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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