Proof convergence of random variables (almost sure)

Question

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 $\epsilon>0$, $\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon\right)<\infty$.

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

Answer 1

Important inequalities

Williams - Probability with Martingales

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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)

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'if'

Suppose $\forall \epsilon>0$,

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon\right)<\infty$$

By BCL1, we have

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

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

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

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

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

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

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

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'only if'

Suppose $\exists \epsilon_0 > 0$ s.t.

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

By BCL2, we have

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

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

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

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

However since

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

we have

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

$$\to \forall \epsilon > 0, \textrm{P}\left([\limsup \left|X_{n}\right|]<\epsilon\right) = 1$$

$$\to \forall \epsilon > 0, \textrm{P}\left(\liminf \left[|X_{n}\right|< \epsilon]\right) = 1$$

Choose $\epsilon = \epsilon_0$ to get:

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

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