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More precisely, is there an $n$-manifold $M$ with an $n$-form $\omega$ such that there is no measure $\nu$ on $M$ satisfying $$\int f \omega = \int f d\mu $$ for all compactly supported smooth functions $f$?

EDIT: More generally, assuming the answer to the above is yes, what if we have a $k$-form $\omega$ ($k\leqslant n$)? Then for every oriented $k$-submanifold $S$ we have the functional $$f\mapsto \int _S f \omega,$$ so for every such $S$ there is a measure such that this is integration wrt it. But is there a single measure for all such $S$? I guess you'd have to say something about the orientation here...

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    $\begingroup$ No. It always comes from a measure. This follows from the Riesz Representation Theorem, since $M$ is locally compact Hausdorff, and $\int (\cdot) \omega$ is a positive linear functional (if it isn't, take $\int - (\cdot) \omega$) . $\endgroup$
    – Aloizio Macedo
    Commented May 17, 2016 at 4:10
  • $\begingroup$ This might be a stupid question, but why is one of them always a positive functional? Is it true that for all $n$-forms $\omega$ either $\omega$ or $-\omega$ positively oriented? $\endgroup$
    – Cronus
    Commented May 19, 2016 at 10:02
  • $\begingroup$ Also, what happens if I take a $k$-form for $k<n$? (The details of the questions are above, I edited the original question). Thanks for your answer in any case! $\endgroup$
    – Cronus
    Commented May 19, 2016 at 10:02
  • $\begingroup$ Related $\endgroup$ Commented May 19, 2016 at 10:12

1 Answer 1

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I am answering my own question because Aloizio Macedo answered it in the comments, and I don't want it to remain unanswered. (I hope this is OK).

$M$ is locally compact and Hausdorff and either $\int (\cdot) \omega$ or $-\int (\cdot) \omega$ is a positive linear functional*, so by Riesz Representation Theorem one of them comes from a measure.

*I'm not entirely sure I understand why one of them is positive, but I think it's probably an elementary fact of differential forms.

I still don't know if there's an answer to the more general question I asked.

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