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