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

## Best Answer

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)<t\} &=& \{\sup_{n\ge 1}\inf_{m\ge n}X_m(\omega) < t\}\\ &=&\bigcap_{n\ge 1}\{\inf_{m\ge n}X_m(\omega) < t\}\\ &=& \bigcap_{n\ge 1}\bigcup_{m\ge n}\{X_m(\omega) < t\}\in \mathcal{B} \end{eqnarray}

Note that if $\sup f_n(\omega)<t$, then $f_n(\omega)\le\sup f_n(\omega)<t\ \forall n\ge 1$, that is the intesection of events $\{f_1(\omega)<t\}$, $\{f_2(\omega)<t\}$ 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 $\{\liminf_{n\ge 1} X_n(\omega)<t\}$ as you can see in the proof.