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Let $\{X_n\}_{n\ge0}$ be a random sequences such that $X_n\to X_{\infty}$ almost surely. Let $\mathcal{T}=\bigcap_{n=1}^{\infty}\sigma(X_n,X_{n+1},...)$ be the tail $\sigma$-algebra. Is $A=\{X_{\infty}\ge 0\}$ a tail event (i.e. $A\in \mathcal{T}$)?. Similar question for $B=\{X_{\infty}>0\}$.

There is a related counterexample when $X_n$ is a sum of random variables $\xi_n$ (however the tail sigma algebra of $(X_n)$ is different from the one of $(\xi_n)$), see the following question showing that $\{ \limsup_{n \to \infty} S_n > 0 \}$ is not a tail $\sigma$ algebra

user2096016
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    Does this answer your question? Show limsup/liminf is in tail field – user10354138 Oct 07 '21 at 04:43
  • @SangchulLee There is a difference between $S_n=X_1+\dots+X_n$ and $X_n$. – user10354138 Oct 07 '21 at 04:43
  • @user10354138 Ah yes, I just realized that. Thank you! I guess I need some sleep to restore my cognitive ability... – Sangchul Lee Oct 07 '21 at 04:43
  • Almost sure convegence is not enough. You require convergence at every point. – Kavi Rama Murthy Oct 07 '21 at 04:56
  • So if I can prove $B$ is not a tail event then $A$ is not too. However, I could not find a counter-example for $B$ assuming that $X_n$ converges almost surely/or surely converges. I am a bit confused now :( – user2096016 Oct 07 '21 at 04:58
  • Complete the tail $\sigma$-algebra with respect to your $\mathbb{P}$, otherwise you could make $X_\infty$ differs from $\lim X_n$ on a null subset $N\subset{\omega:X_n(\omega)\text{ converges}}$ to kill any meaningful statement (e.g., $X_n=1$ but $X_\infty=-1$ on a null set). Then remember the limit of a sequence is positive iff the sequence is eventually positive. – user10354138 Oct 07 '21 at 05:04
  • @user10354138 Thanks. I think that we have ${\lim_n X_{n}\ge 0}=\bigcap_{n\ge1}\bigcup_{k\ge n} {X_n\ge 0}$. Is that what you mean? The case of limsup and $X_n$ does not converges, the identity is certainly not true. – user2096016 Oct 07 '21 at 08:27
  • Oops, I mean bounded away from $0$ not just positive. ${\liminf_n X_n>0}=\bigcup_{q\in\mathbb{Q}, q>0}{\liminf_n X_n\geq q}=\bigcup_{q\in\mathbb{Q}, q>0}\bigcup_n\bigcap_{m>n}{X_m\geq q/2}=\bigcup_{q\in\mathbb{Q}, q>0}\bigcup_n\bigcap_{m>n}{X_m\geq q}$ Similarly $\limsup$, and the nonstrict inequality is just complement up to a null set. – user10354138 Oct 07 '21 at 08:31
  • No, we could identify $\limsup X_n>0$, it is $\bigcup_{q>0}\bigcap_n\bigcup_{m>n}{X_m>q}$. Similarly, we could identify where $X_n$ converge/does not converge, by writing the Cauchy sequences definition with $1/n$s instead of arbitrary $\epsilon$s. So if you define $X_\infty$ the conventional way: $\lim X_n$ if it exists, $0$ otherwise, then $X_\infty$ is measurable with you unaugmented tail $\sigma$-algebra. – user10354138 Oct 07 '21 at 10:01
  • @user10354138. I am a bit confused since there is a counter-example that ${\limsup X_n>0}$ is not tail event when $X_n$ does not converge. – user2096016 Oct 07 '21 at 12:04
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    @user2096016 You are confusing $\limsup S_n>0$ with $\limsup X_n>0$. The counterexample in your link uses the freedom to choose initial $X_1$ in $S_n=X_1+X_2+\dots+X_n$. – user10354138 Oct 09 '21 at 02:53
  • i see, they are different tail sigma-algebras. The tail sigma-algebra generated by $(S_n)$ is different from the tail sigma-algebra generated by its increment. – user2096016 Oct 09 '21 at 09:52

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