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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 too silly for MathOverflow and more suitable for Mathematics Stack Exchange, 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 Mathematics Stack Exchange and on one more forum, but I didn't receive any response.
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    $\begingroup$ 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
    Commented Feb 8, 2012 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
    Commented Feb 8, 2012 at 20:44
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    $\begingroup$ You have to use the area formula in Federer. $\endgroup$
    – Mohan Ramachandran
    Commented Feb 8, 2012 at 21:32
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    $\begingroup$ 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
    Commented Mar 24, 2018 at 17:20
  • $\begingroup$ The referenced MSE question (to which the OP gave an answer four days later, without mentioning that it had been answered here). $\endgroup$
    – LSpice
    Commented Jan 8, 2021 at 18:14

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

Without loss of generality we can assume that $f\geqslant 0$. Fix $\varepsilon>0$ and cover your manifold by $(1\mp\varepsilon)$-Lipschitz charts. Break 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.

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