Show limsup/liminf is in tail field

Given events $X_1, X_2, X_3, ...$, let $\tau = \bigcap_{n\geq1} \sigma(X_n, X_{n+1}, ...)$ be their tail field.

1 How do I show that $\limsup X_n \in \tau$?

What I tried:

$\limsup X_n = \bigcap_{m\geq1} \bigcup_{n\geq m} X_n$

$= (X_1 \cup X_2 \cup ...)$

$\cap (X_2 \cup X_3 \cup ...)$

$\cap ...$

We note that:

$(X_1 \cup X_2 \cup ...) \in \sigma(X_1, X_2, ...)$

$(X_2 \cup X_3 \cup ...) \in \sigma(X_2, X_3, ...)$

and so on.

Now, I am stuck.

If $A \in \scr{F}$ and $B \in \scr{G}$, I don't think it follows that $A \cap B \in \scr{F} \cap \scr{G}$.

2 How do I show that $\liminf X_n \in \tau$ similarly (that is, without noting that $\liminf X_n \subseteq \limsup X_n$)?

Edit: $\liminf X_n \subseteq \limsup X_n$ doesn't cut it. However, if $\limsup >X_n \in \tau$, then $(\limsup X_n)^{C} = \liminf X_n^C \in \tau$

What I tried:

$\liminf X_n = \bigcup_{m\geq1} \bigcap_{n\geq m} X_n$

$= (X_1 \cap X_2 \cap ...)$

$\cup (X_2 \cap X_3 \cap ...)$

$\cup ...$

We note that:

$(X_1 \cap X_2 \cap ...) \in \sigma(X_1, X_2, ...)$

$(X_2 \cap X_3 \cap ...) \in \sigma(X_2, X_3, ...)$

and so on.

I think it is true that if $A \in \scr{F}$ and $B \in \scr{G}$, then $A \cup B \in \scr{F} \cup \scr{G}$ so QED?

Solution 1:

For $\limsup$, notice that for each integer $l$, $$\limsup X_n=\bigcap_{ m\geqslant l}\bigcup_{n\geqslant m}X_n, $$ and the right hand side belongs to $\sigma\left(X_i,i\geqslant l\right)$. Since $l$ is arbitrary you can conclude.

For $\liminf$, your claim in your last sentence is true if $\mathcal F\subset\mathcal G$, which is the case in your context. But this will only show that $\liminf X_n$ is an element of $\sigma\left(X_i,i\geqslant 1\right)$. You can argue similarly as in the case of $\limsup$, or argue with the link between the complement of $\limsup$ and the $\liminf$ of complements.