Theorem

Let $\SS$ be an algebra of sets.

Let $f: \SS \to \overline \R$ be an additive function on $\SS$.

Then $f$ is also strongly additive.

That is:

$\forall A, B \in \SS: \map f {A \cup B} + \map f {A \cap B} = \map f A + \map f B$

Proof

Recall that $\sqcup$ denotes the disjoint union.

First, since:

 $\ds A \cup B$ $=$ $\ds A \cup \paren {B \setminus A}$ Set Difference Union Second Set is Union $\ds$ $=$ $\ds A \sqcup \paren {B \setminus A}$ in view of Definition of Set Difference

we have:

$\paren 1 : \quad \map f {A \cup B} = \map f A + \map f {B \setminus A}$

Secondly, since

 $\ds B$ $=$ $\ds \paren {B \setminus A} \cup \paren {A \cap B}$ Set Difference Union Intersection $\ds$ $=$ $\ds \paren {B \setminus A} \sqcup \paren {A \cap B}$ Set Difference and Intersection are Disjoint

we have:

$\paren 2 : \quad \map f B = \map f {B \setminus A} + \map f {A \cap B}$

On the other hand:

 $\ds A \cup B$ $=$ $\ds \paren {A \cap B} \cup \paren {A \cup B}$ as $A \cap B \subseteq A \cup B$ $\ds$ $=$ $\ds \paren {A \cap B} \cup \paren { \paren {A \cup B} \setminus \paren {A \cap B } }$ Set Difference Union Second Set is Union $\ds$ $=$ $\ds \paren {A \cap B} \sqcup \paren { \paren {A \cup B} \setminus \paren {A \cap B } }$ in view of Definition of Set Difference

so that:

$\map f {A \cup B} = \map f {A \cap B} + \map f {\paren {A \cup B} \setminus \paren {A \cap B} }$

In particular, we have neither:

$\map f {A \cup B} = + \infty \land \map f {A \cap B} = - \infty$

nor:

$\map f {A \cup B} = - \infty \land \map f {A \cap B} = + \infty$

Thus the sum in $\overline {\R}$:

$\map f {A \cup B} + \map f {A \cap B}$

is well-defined.

Finally:

 $\ds \map f {A \cup B} + \map f {A \cap B}$ $=$ $\ds \map f A + \map f {B \setminus A} + \map f {A \cap B}$ by $\paren 1$ $\ds$ $=$ $\ds \map f A + \map f B$ by $\paren 2$

$\blacksquare$