Show that if $\{\mathscr{A}_n\}_{n\in\mathbb{N}}$ is an increasing sequence of $\text{Algebras}$ of subsets of a set $X$. Show by example that even if $\{\mathscr{A}_n\}$ is a $\sigma-\text{Algebras}$ for every ${n\in\mathbb{N}}$ the union still may not be a $\sigma-\text{Algebra}$. Solution:Let $X=\mathbb{N}$, and $\mathscr{A}_n=$ the family of subsets of $\{1,2,3...n\}$ and their complements. Clearly, $\mathscr{A}_n$ is a $\sigma-\text{Algebras}$ and $\mathscr{A}_1\subset\mathscr{A}_2\subset\mathscr{A}_3...$ However $\bigcup_\limits{n\in\mathbb{N}}^{}\mathscr{A}_n$ is the family of all finite and co-finite subsets of $\mathbb{N}$ which is not a $\sigma-\text{Algebra}$$\:\:\blacksquare$

How can $\bigcup_\limits{n\in\mathbb{N}}^{}\mathscr{A}_n$ not be a $\sigma-\text{Algebra}$ if the sequence is increasing? It seems a contradiction to me. Thanks in advance.

  • $\begingroup$ What about the example is eluding you? It seems perfectly clear to me. $\endgroup$ Apr 16, 2017 at 13:35
  • $\begingroup$ All the $\mathscr{A}_n$ are finite algebras, so trivially $\sigma$-algebras. The first time infinite unions can appear at all is after taking the final union. $\endgroup$ Apr 16, 2017 at 13:37
  • $\begingroup$ It is the fact the sequences are increasing which means the union of the sigma-algebras must contain $\mathbb{N} $which countable infinite, that is why I think the union is a sigma-algebra too. What I do not get is the reason behind the family of finite and co-finite subsets of $\mathbb{N} $. $\endgroup$ Apr 16, 2017 at 13:38
  • $\begingroup$ Already $\mathscr{A}_0$ contains $\mathbb{N}$ as the complement of $\emptyset$. Nothing to do with increasingness. $\endgroup$ Apr 16, 2017 at 13:41
  • $\begingroup$ Why does the union is not a sigma-algebra. It must contain$\mathbb{N}$,right? $\endgroup$ Apr 16, 2017 at 13:42

2 Answers 2


A set $A$ is in $\mathscr{A}_n$ iff $A \subseteq \{1,\ldots n\}$ or $X\setminus A \subseteq \{1,\ldots,n\}$. This is just the definition of $\mathscr{A}_n$.

$A \in \mathscr{A}:= \bigcup_n \mathscr{A}_n$ iff $\exists n: A \in \mathscr{A}_n$ iff $\exists n: A \subseteq \{1,\ldots, n\} \text{ or } X\setminus A \subseteq \{1,\ldots,n\}$ iff $A$ is finite or $A$ is cofinite (i.e. has finite complement )

Now, $A_k = \{2k\}$ for $k \in \mathbb{N}$ is in $\mathscr{A}_{2k} \subseteq \mathscr{A}$, but the union of the $A_k$ is the set of even numbers $E$ which is not in $\mathscr{A}$ by the above criterion: it's neither finite nor cofinite. So $\mathscr{A}$ is indeed not a $\sigma$-algebra, exactly for the reason claimed.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Nov 10, 2018 at 0:08

Hint: Find a sequence $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{N}$ such that both $$A := \bigcup_{n \in \mathbb{N}} \{a_n\} \qquad \text{and} \qquad A^c$$ are infinite sets. Show that $A \notin \bigcup_k \mathcal{A}_k$ and conclude that $\bigcup_k \mathcal{A}_k$ is not closed under countable unions, hence not a $\sigma$-algebra.

  • $\begingroup$ How can $A\notin\bigcup_k \mathcal{A}_k$ if $\mathcal{A}_k$, contains $\mathbb{N}$? $\endgroup$ Apr 16, 2017 at 15:16
  • $\begingroup$ @PedroGomes why do you think that $\mathbb{N} \in \mathcal{A}_k$ implies $A \in \bigcup_k \mathcal{A}_k$.....? You have to choose the sequence $(a_n)_n$ in such a way that $A^c$ is infinite (in particular $A \neq \mathbb{N}$ because otherwise $A^c = \emptyset$ wouldn't be an infinite set). $\endgroup$
    – saz
    Apr 16, 2017 at 15:19

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