liminf of sequence of iid random variables if $P(\liminf (X_n>a))=0$, does that mean $\liminf X_n <a$ almost surely? Where $X_n$ are iid random variables. 
 A: Important inequalities
Williams - Probability with Martingales



Deduced similarly:
(iii) If $\liminf x_n > z$, then
$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)
(iv) If $\liminf x_n < z $, then
$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)


if $P(\liminf (X_n>a))=0$, does that mean $\liminf X_n <a$ almost surely?

Note that $P(\liminf (X_n>a))=0 \iff P(\limsup (X_n \le a))=1$
Now it is not true that 
$$P(\limsup (X_n \le a))=1 \to P(\liminf X_n <a) = 1$$
Consider $X_n = a - 1/n$.
However
$$P(\limsup (X_n \le a))=1 \to P(\liminf X_n \le a) = 1$$
The contrapositive is
$$P(\liminf X_n > a) = 1 \to P(\liminf (X_n > a))=1$$
This can proven by proving
$$\liminf X_n > a \subseteq \liminf (X_n > a)$$
Pf: Suppose
$$\sup_{m \ge 1} \inf_{n \ge m} X_n := \beta > a$$
Then $\forall \epsilon > 0, \exists m \ge 1$ s.t.
$$\inf_{n \ge m} X_n := \alpha_m \in (\beta - \epsilon, \beta]$$
Choose $\epsilon > 0$ s.t.
$$\inf_{n \ge m} X_n := \alpha_m > a$$
$$\to \inf (X_m, X_{m+1}, ...) := \alpha_m > a$$
$$\to X_n \ge \alpha_m > a \ \forall n \ge m \ QED$$
