Countably Additive Function of Null Set

Theorem

Let $\overline \R$ denote the extended set of real numbers.

Let $f: \AA \to \overline \R$ be a function be a countably additive function.

Suppose that there exists at least one $A \in \AA$ where $\map f A$ is a finite number.

Then:

$\map f \O = 0$

Suppose that $A \in \AA$ such that $\map f A$ is a finite number.

So, let $\map f A = x$.

Consider the sequence $\sequence {S_i} \subseteq \AA$ defined as:

$\forall i \in \N: S_i = \begin{cases}

A & : i = 0 \\ \O & : i > 0 \end{cases}$

Then:

$\ds \bigcup_{i \mathop \ge 0} S_i = A$

Hence:

$\ds \map f A$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \map f {\bigcup_{i \mathop \ge 0} S_i}$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \sum_{i \mathop \ge 0} \map f {S_i}$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \map f A + \sum_{i \mathop > 0} \map f \O$

$=$

$\ds x$

$\ds \leadsto \ \ $

$\ds \sum_{i \mathop > 0} \map f \O$

$=$

$\ds 0$

It follows directly that:

$\map f \O = 0$

$\blacksquare$

$\blacksquare$