Countably Additive Function of Null Set

Theorem
Let $\mathcal A$ be a $\sigma$-algebra.

Let $f: \mathcal A \to \overline {\R}$ be a real-valued function, where $\overline {\R}$ denotes the set of extended real numbers.

Let $f$ be a countably additive function:
 * $\displaystyle f \left({\bigcup_{i \in \N} A_i}\right) = \sum_{i \in \N} f \left({A_i}\right)$

such that there exists at least one $A \in \mathcal A$ where $f \left({A}\right)$ is a finite number.

Then $f \left({\varnothing}\right) = 0\ $.

Proof
Suppose that $A \in \mathcal A$ such that $f \left({A}\right)$ is a finite number.

So, let $f \left({A}\right) = x$.

Consider the sequence $\left \langle {S_i}\right \rangle \subseteq \mathcal A$ defined as:
 * $\forall i \in \N: S_i = \begin{cases}

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

Then $\displaystyle \bigcup_{i \ge 0} S_i = A$.

Hence:

It follows directly that $\mu \left({\varnothing}\right) = 0$.