Measure is Subadditive

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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:

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


Proof

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

So Additive Nowhere Negative Function is Subadditive‎ applies.


Hence the result directly:

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

$\blacksquare$


Sources