# Tail $\sigma$-algebra for a sequence of sets

1. 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 $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$?

2. 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!

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The tail sigma-algebra is the sigma-algebra of sets $B$ such that, for every integer $N$ one can build $B$ from the sets $A_n$ with $n\ge N$ only. For example the limsup/liminf of $(A_n)_n$ is also the limsup/liminf of $(A_{n+N})_n$ hence the limsup/liminf is in the tail sigma-algebra. There is no measure involved here.
In the independent case with respect to a probability $P$, the tail sigma-algebra is trivial in the sense that it contains only sets of $P$-probability zero or one. For example the limsup/liminf of any sequence which is independent for $P$ has $P$-probability zero or one. This is a property of $P$ in relation with the sigma-algebras considered but definitely not a property of the sigma-algebras alone.
Thanks! For the tail of a sequence of general $\sigma$-algebras $\{\mathcal{F}_n\}$, (1) can I cay a set $B$ belongs to the tail if and only if $B$ can be built from $\{ \mathcal{F}_n: n \geq N \}$ for any integer $N$? (2) what does "$B$ can be built from $\{ \mathcal{F}_n: n \geq N \}$ " mean? Is it that there exists some mapping $f$, such that $B=f(\{ \mathcal{F}_n: n \geq N \})$ ? – Tim May 10 '11 at 14:20
That $B$ belongs to the sigma-algebra generated by the union over $n\ge N$ of the sigma-algebras $F_n$. – Did May 10 '11 at 17:02