Measure is countably subadditive

Theorem

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

Then $\mu$ is $\sigma$-subadditive, that is:

$\ds\forall \sequence {A_n} _{n \in \N} \subseteq \Sigma: \map \mu {\bigcup _{n \in \N} A_n} \le \sum _{n \in \N} \map \mu {A_n}$

For each $n\in\N$, let:

$\ds A'_n := A_n \setminus \bigcup _{i=0} ^{n-1} A_i$

Then, by (SA3) and (SA2'):

$\forall n\in\N : A'_n \in \Sigma$

Furthermore, by construction:

$\bigcup _{n \in \N} A'_n = \bigcup _{n \in \N} A_n$

$\sequence {A'_n} _{n \in \N}$ are pairwise disjoint

$\forall n\in\N : A' _n\subseteq A_n$

Therefore:

$\ds \map \mu {\bigcup _{n \in \N} A_n}$

$=$

$\ds \map \mu {\bigcup _{n \in \N} A'_n}$

$\ds $

$=$

$\ds \sum _{n \in \N} \map \mu {A' _n}$

(2) of definition of measure

$\ds $

$\le$

$\ds \sum _{n \in \N} \map \mu {A_n}$

$\blacksquare$