Additive and Countably Subadditive Function is Countably Additive

Theorem
Let $\Sigma$ be a $\sigma$-algebra over a set $X$.

Let $f: \Sigma \to \overline {\R}_{\ge 0}$ be an additive and countably subadditive function, where $\overline {\R}_{\ge 0}$ denotes the set of positive extended real numbers.

Then $f$ is countably additive.

Proof
Let $\left \langle {S_n}\right \rangle_{n \in \N}$ be a sequence of pairwise disjoint elements of $\Sigma$.

Let $N \in \N$ be any natural number.

By Subset of Union:
 * $\displaystyle \bigcup_{n=0}^N S_n \subseteq \bigcup_{n=0}^\infty S_n$

So we can apply the monotonicity of $f$ to conclude that:
 * $\displaystyle f \left({\bigcup_{n=0}^N S_n}\right) \le f \left({\bigcup_{n=0}^{\infty} S_n}\right)$

Also, it follows by Finite Union of Sets in Additive Function that:
 * $\displaystyle f \left({\bigcup_{n=0}^N S_n}\right) = \sum_{n=0}^N f \left({S_n}\right)$

Hence:
 * $\displaystyle \sum_{n=0}^N f \left({S_n}\right) \le f \left({\bigcup_{n=0}^{\infty} S_n}\right)$

Taking the limit as $N \to \infty$, it follows by Lower and Upper Bounds for Sequences that:
 * $\displaystyle \sum_{n=0}^{\infty} f \left({S_n}\right) \le f \left({\bigcup_{n=0}^{\infty} S_n}\right)$

The reverse inequality holds because of the countable subadditivity of $f$, and thus equality holds.