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

To show the reverse inequality holds, first note that $$f\ $$ is monotonic. If $$A\subseteq B\in X$$, then $$f(B) = f(\big(B\cap A\big) \cup \big(B - A\big)) \geq f(B\cap A) + f(B - A)$$ (by subadditivity) $$= f(A) + f(B - A) \geq f(A)$$.

Next, note that additive functions are finitely additive. So $$f(\bigcup_{i=0}^n A_i) = \sum_{i=0}^n f(A_i)$$ for each $$n$$.

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

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