Suppose $\{A_n, n \in \mathbb{N}\}$ is a sequence of subsets of $\Omega$. Each $A_n$ generates a $\sigma$-algebra as $\mathcal{A}_n:=\{ A_n, A_n^c, \emptyset, \Omega \}$. I was wondering if we can simplify $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$, i.e., the tail $\sigma$-algebra of the sequence of $\sigma$-algebras $\{ \mathcal{A}_n, n \in \mathbb{N} \}$?
I guess $\limsup_{n \rightarrow \infty} A_n:= \cap_{i=1}^{\infty} \cup_{j=i}^{\infty} A_n$ does belong to $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$, and I don't know if $\liminf_{n \rightarrow \infty} A_n:= \cup_{i=1}^{\infty} \cap_{j=i}^{\infty} A_n$ belongs to $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$?
If $\{A_n, n \in \mathbb{N}\}$ are independent events on probability space $(\Omega, \mathcal{F}, P)$, can we further simplify the tail $\sigma$-algebra $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$? The reason I asked this is because I was wondering why the tail σ-algebra is said to be trivial in this independent case and how it looks like to be trivial?
Thanks and regards!