Proof of one of the operations on Random Variables

Question

This is a known operation on Random Variables (r.vs.):

If $(X_n : n \in \mathbb{N})$ is a sequence of r.vs., then $\inf X_n$, $\liminf X_n$, as well as $\sup X_n$ and $\limsup X_n$ are also r.vs.

Proof given in class :

Notice first that $\{\inf X_n \ge c\} = \bigcap_n \{X_n \ge c\} \in \mathcal F$. Hence, $\inf X_n$ is an r.v.

Let $L_n(\omega) := \inf \{X_r(\omega) : r \ge n\}$.

By the previous argument, $L_n$ is an r.v.

However, $L(\omega) := \liminf X_n(\omega) = \sup L_n(\omega)$.

Thus, $\{L \le c\} = \{\sup L_n \le c\} = \bigcap_n \{L_n \le c\} \in \mathcal F$, and $L$ is an r.v.

Now I dont understand how $\{\inf X_n \ge c\} = \bigcap_n \{X_n \ge c\}$ and later on $\{\sup L_n \le c\} = \bigcap_n \{L_n \le c\}$. As a result I do not understand the proof at all. Could someone please explain the proof? I've attached the screenshot of the proof also just in case

Answer 1

To show $\{\inf X_n \ge c\} = \bigcap_n \{X_n \ge c\}$, note the following statements are equivalent for any $\omega$ in the sample space:

• $\omega \in \{\inf X_n \ge c\}$
• $\inf X_n(\omega) \ge c$
• $c$ is a lower bound for $\{X_n(\omega) : n \in \mathbb{N}\}$ (because $\inf X_n(\omega)$ is the greatest such lower bound)
• $X_n(\omega) \ge c$ for all $n$
• $\omega \in \{X_n \ge c\}$ for all $n$
• $\omega \in \bigcap_n \{X_n \ge c\}$.

The case $\{\sup L_n \le c\} = \bigcap_n \{L_n \le c\}$ is very similar.