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

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?

## 1 Answer

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.