Measure is Countably Subadditive

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.


Then $\mu$ is a countably subadditive function.


Proof

Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets in $\Sigma$.

It is required to show that:

$\ds \map \mu {\bigcup_{n \mathop \in \N} E_n} \le \sum_{n \mathop \in \N} \map \mu {E_n}$


Now define the sequence $\sequence {F_n}_{n \mathop \in \N}$ in $\Sigma$ by:

$F_n := \ds \bigcup_{k \mathop = 0}^n E_n$

By Subset of Union, it follows that, for all $n \in \N$, $F_n \subseteq F_{n + 1}$.

Hence, $\sequence {F_n}_{n \mathop \in \N}$ is increasing.

It is immediate that $F_n \uparrow \ds \bigcup_{n \mathop \in \N} E_n$, where $\uparrow$ signifies the limit of an increasing sequence of sets.

Now reason as follows:

\(\ds \map \mu {\bigcup_{n \mathop \in \N} E_n}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \mu {F_n}\) Measure of Limit of Increasing Sequence of Measurable Sets
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \mu {E_0 \cup \cdots \cup E_n}\) Definition of $F_n$
\(\ds \) \(\le\) \(\ds \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n \map \mu {E_k}\) Measure is Subadditive: Corollary
\(\ds \) \(=\) \(\ds \sum_{k \mathop \in \N} \map \mu {E_k}\)


Hence the result.

$\blacksquare$


Sources