Non-Negative Additive Function is Monotone

Theorem
Let $\SS$ be an algebra of sets.

Let $f: \SS \to \overline \R$ be an additive function, that is:
 * $\forall A, B \in \SS: A \cap B = \O \implies \map f {A \cup B} = \map f A + \map f B$

If $\forall A \in \SS: \map f A \ge 0$, then $f$ is monotone, that is:
 * $A \subseteq B \implies \map f A \le \map f B$

Proof
Let $A \subseteq B$, and let $f$ be an additive function.

From Set Difference Union Intersection:
 * $B = \paren {B \setminus A} \cup \paren {A \cap B}$

From Intersection with Subset is Subset we have that:
 * $A \subseteq B \implies A \cap B = A$

So:
 * $A \subseteq B \implies B = \paren {B \setminus A} \cup A$

From Set Difference Intersection with Second Set is Empty Set:
 * $\paren {B \setminus A} \cap A = \O$

So $B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.

So, by the definition of additive function:
 * $\map f B = \map f {B \setminus A} + \map f A$

But as, by hypothesis, $\map f {B \setminus A} \ge 0$:
 * $\map f B \ge \map f A$

hence the result.