Countably Additive Function of Null Set

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.

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

Then:
 * $\map f \O = 0$

Proof
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:

It follows directly that:
 * $\map f \O = 0$

Proof 2
This follows immediately from Countably Additive Function Dichotomy by Empty Set.