# Use of Borel Canteli Lemma for Almost sure convergence, proof verification

I attempted to prove that $$X_{n}$$ converges almost surely to $$0$$ with the use of Borel Cantelli and Chebyshev inequality. \begin{align*} \mathbb{P}(\left \{ \omega \in \Omega:\left | X_{n} \right | \leq \epsilon, \exists \ n_{0} \ s.t. \ \forall \ n \geq n_{0} \right \}) &= \mathbb{P}\left(\bigcap_{n_{0}=1}^{\infty}\bigcup_{n\geq n_{0}}\left \{ \omega \in \Omega:\left | X_{n} \right | \leq \epsilon\right \}\right)\\ &= 1-\mathbb{P}\left(\bigcap_{n_{0}=1}^{\infty}\bigcup_{n\geq n_{0}}\left \{ \omega \in \Omega:\left | X_{n} \right | > \epsilon\right \}\right). \end{align*}

I define as $$A_{n}=\left \{ \omega \in \Omega:\left | X_{n} \right | > \epsilon\right \}.$$ Now from Chebyshev inequality I get

$$\mathbb{P}(\left \{ \omega \in \Omega:\vert X_{n}\vert > \epsilon\right \})\leq \frac{\sigma^{2}}{\epsilon}\Rightarrow \sum_{n}\mathbb{P}(A_{n}) \leq \frac{\sigma^{2}}{\epsilon}.$$

Hence, from the Borel Cantelli lemma https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma

and I get that $$\mathbb{P}\left(\bigcap_{n_{0}=1}^{\infty}\bigcup_{n\geq n_{0}}\left \{ \omega \in \Omega:\left | X_{n} \right | > \epsilon\right \}\right)=0$$. So, $$\mathbb{P}(\left \{ \omega \in \Omega:\left | X_{n} \right | \leq \epsilon, \exists \ n_{0} \ s.t. \ \forall \ n \geq n_{0} \right \})=1.$$

Your application of Chebysev's inequality is not correct. What you can say is $$P(|X_n| >\epsilon) \leq \frac {EX_n^{2}} {\epsilon^{2}}$$. You provided no information about variance of $$X_n$$. If you know that $$\sum EX_n^{2} <\infty$$ then the rest of your proof would be correct.