liminf and limsup in probability 
Consider a sequence of random variable $(X_n)$. Prove the following inequality:
  $$
\mathbb P\left(\liminf\{X_n \leq x\}\right) \leq \liminf\mathbb P\left(\{X_n \leq x\}\right)\leq \limsup\mathbb P\left(\{X_n \leq x\}\right) \leq \mathbb P\left(\limsup\{X_n \leq x\}\right).
$$

I can see that the second $\leq$ is simply the result for real sequence. However, I do not see how to get the first and the last one. It seems that the two are similar. Could anyone help me, please? Thank you!

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 $P(\limsup A_n) \ge \limsup P(A_n)$.
