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.
Best Answer
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?