# Riemannian Measures, Densities and Radon–Nikodym Theorem

If $M$ is a smooth manifold and $\mu$ is a $1$-density thereon then we may define a Borel measure (on Borel sets $A$) on $M$ as: $$\nu(A) = \int_M I_A \mu.$$

My question is does the converse hold also? Is not, then when would it.
That is, if $\nu$ is a measure on $M$ then when does there exist a $1$-density $\mu$ such that $\nu$ and $\mu$ are related as above.

In short is there a version of the Radon–Nikodym theorem relating densities and measures on manifolds?

## 1 Answer

This follows from the usual Radon-Nikodym theorem. Observe that, given two metrics $g_0,g_1$ on $M$ with volume densities $dV_{g_0}$, $dV_{g_1}$ then there exists a positive smooth function $\rho_{10}: M\to (0,\infty)$ such that

$$dV_{g_1}= \rho_{10} dV_{g_0}.$$

Hence, if $B\subset M$ is a Borel subset, then

$$V_{g_0}(B)=0 \Longleftrightarrow V_{g_1}(B) =0.$$

In other words, the concept of negligible set has an intrinsic meaning on a manifold.

You can characterize a density as being a Borel measure $\mu$ on $M$ such that $\mu(S)=0$ for any any negligible Borel subset $S\subset M$.

• I'm not terribly familiar with densities, but are they usually taken to be smooth? In that case the last sentence wouldn't be right. – Nate Eldredge Sep 23 '15 at 16:51
• Densities are sections of a real line bundle. They can be smooth, they can be only continuous, and they can be only measurable. – Liviu Nicolaescu Sep 23 '15 at 22:56