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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 $(\mathcal{F}_n)$.

    Actually @Did has somehow explained it in one earlier question of mine:

    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.

But in a certain sense (1) is almost right. If there is a finite measure here, each element of the tail agrees with such a limsup up to a set of measure zero.

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+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

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