Finitely additive measure

Remarks

Let $\mathcal{A}$ be the algebra 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. Clause 1 holds, and clause 2 holds since two disjoint sets cannot both have finite complement. This $\nu$ is not countably additive, since $\nu(\N) = 1$ while $\nu(\set{n}) = 0$ for each $n$. On the same algebra, the function with $\mu(\varnothing) = 0$ and $\mu(A) = 1$ for $A \ne \varnothing$ is not finitely additive, since clause 2 fails for $A = \set{1}$ and $B = \set{2}$. By induction, clause 2 gives additivity over any finite disjoint union. A finitely additive measure is monotone: $\mu(A) \le \mu(B)$ when $A \subseteq B$. Other names are content and charge; the word content is also used for the same notion on a ring or a semiring of sets.