Given a signed measure $u$ on $(X,A)$ and $\mid u(\cup_{n=1}^{\infty}E_n) \mid < \infty$, where {$E_n: n \in N$} is pairwise disjoint family, I have to show that $\sum_{n=1}^{\infty}u(E_n)$ is absolutely convergent. From the condition $\mid u(\cup_{n=1}^{\infty}E_n) \mid < \infty$, two things I have observed. First, $-\infty < \sum_{n=1}^{\infty}u(E_n) < \infty$ and second, for each n, $-\infty < u(E_n)<\infty$. But how to conclude that $\sum_{n=1}^{\infty}\mid u(E_n) \mid < \infty$. Any hint.
Definition I am using is $u$ is signed measure on $(X,A)$ if
- $u(\{\})=0$,
- u either takes $\infty$ or -$\infty$ but not both.
- $u$ is countably additive.
Denote by $E_+$ the union of all $E_i$'s such that $u(E_i)>0$ and similarly $E_-$ the union of all $E_i$'s with negative measure. As the $E_i$'s are pairwise disjoint, we get by the countable additivity of $u$ $$ u(E_+) = \sum_{i\geq 1} \max\{ u(E_i), 0\} $$ and $$ u(E_-) = \sum_{i\geq 1} \min \{ u(E_i), 0 \}.$$ By point three we get that either $u(E_+)<\infty$ or $u(E_-)>-\infty$. However, we have that $$ \sum_{i\geq 1} u(E_i) = \sum_{i\geq 1} ( \max\{ u(E_i), 0\} + \min\{ u(E_i), 0\}) $$ is convergent. If $u(E_+)$ is convergent, then so is $u(E_-)$ as we have $u(E_-)=\sum_{i\geq 1} u(E_i)-u(E_+).$ The same way we get that $u(E_+)$ is convergent if $u(E_-)$ is. However, if $u(E_\pm)$ are convergent, then they are absolutely convergent, as they have a sign. Thus, also the sum is absolutely convergent, which is what we wanted to prove.