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