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Consider a sequence of $\sigma$-algebras $\mathcal{F}_1,\mathcal{F}_2,\dots$. Is it true that for any event $B$ in the tail $\sigma$-algebra $\mathcal{F_{\text{Tail}}}$, it can be expressed as the $\limsup$ of some sequence of events $A_1,A_2,\dots$ such that $A_i \in \mathcal{F}_i$ for all $i$? Unfortunately, I have absolutely no intuition in this problem, and thus have no idea how to proceed.

Any help provided would be appreciated.

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  • $\begingroup$ How do you define the tail $\sigma$-algebra to begin with - is this $\bigcap_n \sigma(\mathcal F_n, \mathcal F_{n+1}, \dots)$? $\endgroup$
    – R W
    Commented Oct 5, 2019 at 13:27
  • $\begingroup$ Yes, that is correct. $\endgroup$
    – yellowello
    Commented Oct 5, 2019 at 15:04

1 Answer 1

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No, let $\mathcal S_0=\{\{0,1\},\emptyset\}$ and let $\mathcal S_1$ be the powerset of $\{0,1\}$. Let $\mathcal F_n=\mathcal S_0^{n-1}\times\mathcal S_1 \times\mathcal S_0^{\infty}$.

Let us write $X\in\{0,1\}^{\mathbb N}$ as $(X_n)_n$.

Let $B=\{X:X_n=1$ for at most finitely many $n\}$. Suppose $B=$ lim sup $A_n$. Then

$X_n=1$ for at most finitely many $n$ $\iff$ infinitely many $A_n$ occur.

Case 1: for infinitely many $n$, $A_n=\{0,1\}$. Then $X_n=1$ for all $n$ is a counterexample.

I'll let you think about the other cases.

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    $\begingroup$ Which space are you talking about? First you say that all $\mathcal F_n$ are the powerset of $\{0,1\}$, meaning that the state space is just the two-point set $\{0,1\}$. However, in the next sentence you are talking about subsets of the space of binary sequences. $\endgroup$
    – R W
    Commented Oct 5, 2019 at 14:47
  • $\begingroup$ I assume that each $X_n$ denotes the value of the $n^{\text{th}}$ number in the binary sequence. This is my attempt to show that $B$ is not the limsup of any $A_n$: Suppose there are infinitely many $A_n$ such that $1 \in A_n$. Then since $\limsup{A_n}$ contains sequences that have infinitely many 1s, this choice is not possible. This means that for sufficiently large $N$, $A_n = \{0\}$ for $n \geq N$. This however excludes all sequences that have finitely many 1s but contains 1s after the $N^\text{th}$ digit. Therefore, no choice of $A_n$ is possible. Am I correct? $\endgroup$
    – yellowello
    Commented Oct 5, 2019 at 15:11
  • $\begingroup$ @RW Thanks I think it may be fixed now $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Oct 5, 2019 at 18:34
  • $\begingroup$ @yellowello I think you're right but you may want to write it more carefully (than I did) $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Oct 6, 2019 at 0:32

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