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
\(\ds \) \(=\) \(\ds \begin {cases} +\infty & : \map f \O = +\infty \\ -\infty & : \map f \O = -\infty \end {cases}\)

$\blacksquare$