Volume form and measure on manifold

Question

Let $M$ be a compact smooth manifold, and $\omega$ a volume form. I'd like to put a Borel measure on $M $ induced by $\omega$. How do I do that? Can someone point to a reference?

Answer 1

The easiest way to do it is to use the Riesz representation theorem. Given a volume form $\omega \in \Omega^m(M)$ (where $m = \dim M$), we have a linear functional $\Lambda \colon C^{\infty}(M) \rightarrow \mathbb{R}$ given by $\Lambda(f) := \int_M f\omega$. If you follow the development of integration theory on smooth manifolds, you can verify that it also makes sense to integrate not only smooth top forms on $M$ but also continuous top forms (that is, continuous sections of $\Lambda^m(T^{*}M)$). Hence, this functional actually extends to a functional defined on $C^0(M)$ and this is a positive functional. By the Riesz representation theorem there exists a unique regular Borel measure $\mu_{\omega}$ such that

$$ \Lambda(f) = \int_M f\omega = \int_M f d\mu_{\omega} $$

for all $f \in C^0(M)$.

Answer 2

In an $m$-dimensional manifold $M$, the volume form $\omega$ natually induces a Borel measure, which by way of calculation is equivalent to the Hausdorff measure $\mathscr{H}_m$ where $$\mathscr{H}_m (U)= \int_U dV$$

For any $U \subset V$

For a reference see Complete Analytic Sets by Chirka.