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In this paper Tao explains that "The unsigned definite integral generalises to the Lebesgue integral, or more generally to integration on a measure space. Finally, the signed definite integral generalises to the integration of forms, which will be our focus here."

The phrase seems to point to these to kinds of integration being necessarily separate from each other, but is this the case?

On page $100$ of Spivak's Calculus on Manifolds the following definition is made:

If $\omega$ is a $k$-form on $\mathbb{R}^k$, then $\omega = f\ dx_1\land\ldots\land dx_k$ for a unique function $f:\mathbb{R}^k\to\mathbb{R}$. We define $$\int_{[0,1]^k}\omega := \int_{[0,1]^k}f.$$

Can't the right-side be a Lebesgue (as opposed to a Riemann) Integral, allowing us to say we are defining Lebesgue integration on $k$-forms? From what I gather which form of integration we choose (Riemann, Lebesgue, gauge, etc) plays a role in regards to which $k$-forms are integrable.

Sam
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  • The situation is not different from the question if you interpret $\int_a^bf(x),dx$ as Riemann or Lebesgue integral. As you know, unless you deal with somewhat pathological counter examples it does not matter in most practical situations. I prefer Lebesgue always because it offers the better convergence theorems. – Kurt G. Mar 13 '23 at 13:26
  • Integration of differential forms can indeed by defined using Lebesgue’s theory, and I prefer it this way since it doesn’t have strict restrictions on the types of sets involved, or the regularity level of the forms. See my answer here for a more detailed discussion. – peek-a-boo Mar 14 '23 at 23:18
  • Wendell Fleming's FUNCTIONS OF SEVERAL VARIABLES does exactly this. I'm not crazy about this book finding it unnessecerily dry in many places. But it certainly answers your question in the affirmative. – Mathemagician1234 Jul 15 '23 at 03:42

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