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This question has been asked a couple of times before but I have been strugling trying to understand the previous explanations. I want to understand it concisely with an easy counterexample. So far, I have been trying to understand the first answer of the following post:

Example where union of increasing sigma algebras is not a sigma algebra

The idea in this answer is to take $X= \mathbb{N}$, then $\lbrace 1,2,...,n\rbrace$ for each $n \in \mathbb{N}$, an then consider the sigma algebra for each of this sets, i.e $A_{n}= \sigma (\lbrace 1...,n \rbrace)$. Obviously each $A_{n}$ is a sigma algebra and $A_{n} \subseteq A_{n+1}$ for each $n \in \mathbb{N}$ (i.e the sequence is increasing).

So now we want to prove that $\bigcup_{n=1}^{\infty}A_{k}$ is not a sigma algebra. After this things become more difficult to understand. But If I understand right, he takes a set $F_{2n}=\lbrace 2,4,....,2n \rbrace$ and this set belong to $\bigcup_{k=1}^{\infty}A_{n}$ because for example $4 \in F_{2n}$ we have that $4 \in A_{4}=\sigma (\lbrace 1,2,3,4 \rbrace)$ and it belongs in general to every $A_{n}$ for $n \geq 4$. Proving $F_{2n} \subset \bigcup_{k=1}^{\infty}A_{n}$ but by definition of union $F_{2n}$ should belong to some $A_{k}$ and he argues this doesnt happen which I dont understand since $F_{2n}$ is a subset of $A_{2n}$ right? Also I dont understan the scheme of the proof I mean why this proves this is not a sigma algebra?

I would appreciate a nice explanation of the answer there,thanks!

Andrés E. Caicedo
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Cos
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2 Answers2

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The increasing union of $A_n$ captures exactly the subsets of $\Bbb N$ that are either finite or cofinite. But there are plenty of sets (such as the even numbers) that are (a) a countable union of finite sets but (b) are neither finite nor cofinite. Since the set of even numbers is a countable union of sets within $\cup A_n$, but is not itself within that union, it follows that $\cup A_n$ is not a $\sigma-$algebra.

Robert Shore
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Each $A_n$ is a set of subsets (namely for a fixed $n$, $A \in A_n$ iff $A$ is a subset of $\{1,2,3,\ldots,n\}$ or the complement of such a subset). In particular each $A \in A_n$ is finite (the first case), or the complement of a finite set (the second case)

By definition of a union $A \in \bigcup_n A_n$ (a union of set of subsets is again a set of subsets) iff there is some $n$ such that $A \in A_n$.

And for no $n$ can we say that $\bigcup_n F_{2n}$ is in $A_n$, because the union is not finite, and not the complement of a finite set (the complement consists of the set of all odd integers). So it is not in $\bigcup_n A_n$. That $4 \in F_{2n}$ is true, but that's not relevant, the whole set has to be in one of the $A_n$ or it's not in the union of the $A_n$.

Henno Brandsma
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