Additive Function on Empty Set is Zero

Theorem
Let $\mathcal A$ be an algebra of sets.

Let $f: \mathcal A \to \overline {\R}$ be an additive function on $\mathcal A$.

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

Proof
First we check that the statement of the theorem makes sense.

And we have that from Properties of Algebras of Sets we have that $\varnothing \in \mathcal A$.

Now let $X \in \mathcal A$.

From Intersection with Null we have that $X \cap \varnothing = \varnothing$, and so $X$ and $\varnothing$ are disjoint.

As $f$ is additive, it follows by definition that:
 * $f \left({X \cup \varnothing}\right) = f \left({X}\right) + f \left({\varnothing}\right)$

But $X \cup \varnothing = X$ from Union with Null and so:
 * $f \left({X}\right) = f \left({X}\right) + f \left({\varnothing}\right)$

Hence the result:
 * $f \left({\varnothing}\right) = 0$