Let $X_1, X_2, ...$ be an independent sequence of random variables on $(\Omega, \mathscr{F}, \mathbb{P})$.
What I'm trying to prove is:
Prove that $X_1, X_2, ..., X_k$ is independent of $\liminf X_n$.
I've seen a proof of it already, but it was complicated. I am trying to prove it in another way.
This seems that this can be almost immediately just from the definition/s and can be extended to limsup and to any finite collection of the random variables of the sequence.
Based on this, I guess:
$\liminf X_n = \sup_{n \geq 1} \inf_{m \geq n} \ X_m = \sup_{n \geq k+1} \inf_{m \geq n} \ X_m$
I'm not quite sure how to say this precisely, but I think such shows that $\liminf X_n$ has nothing to do with $X_1, ..., X_k$. Perhaps we say something like $\sigma(\liminf X_n) \subseteq \sigma(X_{k+1}, X_{k+2}, ...)$, which is independent of $X_1, ..., X_k$ ?
Similarly, I guess that $\limsup X_n = \inf_{n \geq 1} \sup_{m \geq n} \ X_m = \inf_{n \geq k+1} \sup_{m \geq n} \ X_m$ and $\sigma(\limsup X_n) \subseteq \sigma(X_{k+1}, X_{k+2}, ...)$, which is independent of $X_1, ..., X_k$.
Extending to proving $\{X_k\}_{k \in K}$,where K is finite and $K \subseteq \mathbb{N}$, is independent of $\limsup X_n$ or $\liminf X_n$:
Pick the highest index , call it k* and then
$\liminf X_n = \sup_{n \geq 1} \inf_{m \geq n} \ X_m = \sup_{n \geq k^{*}+1} \inf_{m \geq n} \ X_m$
Pls do not present a different kind of proof if what I am attempting is wrong. Pls just state what is wrong with my attempt.