# 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.

• This follows from the implicit function theorem, no? Jul 23, 2021 at 14:45
• I do not see how. Can you give some details please? Jul 23, 2021 at 14:47

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.

• @Bogdan, $f=max\{f,0\}-(-min\{f,0\})$, so it suffices to prove the result for positive f, for which the proof extends verbatim. Jul 28, 2021 at 20:54
• @Bogdan, indeed it would be probably more accurate to say "inverse function theorem" - for me these are minor variants of the same thing. I added some details. Aug 1, 2021 at 9:22
• @Bogdan, I am not sure I understand the question. I fix the point $x$ and then rotate the coordinate axes so that the first basis vector points in the direction of the gradient of $\nabla \phi(x)$, the second one is any vector orthogonal to it, etc. Aug 16, 2021 at 16:08
• @Bogdan, in these new coordinates, the gradient of $\phi$ has coordinates $(\frac{\partial \phi}{\partial x_1},0,\dots,0)$, so $\frac{\partial \phi}{\partial x_1}$ cannot be zero because we assume the gradient does not vanish. Aug 16, 2021 at 18:18
• @Bogdan, sure, but this is the derivative with respect to the first coordinate in the original coordinates, not the new ones. Aug 16, 2021 at 18:35

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.