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:
 * $\displaystyle f \left({\bigcup_{n \in \N} A_n}\right) \le \sum_{n \in \N} f \left({A_n}\right)$

To show the reverse inequality holds, first note that as $f$ is non-negative, $f$ is also monotonic.

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

for each $n$.

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

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