |\ \phi(x)=0\}\big )\;\; ?$$
Here $\mathcal{H}^1$ denotes the Hausdorff 1-dimensional measure.
In the article of L.Modica -The gradient theory of phase transitions and the minimal interface criterion that can be found here: https://www.math.cmu.edu/~tblass/CNA-PIRE/Modica1987.pdf this property is proved only for the signed distance function (see Lemma 3, at page 8), but from more examples that I take it seems to be valid for many other level functions.
 A: As Leo Moos suggested in the comments, in any dimension $d$, this is a simple consequence of the implicit function theorem.
The implicit function theorem implies that every point $x\in\phi^{-1}(0)$ has a neighborhood $\Omega_x$ that is a diffeomorphic image $\varphi_x(Q_{\delta_x})$ of a box $Q_{\delta_x}=\{|y_i|\leq\delta_x,\;1\leq i \leq d\}$, such that $\phi(\varphi(y))=|\nabla \phi(x)|\cdot y_1$, and such that $\varphi_x(0)=x$ and $D\varphi_x(0)$ is a rotation. (Details: assume wlog that $x=0$, and, by rotating, that $\nabla \phi (0)/|\nabla \phi (0)|=e_1,$ the first basis vector. Consider the function $g:\mathbb{R}^d\to \mathbb{R}^d$ defined by $g(x)=(\phi(x)/|\nabla \phi (0)|,x_2,\dots,x_d)$. Then $Dg(0)$ is the identity, and $\varphi_x$ is the inverse $g^{-1}$ provided by the inverse function theorem.) Given $\varepsilon>0$, by choosing smaller $\delta_x$ if necessary, we can ensure that the restriction of $\varphi_x$ to the leaves $\{y_1=h\}$ distorts the $(d-1)$-area by no more than $1+\varepsilon$, i. e., in $Q_{\delta_x}$,
$$1-\varepsilon<\left(\det_{2\leq i,j\leq d}\left(\partial_{y_i}\varphi\cdot\partial_{y_j}\varphi\right)\right)^\frac12\leq 1+\varepsilon.$$
By compactness, we can choose a finite cover $\Omega_i=\Omega_{x_i}$, $i=1,\dots,n$ of $\phi^{-1}(0)$. Then, for $\epsilon_0>0$ small enough, we have $W_{\epsilon_0}:=\{x:|\phi(x)|\leq\epsilon_0\}\subset \cup_{i=1}^n\Omega_i$. Put $\Omega_0=\Omega\setminus W_{\epsilon_0}$, then $\Omega_0,\Omega_1,\dots,\Omega_n$ is a finite cover of $\Omega$, and we can pick a partition of unity $f_0,\dots,f_n$ subordinate to that cover. We have for $|h|<\epsilon_0$,
$$
\mathcal{H}^{d-1}(\phi^{-1}(h))=\sum_{i=1}^n\int f_i d\mathcal{H}^{d-1}(\phi^{-1}(h)),
$$
hence, by changing the variable,
$$
(1-\varepsilon)\sum_{i=1}^n\int_{y_1=h} f_i\circ\varphi_i\leq \mathcal{H}^{d-1}(\phi^{-1}(h))\leq (1+\varepsilon)\sum_{i=1}^n\int_{y_1=h} f_i\circ\varphi_i.
$$
As $h\to 0$, the terms in the right-hand side tend to
$$
(1+\varepsilon)\int_{y_1=0}f_i\circ\varphi_i\leq (1+\varepsilon)^2\int f_i d\mathcal{H}^{d-1}(\phi^{-1}(0)),
$$
and similarly for the LHS. This means that
$$
(1-\varepsilon)^2\mathcal{H}^{d-1}(\phi^{-1}(0))\leq\liminf_{h\to 0} \mathcal{H}^{d-1}(\phi^{-1}(h)),
$$
$$
\limsup_{h\to 0} \mathcal{H}^{d-1}(\phi^{-1}(h))\leq (1+\varepsilon)^2\mathcal{H}^{d-1}(\phi^{-1}(0)),
$$
and since $\varepsilon>0$ is arbitrary, we are done.
A: Not a complete answer, but just some ideas that are too long for a comment.
Note: Here we use $\mathcal H^n, \mathcal L^n$ to denote the Hausdorff and Lebesgue measure respectively.
Let $\Omega \subset \mathbb R^n$ be an open bounded domain, and let $f: \Omega  \to \mathbb R$ be a $C^1$ function such that $\nabla f(x) \neq 0$ for all $x \in \Omega$,
Define $g: \mathbb R\to \mathbb R$ by $g(y) := \mathcal L^n \{x \in \mathbb \Omega \ | \ f(x) \leq y\}$.
Then an application of the coarea formula gives us the following:
Lemma: $g$ is differentiable at $y$ if and only if $h(x) := \mathbb 1_{\{f(x) = y\}}(x) \frac{1}{|\nabla f(x)|}$ is $\mathcal H^{n-1}$-integrable, and we have
$$g’(y) = \int_{\Omega} \mathbb 1_{\{f(x) = y\}}(x) \frac{1}{|\nabla f(x)|}.$$
Now in the problem given, by compactness we may assume wlog that $\nabla \phi$ is bounded below on $\Omega$ by some absolute constant $c > 0$. Since $\phi$ is smooth, the level sets $E_y := \{\phi(x) = y\}$ are smooth submanifolds of $\Omega$, in particular they have finite $\mathcal H^1$ measure.
Thus $\mathbb 1_{\{\phi(x) = y\}}(x) \frac{1}{|\nabla \phi(x)|}$ is $\mathcal H^1$ integrable, and so, taking $f = \phi$ in the lemma, we have that $g$ is differentiable for all $-\varepsilon < y < \varepsilon$.
Now by the uniform bound on $|\nabla f|$, $g’(y)$ is comparable to $\mathcal H^1 (E_y)$ up to a scalar constant. After some epsilon chasing, we will be done if we can show that $g$ is in addition continuously differentiable. However I am having trouble showing this.
