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.



Polar coordinates, bounded domain with $C^{1}$ boundary
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, $F(x)=d(x,D)-d(x,R^n\setminus D)$).


