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 Subset Equivalences 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 with Intersection 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.