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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!

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  • $\begingroup$ Isn't $L_t$ equal to $\{u=t\}$ rather than $\{u\geq t\}$ ? $\endgroup$
    – Thomas Richard
    Commented May 7, 2013 at 15:13
  • $\begingroup$ What kind of regularity do you have on $u$? $\endgroup$
    – Benoît Kloeckner
    Commented May 7, 2013 at 16:12
  • $\begingroup$ @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? $\endgroup$
    – Ryan O'Donnell
    Commented May 7, 2013 at 23:29
  • $\begingroup$ @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. $\endgroup$
    – Ryan O'Donnell
    Commented Jun 24, 2013 at 10:34
  • $\begingroup$ 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). $\endgroup$
    – Benoît Kloeckner
    Commented Jun 24, 2013 at 10:59

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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.

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  • $\begingroup$ (The link above, ams.org/mathscinet-getitem?mr=1857292, is to the review of Functions of bounded variation and free discontinuity problems, by Ambrosio, Fusco, and Pallara.) $\endgroup$
    – Andrés E. Caicedo
    Commented Jun 25, 2013 at 0:58
  • $\begingroup$ Excellent reference, thanks bezirsk! $\endgroup$
    – Ryan O'Donnell
    Commented Jun 28, 2013 at 1:16

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