Countably Additive Function Dichotomy by Empty Set
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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 function be a countably additive function.
Then the exactly one of the following is true:
- $\paren 1$: $\map f \O = 0$
- $\paren 2$: $\map f \O = + \infty$. Moreover, $\map f A = + \infty$ for all $A \in \AA$.
- $\paren 3$: $\map f \O = - \infty$. Moreover, $\map f A = - \infty$ for all $A \in \AA$.
Proof
Suppose $\map f \O \ne 0$.
Then:
\(\ds \map f \O\) | \(=\) | \(\ds \map f {\bigcup_{n \mathop \in \N} \O}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N} \map f \O\) | Definition of Countably Additive Function | |||||||||||
\(\ds \) | \(\in\) | \(\ds \set {+\infty, -\infty}\) |
$\Box$
Furthermore, for each $A \in \AA$ we have:
\(\ds \map f A\) | \(=\) | \(\ds \map f {A \cup \bigcup_{\N_{>0} }\O}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map f A + \sum_{n \mathop \in \N_{>0} } \map f \O\) | Definition of Countably Additive Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin {cases} +\infty & : \map f \O = +\infty \\ -\infty & : \map f \O = -\infty \end {cases}\) |
$\blacksquare$