Countably additive function Let R be a sigma algebra, and g be a real valued function on R such that for a sequence ($A_{n}$) of disjoint members of R, we have that g($A$) (where $A$ is the countable union) is equal to the series $g(A_{1})$+$g(A_{2})$ + .....
Is it then true that  $g(A_{1})$+$g(A_{2})$ + ..... is absolutely convergent. If so, is there a easy proof? 
 A: I interpret the premises as demanding that for every $A\in R$ and every disjoint sequence $(A_n)_{n\in\mathbb{Z}^+}$ of sets in $R$ with
$$A = \bigcup_{n=1}^\infty A_n,$$
you have
$$g(A) = \lim_{m\to\infty} \sum_{n=1}^m g(A_n).$$
Then, since the union of a sequence of sets is invariant under permutations, for every disjoint sequence $(A_n)$ in $R$, and every permutation $\pi \colon \mathbb{Z}^+ \to \mathbb{Z}^+$, you have that
$$g(A) = \lim_{m\to\infty} \sum_{n=1}^m g(A_{\pi(n)}),$$
and that means precisely that the series
$$\sum_{n=1}^\infty g(A_n)$$
is unconditionally convergent.
For sequences of real numbers (more generally, in finite-dimensional real [or complex] vector spaces), unconditional convergence is equivalent to absolute convergence.
So if I interpreted the premises correctly, indeed $\sum g(A_n)$ is absolutely convergent.
Whether this is an easy proof depends on whether you can just use the equivalence of unconditional and absolute convergence, or need to prove that. [Although that isn't very hard either.]
