The following if Exercise 5.8 in Folland's Real Analysis, 2nd edition:
Let $(X, \mathcal M)$ be a measurable space and let $M(X)$ be the space of complex measures on $(X, \mathcal M)$. Then $||\mu|| = |\mu|(X)$ is a norm on $M(X)$ that makes it into a Banach space.
This question already has an answer here, but I would like help to understand a particular proof by Jonathan Conder I found on the internet, which seems more elegant then those presented in the link above.
The proof of completenes goes as follows:
If $\sum_1^\infty \nu_n$ is an absolutely convergent series in $M(X)$ and $A \in \mathcal M$, then $\sum_1^\infty \nu_n(A)$ also converges absolutely because $|\nu_n(A)| \leq |\nu_n|(A) \leq |\nu_n|(X)$ for all $n \in \Bbb{N}$. Hence we may define $\nu: \mathcal M \longrightarrow \Bbb{C}$ by $\nu(A) = \sum_1^\infty \nu_n(A)$. Clearly $\nu(\emptyset) = 0$. If $(A_k)$ is a sequence of disjoint sets in $\mathcal M$ whose union in $A$, then $$ \sum_{k = 1}^\infty |\nu(A_k)| \leq \sum_{k = 1}^\infty \sum_{n = 1}^\infty |\nu_n(A_k)| = \sum_{n = 1}^\infty \sum_{k = 1}^\infty |\nu_n(A_k)| \leq \sum_{n = 1}^\infty \sum_{k = 1}^\infty |\nu_n|(A_k) = \sum_{n = 1}^\infty |\nu_n|(A) \leq \sum_{n = 1}^\infty ||\nu_n|| < \infty, \quad \quad (1) $$ by Tonelli's theorem, and Fubini's theorem implies that $$ \sum_{k = 1}^\infty \nu(A_k) = \sum_{k = 1}^\infty \sum_{n = 1}^\infty \nu_n(A_k) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \nu_n(A_k) = \sum_{n = 1}^\infty \nu_n(A) = \nu(A). \quad \quad (2) $$ In other words, $\sum_{k = 1}^\infty \nu(A_k)$ converges absolutely to $\nu(A)$, so $\nu \in M(X)$.
Then the author proceeds to show that $\sum_{n = 1} \nu_n$ converges to $\nu$ in $M(X)$, but my question regards the above (I can provide the full file if requested).
Which version of the Fubini and Tonelli theorems is the author using? Is it considering the sums as integrals with the counting measure? I would appreciate if somebody could explain the second passage (the first equality) in (1).
Any help will be the most appreciated.
Thanks in advance and kind regards.
