Let $(X,\Sigma,\mu)$ is a signed measure space, $\mu:\Sigma \to \mathbb R$.
Show that $\nu:\Sigma \to [0,\infty]$, for $A \in \Sigma$ $\quad\nu(A)=\sup\sum\limits_{m=1}^{\infty}|\mu(A_m)| \ $ where $A=\bigcup\limits_{m=1}^{\infty}A_m \ $ and for $\ n\neq m \quad A_n\cap A_m= \emptyset $. Supremum is being taking over $A_n\in \Sigma$ sets in union.
I couldn't show its $\sigma$-additiveness.
I've written $A_n=\bigcup\limits_{k=1}^{\infty}B_{nk} \ $ for all $n \in \mathbb N\ $ where $\ B_{nk}$ sets are disjoint.
$\nu(\bigcup\limits_{n=1}^{\infty}A_n)=\sup_{A_n}\sum\limits_{n=1}^{\infty}|\mu(A_n)|=\sup_{A_n}\sum\limits_{n=1}^{\infty}\left|\sum\limits_{k=1}^{\infty}\mu(B_{nk})\right| \qquad (1)$
$\sum\limits_{n=1}^{\infty}\nu(A_n) =\sum\limits_{n=1}^{\infty}\sup_{B_{nk}}\sum\limits_{k=1}^{\infty}|\mu(B_{nk})| \qquad (2)$
I've tried to shoe that $(1)$ and $(2)$ are equal. I've tried to use supremum and triangle inequality for showing $(1)\leq(2)\ $ and $\ (2)\leq(1)$ but I couldn't obtain. I think there are some mistakes.
Sorry in advance if it is so silly.
Thanks for any help
Hint: Let $A$ be disjoint union of $A_n$'s. If $(C_j)$ is a disjoint sequence with union $A$ then, for each $n$, $C_j \cap A_n, j\geq 1$ gives disjoint sets with union $A_n$ so $\nu (A_n) \geq \sum_j |\mu(C_j \cap A_n|$. Sum this over $n$ and show that $\nu (A) \leq \sum \nu(A_n)$.
The reverse inequality comes form the following: If, for each $n$, $(B_{nk})_k$ is a disjoint sequence with union $A_n$ then the collection $\{B_{nk}: n,k \geq 1\}$ is countable disjoint collection of sets with union $A$.