Countably Additive Function also Finitely Additive
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:
- $\displaystyle f \left({\bigcup_{i \in \N} A_i}\right) = \sum_{i \in \N} f \left({A_i}\right)$
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:
- $\displaystyle f \left({\bigcup_{i \ge 1} A_i}\right) = \sum_{i \ge 1} f \left({A_i}\right)$
where $\left \langle {A_i} \right \rangle$ is any sequence of pairwise disjoint elements of $\mathcal A$.
We need to show that:
- $\displaystyle \forall n \in \N: f \left({\bigcup_{i=1}^n A_i}\right) = \sum_{i=1}^n f \left({A_i}\right)$
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:
- $\displaystyle \bigcup_{i \ge 1} B_i = \bigcup_{i = 1}^n A_i$
Thus
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \mu \left({\bigcup_{i = 1}^n A_i}\right)\) | \(=\) | \(\displaystyle \mu \left({\bigcup_{i \ge 1} B_i}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \ge 1} f \left({B_i}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i=1}^n f \left({A_i}\right) + \sum_{i>n} f \left({\varnothing}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i=1}^n f \left({A_i}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | from Countably Additive Function of Null Set |
Hence the result.
$\blacksquare$