(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.


1 Answer 1


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.

  • 1
    $\begingroup$ Oh... I see.... $\endgroup$
    – Hermi
    Commented Dec 12, 2019 at 4:51

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