The union of a NOT strictly increasing sequence of $\sigma$-algebras is not a $\sigma$-algebra

Question

Im refering to

The Problem with strictly increasing sequences

What will change if I have not a strictly increasing sequence but just an increasing one:

$A_1\subseteq A_2 \subseteq A_3 \subseteq \cdots $?

Answer 1

If there are infinitely any strict inclusions $A_n\subsetneq A_{n+1}$, then by discarding all equalities in the chain, you get a strictly increasing chain which has the same union, and hence this union is not a $\sigma$-algebra by the problem you've linked to.

On the other hand, if there are only finitely many strict inclusions, then for some $n$ we have $A_n=A_{n+1}=\cdots$, and hence $\cup_i A_i=A_n$, which is a $\sigma$-algebra by assumption.