Another question on almost sure and convergence in probability Convergence in probability implies convergence on a subsequence almost surely.
But this means we fix a subsequence, such that $X_{n_k}$ converges for almost every $\omega$, right? The subsequence we pick does not depend on the $\omega$ right? 
 A: Also, you can directly apply Borel Cantelli to the sequence of events $|X_{n_k}-X|>2^{-k}$.
A: Yes, we can take a sequence $\{n_k\}$ which works for almost all $\omega$. To see that, fix a subsequence $\{n_k\}$ such that for each $k$,
$$ \Pr\left(|X_{n_k}-X|>2^{-k}\right)\leqslant 2^{-k}.$$
This one can be constructed by induction. Indeed, we first use the definition of convergence in probability with $\varepsilon=1/2$. We know that $\Pr\left(|X_{n}-X|>2^{-1}\right)$ goes to zero as $n$ goes to infinity, hence we are sure that for some $n_1$, $\Pr\left(|X_{n_1}-X|>2^{-1}\right)\leqslant 2^{-1}$. Now assume that we constructed $n_1<n_2<\dots<n_{k-1}$ such that for all $j\in\{1,\dots,k-1\}$,
$$ \Pr\left(|X_{n_j}-X|>2^{-j}\right)\leqslant 2^{-j}.$$
We now use the definition of convergence in probability with $\varepsilon=2^{-k}$. We know that $\Pr\left(|X_{n}-X|>2^{-k}\right)$ goes to zero as $n$ goes to infinity, hence we are sure that there is some $N$ such that for all $n\geqslant N$, $\Pr\left(|X_{n}-X|>2^{-k}\right)\leqslant 2^{-k}$. Consequently, a $n_k$ bigger than $n_{k-1}$ and $N$ does the job.
By the Borel-Cantelli lemma applied to $A_k:=\left\{|X_{n_k}-X|>2^{-k}\right\}$,
$$P\left(\limsup_{k\to+\infty}\left\{|X_{n_k}-X|>2^{-k}\right\}\right)=0.$$
This proves convergence almost everywhere of $\{X_{n_k}\}$ to $X$.
A: If $n_k$ is a subsequence with $P(|X_{n_k}-X|>\frac{1}{2^k}) \le \frac{1}{3^k}$
then
$$
E[\sum_{k\ge 0} \min(1,|X_{n_k}-X|) ] \le \sum_{k\ge 0} \frac{1}{2^k} + \frac{1}{3^k} <+\infty.
$$
The non-negative random variable $\sum_{k\ge 0} \min(1,|X_{n_k}-X|)$ is thus finite with probability 1 and in this event the summand must converge to 0, that is,  $\min(1,|X_{n_k}-X|)\to 0$, which implies $|X_{n_k}-X|\to 0$.
