Riemannian measure and Hausdorff measure in a general Riemannian Manifold

Let $M$ be a Riemannian manifold and let $\mu$ be its Riemannian measure. This is the measure obtained by Riesz reprersentation theorem such that for every continuous function with compact support $f$

$\int_M f d\mu = \sum_i^n \int_{U_{i}}(\rho_i\sqrt{G_i}f)\circ \phi_i^{-1}dx$

where $(U_i,\phi_i )$ is a finite covering of $supp f$, $\rho_i$ is a partition of unity subordinated to $U_i$ and $G_i$ is the determinant of the metric of $M$ in the $\phi_i$ -coordinates.

If $M=R^n$ is the standard euclidean space then $\mu = L^n$, where $L^n$ is the Lebesgue measure.

On the other hand on $M$ (not necessarily equal to $R^n$) we can define the $n-$dimensional Hausdorff measure $H^n$. It is a standard result that if $M= R^n$ then $\mu=L^n=H^n$.

Now the question: is it true that $\mu=H^n$ for every Riemannian manifold $M$?

Thanks

• I would guess they are equal. You only need to show that they coincide on chart domains (being $\sigma$-finite measures, and using the "monotone class" theorem). – Olivier Bégassat Apr 27 '13 at 16:57
• We know that $H^n$ is Borel regular (actually it holds in every metric space). If we prove that $H^n$ is also finite on compact sets we conclude that $H^n$ is $\sigma$-finite. But i'm not sure that it is easy.... – user55449 Apr 27 '13 at 17:07
• Just take a chart domain $U$, and use the fact that the distance function induced by the riemannian metric (on any compact subset $K$) will be bounded above and below by the standard euclidean distance: $$c d_{std}\leq d_g\leq Cd_{std}$$ where $0<c<C$ depend on the compact subet $K\subset U$. Using the definition of the Hausdorff measure we have that the Hausdorff measure of $K$ with the metric induced by $g$ on $U$ is at most $C$ times the standard Lebesgue measure of $K$. Since $M$ is covered by a sequence of compact subsets, the Hausdorff measure will be $\sigma$-finite. – Olivier Bégassat Apr 27 '13 at 17:14
• Actually I'm not sure they are equal, but to tackle the problem the above approach should work. – Olivier Bégassat Apr 27 '13 at 17:17

Yes, you can reduce the problem to the Euclidean case using normal coordinates, working along the lines of the comments by Olivier Bégassat, but with $c\approx C\approx 1$. Precisely, given $\epsilon>0$, you can use the normal coordinates to cover the manifold by patches $(U_i,\phi_i)$ such that
1. the metric tensor in these coordinates is $\delta_{ij}+O(\epsilon)$
2. the pushforward of the Lebesgue measure under $\phi_i^{-1}$ is comparable to $\mu$ to within the factor $1+O(\epsilon)$. (This is a consequence of 1.)
Property 1 allows you to compare the Hausdorff measures on $M$ and on $\mathbb R^n$, while property 2 compares Riemannian and Lebesgue measures. It follows that within each patch on $M$ the Hausdorff and Riemannian measures agree up to a factor of $1+O(\epsilon)$. Since $\epsilon$ was arbitrary, the measures are equal.