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Let $M$ be a differentiable manifold. Let $\mu$ be a (probability) measure on $M$.

What are the conditions under which $\mu$ is given by a differential form on $M$? I imagine some sort of compatibility of the topology or the differentiable structure of $M$ with the $\sigma$-algebra of $\mu$ would be required.

(Apologies if the question is too elementary for this forum. A pointer to the relevant result in the literature would suffice.)

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    $\begingroup$ If you mean by a top degree differential form (so assume $M$ orientable) then I’d look at absolute continuity with respect to the manifold’s natural measure class. $\endgroup$
    – Francois Ziegler
    Commented Sep 23, 2018 at 5:24
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    $\begingroup$ The answer to the question will vary depending on whether we want the (top degree) differential form to be smooth, differentiable, continuous, or merely measurable, whether or not we demand that its integral be finite, etc. $\endgroup$
    – John Baez
    Commented Sep 23, 2018 at 5:50
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    $\begingroup$ Your measure will be a continuous functional on $C^\infty(M)$, hence it is by definition a distributional section of the $\Lambda^{top}M$ bundle (aka a distribution). Now you just need a condition under which this distribution is represented by a smooth function. That really depends on what information you have available. Sledgehammer approach: check that the singular support (or the wave front set) is empty. $\endgroup$
    – Igor Khavkine
    Commented Sep 23, 2018 at 6:53
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    $\begingroup$ For the sake of precision in what @IgorKhavkine describes, a measure is not intrinsically a section of $\Lambda^{top} M$, but instead one can make an identification depending on a choice of generator for $\Lambda^{top} M$. I.e. given a measure $\mu$ and a volume form $\omega$ you are talking about the operation $\alpha \mapsto \int \frac{\alpha}{\omega} \mathrm{d}\mu$ $\endgroup$
    – user13113
    Commented Sep 23, 2018 at 7:38
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    $\begingroup$ A "distributional section", as per the comment by Igor Khavkine, is also known as a (top degree / lowest dimension) current on $M$ ( en.wikipedia.org/wiki/Current_(mathematics) ). The theory of currents is precisely a common generalization of measure theory and differential forms on a manifold. $\endgroup$
    – Qfwfq
    Commented Sep 23, 2018 at 11:50

2 Answers 2

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I assume that $\mu$ is a measure defined on the $\sigma$-algebra of Borel sets. First, on any manifold the notion of negligible set is well defined.

If $M$ is orientable and $\mu(N)=0$ for any negligible Borel set then the Radon-Nikodym theorem implies that, for any smooth volume form $\omega$ on $M$, there is a positive measurable function $\rho_\omega\colon M\to\mathbb{R}$ such that

$$\mu(U)=\int_U \rho_\omega \omega, $$

for any open set $U\subset M$.

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  • $\begingroup$ @BenMcKay You are right. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 23, 2018 at 15:38
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Any smooth manifold has a canonical σ-ideal of negligible subsets, and μ must vanish on these.

Apart from that, the Lie derivative of μ with respect to any smooth vector field must exist.

This is how smooth measures are defined by Ramanan in Definition 1.9 of Chapter 3 of Global Calculus, for example.

Remark 2.8 in Chapter 8 there explains how this definition is equivalent to the traditional definition of a smooth measure as a smooth section of the line bundle of densities.

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