Countably Additive Function Dichotomy by Empty Set

Theorem
Let $\AA$ be a $\sigma$-algebra.

Let $\overline \R$ denote the extended set of real numbers.

Let $f: \AA \to \overline \R$ be a function be a countably additive function.

Then the exactly one of the following is true:
 * $\paren 1$: $\map f \O = 0$
 * $\paren 2$: $\map f \O = + \infty$. Moreover, $\map f A = + \infty$ for all $A \in \AA$.
 * $\paren 3$: $\map f \O = - \infty$. Moreover, $\map f A = - \infty$ for all $A \in \AA$.

Proof
Suppose $\map f \O \ne 0$.

Then:

Furthermore, for each $A \in \AA$ we have: