Lemma 13.1.5. Let $X$ be a set and $\alg \subset 2^{X}$ be an algebra, then the following are equivalent:
For any $\seq{A_n}\subset \alg$, $\bigcup_{n \in \nat^+}A_{n} \in \alg$.
For any $\seq{A_n}\subset \alg$ with $A_{n} \subset A_{n+1}$ for all $n \in \natp$, $\bigcup_{n \in \natp}A_{n} \in \alg$.
For any $\seq{A_n}\subset \alg$ pairwise disjoint, $\bigsqcup_{n \in \natp}A_{n} \in \alg$.
Proof. (2) $\Rightarrow$ (1): For each $n \in \nat$, let $B_{n} = \bigcup_{k = 1}^{n} A_{n} \in \alg$, then $\bigcup_{n \in \natp}A_{n} = \bigcup_{n \in \nat}B_{n} \in \alg$.
(3) $\Rightarrow$ (2): Denote $A_{0} = \emptyset$ and $B_{n} = A_{n} \setminus A_{n-1}$, then $\seq{B_n}\subset \alg$ is pairwise disjoint and $\bigcup_{n \in \nat}A_{n} = \bigsqcup_{n \in \nat}B_{n} \in \alg$.$\square$