This question is a sequel of this one : commute sum and integral when definig integral of differential manifold

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 $\sigma_{i}$.

We have to prove that $$ \sum_{\alpha\in\Lambda}\int_{\varphi_{\alpha}(U_{\alpha})}\rho_{\alpha}\omega =\sum_{i\in I}\int_{\phi_{i}}\sigma_{i}\omega $$

At some point, every single course I read make the following step:

$$ \sum_{\alpha\in\Lambda}\sum_{i\in I}\int_{\phi_i(V_i)}\rho_{\alpha}\sigma_i\omega= \sum_{i\in I}\sum_{\alpha\in\Lambda}\int_{\phi_i(V_i)}\rho_{\alpha}\sigma_i\omega $$ In fact, this commutation of sums if almost always hidden in a notation like «$\sum_{i,\alpha}$».

MY QUESTION IS : How do we commute these two sums ?


In fact we have to split the definition of $\int_M\omega$ in two parts. First we suppose $\omega$ to be positive on the charts $\varphi_{\alpha}$ (then it is positive on the charts $\phi_i$ because they are oriented the same way). And then we make the general case by splitting the positive and negative parts.

The reason of that idea is the following. If I know that $\omega$ is positive, I can commute the sums by invoking Fubini theorem by seeing the sums as integrals with respect to the counting measure on $\Lambda$ and $I$ of the function $f: \Lambda\times I \to \mathbb{R}$

$$ (\alpha,i) \mapsto \int_{\phi_i(V_i)}\rho_{\alpha}\sigma_i\omega. $$

In order to check the hypothesis of Fubini, I need $f\in L^1(\Lambda\times I)$. And for that I need to check

$$ \sum_i\left[\sum_{\alpha} \big| \int_{\phi_i(V_i)}\rho_{\alpha}\sigma_i\omega \big| \right]<\infty. $$

Is it the right way to define the integral $\int_M\omega$ ?

  • $\begingroup$ The sums are done AFTER the integral. What I can do is to restrict the sum over $i$ to a finite part $I_{\alpha}\subset I$ because the support of $\rho_{\alpha}$ is compact. But then I remain with $\sum_{\alpha}\sum_{i\in I_{\alpha}}f(\alpha, i)$. $\endgroup$ Commented May 17 at 4:34


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