Non-Negative Additive Function is Monotone

Theorem
Let $\mathcal S$ be an algebra of sets.

Let $f: \mathcal S \to \overline {\R}$ be an additive function, i.e.:
 * $\forall A, B \in \mathcal S: A \cap B = \varnothing \implies f \left({A \cup B}\right) = f \left({A}\right) + f \left({B}\right)$

If $\forall A \in \mathcal S: f \left({A}\right) \ge 0$, then $f$ is monotone, i.e.:
 * $A \subseteq B \implies f \left({A}\right) \le f \left({B}\right)$

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

From Set Difference Union Intersection we have that:
 * $B = \left({B \setminus A}\right) \cup \left({A \cap B}\right)$

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

So:
 * $A \subseteq B \implies B = \left({B \setminus A}\right) \cup A$

From Set Difference Intersection Second Set is Empty Set we have that:
 * $\left({B \setminus A}\right) \cap A = \varnothing$

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

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

But as, by hypothesis, $f \left({B \setminus A}\right) \ge 0$, it follows that:
 * $f \left({B}\right) \ge f \left({A}\right)$

hence the result.