Equivalence of limit supremum of sequence of sets and sequence of functions

Question

Let $X_1,X_2,\dots$ be a sequence of real-valued random variables where $X_n:\Omega \to \mathbb{R}$.

For any sequence of events $A_n \subset \Omega$ define $\limsup_nA_n := \bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}A_m$.

Moreover, define the (extended-real) random variable $Y \equiv (\limsup_nX_n)$ by $Y(\omega) := \limsup_n\{X_n(\omega) \mid n=1,2,\dots\}$.

How would you establish that

$$ \limsup_{n}\{\omega \mid X_n(\omega) \in B\} = \{\omega \mid (\limsup_nX_n)(\omega)\ \in B \}, \qquad (*) $$

for any Borel set $B$? Is the proposition $(*)$ even true?

Answer 1

After further thought I believe that $(*)$ is in general false.

Let $B = \{ b\}$ be a singleton set. Observe that

1. $\omega \in \{ \omega' \mid (\limsup_nX_n)(\omega') = b \}$ if and only if $\limsup\{X_1(\omega), X_2(\omega), \dots \} = b$.

2. $\omega \in \limsup_n\{ \omega' \mid X_n(\omega') = b \}$ if and only if for all $n$ there exists $m>n$ such that $X_m(\omega)=b$, or in other words $X_n(\omega) = b$ infinitely often.

Let $\omega$ be such that the sequence $(X_n(\omega))_{n=1}^{\infty} = (0,1,0,1,0,1,0,1,\dots)$ and put $b=0$. Then $(\limsup_nX_n)(\omega) = 1$, however $X_n(\omega) = 0$ occurs infinitely often.