Countably Additive Function also Finitely Additive
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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 if and only if:
- $\ds \map f {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map f {A_i}$
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:
\(\ds \map f {\bigcup_{i \mathop = 1}^n A_i}\) | \(=\) | \(\ds \map f {\bigcup_{i \mathop \ge 1} B_i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop \ge 1} \map f {B_i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map f {A_i} + \sum_{i \mathop > n} \map f \O\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map f {A_i}\) | as $\map f \O = 0$ |
Hence the result.
$\blacksquare$