liminf and limsup in probability

Let $A_n = \{X_n \le x\}$. Then

$$\liminf A_n = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k$$

The sequence $\{\cap_{k\ge n} A_k\}_{n = 1}^\infty$ is a sequence which increases to $\liminf A_n$,

$$P(\liminf A_n) = \lim_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right)\tag{1}$$

by continuity of $P$ from below. Since $\cap_{k\ge n} A_k \subseteq A_n$ for all $n$, $P(\cap_{k\ge n} A_k) \le P(A_n)$ for all $n$. Hence

$$\liminf_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right) \le \liminf_{n\to \infty} P(A_n)\tag{2}$$

By $(1)$ and $(2)$,

$$P(\liminf_{n\to \infty} A_n) \le \liminf_{n\to \infty} P(A_n).$$

For the last equality, let $\Omega$ be the underlying sample space. Apply the lim inf inequality to the sequence $B_n = \Omega \setminus A_n$ to get $P(\limsup A_n) \ge \limsup P(A_n)$.

This has nothing to do with random variables. Just let $A_n = \{X_n \leq x\}$

$$ \mathbb P\left(\liminf A_n\right) \leq \liminf\mathbb P\left(A_n\right)\leq \limsup\mathbb P\left(A_n\right) \leq \mathbb P\left(\limsup A_n\right). $$

The inequalities are:

1. Fatou's Lemma
2. A real analysis proposition
3. Reverse Fatou's Lemma