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Let $(M,g)$ be a compact Riemannian manifold and take $f\in L^1(M, dV_g)$ for the Riemannian volume form $V_g$. If we define $m(B):=\int_Bf(x)dV_g(x)$, which sets are $m$-measurable? Heck, maybe the real questions is that what sets are $dV_g$-measurable? I'm inclined to believe that the Riemannian volume form behaves like a Lebesgue measure, so e.g. all open subsets of $M$ are $dV_g$-measurable. But I have had hard time finding a reference which considers measures on Riemannian manifolds.

Cartesian Bear
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  • The reason you don't find references specific to Riemannian manifolds is because this question is in no way specific to Riemannian geometry. You are given a measure space $(\Omega, \mathcal{A},\mu)$, and your are looking for the measurable subsets of an associated absolutely continuous measure $fd\mu$. In charts, the Riemannian measure is absolutely continuous with respect to the Lebesgue measure (the absolutely continuous factor being $\sqrt{\det g}$). The measurable subsets for the Riemannian measure are exactly those of the Lebesgue measure. – Didier Jul 25 '23 at 17:17
  • I’ve written several answers on this topic by now (see the various links in the comments here). You can without much thought define the measure $dV_g$ on the Borel $\sigma$-algebra defined by the underlying topological space of $M$. But, using local charts, you can also elevate this to the “Lebesgue $\sigma$-algebra” of $M$ (see the links for the definition). – peek-a-boo Jul 25 '23 at 23:52

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