There are two tools, generalizing a concept of a volume to the case of submanifolds in $\mathbb{R}^n$, namely the Hausdorff measure $H^k$ and the volume form. The question is how to show that if $M$ is an orientable $k$-submanifold in $\mathbb{R}^n$ with a volume form $dV$ then $$ \int\limits_{M} f(x) dV = \int\limits_{M} f(x) H^k(dx) $$ if the integrals exist.

P.S. Maybe the question is sufficiently silly for mathoverflow.com and a more suitable for math.stackexchange.com, but I have 2 reasons to post it here:

  1. In books on geometric integration theory (Krantz, Parks; Federer) I failed to find an answer.
  2. I've already posted the question on math.stackexchange.com and on one more forum, but I didn't receive any response.

If this question is very silly, I will delete it.

Thank you.

  • $\begingroup$ It's not a silly question, but have you tried to work it out yourself for a smooth manifold using just the definitions and maybe the implicit function theorem? $\endgroup$ – Deane Yang Feb 8 '12 at 20:26
  • $\begingroup$ Yes, in the case of $n=3$, $k=2$ it is the theorem about equality of the surface integrals of first and second kinds and there is no problem. In the case of arbitrary $n$ and $k$ I can manipulate with the integral on the left and I can receive some different representations of this integral (using local coordinates), but I don't know how to reduce the integral on the right to some suitable form. $\endgroup$ – Appliqué Feb 8 '12 at 20:44
  • 3
    $\begingroup$ You have to use the area formula in Federer. $\endgroup$ – Mohan Ramachandran Feb 8 '12 at 21:32
  • $\begingroup$ This is one of the folklore results that is well known and hard to find. It follows from Federer's area formula, but Federer's theorem is a way to difficult for this fact. The right proof has been sketched below by Anton Petrunin. Unfortunately it is not easy to find this argument in the literature. $\endgroup$ – Piotr Hajlasz Mar 24 '18 at 17:20

I guess you know that it is true in $\mathbb R^k$.

Without loss of generality we can assume that $f\ge 0$. Fix $\varepsilon>0$ and cover your manifold by $(1\mp\varepsilon)$-Lipschitz charts. Brake your integrals into pieces using subordinate partition of unity and put these pieces back together. Since in $\mathbb R^k$ you have equality, you will get $$ \int\limits_{M} f(x)\cdot dV\ \ \lessgtr\ \ (1\pm \varepsilon)^k\cdot\int\limits_{M} f(x)\cdot H^k(dx) $$ Since $\varepsilon>0$ is arbitrary your statement follows.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.