# Volume form induces Borel measure: proof verification

by TheGeekGreek   Last Updated August 14, 2019 15:20 PM

Disclaimer: I asked this question already on the regular mathematics site here, but to no avail, even with a bounty. I think answering said question is still of value.

Proposition. Let $$M$$ be a compact smooth manifold of positive dimension and $$\omega$$ a volume form on $$M$$. Suppose that $$F \in \operatorname{Diff}(M)$$ such that $$F^*\omega = \omega$$. Then there exists a finite $$F$$-invariant regular Borel measure on $$M$$.

Proof. Define $$I \in (C^\infty(M))^*$$ by $$I(f) := \int_M f\omega.$$ Because $$C^\infty(M)$$ is dense in $$C^0(M)$$ under the sup-norm, $$I$$ extends uniquely to an element of $$(C^0(M))^*$$. By the Riesz representation theorem, there exists a unique finite regular Borel measure $$\mu$$ such that $$I(f) = \int_M fd\mu$$ for all $$f \in C^0(M)$$. Thus left to show is $$F$$-invariance, that is $$\mu(F^{-1}(A)) = \mu(A)$$ for all measurable $$A \subseteq M$$ or equivalently, $$F_* \mu = \mu$$, where $$F_*\mu$$ denotes the pushforward-measure of $$\mu$$ under $$F$$. By assumption, we have that $$I(f) = \int_M f\omega \overset{\dagger}{=} \int_M F^*(f\omega) = \int_M (f \circ F)\omega = I(f \circ F)$$ for all $$f \in C^\infty(M)$$ since $$F^*\omega = \omega$$ implies that $$F$$ is orientation-preserving. Thus also $$\int_M f d\mu = \int_M(f \circ F)d\mu$$ for all $$f \in C^0(M)$$. The latter is exactly the integral $$\int_M f d(F_*\mu).$$ But then also $$I(f) = \int_M fd(F_*\mu)$$ for all $$f \in C^0(M)$$ which implies by uniqueness that $$F_*\mu = \mu$$ because $$F_*\mu$$ is also regular as $$\mu$$ is. $$\square$$

Is that proof correct? Is there a simpler argument?

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