Measure is Subadditive

Theorem
Let $\left({X, \mathcal A, \mu}\right)$ be a measure space.

Then $\mu$ is subadditive, that is:
 * $\forall A, B \in \mathcal A: \mu \left({A \cup B}\right) \le \mu \left({A}\right) + \mu \left({B}\right)$

Corollary
Let $A_1, \ldots, A_n \in \mathcal A$.

Then $\displaystyle \mu \left({\bigcup_{k = 1}^n A_k}\right) \le \sum_{k = 1}^n \mu \left({A_k}\right)$.

Proof
A measure is an additive function, and, by definition, nowhere negative.

So Additive Nowhere Negative Function is Subadditive‎ applies.

Hence the result directly:
 * $\mu \left({A \cup B}\right) \le \mu \left({A}\right) + \mu \left({B}\right)$

The corollary is an application of Finite Union of Sets in Subadditive Function.