# Proof explanation about why $\liminf_{n}X_{n}$ is a random variable

borel-setslimsup-and-liminfprobability theoryrandom variables

Definition 1: $$X: \varOmega \to \varOmega'$$ is called a measurable map from $$(\varOmega ,\mathcal{F}) \to (\varOmega' , \mathcal{F}')$$ if for all $$A'\in\mathcal{F}' \implies X^{-1}(A') \in\mathcal{F}$$ where $$\mathcal{F},\mathcal{F}'$$ is a sigma-algebra over $$\varOmega ,\varOmega'$$ respectively

Definition 2: A random variable on a probability space $$(\varOmega , \mathcal{F} ,\mathbb P )$$ is a measurable function $$X : (\varOmega,\mathcal{F}) \to (\mathbb R, \mathcal{B})$$

The question is: Assume that $$(X_{n})_{n}$$ is a sequence of random variables. I need to prove that $$\liminf_{n}X_{n}$$ is a random variable

Since $$\mathcal{B}=\sigma((-\infty,t]:t\in\mathbb{R})$$ its enough to show that for any $$t \in \mathbb{R}$$ $$\liminf_{n}X_{n} \leq t$$ is an event

My professor solved that like that:

Let $$W = \liminf_{n}X_{n}$$. Note that for any $$t \in \mathbb R$$,
$$\left\{W > t \right\} = \{\exists n, \forall k\ge n , X_{k}>t\} = \bigcup_{n}\bigcap_{k\geq n}\{X_{k}>t\}=\liminf_{n}\{X_{n}>t\}$$ which is an event as a countable union of a countable intersection of events.Thus $$\left\{W \leq t\right\} = \left\{W > t\right\}^c$$ is an event. So $$W$$ is measutable and thus a random variable.

I don't understand the logic behind this, from the answer I can see that
$$\left\{W \leq t\right\} = \bigcap_{n}\bigcup_{k\geq n}\left\{X_{k}\leq t\right\} = \limsup_{n}\left\{X_{n}\leq t \right\}$$

but the definition of $$\liminf_{n}X_{n}$$ is that $$X_{n}$$ will occur after some large enough $$n$$ onward. that is ,eventually all $$X_{n}$$ occur.
so if we say that $$\left\{W > t\right\}$$ its mean that after large enough $$n$$ we will have that for all $$k \geq n$$ $$X_{k}>t$$ so indeed $$\left\{W > t\right\} = \liminf_{n}\left\{X_{n}>t\right\}$$

but why can't I say the same with
$$\left\{W \leq t\right\}$$ ? I mean why cant I use the same logic that after large enough $$n$$ we will have that for all $$k \geq n$$ $$X_{k}\leq t$$ so we will have that $$\left\{W \leq t\right\} = \liminf_{n}\left\{X_{n}\leq t\right\}$$?

I saw the question here (Prove that $\liminf_{n}X_{n}$ is a random variable) but this is not answering my question

Let be $$f_n(\omega)=\inf_{m\ge n} X_m(\omega)$$, then $$W=\sup_n f_n(\omega)\le t$$ implies $$f_n(\omega)\le t$$, $$\forall n\ge 1$$, so your question is equivalent to say that for some large $$n$$ inequality $$f_n(\omega)=\inf_{m\ge n} X_m(\omega)\le t$$ implies $$X_m\le t\quad\forall m\ge n$$, wich is not true if you take by example $$X_m(\omega) = m$$. So $$f_n(\omega)\le t$$ ensure only existence of some $$m$$ such that $$X_m\le t$$.
Edit: The proof step by step: $$\begin{eqnarray} \{\liminf_{n\ge 1} X_n(\omega)
Note that if $$\sup f_n(\omega), then $$f_n(\omega)\le\sup f_n(\omega), that is the intesection of events $$\{f_1(\omega), $$\{f_2(\omega) and so on.
The reason for taking $$<$$ instead of $$\le$$ is because of the event $$A=\{\inf_{m\ge n} X_m\ge t\}=\bigcap_{m\ge n}\{X_m\ge t\}$$ by reasoning as we did above with $$\sup f_n$$, so we can use the complement of $$A$$ to derivate the form of $$\{\liminf_{n\ge 1} X_n(\omega) as you can see in the proof.