# Surface integral approximation

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!

• 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.
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.