liminf of sequence of iid random variables if $P(\liminf (X_n&gt;a))=0$, does that mean $\liminf X_n &lt;a$ almost surely? Where $X_n$ are iid random variables.
Important inequalities

Williams - Probability with Martingales







Deduced similarly:

(iii) If $\liminf x_n &gt; z$, then

$ \ \ \ \ \ \ \ (x_n &gt; z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n &lt; z $, then

$ \ \ \ \ \ \ \ (x_n &lt; z)$ infinitely often (that is, for infinitely many n)




  if $P(\liminf (X_n&gt;a))=0$, does that mean $\liminf X_n &lt;a$ almost surely?


Note that $P(\liminf (X_n&gt;a))=0 \iff P(\limsup (X_n \le a))=1$

Now it is not true that 

$$P(\limsup (X_n \le a))=1 \to P(\liminf X_n &lt;a) = 1$$

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

$$P(\liminf X_n &gt; a) = 1 \to P(\liminf (X_n &gt; a))=1$$

This can proven by proving

$$\liminf X_n &gt; a \subseteq \liminf (X_n &gt; a)$$

Pf: Suppose

$$\sup_{m \ge 1} \inf_{n \ge m} X_n := \beta &gt; a$$

Then $\forall \epsilon &gt; 0, \exists m \ge 1$ s.t.

$$\inf_{n \ge m} X_n := \alpha_m \in (\beta - \epsilon, \beta]$$

Choose $\epsilon &gt; 0$ s.t.

$$\inf_{n \ge m} X_n := \alpha_m &gt; a$$

$$\to \inf (X_m, X_{m+1}, ...) := \alpha_m &gt; a$$

$$\to X_n \ge \alpha_m &gt; a \ \forall n \ge m \ QED$$
Important inequalities

Williams - Probability with Martingales







Deduced similarly:

(iii) If $\liminf x_n &gt; z$, then

$ \ \ \ \ \ \ \ (x_n &gt; z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n &lt; z $, then

$ \ \ \ \ \ \ \ (x_n &lt; z)$ infinitely often (that is, for infinitely many n)




  if $P(\liminf (X_n&gt;a))=0$, does that mean $\liminf X_n &lt;a$ almost surely?


Note that $P(\liminf (X_n&gt;a))=0 \iff P(\limsup (X_n \le a))=1$

Now it is not true that 

$$P(\limsup (X_n \le a))=1 \to P(\liminf X_n &lt;a) = 1$$

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

$$P(\liminf X_n &gt; a) = 1 \to P(\liminf (X_n &gt; a))=1$$

This can proven by proving

$$\liminf X_n &gt; a \subseteq \liminf (X_n &gt; a)$$

Pf: Suppose

$$\sup_{m \ge 1} \inf_{n \ge m} X_n := \beta &gt; a$$

Then $\forall \epsilon &gt; 0, \exists m \ge 1$ s.t.

$$\inf_{n \ge m} X_n := \alpha_m \in (\beta - \epsilon, \beta]$$

Choose $\epsilon &gt; 0$ s.t.

$$\inf_{n \ge m} X_n := \alpha_m &gt; a$$

$$\to \inf (X_m, X_{m+1}, ...) := \alpha_m &gt; a$$

$$\to X_n \ge \alpha_m &gt; a \ \forall n \ge m \ QED$$