I have a question about coarea formula.
Last day, I found the following assertion in a paper:
"Since $f \in C(\bar{D})$ and $D$ is a bounded $C^1$-domain,
\begin{align*}
(1)\quad
\lim_{\varepsilon \to 0}\frac{1}{\varepsilon}\int_{D_{\varepsilon}}f\,dx=\int_{\partial D}f\,d\sigma,
\end{align*}
where $D_{\varepsilon}=\{x \in \bar{D}:d(x,\partial D) \le \varepsilon\}$ and $\sigma$ is the surface measure on $\partial D$ ($(d-1)$-dim Hausdorff measure on $\partial D$)."
Since the author state this assertion without proof, I want to know how to prove and generalize this assertion.
My attempt
Let $D$ be a bounded domain on $\mathbb{R}^d$. That is, $D$ is a connected bounded open subset of $\mathbb{R}^d$. We assume $m(\partial D)=0$ ($m$ is the $d$-dim Lebesgue measure).
-
Define $F(x)=\inf_{y \in \partial D}|x-y|$.
Note that $F$ is a Lipschitz continuous function on $\mathbb{R}^d$ and $|\nabla F|=1$ $m$-a.e. -
Take $f \in L^{1}(\bar{D},m)$ and extend it on $\mathbb{R}^d$ by putting $f=0$ on $\mathbb{R^d}\setminus \bar{D}$.
Using coarea formula, we have $\int_{\mathbb{R}} \int_{\{F=s\}}|f|\,d\sigma\,ds<\infty $ and
\begin{align*}
\int_{\{F>t\}}f\,dx=\int_{t}^{\infty}\left(\int_{\{F=s\}}f\,d\sigma \right)\,ds,\quad t \in \mathbb{R}.
\end{align*}
Therefore, we have
\begin{align*}
\frac{1}{\varepsilon}\int_{\{0<F \le \varepsilon\}}f\,dx=\frac{1}{\varepsilon} \int_{0}^{\varepsilon} \left(\int_{\{F=s\}}f\,d\sigma \right)\,ds.
\end{align*}
Since $f=0$ on $\mathbb{R}^d \setminus \bar{D}$,
\begin{align*}
\frac{1}{\varepsilon}\int_{D_{\varepsilon}}f\,dx=\frac{1}{\varepsilon} \int_{0}^{\varepsilon} \left(\int_{\{F=s\}}f\,d\sigma \right)\,ds.
\end{align*}
However, is $s \mapsto \int_{\{F=s\}}f\,d\sigma$ continuous?
My question
When is $s \mapsto \int_{\{F=s\}}f\,d\sigma$ continuous? Even if this is not true, under what conditions do we still have
$$
\frac{1}{\varepsilon} \int_{0}^{\varepsilon} \left(\int_{\{F=s\}}f\,d\sigma \right)\,ds \to \int_{\{F=0\}}f\,d\sigma \, ?
$$
Even if $D$ is Lipschitz domain and $f$ is continuous, (1) must be valid.
Thanks for any information.
Information for (1)
Recently, I found an extension of (1) in this paper: enter link
description here. If you are interested, please look at Lemma 7.1
in this paper.
Best Answer
Disintegration of the volume w.r.t. a surface measure on $\Sigma_s=\{F=s\}\cap D$ and the parameter $s$ is difficult when $\Sigma_0=\partial D$ is only $C^1$. $\Sigma_s$ is in general only Lipschitz, although with a Lipschitz constant that goes to 1 as $s\rightarrow 0$. I guess there is still a well-defined surface measure on $\Sigma_s$ but I wouldn't know how to use it in calculations. It is much easier to work directly with the volume form:
In the following, write $n_p$ for the unit vector, normal to the tangent space of $\Sigma_0$ at $p$ and pointing inwards in $D$.
First a geometric estimate: For $\delta>0$ consider the quantity: $$ \alpha(\delta) = \{ n_p \cdot v : p\in \Sigma_0, v \in T_q \Sigma_0, |v|=1, d(p,q)\leq \delta\}.$$ The dot-product is the cosine of the angle between the normal vector $n_p$ at $p$ and a tangent vector at $q$ when $p,q\in \Sigma_0$ are $\delta$-close. $\alpha$ estimates the local 'flatness' of the surface. We have $\alpha(0)=0$ and $\lim_{\delta\rightarrow 0} \alpha(\delta)=0$ because $\Sigma_0$ is assumed $C^1$ and compact.
Let $f$ be a continuous function with support in a small ball $B(p,\delta/2)$ around $p\in \Sigma_0$. We will show that for $\delta$ small $$ \lim_{s\rightarrow 0 } \frac{1}{s} \int_{\{0\leq F \leq s\}\cap D} f\; d^d x = \int_{\Sigma_0} f\; d\sigma_0$$ where $d\sigma_0$ is the area measure on the surface (well-defined as $\Sigma_0$ is $C^1$). This suffices since for general $f$ you may use a partition of unity to reduce to this case.
Local coordinates: Choosing $\delta>0$ small enough we may ensure that $B(p,\delta)\cap \Sigma_0$ is the (embedded) image of a $C^1$ map: $$ j : U \rightarrow \Sigma_0 \subset {\Bbb R}^d, \; 0\in U \subset {\Bbb R}^{d-1}, j(0)=p, $$ where $j'(u)$ is injective for all $u\in U$ and its image is the tangent space of $\Sigma_0$ at $j(u)$. Write $n(u) = n_{j(u)}$ and consider now the map: $$ \Phi(u,s) = j(u) + s n(u) , u\in U, s\in [0,\delta/2]$$
The Jacobian determinant, $J(u,s) = \det \Phi' = \det ( j'(u) + s n'(u), n(u))$ is non-zero at $(u,s)=(0,0)$ and since continuous in $u$ and $s$ we may by the inverse function theorem (possibly choosing $\delta$ smaller) assume that $\Phi$ is a bijection onto its image. We then have an expression for the volume element $d^d x = J(u,s) d^{d-1}u ds$ in local coordinates. Furthermore, when restricted to $s=0$ it gives the $(d-1)$-dim area element on the surface (because $n$ is a unit normal to the tangent space): $$ d\sigma_0 = J(u,0) d^{d-1} u $$
The distance from $\Phi(u,s)$ to $\Sigma_0$ is clearly at most $s$ but a geometric consideration (make a drawing) using the definition of $\alpha$ shows that it is not smaller than $s \sqrt{1-\alpha(\delta)^2}$ (the last factor is an upper bound for the sine of the angle between two normal vectors in $\Sigma_0\cap B(p,\delta)$). It follows that for $0\leq s \leq \delta/2$ and restricting to the set $B(p,\delta/2)$ we have $$ \{x\in D: 0\leq F\leq s \sqrt{1-\alpha(\delta)^2}\} \subset \Phi(U,s) \subset \{x\in D: 0\leq F\leq s\} . $$ Therefore, $$ \frac{1}{s} \int_{0\leq F\leq s \sqrt{1-\alpha(\delta)^2} } f \; d^d x \leq \frac{1}{s} \int_0^s \left( \int_U f(u,s) J(u,s) d^{d-1}u \right) ds \leq \frac{1}{s}\int_{0\leq F\leq s} f \; d^d x $$ As $s\rightarrow 0^+$ the middle term goes to $\int_{\Sigma_0} f\; d\sigma_0$ by simple continuity, and the outer terms (limsup and liminf) agree within a factor $ \sqrt{1-\alpha(\delta)^2}$. Patching things together we may finally let $\delta$ go to zero and conclude since this factor goes to one.