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Let $D$ be a bounded Lipschitz domain and $f$ is continuous up to $\partial D$. Is it true that $$\int_{\partial D}f(x)d\sigma(x) = \lim_{\epsilon\to 0}\frac{1}{\epsilon}\int_{D^{\epsilon}}f(x)dx$$ where $D^{\epsilon}=\{y\in D: d(y,\partial D)<\epsilon\}$?

When $\partial D$ is $C^2$, we can parametrize each point of $D^{\epsilon}$ by a point $(\xi,\delta)\in \partial D\times [0,\epsilon)$ bijectively, when $\epsilon$ is small enough. But for Lipschitz it is not clear to me. Thanks!

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up vote 0 down vote accepted

This is really a question about Minkowski content. If you look at the cited article (under properties), this seems to indicate that under your Lipschitz condition, the answer to your question is YES (since the full-dimensional Hausdorff measure is the Lebesgue measure).

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Thanks for pointing me to the concept of Minkowski content! I found that Theorem 3.2.39 of "Geometric measure theory" by Federer provides an answer. – Fantastic Sep 6 '12 at 6:24

There might be a more abstract/direct way to prove it, but I would try to prove the statement by using:

  • Almost everywhere differentiability of Lipschitz mappings (giving that almost everywhere the boundary is like a hyperplane on small scales).
  • The fact that we have a Lipschitz domain so that for the parts where the boundary is not sufficiently close to a hyperplane we have bounded estimates between the measure of the boundary and the corresponding measure of its $\epsilon$-neighbourhood.
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Let us assume $D$ is $C^1$: it means that for any point of $x\in\partial D$, there exists an open neighborhood $V$ of $x$ and a $C^1$ function $\rho:V\longrightarrow\mathbb R$ with $d\rho\not=0$ such that $$ D\cap V\equiv \rho(x)<0. $$ Using a partition of unity, we may assume that $\rho$ is $C^1$ defined globally with $d\rho\not=0$ at $\partial D$. Then to define the Euclidean surface measure on $\partial D$, we set formally $$ \int_{\partial D} f d\sigma=\int_{\mathbb R^n}f(x)\delta_0(\rho(x))\Vert d\rho(x)\Vert dx, $$ where $\delta_0$ is the one-dimensional Dirac mass. To justify this, we take $\chi\in C_c^\infty(\mathbb R)$ with $\int \chi(t) dt =1$ and define $$ \int_{\partial D} f d\sigma=\lim_{\epsilon\rightarrow 0_+}\int_{\mathbb R^n}f(x)\chi(\rho(x)\epsilon^{-1})\Vert d\rho(x)\Vert \epsilon^{-1}dx. $$ We obtain a Radon measure supported on $\partial D$ which is independent of the choice of $\chi$ as above and is the Euclidean surface measure on $\partial D$. That construction should extend to the case where $\rho$ is Lipschitz continuous.

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