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 $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint elements of $\Sigma$.

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

By Set is Subset of Union:
 * $\ds \bigcup_{n \mathop = 0}^N S_n \subseteq \bigcup_{n \mathop = 0}^\infty S_n$

From Non-Negative Additive Function is Monotone:
 * $\ds \map f {\bigcup_{n \mathop = 0}^N S_n} \le \map f {\bigcup_{n \mathop = 0}^\infty S_n}$

Also, from Finite Union of Sets in Additive Function:
 * $\ds \map f {\bigcup_{n \mathop = 0}^N S_n} = \sum_{n \mathop = 0}^N \map f {S_n}$

Hence:
 * $\ds \sum_{n \mathop = 0}^N \map f {S_n} \le \map f {\bigcup_{n \mathop = 0}^\infty S_n}$

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

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