Additive and Countably Subadditive Function is Countably Additive

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

Proof
Let $$\left({A_n}\right)_{n \in \N}$$ be a sequence of disjoint sets in $$X\ $$. By countably subadditivity, $$f\left({\bigcup_{n \in \N} A_n}\right) \leq \sum_{n \in \N} f \left({A_n}\right)$$ automatically.

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

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

$$ $$ $$ $$

Next, note that additive functions are 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.