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 is Strongly Additive:
 * $\forall A, B \in \mathcal A: f \left({A \cup B}\right) = f \left({A}\right) + f \left({B}\right) - f \left({A \cap B}\right)$

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