Union of ascending chain of $\sigma$-algebras is not necessarily a $\sigma$-algebra, need help concluding.

Suppose $\mathcal{A}_1 \subset \mathcal{A}_2 \subset \ldots$ are $\sigma$-algebras consisting of subsets of a set $X$. Is $\cup_{i = 1}^\infty \mathcal{A}_i$ a necessarily a $\sigma$-algebra? If not, give a counterexample.

Let $\sigma(A_1, \ldots, A_n)$ denote the smallest $\sigma$-algebra containing the sets $A_1, \ldots, A_n$. Consider the following family of $\sigma$-algebras $(\mathscr{A}_n)_{n = 0}^\infty$ on the natural numbers $\mathbb{N} = \{0, 1, 2, \ldots\}$ defined by$$\mathscr{A}_n = \sigma(\{0\}, \ldots, \{n\}).$$Clearly, $\mathscr{A}_1 \subset \mathscr{A}_2 \subset \ldots.$

Now consider the sets $A_n \in \mathscr{A}_n$ defined by$$A_n = \{0\} \cup \{2\} \cup \ldots \cup \{n - 2\} \cup \{n\} \text{ if }n\text{ is even,}$$$$A_n = \{0\} \cup \{2\} \cup \ldots \cup \{n - 3\} \cup \{n - 1\} \text{ if }n\text{ is odd.}$$Question. Why does it follow that$$A = \bigcup_{k = 0}^\infty A_k \notin \mathscr{A}_n$$for every $n$?

• Well, what is $A$? – Did Sep 13 '16 at 21:17

$A$, the set of nonnegative even numbers, is an infinite set with infinite complement. For all $n$, the members of $\mathscr{A}_{n}$ are either finite or have finite complements. As $A$ satisfies neither of these conditions, it's not in $\mathscr{A}_{n}$.