# Hausdorff measure and the volume form

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.

• 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? – Deane Yang Feb 8 '12 at 20:26
• 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. – Appliqué Feb 8 '12 at 20:44
• You have to use the area formula in Federer. – Mohan Ramachandran Feb 8 '12 at 21:32
• 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. – 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.