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