Additive and Countably Subadditive Function is Countably Additive

Theorem
Let $$f: X \to Y$$ be an additive and countably subadditive set function which is nonnegative everywhere on its domain.

Then $$f\ $$ is countably additive.

Proof
Let $$\left \langle {A_n}\right \rangle_{n \in \N}$$ be a sequence of pairwise disjoint sets in $$X\ $$.

By countable subadditivity, it follows that:
 * $$f \left({\bigcup_{n \in \N} A_n}\right) \le \sum_{n \in \N} f \left({A_n}\right)$$

To show the reverse inequality holds, first note that $$f\ $$ is monotonic.

If $$A \subseteq B \in X$$, then:

$$ $$ $$ $$

Next, note that a countably additive function is also finitely additive. So:
 * $$f \left({\bigcup_{i=0}^n A_i}\right) = \sum_{i=0}^n f \left({A_i}\right)$$

for each $$n$$.

But by monotonicity, $$f \left({\bigcup_{i=0}^n A_i}\right) \leq f \left({\bigcup_{n \in \N} A_n}\right)$$ for each $$n$$.

Hence $$\sum_{i=0}^n f \left({A_i}\right) \leq f \left({\bigcup_{n \in \N} A_n}\right)$$ for each $$n$$, and so the inequality holds in the limit.