Measure is Countably Subadditive

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Theorem

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


Then $\mu$ is a countably subadditive function.


Proof

Let $\left({E_n}\right)_{n \in \N}$ be a sequence of sets in $\Sigma$.

It is required to show that:

$\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E_n}\right) \le \sum_{n \mathop \in \N} \mu \left({E_n}\right)$


Now define the sequence $\left({F_n}\right)_{n\in\N}$ in $\Sigma$ by:

$F_n := \displaystyle \bigcup_{k \mathop = 1}^n E_n$

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

Hence, $\left({F_n}\right)_{n\in\N}$ is increasing.

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

Now reason as follows:

\(\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E_n}\right)\) \(=\) \(\displaystyle \lim_{n \to \infty} \mu \left({F_n}\right)\) Characterization of Measures, $(3)$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \to \infty} \mu \left({E_1 \cup \cdots \cup E_n}\right)\) Definition of $F_n$
\(\displaystyle \) \(\le\) \(\displaystyle \lim_{n \to \infty} \sum_{k \mathop = 1}^n \mu \left({E_k}\right)\) Measure is Subadditive: Corollary
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop \in \N} \mu \left({E_k}\right)\)


Hence the result.

$\blacksquare$


Sources