Countable additivity

Remarks

On the power set of $\N$, the function $\mu(A) = \sum_{n \in A} 2^{-n}$ is countably additive. The series has non-negative terms, so its sum does not depend on how the terms are grouped. On the family $\mathcal{F}$ of subsets $A \subseteq \N$ such that $A$ or $\N \setminus A$ is finite, let $\nu(A) = 0$ for finite $A$ and $\nu(A) = 1$ otherwise. This $\nu$ is finitely additive but not countably additive. Indeed $\N = \bigcup_{n} \set{n}$ is a disjoint union of sets in $\mathcal{F}$, but $\nu(\N) = 1$ and $\sum_n \nu(\set{n}) = 0$. If $\varnothing \in \mathcal{F}$ and $\mu$ takes at least one finite value, then countable additivity forces $\mu(\varnothing) = 0$: take $A_1 = A$ with $\mu(A) < \infty$ and $A_n = \varnothing$ for $n \ge 2$. The same definition is used for functions with values in $\R$, $\C$ or a Banach space; then the series must converge.