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I have a question about a integral on a surface.

It is well known that for any Integrable function $f$ defined on $\mathbb{R}^{n}$, it holds that \begin{equation} (1) \quad \frac{d}{dr} \int_{B(0,r)}f\,dm=\int_{\partial B(0,r)}f\,d \sigma \quad m\text{-a.e. }r. \end{equation} Here and hereafter $m$ denotes the $n$-dim Lebesgue measure, $\sigma$ the $(n-1)$ dim Hausdorff measure (surface measure) and $B(0,r)$ the open ball of radius $r$ centered at origin.

Question

Let $D \subset \mathbb{R}^{n}$ be a bounded domain with $C^{1}$ boundary. Set \begin{align} D_{\epsilon}=\left\{ x \in \bar{D} : d\left(x,\partial D \right) \leq \epsilon \right\} \end{align} Can we show the following equation? : \begin{align} \lim_{\epsilon \to 0} \frac{1}{\epsilon}\int_{D_{\epsilon}}f\,dm=\int_{\partial D}f\,d\sigma ,\quad (f \in C(\bar{D})) \end{align}

This is a generalization of $(1)$.

If you know how to prove this equation or helpful references, please let me know.

Thank you in advance.

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I think the cleanest proof is based on the coarea formula, which holds for pretty rough functions. Describe your set $D$ as the level set $\{F(x)\le t\}$ for some suitable function $F:R^n\to R$. By coarea formula you can write $$ \int _ {t-\epsilon<F(x)\le t}f(x)dx= \int_{t-\epsilon}^{t} \int_{F(x)=s} \frac{f(x)}{|\nabla F(x)|}dH_{n-1}ds $$ where $dH_{n-1}$ is the surface measure on the set $\{F(x)=s\}$. This gives $$ \epsilon^{-1}\int _ {t-\epsilon<F(x)\le t}f(x)dx \to \int_{F(x)=t} \frac{f(x)}{|\nabla F(x)|}dH_{n-1}. $$ To obtain your formula, just choose $F(x)=d(x,\partial D)$ (or if you want it smoother, $F(x)=d(x,D)-d(x,R^n\setminus D)$).

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Using a partition of unity, you can reduce the problem to the case when $f$ has compact support and $D$ is a subgraph of Lipshitz function with arbitrary small Lipschitz constant, say $\varepsilon>0$.

In the latter case it is straightforward to prove that your equality holds up to $e^{\pm\varepsilon}$. Since $\varepsilon$ is arbitrary, the statement follows.

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