Measure is countably subadditive
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Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Then $\mu$ is $\sigma$-subadditive, that is:
- $\ds\forall \sequence {A_n} _{n \in \N} \subseteq \Sigma: \map \mu {\bigcup _{n \in \N} A_n} \le \sum _{n \in \N} \map \mu {A_n}$
Proof
For each $n\in\N$, let:
- $\ds A'_n := A_n \setminus \bigcup _{i=0} ^{n-1} A_i$
- $\forall n\in\N : A'_n \in \Sigma$
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 | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum _{n \in \N} \map \mu {A_n}\) | Measure is Monotone |
$\blacksquare$