Edit: Let us define the tail sigma-algebra as $T = \cap_{i \in \mathbb{N}}T_i, T_i = \sigma((X_{n})_{n > i})$ of a sequence of random variables $X_n$.
All the examples I have seen so far about events belonging/not belonging to the tail sigma-algebra fall essentially to either i.) decomposing the event to a relationship with the $\lim\inf$ and $\lim\sup$ and using intersections, unions and complements to argue why an event belongs to the tail sigma-algebra or ii.) constructing such a sequence $X_1,X_2,\dots$ of r.v.s that some finite random variables $X_{i_1},\dots,X_{i_n}$ define the event in question, so that one can conclude "Since this event depends on some finite collection of the elements of the random sequence, the event does not belong to the tail sigma-algebra".
But what if you'd want to argue about an event no belonging to the tail sigma-algebra in a more rigorous way? E.g. let $X_1,X_2,\dots$ be real valued r.v.s on the probability space $(\Omega, F, P)$ and let $E = \{\omega \in \Omega\mid X_1 > 0\}$. Evidently this event does not belong to the tail sigma-algebra, but how would you argue about it with only the algebraic/set theoric definitions?
Best Answer
As usual, to show that an element is not in intersection, you need to show that it is not in some element.
For example, if $X_i$ are independent, then any event from $\sigma(X_i)$ if $i > 1$ is independent from any event from $\sigma(X_1)$. In particular, your $E \in \sigma(X_1)$ is independent from any event in $T_2$ and so is independent from $T$. As any event with non-zero and non-one probability is not independent from itself, it implies that if $P(E) \neq 0, 1$ then $E \notin T_2$ and so $E \notin T$.
If, for example, $E$ is empty ($X_1$ is always negative) then $E$, of course, is in $T$.
And if $X_i$ are not independent, then, in general, $E$ also can be in $T$. For example, if $X_1 = X_2 = \ldots$ then $T = T_i = \sigma(X_1)$, and so of course $E \in T$ in such case.