Countably Additive Function of Null Set

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:
 * $$f \left({\bigcup_{i \in \N} A_i}\right) = \sum_{i \in \N} f \left({A_i}\right)$$

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

Hence:

$$ $$ $$ $$ $$

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