## Theorem

Let $\AA$ be an algebra of sets.

Let $f: \AA \to \overline {\R}$ be an additive function such that:

$\forall A \in \AA: \map f A \ge 0$

Then $f$ is subadditive.

## Proof

If $f$ is additive then by Additive Function is Strongly Additive:

$\forall A, B \in \AA: \map f {A \cup B} = \map f A + \map f B - \map f {A \cap B}$

As $\map f {A \cap B} \ge 0$, the result follows by definition of subadditive:

$\forall A, B \in \AA: \map f {A \cup B} \le \map f A + \map f B$

$\blacksquare$