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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

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    $\begingroup$ 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). $\endgroup$ – Olivier Bégassat Apr 27 '13 at 16:57
  • $\begingroup$ 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.... $\endgroup$ – user55449 Apr 27 '13 at 17:07
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    $\begingroup$ 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. $\endgroup$ – Olivier Bégassat Apr 27 '13 at 17:14
  • $\begingroup$ Actually I'm not sure they are equal, but to tackle the problem the above approach should work. $\endgroup$ – Olivier Bégassat Apr 27 '13 at 17:17
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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.

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