Countably Additive Function also Finitely Additive

Theorem
Let $\AA$ be a $\sigma$-algebra.

Let $\overline \R$ denote the extended set of real numbers.

Let $f: \AA \to \overline \R$ be a countably additive function.

Then $f$ is a finitely additive function.

Proof
We have that $f$ is defined as countably additive :


 * $\ds \map f {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map f {A_i}$

where $\sequence {A_i}$ is any sequence of pairwise disjoint elements of $\AA$.

We need to show that:


 * $\ds \forall n \in \N: \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$

The claim is trivial if $f=+\infty$ or $f=-\infty$, constantly.

Thus we assume $\map f \O = 0$ by Countably Additive Function Dichotomy by Empty Set.

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

Let $\sequence {B_i}$ be the sequence of pairwise disjoint elements of $\AA$ defined as:


 * $B_i = \begin{cases}

A_i & : i \le n \\ \O & : i > n \end{cases}$

It follows that:


 * $\ds \bigcup_{i \mathop \ge 1} B_i = \bigcup_{i \mathop = 1}^n A_i$

Thus:

Hence the result.