Equivalence of inequalities of limsup and liminf 
Prove the following equivalence relationship: $$P(\lim\inf A_n)\leq \lim\inf P(A_n) \Leftrightarrow \ P(\lim \sup A_n) \geq \lim \sup P(A_n).$$

My attempt: I could not see how to prove this relationship yet, except for my hypothesis, based on the definition of $\lim \inf$ and $\lim \sup$ that we could rewrite $\lim \sup A_n = -\lim \inf(-A_n)$?? But then how to show that $P(-\lim \inf(-A_n)) = - P(\lim \sup (A_n))$?
Could anyone please tell me if I'm on the right track help me with this problem? Really appreciate your input.
 A: We show that the $\liminf$ inequality implies the $\limsup$ inequality (the other way is similar).
Consider the sequence $\{A_n^c\}_{n\geq 1}$. We have that by De Morgan's laws:
$$\liminf_{n\to \infty} A_n^c = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k^c=\bigcup_{n \ge 1}\left( \bigcup_{k\ge n} A_k\right)^c
=\left(\bigcap_{n \ge 1} \bigcup_{k\ge n} A_k\right)^c=\left(\limsup_{n\to \infty} A_n\right)^c.$$
Therefore
$$1-\limsup_{n\to \infty} P(A_n)=\liminf_{n\to \infty} (1-P(A_n))\\=\liminf_{n\to \infty} P(A_n^c)\geq P(\liminf_{n\to \infty} A_n^c)=1-P(\limsup_{n\to \infty} A_n)$$
which implies $$\limsup_{n\to \infty} P(A_n)\leq P(\limsup_{n\to \infty} A_n).$$
P.S. We include a proof of the $\liminf$ inequality (my first answer where a miss OP's request about equivalence).
By definition
$$\liminf_{n\to \infty} A_n = \bigcup_{n \ge 1} \bigcap_{k\ge n} A_k.$$
Since for any $n$, $\cap_{k\ge n} A_k \subseteq A_n$, it follows that $P(\cap_{k\ge n} A_k) \le P(A_n)$. Hence
$$\liminf_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right) \le \liminf_{n\to \infty} P(A_n)\tag{2}$$
Moreover the sequence $\{\cap_{k\ge n} A_k\}_{n\geq 1}$  is increasing and by continuity of $P$ from below,
$$P(\liminf A_n) = \lim_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right)=\liminf_{n\to \infty} P\left(\bigcap_{k \ge n} A_k\right).$$
A: I think the correct statement should be 
$$P(\liminf A_n) \leq \liminf P(A_n) \Leftrightarrow P(\limsup A_n^c) \geq \limsup P(A_n^c) \tag{1}$$
instead of the one you stated. 
To prove the $(1)$, it suffices to note that
$$\left(\liminf A_n\right)^c = \limsup A_n^c,$$
which can be easily verified by definitions of upper/lower limit sets and De Morgan law.
Now the proof can be completed by just using $P(E^c) = 1 - P(E)$ for any event $E$ and $\liminf(1 - a_n) = 1 - \limsup a_n $ for any real sequence $\{a_n\}$.
For example, if you are given $P(\liminf A_n) \leq \liminf P(A_n)$, then it implies
$$1 - P(\liminf A_n) \geq 1 - \liminf P(A_n),$$
which is equivalent to
$$P((\liminf A_n)^c) \geq \limsup P(A_n^c),$$
or
$$P(\limsup A_n^c) \geq \limsup P(A_n^c).$$
