I admit that my measure theoretic background is not quite sharp, and now I'm studying advanced probability theory.
In this "easy" exercise, I have to show that if $a_n$ is a sequence decreasing towards zero, then the random variables $X_n = a_nX$ are converging almost sure to zero.
My argument goes as follows:
For fixed $\epsilon >0$ and fixed $\omega \in \Omega$, we can find $N(\omega)$ such that for all $m \geq N(\omega)$: $|a_mX(\omega)| \leq \epsilon$
Now let $N^{\ast}$ = $\max_{\omega \in \Omega} \{N(\omega) \, : \, |a_mX(\omega)| \leq \epsilon \, \text{ for } \, m \geq N(\omega) \}$
Then we have $$\sum_{n=1}^{\infty} P(|aX_n| \geq \epsilon) = \sum_{n=1}^{N^{\ast}} P(|aX_n| \geq \epsilon) < \infty $$ and from Borel-Cantellis lemma we conclude that $aX_n \longrightarrow 0$ a.s.
Is this a correct argument, or is it flawed? Could we conclude the almost sure convergence in a simpler way?
thanks!
Your first sentence completes the proof. Nothing more to be done! $\{\omega : a_nX(\omega) \to 0\}=\Omega$ which prove that $a_n X \to 0$ a.s. since $P(\Omega)=1$.