# Almost sure convergence problem

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}

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*}

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.