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so today I was looking at the Generalized Stokes' Theorem: \begin{align} \intop_{\Omega} d\omega=\intop_{\partial\Omega}\omega\ \ , \end{align} where $\Omega$ is some region, and $\omega$ is a differential form. What I would like to know is how to convert a differential form into a measure. For example, I read in Choquet-Bruhat's Analysis, Manifolds, and Physics that there was a way to do this, but I am unsure on how to apply it in a more abstract setting. Any advice or help you can provide would be helpful, particularly in how it applies to the Generalized Stokes' Theorem.

Sora8DTL
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    From the Wikipedia article on Differential forms: "Formally, in the presence of an orientation, one may identify n-forms with densities on a manifold; densities in turn define a measure, and thus can be integrated (Folland 1999, Section 11.4, pp. 361–362)." – M Turgeon Aug 23 '23 at 20:32
  • That is really cool! Do you happen to have a formula for how to compute these densities? I know that a differential form is different than a tensor density if that is what the statement is saying, so an example of this identification would be useful for computation. For example $dx \wedge dy = f("some\ density")$. – Sora8DTL Aug 23 '23 at 20:53
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    One way to get a measure from the usual DG definition of integration (at least in the embedded case) is by thinking of the orientation as a vector in the dual to the top exterior power, and the measure in the Hausdorff measure. In symbols, if $M^n\subset \mathbb{R}^{n+k}$, then $\int_M \omega=\int_M \langle \omega,\xi\rangle d\mathscr{H}^n.$ See the discussion of currents in chapter 6.2 of Leon Simon's GMT notes. But you're not "turning $\omega$ into a measure" in this method. – Mr. Brown Aug 23 '23 at 23:51
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    @Sora8DTL Unfortunately I don't! But it shouldn't be too difficult to track down a copy of Folland, it is (was?) a standard textbook for Real Analysis/Measure Theory. The title, if you're looking for it, is Real Analysis: Modern Techniques and Their Applications. – M Turgeon Aug 24 '23 at 16:15
  • I think that it's as easy as $\mu(E) = \int_E \omega.$ This however gives a signed measure, or even a complex valued measure, since the differential form might be non-positive. – md2perpe Aug 25 '23 at 13:26
  • That's a bit glib, @md2perpe. What degree form is $\omega$? What restrictions are you imposing on measurable subsets $E$? How are you orienting them in order to integrate? Etc. – Ted Shifrin Aug 25 '23 at 19:06
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    See Are differential forms lebesgue measures and the two links within (especially the first one). – peek-a-boo Sep 18 '23 at 11:09

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