# if $X_n \xrightarrow{P} X$, then there is a subsequence $\left(X_{n_k}\right)$ such that $X_{n_k} \xrightarrow{a.s}X$

I would like to show that if $X_n \xrightarrow{P} X$, then there is a subsequence $\left(X_{n_k}\right)$ such that $X_{n_k}\xrightarrow{a.s}X$.

I will enumerate some knowledge facts that I would to use. Moreover, I would like to know if these facts are well enunciated for a correct demonstration:

Fact 1.- If $\lim_{n \to \infty}P(|X_n - X|>\epsilon) = 0$, $\forall \epsilon>0$, then there is a subsequence $\left(n_k\right)$ such that $\Sigma_{k \in \mathbb{N}}P(|X_{n_k} - X|>\epsilon) < \infty$, $\forall \epsilon>0$.

Fact 2.- If $\Sigma_{n \in \mathbb{N}}P(|X_n - X|>\epsilon) < \infty$ $\forall \epsilon>0$, then $X_{n}\xrightarrow{a.s}X$

Note that with the two facts, I can prove my problem. In other words:

$X_n \xrightarrow{P} X \implies \hbox{hypothesis of Fact 1} \implies \hbox{hypothesis of Fact 2} \implies X_{n_k} \xrightarrow{a.s}X$

So, I would like to prove the Fact 1. I other words, for me, it is sufficient to prove that

if $\lim_{n\to \infty} a_n = 0$, then there is $(a_{n_k})$ subequence that $\Sigma_{k}a_{n_k} < \infty$. Is this fact true?

• as written, fact 1 is trivially true, fact 2 is trivially false – user363464 Jun 8 '17 at 3:59
• The fact 2 is a implication of Borel Cantelli. – orrillo Jun 8 '17 at 4:01
• You need to take $\epsilon_k \rightarrow 0$ for this theorem – clark Jun 8 '17 at 4:02
• @orrillo for fact 2 you need strict $<\infty$ – user363464 Jun 8 '17 at 4:03
• Sorry, You are right. Now, How the fact 1 is trivial? – orrillo Jun 8 '17 at 4:13