Working with the level set method introduced by *Osher & Sethian* in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here [Lebesgue measure of a neighbourhood of a curve](https://mathoverflow.net/questions/385144/lebesgue-measure-of-a-neighbourhood-of-a-curve)). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.


Here is my question:


Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi\_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:


$\bullet\ \phi\_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits\_{n\to\infty} \sup\_{x\in\overline{\Omega}}|\phi\_n(x)-\phi(x)|+\sup\_{x\in\overline{\Omega}} |\nabla\phi\_n(x)-\nabla\phi(x)|=0$


$\bullet$ $\nabla\phi\_n\neq 0$ on $\phi\_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$


$\bullet$ $\{\phi=0\}\Subset\Omega$


Then how can we prove that:


$$\lim\limits\_{n\to\infty} \int\_{\phi\_n=0} 1 \ d\sigma=\int\_{\phi=0} 1\ d\sigma$$


?


For the continuity of the perimeter on level sets with respect to the level you can see other posts here: [Continuity of Hausdorff measure on level sets](https://mathoverflow.net/questions/398167/continuity-of-hausdorff-measure-on-level-sets), [Continuity of surface integrals on level sets](https://mathoverflow.net/questions/401707/continuity-of-surface-integrals-on-level-sets)



Continuity of the perimeter of level sets w.r.t. level function
As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi\_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi\_n(x)=0\}$ is empty. The measure of *interior* zeros is neither continuous: if $\psi\_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi\_n(x)=0\}$ is a circle of radius $1-o(1).$


You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi\_n$ by $N\ge0$ closed curves $\{\gamma\_{j,n}:\mathbb{S}^1\to\Omega\}\_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma\_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$ and $\phi\_n$, which is finite and definitively constant wrto $n$. Then the length of each component $\int\_{\mathbb{S}^1}|\dot\gamma\_{j,n}(s)|ds$ passes to the limit.


