Sum of Additive Function Values is Well-Defined
Theorem
Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Let $A, B \in \SS$.
Then the sum
- $\map f A + \map f B$
is well-defined in the extended real numbers $\overline \R$.
Proof
Suppose the sum $\map f A + \map f B$ is void.
By Definition of Extended Real Addition, this happens when the sum is $\paren{ +\infty } + \paren{ -\infty }$, or $\paren{ -\infty } + \paren{ +\infty }$.
Without loss of generality, assume that $\map f A = +\infty$, and $\map f B = -\infty$.
Set Difference and Intersection are Disjoint shows that $A \setminus B$ and $A \cap B$ are disjoint.
Set Difference and Intersection are Disjoint shows that $B \setminus A$ and $A \cap B$ are disjoint.
By definition of additive function and Set Difference Union Intersection, it follows that:
- $ \map f { A \setminus B } + \map f { A \cap B } = \map f A = +\infty$
- $ \map f { B \setminus A } + \map f { A \cap B } = \map f B = -\infty$
Suppose $\map f { A \cap B } = +\infty$.
It follows that:
- $\map f { B \setminus A } + \paren{ +\infty } = -\infty$
which is a contradiction.
Suppose $\map f { A \cap B } = -\infty$.
It follows that:
- $\map f { A \setminus B } + \paren{ -\infty } = +\infty$
which is a contradiction.
Finally, suppose $\map f { A \cap B } \in \R$.
As $\map f { A \setminus B } + \map f { A \cap B } = +\infty$, it follows that $\map f { A \setminus B } = +\infty$.
As $\map f { B \setminus A } + \map f { A \cap B } = -\infty$, it follows that $\map f { B \setminus A } = -\infty$.
By definition of additive function and Set Difference is Disjoint with Reverse, it follows that the sum:
- $\map f { A \setminus B } + \map f { B \setminus A }$
is well-defined.
As this sum is equal to $\paren{ +\infty } + \paren{ -\infty}$, this is a contradiction.
$\blacksquare$