# Technical question on perimeter of level sets

Sorry for asking such a basic question, but this is not my area of expertise.

In my work I'm using the coarea formula: for $\Omega \subseteq \mathbb{R}^n$ open and bounded, and $u : \Omega \to \mathbb{R}$ Lipschitz,

$\int_{\Omega} |\nabla u| = \int_{-\infty}^\infty \mathcal{H}_{n-1}(u^{-1}(t))dt$.

I can calculate the LHS and want to use this to reason about the RHS.

Question: Is it true that there is a "standard definition" of surface area such that for almost all $t$, the set $L_t = \{x \in \Omega : u(x) \geq t\}$ is "nice enough" that SurfaceArea$(L_t) = \mathcal{H}_{n-1}(u^{-1}(t))$? And is there a reasonable reference for this?

If it makes a difference, my $u$ is actually defined on the torus $\mathbb{R}^n/\mathbb{Z}^n$, which may make things cleaner.

Thanks!

• Isn't $L_t$ equal to $\{u=t\}$ rather than $\{u\geq t\}$ ? May 7, 2013 at 15:13
• What kind of regularity do you have on $u$? May 7, 2013 at 16:12
• @Thomas Richard: no, intuitively one would expect that the boundary of $\{u \geq t\}$ would be $\{u = t\}$. @Benoit Kloekcner: well... suppose it's merely Lipschitz -- am I out of luck? May 7, 2013 at 23:29
• @Benoit Kloekcner: Actually, in my application with a little work I can make $u$ piecewise-affine, in which case everything is completely fine. However, I would prefer if I could just leave it Lipschitz and cite something, rather than throw in an extra approximation argument. Jun 24, 2013 at 10:34
• I do not have an answer, but this seems plausible for Lipschitz functions. You could have a look in Federer's book or on functional inequalities where this kind of thing are used (sample keywords: Faber-Krahn, Schwarz symmetrization). Jun 24, 2013 at 10:59

The answer to your first question is yes. You can take as surface area of a set the (total) variation of its characteristic function. This is the "standard definition" of perimeter for general measurable sets in $\mathbb{R}^n$. See, for example, this book, in particular Definition 3.35 and Theorem 3.40.