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I'm trying to show that $\{ \limsup_{n \to \infty} S_n > 0 \}$ is not a tail $\sigma$ algebra.

We can rewrite the set as $\{ \omega \in \Omega: \limsup_{n \to \infty} X_1(\omega) + X_2(\omega) + \dots > 0 \}$. In order to show that it is not in tail $\sigma$-algebra, I need to construct $X_n$ such that $\omega \in \sigma(X_1)$ but $\omega \not\in \sigma(X_n)\; \forall n > 1$.

I tried to come up with a trivial example: $\Omega = \{ 1,2,3,4\}$, $$X_1 = \begin{cases}1 & \text{if $\omega \in \{1,2\}$} \\ 0 & \text{otherwise}\end{cases}, \quad X_n(\omega) = 0, \; \forall \omega \in \Omega, \; n > 1$$ But $\sigma(X_1) = \sigma(X_2)$ so this didn't work out. In fact, I'm having trouble defining $X_1$ and $X_n$ where $\sigma(X_1) \neq \sigma(X_n)$ for $n > 1$. Any hints would be appreciated.

MoneyBall
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Let $X_1=I_A$ and $X_n=I_B$ for all $n \geq 2$. Then you can check that $\lim \sup S_n(\omega) >0$ if and only if $\omega \in A \cup B$. But if $A \cup B$ belongs to the tail sigma field then it belongs to $\sigma (B)=\{\emptyset, B, B^{c},\Omega\}$. Can find an example where this fails?

Kavi Rama Murthy
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  • when $A$ and $B$ are disjoint, then $A \not\in \sigma(B)$, but isn't $\sigma(X_n)$ different from $\sigma(B)$? While $\sigma(B) = {\emptyset, B, B^c, \Omega }$, $\sigma(X_n) = { { \omega \in \Omega: X(\omega) \in B }: B \in { {1}, {0}, {1,0}, \emptyset } $? – MoneyBall Oct 13 '20 at 08:37
  • Take non-empty disjoint sets $A,B$ such that $A \cup B \neq \Omega$. Then $A \cup B \notin \sigma (B)$. @MoneyBall – Kavi Rama Murthy Oct 13 '20 at 08:41
  • $\sigma (I_B)= {\emptyset, \Omega, B, B^{c}}$. When you wrote $\sigma (X_n)$ you forgot to take inverse images. You have consider $X_n^{-1}({1}), X_n^{-1}({0})$ etc. – Kavi Rama Murthy Oct 13 '20 at 08:43
  • Ah right, it is the smallest set that makes the $I_B$ measurable. Thank you. – MoneyBall Oct 13 '20 at 08:49