Measure is Countably Subadditive

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

Then $\mu$ is a countably subadditive function.

Proof
Let $\left({A_n}\right)_{n \in \N}$ be a sequence of sets in $\mathcal A$.

It is required to show that:


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

Now define the sequence $\left({B_n}\right)_{n\in\N}$ in $\mathcal A$ by:


 * $B_n := \displaystyle \bigcup_{k=1}^n A_n$

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

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

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

Now reason as follows:

Hence the result.