# 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 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.
• 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? Feb 8, 2012 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. Feb 8, 2012 at 20:44
• You have to use the area formula in Federer. Feb 8, 2012 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. Mar 24, 2018 at 17:20
• The referenced MSE question (to which the OP gave an answer four days later, without mentioning that it had been answered here). Jan 8, 2021 at 18:14

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. 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.