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](https://www.math.cmu.edu/%7Etblass/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.



Continuity of Hausdorff measure on level sets
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.


