The coarea formula for BV functions

Today, we will discuss the coarea formula for $BV$ functions, established by Fleming and Rishel in 1960 (W. Fleming, R. Rishel, An integral formula for total gradient variation, Arch. Math. 11 (1960), 218-222.)

Let us first recall the lower semicontinuity property of the total variation with respect to the strong convergence in $L^1(\Omega)$:
Let $\{u_n\}_{n\in\Omega}$ be a sequence in $BV(\Omega)$ strongly converging to some $u$ in $L^1(\Omega)$ and satisfying $|u_{n}|_{BV} < \infty$. Then $u\in BV(\Omega)$ and $$|u|_{BV(\Omega)} \leq \lim\inf_{n\rightarrow \infty} |u_n|_{BV(\Omega)}.$$ The coarea formula is as stated as follows:

Coarea formula: Let $u$ be a given function in $BV(\Omega)$. Then for almost every $t$ in $\mathbb{R}$, the level set $E_t=\{ x\in \Omega \subset \mathbb{R}^N: u(x) > t\}$ of $u$ is a set of finite perimeter in $\Omega$, and
$$\begin{align} &Du=\int_{-\infty}^{\infty} D\chi_{E_t}\, dt,\\ &|u|_{BV(\Omega)}=|Du|(\Omega)=\int_{-\infty}^{\infty} \int_{\Omega}|D\chi_{E_t}|\, dt. \end{align}$$
The second assertion above is often written in terms of the perimeter of level sets as follows:
$$\begin{align} |u|_{BV(\Omega)}=\int_{-\infty}^{\infty} Per(E_t, \Omega) dt, \end{align}$$
where $Per(E_t, \Omega)=\mathcal{H}^{N-1}(\Omega \cap \partial E_t)$ is the perimeter of the level set $E_t$.

Proof of the coarea formula: We follow the treatment of Variational Analysis in Sobolev and BV spaces by Attouch et al.

Part I: Proof of $|Du|(\Omega) \leq \int_{-\infty}^{\infty}\int_{\Omega}|D\chi_{E_t}|\, dt$:

Let us assume that $D\chi_{E_t}$ belongs to $\mathbf{M}(\Omega, \mathbb{R}^N)$, the set of all $\mathbb{R}^N-$ valued Borel measures which is the set of all the $\sigma-$ additive set functions $\mu:\mathcal{B}(\Omega)\rightarrow \mathbb{R}^N$, with $\mu(\emptyset)=0$. For all $t$ in $\mathbb{R}$ set
$$\begin{equation} f_t=\left\{ \begin{array}{ll} \chi_{E_t} & \mbox{if } t \geq 0 \\ -\chi_{\Omega \backslash E_t} & \mbox{if } t < 0 \end{array} \right. \end{equation}$$ It is easy to see that $u(x)=\int_{-\infty}^{\infty}f_t(x)\,dt$ for all $x\in \Omega$. For all $\bar{\phi} \in C_c^1(\Omega, \mathbb{R}^N)$ we have $$\begin{align*} &\langle Du, \bar{\phi} \rangle \\ &= -\int_{\Omega} u \, \mbox{div} \,\bar{\phi} \, dx \\ & = - \int_{\Omega} \int_{-\infty}^{\infty} f_t \, \mbox{div} \,\bar{\phi} \, dt\,dx\\ &= - \int_{\Omega} \int_{-\infty}^{0} f_t \, \mbox{div} \,\bar{\phi} \, dt\,dx -\int_{\Omega} \int_{0}^{\infty} f_t \, \mbox{div} \,\bar{\phi} \, dt\,dx \\ &= \int_{\Omega} \int_{-\infty}^{0} \chi_{\Omega \backslash E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx -\int_{\Omega} \int_{0}^{\infty} \chi_{E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx \\ &=\int_{\Omega} \int_{-\infty}^{0} (1-\chi_{E_t}) \, \mbox{div} \,\bar{\phi} \, dt\,dx -\int_{\Omega} \int_{0}^{\infty} \chi_{E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx \\ &=\int_{\Omega} \int_{-\infty}^{0} \mbox{div} \,\bar{\phi} \, dt\,dx - \int_{\Omega} \int_{-\infty}^{0} \chi_{E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx-\int_{\Omega} \int_{0}^{\infty} \chi_{E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx \\ &= \int_{\Omega} \int_{-\infty}^{0} \mbox{div} \,\bar{\phi} \, dt\,dx-\int_{\Omega} \int_{-\infty}^{\infty} \chi_{E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx \\ &= 0-\int_{\Omega} \int_{-\infty}^{\infty} \chi_{E_t} \, \mbox{div} \,\bar{\phi} \, dt\,dx \\ &= \int_{-\infty}^{\infty} \langle D\chi_{E_t}, \bar{\phi}\rangle \,dt \end{align*}$$ Hence, $Du=\int_{-\infty}^{\infty} D\chi_{E_t}\, dt$, and $|Du|(\Omega) \leq \int_{-\infty}^{\infty}\int_{\Omega}|D\chi_{E_t}|\, dx dt$.

Part II: (coverse) Proof of $\int_{-\infty}^{\infty}\int_{\Omega}|D\chi_{E_t}|\, dxdt\leq |Du|(\Omega)$:


Step i. We first assume that $u$ belongs to the space $\mathcal{A}(\Omega)$ of piecewise linear and continuous functions in $\Omega$. By linearity, one can assume that $u=a\cdot x+b$ with $a\in \mathbb{R}^N$ and $b\in \mathbb{R}$, so that

$$\begin{align*} &\int_{\Omega} |D\chi_{E_t}|\\ &=\mathcal{H}(\Omega \cap \partial E_t)\\ &=\mathcal{H}(\Omega \cap \{x| a\cdot x +b=t\})\\ &=\int_{\Omega \cap \{x| a\cdot x +b=t\}} d\mathcal{H}^{N-1}(x)\\ &=\int_{\{x| a\cdot x +b=t\}}\chi_{\Omega}(x)\, d\mathcal{H}^{N-1}(x) \end{align*}$$
Thus we get,
$$\begin{align*} &\int_{-\infty}^{\infty} \int_{\Omega} |D\chi_{E_t}|\, dx dt\\ &=|a|\mathcal{L}(\Omega)\\ &=\int_{\{x| a\cdot x +b=t\}}\chi_{\Omega}(x)\, d\mathcal{H}^{N-1}(x)\\ &=\int_{\Omega} |Du| \end{align*}$$
Hence we have,
$$\begin{align*} \int_{-\infty}^{\infty} \int_{\Omega} |D\chi_{E_t}|\, dx dt=|Du|(\Omega) \dots \mbox{ if } u=a\cdot x+b. \end{align*}$$

Step ii. Now we prove that $\int_{-\infty}^{\infty} \int_{\Omega} |D\chi_{E_t}|\, dx dt\leq|Du|(\Omega)$ for all $u\in BV(\Omega)$. We know that the space of piecewise linear and continuous functions $\mathcal{A}(\Omega)$ is dense in $W^{1, 1}(\Omega)$ when equipped with the strong topology. We also know that the space $C^{\infty}\cap W^{1, 1}(\Omega)$ is dense in $BV$ when equipped with the intermediate convergence. Thus, there exists a sequence $\{u_n\}_{n\in \mathbb{N}} \in \mathcal{A}(\Omega)$ such that $u_n\rightharpoonup u$ for the intermediate convergence. For each of the functions $u_n$ we set $E_{n, t}:=\{x\in \Omega: u_n(x) > t\}$. Due to the intermediate convergence, we have
$$\begin{align*} &\int_{\Omega}|Du|=\lim_{n\rightarrow \infty}\int_{\Omega}|Du_n| \dots \mbox{the intermediate convergence}\\ &=\lim_{n\rightarrow \infty} \int_{-\infty}^{\infty} \int_{\Omega} |D\chi_{E_{n,t}}|\, dx dt \dots \mbox{from step i}\\ &\geq \int_{-\infty}^{\infty} \lim\inf_{n\rightarrow \infty} \int_{\Omega} |D\chi_{E_{n,t}}|\, dx dt \dots \mbox{Fatou's lemma}\\ \end{align*}$$
From the intermediate convergence we also have $u_n\rightarrow u$ in $L^1(\Omega)$. i.e.
$$\begin{align*} \int_{\Omega} |u_n-u|=\int_{\Omega} \int_{-\infty}^{\infty} |\chi_{E_{n,t}}-\chi_{E_t}|\ dt dx = \int_{-\infty}^{\infty} \Big(\int_{\Omega} |\chi_{E_{n,t}}-\chi_{E_t}|\ dx \Big) dt=0. \end{align*}$$
Thus, for a subsequence $E_{n_m, t}$, and for almost all $t \in \mathbb{R}$, $\chi_{E_{n_m, t}}\rightarrow \chi_{E_{n, t}}$ strongly in $L^1(\Omega)$. The lower semicontinuity of the total variation with respect to the strong convergence in $L^1(\Omega)$ we get $\lim\inf_{n\rightarrow \infty} \int_{\Omega} |D\chi_{E_{n_m,t}}|\, dx \geq \int_{\Omega} |D\chi_{E_{t}}|\, dx$. Thus, we get,
$$\int_{\Omega}|Du| \geq \int_{-\infty}^{\infty} \int_{\Omega} |D\chi_{E_{t}}|\, dx dt.$$
This completes the proof!

(By the way, step ii also proves that $D\chi_{E_t} \in \mathbf{M}(\Omega, \mathbb{R}^N)$ for a.e. $t\in \mathbb{R}$, something that we used in Part I of the proof.)

Hausdorff measure and dimension

Let us talk about Hausdorff measure $\mathcal{H}^s$ named after Felix Hausdorff, which comes up in image processing often. These measures allow us to measure very small subsets of $\mathbb{R}^n$. The idea is that $A$ is an $s-$ dimensional subset of $\mathbb{R}^n$ if $0<\mathcal{H}^s(A)<\infty$. We will follow the notation and treatment of Evans and Gariepy.

Felix Hausdorff

Definitions:
(i) Let $A\subset \mathbb{R}^n$, $0\leq s < \infty$, $0<\delta\leq \infty$. Define $$\mathcal{H}^s_{\delta}(A)=\inf \Big\{ \sum_{j=1}^{\infty} \alpha(s) \Big( \frac{\mbox{diam}(C_j)}{2} \Big)^s \, \Big\vert \,A\subset \cup_{j=1}^{\infty}C_j, \mbox{diam}(C_j)\leq \delta \Big\},$$ where $\{C_j\}\subset \mathbb{R}^n$ is the covering of the set $A$ and $\alpha(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma(\frac{s}{2}+1)}.$ Here $\Gamma(s)=\int_0^{\infty}e^{-x}x^{s-1}\,dx,\,\,(0< s<\infty),$ is the usual gamma function. (ii) The $s-$dimensional Hausdorff measure of $A$ is denoted by $\mathcal{H}^s(A)$ and it is defined as the limit:
$$\mathcal{H}^s(A)=\lim_{\delta\rightarrow 0}\mathcal{H}^s_{\delta}(A)=\sup_{\delta >0} \mathcal{H}^s_{\delta}(A).$$

The book: 'Measure theory and fine properties of functions' by Evans and Gariepy includes the normalizing constant $\alpha(s)$ in the definition, whereas some other authors do not include this constant. The constant $\alpha(s)$ is included because if $s$ is an integer, then $\mathcal{H}^s(A)$ is equal to the $s-$dimensional surface area for nice sets (graph of Lipschitz function).


Some properties of Hausdorff measure:
1.$\mathcal{H}^s$ is a Borel regular measure. Refer to E.G. for the proof.

2. Let us consider a Hausdorff measure $\mathcal{H}^0$ of a singleton $\{a\}\in \mathbb{R}^n$. First note that $\alpha(0)=1$. So, $\mathcal{H}^0(\{a\})=\mathcal{H}^0_{\delta}(\{a\})=1$. Thus, $\mathcal{H}^0$ is a counting measure.

3. $\mathcal{H}^1=\mathcal{L}^1$ on $\mathbb{R}^1$, i.e. $\mathcal{H}^1$ measure of a line segment is its length.

4. Since $\mathbb{R}^n$ is not $\sigma-$finite, $\mathcal{H}^s$ is not a Radon measure for $0\leq s < n$.

5. If $s > n$ then $\mathcal{H}^s\equiv 0$ on $\mathbb{R}^n$.
To see this is easy. Let us consider a unit cube $Q$ in $\mathbb{R}^n$, fix an integer $m \geq 1$. The unit cube can be decomposed into $m^n$ cubes with sides $1/m$ and diameter ${n^{1/2}}/{m}$. Therefore,
$$\mathcal{H}^s_{{\sqrt{n}}/{m}}(Q)\leq \sum_{i=1}^{m^n} \alpha(s) (n^{1/2}/m)^s = \alpha(s) n^{s/2} m^{n-s}.$$ If $s >n$, then $m^{n-s}\rightarrow 0$ as $m\rightarrow \infty$ i.e. as the size of the covering vanishes. Thus, $\mathcal{H}^s(Q)=0$, so $\mathcal{H}^s(\mathbb{R}^n)=0$ for $s >n$.

6. Scaling: $\mathcal{H}^s(\lambda A)=\lambda^s\mathcal{H}^s(A)$ for all $\lambda > 0$, $A \subset \mathbb{R}^n$. (Easy to prove.)

7. Invariance under affine transformation: $\mathcal{H}^s(L(A))=\mathcal{H}^s(A)$ for each affine isometry $L:\mathbb{R}^n\rightarrow \mathbb{R}^n, \, A \subset \mathbb{R}^n$.

8. $\mathcal{H}^n=\mathcal{L}^n$ on $\mathbb{R}^n$.


Hausdorff-Besicovitch dimension:
The hausdorff dimension of a set $A\subset \mathbb{R}^n$ is defined to be

$$\mathcal{H}_{\mbox{dim}}(A)=\inf\{ 0\leq s < \infty | \mathcal{H}^s(A)=0 \}.$$

The following assertions follow from the definition:
Countable sets have Hausdorff dimension $0$. The Euclidean space has Hausdorff dimension $n$.
The circle has Hausdorff dimension $1$.
The Cantor set (a zero-dimensional topological space) is a union of two copies of itself, each copy shrunk by a factor 1/3; this fact can be used to prove that its Hausdorff dimension is ${\ln 2}/{\ln 3}$.