Define the limit inferior of a sequence of sets as $$\liminf A_n = \bigcup_{N \geq 0} \bigcap_{n \geq N} A_n , $$ and the limit inferior of a sequence as $$\liminf X_n = \sup_{N \geq 0} \inf_{n \geq N} X_n, $$
Is it true that $ \sup (\liminf A_n) = \liminf( \sup A_n) $? If it is not, under what conditions is it?
Now for a secondary point, let $ L^\infty_t = (\Omega , \mathcal{F}_t, \mathbb{P}) $ be a probability space, and for each $t \in [0,T]$ we have a probability space ($\mathfrak{F}= (\mathcal{F}_t)_t \small{\in[0,T]} $ is a filtration). How does the above question change if, instead of sup we use the essential supremum of a family of random variables random variables, i.e. for $X \in L^\infty_s, t \leq s, \text{ess} \sup_t X $ is the smaller $\mathcal{F}_t$-measurable random variable that is greater than $X$.
I can give you a counterexample for the first part of the question.
$$A_n = \cases{ [0;2] & if $n$ is even \\ [0,1] \cup [3,4]& if $n$ is odd}$$ Then $$\liminf A_n = [0,1]$$ $$\sup (\liminf A_n)=1$$ on the other hand $$\sup A_n = \cases{ 2 & if $n$ is even \\ 4& if $n$ is odd}$$ $$\liminf(\sup A_n) = 2$$ Remark: I suspect that with some effort one could show that $$ \sup (\liminf A_n) \le \liminf (\sup A_n)$$ but the converse inequality is clearly false.