Additive Nowhere Negative Function is Subadditive

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

Let $$f: \mathcal A \to \overline {\R}$$ be an additive function such that:
 * $$\forall A \in \mathcal A: f \left({A}\right) \ge 0$$

Then $$f$$ is subadditive.

Proof
If $$f$$ is additive then by Additive Function on Union of Sets:
 * $$\forall A, B \in \mathcal A: f \left({A \cup B}\right) = f \left({A}\right) + f \left({B}\right) - f \left({A \cap B}\right)$$

As $$f \left({A \cap B}\right) \ge 0$$, the result follows by definition of subadditive:
 * $$\forall A, B \in \mathcal A: f \left({A \cup B}\right) \le f \left({A}\right) + f \left({B}\right)$$