Almost sure convergence problem

Question

I'm working on a problem in which I should prove "almost sure" convergence for a sequence of random variables. I'm using Borel-Cantelli lemma to prove it. Here is the question and my solution - I appreciate if you can give me feedback and whether you have any other solution to this problem.

Assumptions:

• $X_n$ is a sequence of random variables
• $\mathrm{Pr}(|X_n| > \epsilon)\le c^n$ for sufficiently large n, where $ 0 < c < 1$ and for $\forall \epsilon > 0$

Goal: to show:

$$X_n \overset{\text{a.s.}}{\rightarrow} 0 $$ where "$\overset{\text{a.s.}}{\rightarrow}$" means almost sure convergence}

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My solution:

As $n \rightarrow \infty$, we can write:

$$\sum_{j = 1}^{\infty} \mathrm{Pr}(|X_n| > \epsilon) \le \frac{c}{1-c} < \infty$$

Then by the application of Borel-Cantelli lemma, we can conclude:

\begin{align*} && \mathrm{Pr}\{|X_n| > \epsilon, i.o\} &= 0 \\ &\Rightarrow & 1 - \mathrm{Pr}\{|X_n| > \epsilon, i.o\} &= \mathrm{Pr}\{|X_n| \le \epsilon, i.o\} \\ & \Rightarrow & \mathrm{Pr}\{|X_n| \le \epsilon, i.o\} &= 1 \\ & \Rightarrow & X_n &\overset{\text{a.s.}}{\rightarrow} 0 \end{align*}

What do you think about this solution?

Answer 1

Careful, the complement of the set $\{|X_n|\leqslant\varepsilon, \mbox{i.o.}\}$ is not $\{|X_n|>\varepsilon, \mbox{i.o.}\}$ (it is $\{\exists m\mid \forall n\geqslant m,|X_n|>\varepsilon\}$.

So the application of Borel-Cantelli's lemma yields a set $\Omega_{\varepsilon}\subset \Omega$ of measure $1$ for which if $\omega\in\Omega'$, then there is an integer $N=N(\omega)$ such that $|X_n|\leqslant \varepsilon$ whenever $n\geqslant N$.

Take $\Omega':=\bigcap_{q\in\Bbb Q,q>0}\Omega_q$: there is pointwise convergence on $\Omega'$ to $0$ and $\mathbb P(\Omega')=1$.

Since we just need the convergence of $\mathbb P(|X_n|>\varepsilon)$, we actually proved that

If for each $\varepsilon>0$, the series $\sum_{n=1}^{+\infty}\mathbb P(|X_n|>\varepsilon)$ is convergent, then $X_n\to 0$ almost surely.