Additive Function is Strongly Additive

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

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

Then:
 * $\forall A, B \in \mathcal S: f \left({A \cup B}\right) = f \left({A}\right) + f \left({B}\right) \setminus f \left({A \cap B}\right)$

Proof
From Set Difference and Intersection form Partition, we have that:


 * $A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$;
 * $B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.

So, by the definition of additive function:
 * $f \left({A}\right) = f \left({A \setminus B}\right) + f \left({A \cap B}\right)$
 * $f \left({B}\right) = f \left({B \setminus A}\right) + f \left({A \cap B}\right)$

We also have from Set Difference Disjoint with Reverse that $\left({A \setminus B}\right) \cap \left({B \setminus A}\right) = \varnothing$.

Hence:

Hence the result.