Additive Function on Union of Sets

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f \left({A}\right) + f \left({B}\right)\)	\(=\)	\(\displaystyle f \left({A \setminus B}\right) + 2 f \left({A \cap B}\right) + f \left({B \setminus A}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f \left({\left({A \setminus B}\right) \cup \left({A \cap B}\right) \cup \left({B \setminus A}\right)}\right) + f \left({A \cap B}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Difference and Intersection form Partition: Corollary
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle f \left({A \cup B}\right) + f \left({A \cap B}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Hence the result.

$\blacksquare$