Proof convergence of random variables (almost sure) Let $\left\{ X_{n}\right\} _{n\in\mathbb{N}}$ be a sequence of independent random variables. Prove that $X_{n}\overset{a.e.}{\rightarrow}0$ if and only if, for all $\epsilon&gt;0$, $\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon\right)&lt;\infty$. 

I guess I have to use the Borel Cantelli Lemma, right? Thanks.
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'

Suppose $\forall \epsilon&gt;0$,

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon\right)&lt;\infty$$

By BCL1, we have

$$\textrm{P}\left(\limsup [\left|X_{n}\right|&gt;\epsilon\right]) = 0$$

$$\to \textrm{P}\left(\liminf [\left|X_{n}\right|\le\epsilon\right]) = 1$$

$$\to \textrm{P}\left([\liminf \left|X_{n}\right|]\le\epsilon\right) = 1 \ \text{and} \ \textrm{P}\left([\limsup \left|X_{n}\right|]\le\epsilon\right) = 1$$

$$\to \textrm{P}\left([\liminf \left|X_{n}\right|]=0\right) = 1 \ \text{and} \ \textrm{P}\left([\limsup \left|X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\lim \left|X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\left|\lim  X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\lim X_{n}]=0\right) = 1$$



'only if'

Suppose $\exists \epsilon_0 &gt; 0$ s.t.

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon_0\right)=\infty$$

By BCL2, we have

$$\textrm{P}\left(\limsup [\left|X_{n}\right|&gt;\epsilon_0\right]) = 1$$

$$\to \textrm{P}\left([\limsup \left|X_{n}\right|]&gt;\epsilon_0\right) = 1$$

$$\to \textrm{P}\left([\limsup \left|X_{n}\right|]\le\epsilon_0\right) = 0$$

$$\to \textrm{P}\left(\liminf \left[|X_{n}\right|\le\epsilon_0]\right) = 0$$

However since

$$\textrm{P}\left([\lim X_{n}]=0\right) = 1,$$

we have

$$\textrm{P}\left([\limsup \left|X_{n}\right|]=0\right) = 1$$

$$\to \forall \epsilon &gt; 0, \textrm{P}\left([\limsup \left|X_{n}\right|]&lt;\epsilon\right) = 1$$

$$\to \forall \epsilon &gt; 0, \textrm{P}\left(\liminf \left[|X_{n}\right|&lt; \epsilon]\right) = 1$$

Choose $\epsilon = \epsilon_0$ to get:

$$\textrm{P}\left(\liminf \left|[X_{n}\right|&lt; \epsilon_0]\right) = 1 ↯$$

$\therefore, \forall \epsilon&gt;0, \sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon\right)&lt;\infty$ 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'

Suppose $\forall \epsilon&gt;0$,

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon\right)&lt;\infty$$

By BCL1, we have

$$\textrm{P}\left(\limsup [\left|X_{n}\right|&gt;\epsilon\right]) = 0$$

$$\to \textrm{P}\left(\liminf [\left|X_{n}\right|\le\epsilon\right]) = 1$$

$$\to \textrm{P}\left([\liminf \left|X_{n}\right|]\le\epsilon\right) = 1 \ \text{and} \ \textrm{P}\left([\limsup \left|X_{n}\right|]\le\epsilon\right) = 1$$

$$\to \textrm{P}\left([\liminf \left|X_{n}\right|]=0\right) = 1 \ \text{and} \ \textrm{P}\left([\limsup \left|X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\lim \left|X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\left|\lim  X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\lim X_{n}]=0\right) = 1$$



'only if'

Suppose $\exists \epsilon_0 &gt; 0$ s.t.

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon_0\right)=\infty$$

By BCL2, we have

$$\textrm{P}\left(\limsup [\left|X_{n}\right|&gt;\epsilon_0\right]) = 1$$

$$\to \textrm{P}\left([\limsup \left|X_{n}\right|]&gt;\epsilon_0\right) = 1$$

$$\to \textrm{P}\left([\limsup \left|X_{n}\right|]\le\epsilon_0\right) = 0$$

$$\to \textrm{P}\left(\liminf \left[|X_{n}\right|\le\epsilon_0]\right) = 0$$

However since

$$\textrm{P}\left([\lim X_{n}]=0\right) = 1,$$

we have

$$\textrm{P}\left([\limsup \left|X_{n}\right|]=0\right) = 1$$

$$\to \forall \epsilon &gt; 0, \textrm{P}\left([\limsup \left|X_{n}\right|]&lt;\epsilon\right) = 1$$

$$\to \forall \epsilon &gt; 0, \textrm{P}\left(\liminf \left[|X_{n}\right|&lt; \epsilon]\right) = 1$$

Choose $\epsilon = \epsilon_0$ to get:

$$\textrm{P}\left(\liminf \left|[X_{n}\right|&lt; \epsilon_0]\right) = 1 ↯$$

$\therefore, \forall \epsilon&gt;0, \sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|&gt;\epsilon\right)&lt;\infty$ QED