Countably Additive Function also Finitely Additive

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Theorem

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

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


Let $f$ be a countably additive function:

$\displaystyle f \left({\bigcup_{i \mathop \in \N} A_i}\right) = \sum_{i \mathop \in \N} f \left({A_i}\right)$

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


Then $f$ is a finitely additive function.


Proof

We have that $f$ is defined as countably additive if and only if:

$\displaystyle f \left({\bigcup_{i \mathop \ge 1} A_i}\right) = \sum_{i \mathop \ge 1} f \left({A_i}\right)$

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


We need to show that:

$\displaystyle \forall n \in \N: f \left({\bigcup_{i \mathop = 1}^n A_i}\right) = \sum_{i \mathop = 1}^n f \left({A_i}\right)$


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:

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


Thus:

\(\displaystyle f \left({\bigcup_{i \mathop = 1}^n A_i}\right)\) \(=\) \(\displaystyle f \left({\bigcup_{i \mathop \ge 1} B_i}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop \ge 1} f \left({B_i}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n f \left({A_i}\right) + \sum_{i \mathop > n} f \left({\varnothing}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n f \left({A_i}\right)\) from Countably Additive Function of Null Set

Hence the result.

$\blacksquare$