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, this happens when the sum is $\paren{ +\infty } + \paren{ -\infty }$, or $\paren{ -\infty } + \paren{ +\infty }$.

, 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.