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!
Best Answer
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.