# Find two counterexamples

(1)Suppose $$\mathcal{A}_1\subset \mathcal{A}_2 \subset \mathcal{A}_3 ...$$ are $$\sigma$$-algebra consisting of subsets of a set $$X$$. Is $$\cup_{i}^{\infty} \mathcal{A}_i$$ necessarily a $$\sigma$$-algebra? If not, give a counterexample.

I find a counterexample of (1): 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$$A_n = \{0\} \cup \{2\} \cup \ldots \cup \{n - 2\} \cup \{n\} \text{ if }n\text{ is even,}$$$$A_n = \{0\} \cup \{2\} \cup \ldots \cup \{n - 3\} \cup \{n - 1\} \text{ if }n\text{ is odd.}$$ $$A = \bigcup_{k = 0}^\infty A_k \notin \mathscr{A}_n$$for every $$n$$

(2) Suppose $$\mathcal{M}_1\subset \mathcal{M}_2 \subset \mathcal{M}_3 ...$$ are monotone classes. Let $$\mathcal{M}=\cup_{i}^{\infty} \mathcal{M}_i$$. Suppose $$A_j\to A$$ and each $$A_j\in \mathcal{M}$$. Is $$A$$ necessarily in $$\mathcal{M}$$? If not, give a counterexample.

I feel like two questions are not. Thanks for any hints.

A funny thing is that, actually you have answered your own question: For $$\sigma$$-algebra is certainly a monotone class, and you have exhibited a set, the so-called $$A$$ such that $$A\notin\mathscr{A}_{n}$$ for each $$n$$, certainly $$A$$ is such that $$B_{n}:=\displaystyle\bigcup_{k=1}^{n}A_{k}\rightarrow A$$, as $$(B_{n})$$ is increasing. Note that $$B_{n}\in\mathscr{A}_{n}$$ as well.