Prove that $ \liminf_{n}X_{n} $ is a random variable

Question

We define random variable $ X: \varOmega \to \mathbb{R} $ as a function such that for and Borel set $ \forall B\in\mathbb{\mathcal{B}} $ it follows that $ X^{-1}\left(B\right)\in\mathcal{F} $ where $ \mathcal{F} $ is a sigma-algebra over $ \varOmega $

Now assume that $ \left(X_{n}\right)_{n} $ is a sequence of random variables. Is it true that the function $ Y:\varOmega \to \mathbb{R} $ by :

$ Y\left(\omega\right)=\liminf_{n}X_{n}\left(\omega\right) $ is a random variable?

Here's what I tried.

I want to prove that indeed it is a random variable. Since $ \mathcal{B}=\sigma\left((-\infty,t]:t\in\mathbb{R}\right) $ its enough to show that for any $ t \in \mathbb{R} $ it follows that $ Y^{-1}\left((-\infty,t]\right)\in\mathcal{F} $

But

$ Y^{-1}\left((-\infty,t]\right)=\left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} $

Now if I'll prove that

$ \left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} =\bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $

Then it will end the proof because the latter is an event in the sigma-algebra. I already proved that $ \left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} \supseteq\bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $

But Im having trouble to prove that $ \left\{ \omega\in\varOmega:\liminf_{n}X_{n}\left(w\right)\in(-\infty,t]\right\} \subseteq\bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $

Because if I'll take $ \omega $ such that $ \liminf_{n}X_{n}\left(w\right)\in(-\infty,t] =t $, thenI cannot promise that this $ \omega $ is in $ \bigcap_{k}^{\infty}\bigcup_{n\geq k}X_{n}^{-1}\left((-\infty,t]\right) $.

Any ideas?

Answer 1

We'll instead prove that $$ \{ \liminf X_n \leq t \} = \bigcap_{r \geq 1} \bigcap_{k \geq 1} \bigcup_{n \geq k} \left\{ X_n < t + \frac1r \right\} =: \bigcap_{r \geq 1} E_r $$ Adding the wiggle room of "$+1/r$" fixes the problem you were having.

For the $\supseteq$ direction, if $\omega \in E_r$ then $X_n(\omega) < t + 1/r$ infinitely often, which implies $\liminf X_n < t+ 1/r$. So if this holds for every $r \geq 1$ then $\liminf X_n \leq t$.

For the $\subseteq$ direction, $\liminf X_n(\omega) \leq t$ implies that for every $\epsilon > 0$, $X_n(\omega) < t+\epsilon$ infinitely often, which is equivalent to the statement that $\omega \in E_r$ for every $r$.