# Tail sigma algebra and $\limsup$ of a sequence of subsets

On a set $\Omega$, there is a sequence of sigma algebras $(\mathcal{F}_n)_{n \in \mathbb{N}}$. The tail sigma algebra of $(\mathcal{F}_n)$ is defined to be $\cap_{n=1}^{\infty} \sigma(\cup_{m=n}^\infty \mathcal{F}_m )$. I was wondering if the following two statements are true:

1. $\forall B$ in the tail sigma algebra of $(\mathcal{F}_n)$, there exist $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$, such that $\limsup_n A_n = B$.
2. $\forall A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$, $\limsup_n A_n$ is in the tail sigma algebra of $(\mathcal{F}_n)$.

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.

But I don't quite understand it well in the hindsight:

• for any subset $B$ in the tail sigma algebra and any integer $N$, how does one build $B$ from the sets $A_n \in \mathcal{F}_n$ with $n\ge N$?
• how does the above explain that $\limsup_n A_n$ is in the tail sigma algebra of $(\mathcal{F}_n)$?
3. Similar questions to the above two for relations:

between $\sigma(\cup_{n=1}^{\infty} \cap_{m=n}^\infty \mathcal{F}_m )$ and $\liminf_n A_n$ for $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$?

between $\cap_{n=1}^{\infty} \sigma(\cup_{m=n}^\infty \mathcal{F}_m )$ and $\liminf_n A_n$ for $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$?

between $\sigma(\cup_{n=1}^{\infty} \cap_{m=n}^\infty \mathcal{F}_m )$ and $\limsup_n A_n$ for $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$?

Thanks and regards!

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(2). $\limsup_n A_n$ is in the tail sigma-algebra.
It suffices to show: for any $p$, we have $\limsup_n A_n \in \sigma(\bigcup_{m=p}^\infty \mathcal F_m)$. And, as Didier said, this is because we can write $\limsup_n A_n$ as $\limsup_n A_{n+p}$. Indeed, each $A_{n+p}$ is in $\sigma(\bigcup_{m=p}^\infty \mathcal F_m)$, so their countable combinations (such as limsup) are also in $\sigma(\bigcup_{m=p}^\infty \mathcal F_m)$.
(1). is not right. We could take $\mathcal F_n$ finite sigma-algebras (consisting of finite unions of intervals) such that $\bigcup_{n=1}^\infty \mathcal F_n$ contains all intervals with rational endpoints, so $\sigma(\bigcup_{n=p}^\infty \mathcal F_n)$ is the Borel sigma algebra $\mathcal B$, so $\bigcap_p \sigma(\bigcup_{n=p}^\infty \mathcal F_n)) = \mathcal B$. But there are elements in $\mathcal B$ that are more complicated than limsups of finite unions of intervals.
+1. Thanks! (1) $\liminf_n A_n = \liminf_n A_{n+p}$ is true similar to the $\limsup$ case, isn't it? (2) I wonder if there is something to say about part 3? (Or we may disregard it if it is not meaningful.) (3) Have you seen $\sigma(\cup_{n=1}^{\infty} \cap_{m=n}^\infty \mathcal{F}_m )$ having been considered? (Maybe less than the tail sigma algebra?) –  Tim Mar 12 '12 at 0:56
Consider the example: in $[0,1)$ let $\mathcal F_m$ be the sets that depend only on the $m$th binary digit. Compute your sigma-algebras in that case and see how they compare. –  GEdgar Mar 12 '12 at 14:54
GEdgar, how do you know, we can just write limsup An as $limsup A_{n+p}$ ? I thought the definition of limsup holds only for p= 0? upload.wikimedia.org/math/6/2/0/… I have this strange feeling that you can take away the first few things to intersect similar to this one (you can take away the first few random variables in the sequence) math.stackexchange.com/a/546801/140308 –  BCLC May 14 at 19:53