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

Solution 1:

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.