Prove that if $\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty$ for $\forall \delta > 0$, then $X_n \to 0$ almost surely.

Question

Let $\{X_n\}$ be a sequence of independent random variables. Assume that $\forall \delta > 0$ the following sum converges: $$\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty.$$

How can I prove that $X_n \to 0$ almost surely? In other words, I want to prove that

$$\mathbb{P}(\lim\limits_{n\to +\infty} X_n = 0) = 1.$$

To prove that, I think I need to use Borel–Cantelli lemma. It states that for a sequence of events $\{A_n\}_{n=1}^{\infty}$ such that $A = \bigcap\limits_{N=1}^{\infty} \bigcup\limits_{n > N} A_n $ and $\sum\limits_{n=1}^{\infty}\mathbb{P}\left(A_n\right) < \infty$, it is true that $P(A) = 0$.

From problem statement, I have $A_n = \{|X_n| \geq \delta\}$ and in order to solve the problem I need to put $A = \{\lim\limits_{n\to +\infty} X_n \neq 0\}$. So I need to make sure that this equation holds:

$$\{\lim\limits_{n\to +\infty} X_n \neq 0\} = \bigcap\limits_{N=1}^{\infty} \bigcup\limits_{n > N} \{|X_n| \geq \delta\} $$

I would say the equation above is true if I put $\delta := \frac{1}{N}$, for example, but I'm not allowed to make $A_n$ dependent on $N$, right? I don't know how to use the fact that $\sum\limits_{n=1}^{+\infty} \mathbb{P}(|X_n| \geq \delta) < \infty$ for every $\delta > 0$.

Any clue how to finish the proof? I would appreciate any help to proceed from here.

Answer 1

Just to present an alternative with the notation you used, note that by negating the limit definition, $X_n \not\to 0$ if and only if there exists a $m \in \mathbb{Z}_+$ such that for all $N \in \mathbb{Z}_+$ there exists an integer $n \geq N$ such that $|X_n| > \frac{1}{m}$.

Let's define the event $A_n(\epsilon) = \{ |X_n| > \epsilon\}$. By assumption $\sum_{n \geq 1} \mathbb{P}(A_n(\epsilon)) < \infty$ for all $\epsilon > 0$, and therefore by the Borel-Cantelli lemma $\mathbb{P}(\limsup A_n(\epsilon)) = 0$. (We can also say $A_n(\epsilon)$ occurs infinitely often or i.o., which intuitively says that almost surely $|X_n|$ exceeds $\epsilon$ only finitely many times.)

Therefore, $$ A = \{ \lim_{n \to \infty} X_n \ne 0 \} = \bigcup_{m \geq 1} \bigcap_{N \geq 1} \bigcup_{n \geq N} \{ |X_n| > \frac{1}{m} \} = \bigcup_{m \geq 1} \limsup A_n(1/m) . $$ Thus $\mathbb{P}(A) \leq \sum_{m \geq 1} \mathbb{P}(\limsup A_n(1/m))) = 0$, and we conclude $\mathbb{P}(A^c) = 1$, or equivalently $X_n \to 0$ almost surely.

Answer 2

Let $E_\delta$ be the set of $\omega$ for which $\sum_n 1_{[|X_n| \ge \delta]}(\omega)$ is unbounded. You know that $P E_\delta = 0$ for all $\delta >0$. Let $E = \cup_m E_{1 \over m}$ and note that $P E = 0$.

Suppose $\epsilon>0$ and $\omega \notin E$ then choose ${1 \over m} < \epsilon$ and so $\omega \notin E_{1 \over m}$. Then $\sum_n 1_{[|X_n| \ge {1 \over m}]}(\omega) < \infty$ and so there is some $M$ such that if $m \ge M$ we have $|X_n(\omega)| < { 1\over m} < \epsilon$.