In the book of *S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces*, at page 15, is stated without proof, a formula like that:
$$\text{Per}\_{\Omega}(\omega)=\lim\_{\varepsilon\to 0} \int\limits\_{\Omega}\delta\_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$


where


* $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
* $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
* $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
* $\text{Per}\_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the *relative* perimeter).
* $\delta\_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely
$$
\delta\_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;.
$$
Now, denoting by $\bigvee\_{\Omega} f$ the total variation of $f$ in $\Omega$, here is **my question:**



> 
> **Is it true that for any $\chi\_{\omega}\in BV(\Omega)$ ($\chi\_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi\_n\to \chi\_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi\_n\leq 1, \chi\_n\in W^{1,1}(\Omega)$ for each $n$, we have that:**
> $$\bigvee\_{\Omega} \chi\_{\omega}=\lim\limits\_{n\to\infty} \bigvee\_{\Omega}\chi\_n(x)\;\;\boldsymbol{?}$$
> 
> 
> 




---


**Notes**


* It is known that:
$$\bigvee\_{\Omega} f = \sup\left\{\,\int\limits\_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}\_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert\_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$
* It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that:
$$\bigvee\_{\Omega} \chi=\int\limits\_{\Omega} |\nabla\chi(x)|\ dx.$$
* We have that $\text{Per}\_{\Omega}(\omega)=\bigvee\_{\Omega}\chi\_{\omega}$ for any $\omega\subseteq\Omega$.
* Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so:
$$\bigvee\_{\Omega} \chi\_{\omega}\leq\liminf\limits\_{n\to\infty} \bigvee\_{\Omega}\chi\_n$$
* In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}\_c$ functions. Maybe it is useful.
* I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.


**Motivation**


I'm interested in this type of formulas because if it is indeed true we will have that:
$$\text{Per}\_{\Omega}(\omega)=\lim\limits\_{\varepsilon\to 0}\int\_{\Omega} H'\_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx,
$$
for any function $H\_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H\_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the *Heaviside function*
$$
H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}.
$$ In the above example,
$$
H\_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$
for each $\varepsilon>0$ and $\delta\_{\varepsilon}(x)=H'\_{\varepsilon}(x)$.



Perimeter continuity of $BV$ sets on any sequence from $W^{1,1}$
This seems too strong, regardless of which set $\omega \subset \Omega$ one works with. We suppose that $\omega$ has bounded perimeter, so that $\chi\_\omega \in BV(\Omega)$.


Let $(f\_n \mid n \in \mathbf{N})$ be a sequence of functions in $W^{1,1}(\Omega)$ with
\begin{equation}
\lvert f\_n \rvert\_{L^1} \to 0
\text{ and }
V(f\_n) \geq 3 \, \mathrm{Per}(\omega).
\end{equation}
Let moreover $g\_n \in W^{1,1}(\Omega)$ be a sequence of functions
so that
\begin{equation}
\lvert g\_n - \chi\_\omega \rvert\_{BV} \to 0;
\end{equation}
for example $g\_n$ can be obtained by a mollification argument. Then the two sequences combined have
\begin{equation}
\lvert f\_n + g\_n - \chi\_\omega \rvert\_{L^1} \to 0 
\text{ but }
V(f\_n + g\_n) \geq V(f\_n) - V(g\_n) \geq 3/2 \mathrm{Per}(\omega).
\end{equation}


We construct Lipschitz functions $f\_n$ on $\Omega$ that are independent of $y$. Specifically let, given $A > 0$,
\begin{equation}
f\_n(x,y) =\begin{cases}
Ax & \text{ on $[0,\frac{1}{2n}]$} \\
A/n - Ax & \text{ on $[\frac{1}{2n},\frac{1}{n}]$}
\end{cases}
\end{equation}
and extend this to be $\frac{1}{n}$-periodic in $x$.


Then $\lvert f\_n \rvert\_{L^\infty} \leq \frac{A}{2n} \to 0$ but the total variation is constant equal to $\int\_\Omega \lvert D f\_n \rvert = A \lvert \Omega \rvert$. To conclude it only remains to take $A > 3 \mathrm{Per}(\omega)/\lvert \Omega \rvert$.


