Measure is Subadditive

Theorem

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

Then $\mu$ is subadditive, that is:

$\forall E, F \in \Sigma: \mu \left({E \cup F}\right) \le \mu \left({E}\right) + \mu \left({F}\right)$

Corollary

Let $E_1, \ldots, E_n \in \Sigma$.

Then:

$\ds \map \mu {\bigcup_{k \mathop = 1}^n E_k} \le \sum_{k \mathop = 1}^n \map \mu {E_k}$.

Proof

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

So Additive Nowhere Negative Function is Subadditive‎ applies.

Hence the result directly:

$\mu \left({E \cup F}\right) \le \mu \left({E}\right) + \mu \left({F}\right)$

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $4.3 \ \text{(v)}$