Countably Additive Function also Finitely Additive

Theorem
Let $$\mathcal A$$ be a $\sigma$-algebra.

Let $$f: \mathcal A \to \overline {\R}$$ be a real-valued function, where $$\overline {\R}$$ denotes the set of extended real numbers.

Let $$f$$ be a countably additive function:
 * $$f \left({\bigcup_{i \in \N} A_i}\right) = \sum_{i \in \N} f \left({A_i}\right)$$

such that there exists at least one $$A \in \mathcal A$$ where $$f \left({A}\right)$$ is a finite number.

Then $$f$$ is a finitely additive function.

Proof
We have that $$f$$ is defined as countably additive iff:
 * $$f \left({\bigcup_{i \ge 1} A_i}\right) = \sum_{i \ge 1} f \left({A_i}\right)$$

where $$\left \langle {A_i} \right \rangle$$ is any sequence of pairwise disjoint elements of $$\mathcal A$$.

We need to show that:
 * $$\forall n \in \N: f \left({\bigcup_{i=1}^n A_i}\right) = \sum_{i=1}^n f \left({A_i}\right)$$

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

Let $$\left \langle {B_i}\right \rangle$$ be the sequence of pairwise disjoint elements of $$\mathcal A$$ defined as:
 * $$B_i = \begin{cases}

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

It follows that:
 * $$\bigcup_{i \ge 1} B_i = \bigcup_{i = 1}^n A_i$$

Thus

$$ $$ $$ $$

Hence the result.