Prove this liminf is a tail events Let $A_{k}$,$k\geq1$ be [0,$\infty$)-valued random variables on a common probability space. 
I want to prove the following events are in/not in tail $\sigma$-field T($A_{k}$:$k\geq1$).
First, event $${lim_{k\to\infty}inf(A_{k})<{\infty}}$$
We know that ${lim_{k\to\infty}sup(A_{k})}$ and ${lim_{k\to\infty}inf(A_{k})}$ are tail events. 
For ${lim_{k\to\infty}inf(A_{k})}$ we can argue that elements of the σ-algebra generated by the $A_{k}$ with k≥n are of the form $\bigcap_{k\geq n} A_k$. And if there is some n and points x,y that cannot be separated by the $A_k$ with k≥n for some n, they have to be the same element(I take the argument from this post Atoms in a tail $\sigma$-algebra as $\liminf C_n$). So we use the union to pool the elements from all tails together:  $\bigcup_{n\geq 1}\bigcap_{k\geq n} A_k$ = ${lim_{k\to\infty}inf(A_{k})}$. 
I am a little confused, with the event be defined here using the condition <${\infty}$, how is it different with ${lim_{k\to\infty}inf(A_{k})}$? The intuition of a tail event is, it does not depend on the first n instances. And as the liminf can be think as "ultimately all of them comes up",  whether ultimately $A_k$ is finite (is this the correct intuition of this event?) certainly does not depend on the first n instances, so this event is a tail event.
Second, an event $${inf_{k\geq 1}A_k{\leq 1}}$$
If in the first k instances, we have $A_i$ > $inf_{k\geq 1}$$A_k$, then the probability of this event happening does depend on the first instances, so it is not a tail event, is this argument correct? I certainly know it is not rigorous, can you give any hints on how to do that? I think we do not need to use measure or borel fields though.
 A: $$\liminf A_k = \bigcup_{n\in \Bbb N} \bigcap_{k≥n} A_k \Rightarrow \bigcup_{n\in \Bbb N} \bigcap_{k≥n}A_k\in \mathcal T \iff \bigcup_{n ≥ m} \bigcap_{k≥n} A_k \in \mathcal T \space \forall m\in \Bbb N$$
Since 
$$\liminf A_k, k≥1 = \liminf A_j, j≥m$$ because, if $a\in \liminf A_k, k≥1, \space then\space a\in \bigcup_{n\in \Bbb N} \bigcap_{k≥n} A_k \Rightarrow a\in \bigcup_{n ≥ m} \bigcap_{k≥n} A_k \space and \space a\in \bigcup_{n ≥ m} \bigcap_{k≥n} A_k \Rightarrow a\in \bigcup_{n\in \Bbb N} \bigcap_{k≥n} A_k$
These arrive from the definitions of the set operations.
To convince yourself of the previous, remember $(\bigcap_{k≥n} A_k)_n$  is an increasing sequence of events. Let $B_n = \bigcap_{k≥n} A_k \Rightarrow B_n \subset B_{n+1} \subset \dots \space so \space \bigcup_{n=1}^m B_n = \bigcup_{n=k}^m B_n = B_m , k>1 $
Thus, the event {$\liminf A_k, k\in \Bbb N$} only depends on those $k≥m$
Hence, it is a tail event.
Therefore, $\liminf A_k<\infty, \space k≥1 \iff \liminf A_k <\infty, \space k≥m$
So, the first event is a tail event.
For the second event, 
Let $\inf A_k ≤1, k≥1$ since $\inf A_k$ is an increasing sequence of events, It could be the case that there is an element bigger than 1 in all $A_k, k≥2$ Thus, $\inf A_k\not <1, k≥2$ Hence, $\inf A_k ≤ 1$ can't be a tail event. 
