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I am trying to prove that the definition of the integral of a differential form does not depend on the choice of (oriented) atlas and partition of unity.

We consider charts $\varphi_{\alpha}:U_{\alpha}\to M$ with a partition of unity $\rho_{\alpha}$.

And an other choice of charts $\phi_{i}:V_{i}\to M$ with a partition of unity $\sigma_{i}$.

I see (in many courses) the following computation :

$$ \int_{\varphi_{\alpha}(U_{\alpha})}\rho_{\alpha}\omega=\int_{\varphi_{\alpha}(U_{\alpha})}\sum_i\sigma_i\rho_{\alpha}\omega=\sum_i\int_{\varphi_{\alpha}(U_{\alpha})}\sigma_i\rho_{\alpha}\omega $$

MY QUESTION IS : How do we permute the integral and the sum over $i$ ?

EDIT

As far as the convergence of the integral is concerned, we assume that $\int_{\varphi_{\alpha}}(U_{\alpha})\rho_{\alpha}\omega$ converges for each $\alpha$. And $$ \sum_{\alpha\in\Lambda}\int_{\varphi_{\alpha}}(U_{\alpha})\rho_{\alpha}\omega $$ converges. The final purposes is to prove that $$ \sum_{i\in I}\int_{\phi_{i}}(V_{i})\sigma_{i}\omega $$ converges and has the same value. (this is a purpose slightly after the current question -- see commute two sums when definig integral of differential manifold)

END EDIT

As far as I understand, people do not assume $U_{\alpha}$ to be bounded or $\omega$ to be of constant sign.

Maybe the sum over $i$ is finite for some reason ? Maybe because the partition of unity is locally finite ?

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  • $\begingroup$ It probably depends on the definition of a partition of unity, but I know the definition, where in each neighborhood all but a finite number of the functions are equal to $0$. Then, this integral is just an integral of a finite sum of functions and you can change sum and integral, since $\int 0 + \ldots + \int 0 = \int (0+\ldots +0)$ even over an infinite index set. $\endgroup$
    – user408858
    May 15 at 5:52
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    $\begingroup$ The problem with that explanation is that only a finite number of $\sigma_i$ is non vanishing on each $\phi_i(V_i)$. Here the integral is on $\varphi_{\alpha}(U_{\alpha})$. Since the covering with "$i$" and the covering with "$\alpha$" are independent, I cannot see why restricting the integral over the support of $\rho_{\alpha}$ makes the sum over $i$ finite. $\endgroup$ May 15 at 6:07
  • $\begingroup$ Can't you just write $\int_N f=\int_N\sum_i \sigma_i f=\sum_i \int_N\sigma_i f$ for any subset $N\subset M$ and function $f:M\rightarrow \mathbb{R}$? It is only the special case for $N=\varphi_\alpha(U_\alpha)$ and $f=\rho_\alpha\omega$. I'll give it a try later. $\endgroup$
    – user408858
    May 15 at 6:33
  • $\begingroup$ I cannot because of the permutation of the integral with the sum. As far as I know, the sum is countable, $N$ is not bounded and the function has not a constant sign. One point I see is that the support of $\rho_{\alpha}$ is compact. Thus, the integration is the integral of continuous functions over a compact set. I'll see if I can find a commutation theorem in that case. $\endgroup$ May 15 at 6:49
  • $\begingroup$ Its the central theorem of Lebesgue integrals over measurable functions and measurable sets: All sums and limits of sums are independent of the the order. If both sides converge, the value is the same by the orderless definition of the integral. The proof by measure theory and calculus on manifolds is using transfinite methods, especially, in order to construct non measurable partitions of a set Zorn's Lemma has to be used. $\endgroup$
    – Roland F
    May 15 at 8:18

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Answer adapted from user408858'hint.

We can restrict the domain from $ \varphi_{\alpha}(U_{\alpha})$ to $ supp(\rho_{\alpha})$. Since $ \{ \sigma_i \}$ is locally finite, for each $ m\in supp(\rho_{\alpha})$ we choose

  • $O_m$ : open containing $ m$
  • $ I_m$ : finite subset of $ I$ such that $\sigma_i=0$ on $O_m$ when $i\not\in I_m$.

This is an open covering of the compact $supp(\rho_{\alpha})$. We extract a finite covering. Let $ m_1,\ldots,m_n\in supp(\rho_{\alpha})$ be such that $supp(\rho_{\alpha})\subset \bigcup_{k=1}^nO_{m_k}$.

We set $J=\bigcup_{k=1}^nI_{m_k}$. This is a finite set as finite union of finite sets.

We have $\sigma_i=0$ on $supp(\rho_{\alpha})$ when $i\not\in J$. Thus

$$ \int_{\varphi_{\alpha}(U_{\alpha})}\sum_{i\in I}\sigma_i\rho_{\alpha}\omega= \int_{\varphi_{\alpha}(U_{\alpha})}\sum_{i\in J}\sigma_i\rho_{\alpha}\omega $$ Now the sum is finite and we can do whatever we want.

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