# commute two sums when defining integral of differential manifold

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 ?

MY GUESS.

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 $$\int_M\omega$$ ?

• 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 $\sum_{\alpha}\sum_{i\in I_{\alpha}}f(\alpha, i)$. Commented May 17 at 4:34