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.)

More on intermediate convergence

According to the vectorial version of the Riesz-Alexandroff representation theorem, the dual norm $\vert u \vert_{BV}=\Vert Du \Vert=\sup \{\int_{\Omega} u \,\mbox{ div } \mathbf{\phi}: \mathbf{\phi} \in C_c^1(\Omega, \mathbb{R}^N) \mbox{ and } \Vert \mathbf{\phi} \Vert_{\infty}\leq 1\}$ is also the total mass $|Du|(\Omega)=\int_{\Omega}|Du|$ of the total variation $|Du|$ of the measure $Du$.

Moreover, from classical integration theory, the integral $\int_{\Omega}\phi \,d(Du)$ is well defined for all $Du$ integrable functions $\phi$ from $\Omega$ into $\mathbb{R}^N$, e.g. for functions in $C_b(\Omega, \mathbb{R}^N)$. For the same reasons, $\int_{\Omega}\phi\, d|Du|$ is well defined for all $|Du|$ integrable functions $\phi$, in particular for functions in $C_b(\Omega)$.

Thus, we can say that $|Du_n|\rightharpoonup |Du|$ narrowly in $M(\Omega)$ if $\int_{\Omega}\phi \, d|Du_n|\rightarrow \int_{\Omega}\phi\,d|Du|$, for all $\phi \in C_b(\Omega)$. If we choose $\phi\equiv 1$ on $\Omega$ then $\Vert Du_n \Vert \rightarrow \Vert Du \Vert$ i.e. $|u_n|_{BV}\rightarrow |u|_{BV}$.

Hence, $u_n\rightarrow u$ in $L^1(\Omega)$ and the narrow convergence of measures, $|Du_n|\rightharpoonup |Du|$ in $M(\Omega)$ implies the intermediate convergence, $u_n\rightharpoonup u$ (i.e. $u_n\rightarrow u$ in $L^1(\Omega)$ and $|u_n|_{BV}\rightarrow |u|_{BV}$).


We have the following proposition regarding the intermediate convergence.

Proposition
The following three assertions are equivalent
(i) $u_n\rightharpoonup u$ in the sense of intermediate convergence.
(ii) $u_n\rightharpoonup u$ in $BV(\Omega)$ and $\vert u_n \vert_{BV}\rightarrow \vert u \vert_{BV}$.
(iii) $u_n\rightarrow u$ strongly in $L^1(\Omega)$ and $\vert Du_n \vert\rightharpoonup \vert Du \vert$ narrowly in $M(\Omega)$.

Proof
The equivalence (i) to (ii) is a consequence of a proposition from the previous post. We have just seen above that (i) and (iii) are equivalent. We only need to prove that (ii) implies (iii).

Recall that $u_n\rightharpoonup u$ weakly in $BV(\Omega)$ means that
(a) $u_n\rightarrow u$ strongly in $L^1(\Omega)$ and
(b) the measures $Du_n\rightharpoonup Du$ weakly in $M(\Omega, \mathbb{R}^N)$.

Also, recall that for non-negative Borel measures $\mu_n$ and $\mu$ in $M(\Omega)$ the following statements are equivalent:
(c) $\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu(U)\leq \lim\inf_{n\rightarrow\infty}\mu_n(U)$ for all open subsets $U\subset \Omega$.
(d) $\mu_n\rightharpoonup \mu$ narrowly in $M(\Omega)$.

Let, $\mu_n=|Du_n|$ and $\mu=|Du|$. We have been given that $|Du_n|(\Omega)\rightarrow |Du|(\Omega)$ i.e. we have $\mu_n(\Omega)\rightarrow \mu(\Omega)$. For showing the narrow convergence $\mu_n\rightharpoonup \mu$ we just need to prove the lower-semicontinuity property on $U\subset\Omega$,
i.e. we need to show that for $\mu(U)\leq \lim\inf_{n\rightarrow\infty}\mu_n(U)$ for all open subsets $U\subset \Omega$,
i.e. we need to show that for $|Du|(U)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(U)$ for all open subsets $U\subset \Omega$.

To this effect we use the proposition from the previous blog which says that if $u_n\in BV(\Omega)$ with $|u_n|_{BV(\Omega)}$ bounded, with $u_n\rightarrow u$ in $L^1(\Omega)$, then we have the lower semicontinuity property i.e. $|Du|(\Omega)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(\Omega)$, and $u\in BV(\Omega)$.

Let $U\subset\Omega$ be any open subset. We have $u_n\in BV(\Omega)$, thus as $U\subset\Omega$, we have $u_n\in BV(U)$. Thus, we have $\sup_n |Du_n|(U)\leq \sup_n |Du_n|(\Omega)<\infty$.

The weak convergence $u_n\rightharpoonup u$ in $L^1(\Omega)$ implies the strong convergence $u_n\rightarrow u$ in $L^1(\Omega)$ and thus the strong convergence $u_n\rightarrow u$ in $L^1(U)$ follows as $U\subset \Omega$.

 Now that we have the conditions in the proposition: $u_n\in BV(U)$, $\sup_n |Du_n|(U)<\infty$, and $u_n\rightarrow u$ in $L^1(U)$ we have the lower semicontinuity property: $|Du|(U)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(U)$ for all open subsets $U\subset \Omega$. 

As we have $|Du_n|(\Omega)\rightarrow |Du|(\Omega)$ and $|Du|(U)\leq \lim\inf_{n\rightarrow\infty}|Du_n|(U)$ for all open subsets $U\subset \Omega$, from the equivalence of (c) and (d) we get the narrow convergence: $|Du_n|\rightharpoonup |Du_n|$ narrowly in $M(\Omega). \,\,\hspace{25pt} \square$.

In conclusion, intermediate convergence implies narrow convergence of $|Du_n|$.

Narrow convergence of measures and intermediate convergence of BV functions

Recall By the Riesz-Alexandroff representation theorem the dual of the space $C_0(\Omega,\mathbb{R}^N)$ (and thus of $C_c(\Omega,\mathbb{R}^N)$ can be isometrically identified with $M(\Omega,\mathbb{R}^N)$. i.e. any Borel measure $\mu$ is a bounded linear functional (continuous linear form if you like that) on $C_0(\Omega,\mathbb{R}^N)$ or $C_c(\Omega,\mathbb{R}^N)$. Thus, the weak convergence of a sequence of Borel measures $\mu_n$ is defined as follows.

Weak convergence of Borel measures: We say that a sequence of Borel measures $\mu_n$ converges weakly to $\mu$ i.e. $\mu_n\rightharpoonup \mu$ if $\langle\mu_n, \phi \rangle \rightarrow \langle \mu, \phi\rangle $ for all $\phi \in C_0(\Omega,\mathbb{R}^N)$.

Here, $\langle \mu, \,\phi\rangle\equiv\mu(\phi)\equiv\int_{\Omega} \phi \, d\mu \equiv \sum_{i=1}^N\int_{\Omega} \phi_i d\mu_i$.

Narrow convergence: A sequence $\{\mu_n\}$ in $M(\Omega, \mathbb{R}^N)$ narrowly converges to $\mu$ ($\mu_n\rightharpoonup \mu$ narrowly) in $M(\Omega, \mathbb{R}^N)$ if and only if $$\int_{\Omega} f\, d\mu_n \rightarrow \int_{\Omega} f d\mu,$$ for all $f \in C_b(\Omega, \mathbb{R}^N)$.

Narrow convergence is stronger than weak convergence.
If $\Omega=(0, 1)$ and $\mu_n=\delta_{\frac{1}{n}}$ then $\mu_n\rightharpoonup 0$ weakly. This is because for all $\phi \in C_0(\Omega)$, we have, $\mu_n(\phi)=\phi(\frac{1}{n})\rightarrow \phi(0)=0=\mu(\phi)$.

On the other hand, if $\phi\equiv1$ on $\Omega$, (note that $\phi \not\in C_0(\Omega)$ but $\phi \in C_b(\Omega)$), then we see that $\int_{\Omega} d\mu_n=\mu_n(\Omega)=1$ for all $n$. Essentially what happened with $\mu_n=\delta_{\frac{1}{n}}$ is that it converges weakly to $\mu=0$, but $\mu_n(\Omega)$, the size of $\Omega$ measured in $\mu_n$ does not converge to the size of $\Omega$ in measure $\mu(=0)$, that is $\mu_n\rightharpoonup \mu$ but $\mu_n(\Omega)\nrightarrow \mu(\Omega)$.

We have the following compactness result for the narrow topology.

PROKHOROV's sequential compactness theorem
If for a bounded subset $\mathcal{H}$ of $M^+(\Omega)$ it is true that for all $\epsilon$ there exists a compact subset $K_{\epsilon}\subset \Omega$ such that $\sup\{\mu(\Omega\ K_{\epsilon}):\mu\in \mathcal{H}\}\leq \epsilon$, then $\mathcal{H}$ is sequentially compact for the narrow topology.

The following equivalence proposition says that the convergence $\mu_n(\Omega)\rightarrow \mu(\Omega)$ is precisely what we need to ensure narrow convergence of measures.

Proposition
Let $\mu_n$ and $\mu$ be in $M^+(\Omega)$, non-negative Borel measures, then the following assertions are equivalent:
(i) $\mu_n\rightharpoonup \mu$ narrowly.
(ii)$\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu \rightharpoonup \mu$ weakly.

Proof
Note that as narrow convergence is stronger than weak convergence, it is clear that if $\mu_n\rightarrow \mu$ narrowly then $\mu \rightharpoonup \mu$. Also, $\mu_n(\Omega)\rightarrow \mu(\Omega)$ follows by taking $f\equiv 1$.

So, we need to only prove that if $\mu \rightharpoonup \mu$ weakly and $\mu_n(\Omega)\rightarrow \mu(\Omega)$ then $\mu_n\rightharpoonup \mu$ narrowly. That is, we need to show that for any function $f$ in $C_b(\Omega, \mathbb{R}^N)$ we have $\big| \int_{\Omega} fd\mu_n - \int_{\Omega} fd\mu \big|\rightarrow 0$. To do this we add and subtract $\int_{\Omega} f\,\phi d\mu_n$ and $\int_{\Omega} f\,\phi d\mu$ where $\phi$ is continuous function that is compactly supported in $\Omega$. The result follows as $f$ is bounded i.e. $\Vert f \Vert_{\infty}$ is a finite number and $\int_{\Omega} f\phi d\mu_n \rightarrow \int_{\Omega} f\phi d\mu$ due to compact support of $\phi$ in $\Omega . \hspace{25pt}\square$

More equivalence proposition
Let $\mu_n$ and $\mu$ be in $M^+(\Omega)$, non-negative Borel measures, then the following assertions are equivalent:
(i) $\mu_n\rightharpoonup \mu$ narrowly.
(ii) $\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu(U)\leq \lim\inf_{n\rightarrow \infty} \mu_n(U)$ for all open subsets $U\subset \Omega$.
(iii) $\mu_n(\Omega)\rightarrow \mu(\Omega)$ and $\mu(F)\geq \lim\inf_{n\rightarrow \infty} \mu_n(F)$ for all closed subsets $F\subset \Omega$.
(iv) $\mu_n(B)\rightarrow \mu(B)$ for all Borel subsets $B\subset\Omega.\,\,\hspace{25pt}\square$

Ok, so far so good. But images are not necessarily continuous functions. They might have jump discontinuities, i.e. edges. For this the following result due to Marle (Measure et probabilities) is helpful:

Proposition
Let $\mu_n$ and $\mu$ be non-negative Borel measures $M^+(\Omega)$, with $\mu_n\rightharpoonup \mu$ narrowly and let $f$ be a $\mu_n$ measurable bounded function for all $n$, such that it is discontinuous only on a set of null $\mu-$ measurable set then
$$\hspace{25pt}\int_{\Omega} f\, d\mu_n \rightarrow \int_{\Omega} f\, d\mu \hspace{25pt}\square$$

Now, that we have seen a notion of convergence for sequence of measures that is stronger than weak convergence, let's look at the a notion of convergence for sequence of functions in $BV$ space that is stronger than weak convergence.

Intermediate convergence
Let $\{u_n\}$ be a sequence in $BV(\Omega)$ and $u\in BV(\Omega)$. we say that $u_n\rightharpoonup u$ intermediately (or in the sense of intermediate convergence) if and only if
$$u_n\rightarrow u \mbox{ in } L^1(\Omega),$$
$$\hspace{25pt}\vert u_n\vert_{BV}\rightarrow \vert u\vert_{BV}\,\,\hspace{25pt}\square$$

Recall, from the previous post that for $\{u_n\}\in BV(\Omega)$ with bounded TV mass and $u_n\rightarrow u$ strongly in $L^1(\Omega)$, then we have $u_n\rightharpoonup u$ in $BV$, but $\vert u \vert_{BV}\leq \lim\inf_{n\rightarrow \infty}\vert u_n \vert_{BV}$ i.e. some of the mass is lost.

Ok, now I am hungry, so more about this tomorrow.