Measure is Finitely Additive Function

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

Let $$f: \mathcal A \to \overline {\R}$$ be an countably additive function.

Then $$f$$ is also finitely additive.

Proof
From the definition of countably additive function, for any sequence $$\left \langle {A_i} \right \rangle$$ of pairwise disjoint elements of $$\mathcal A$$:
 * $$f \left({\bigcup_{i \ge 1} A_i}\right) = \sum_{i \ge 1} f \left({A_i}\right)$$

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

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