Counting measure proof Can somebody tell me why the counting measure (so, if $S=P(X)$, then $\mu(A)$=infinity if $A$ isn't finite and $\mu(A)=$#$A$ if $A$ is finite) is a measure? (The second property of a measure isn't clear for me).
 A: To see the $\sigma$-additivity of the counting measure, consider a sequence $(A_k)_{k\in\mathbb{N}}$ of pairwise disjoint sets. If one of the $A_k$ is infinite, say $A_i$, then $\bigcup A_k \supset A_i$ is also infinite, and
$$\sum_{k\in\mathbb{N}} \mu(A_k) = \underbrace{\sum_{k < i}\mu(A_k)}_{0 \leqslant \sum \leqslant \infty} + \underbrace{\mu(A_i)}_{\infty} + \underbrace{\sum_{k > i} \mu(A_k)}_{0\leqslant \sum \leqslant \infty} = \infty = \mu\left(\bigcup_{k\in\mathbb{N}} A_k\right).$$
If infinitely many of the $A_k$ are nonempty, since they are disjoint, the union $\bigcup A_k$ is infinite, and
$$\sum_{k\in\mathbb{N}} \mu(A_k) = \sum_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} \mu(A_k) \geqslant \sum_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} 1 = \infty = \mu\left(\bigcup_{k\in\mathbb{N}} A_k\right).$$
If only finitely many $A_k$ are nonempty and all $A_k$ are finite, then the union is also finite, and the number of its elements is the sum of the number of elements of the nonempty $A_k$ by disjointness, hence
$$\sum_{k\in\mathbb{N}} \mu(A_k) = \sum_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} \mu(A_k) = \mu\left(\bigcup_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} A_k\right)= \mu\left(\bigcup_{k\in\mathbb{N}} A_k\right)$$
holds in that last case too.
