Convergence in probability vs convergence a.s. Let $(X_n)$ be a sequence of random variables such that $(X_n)$ converges to $X$ in probability. Suppose that there exists a subsequence $(X_{n_j})$ that converges to $Y$ almost surely. How can we show that $X=Y$ almost surely? 
I know that if $(X_n)$ converges to $X$ in probability, then there exists a subsequence $(X_{n_k})$ that converges to $X$ almost surely. But what if the subsequences $(X_{n_k})$ and $(X_{n_j})$ do not coincide?
 A: Since $X_n$ converges in probability to $X$, it follows that the subsequence $X_{n_j}$ also converges in probability to $X$. Consequently, we can choose a subsequence $X_{n_{j_k}}$ which converges almost surely to $X$.
On the other hand, by your assumptions on $X_{n_j}$, we have that $X_{n_j}$ converges almost surely to $Y$. In particular, the subsequence $X_{n_{j_k}}$ also converges to $Y$ almost surely.
Since almost sure limits are unique (up to a null set), we conclude that $X=Y$ almost surely.
A: Lemma: If $Z_{j}\stackrel{P}{\to}Z$ and $Z_{j}\stackrel{P}{\to}0$
then $Z=0$ a.s.
Proof: 
For $\epsilon>0$ we find $\left|Z\right|>\epsilon\implies\left|Z-Z_{j}\right|>\frac{1}{2}\epsilon\text{ or }\left|Z_{j}\right|>\frac{1}{2}$
for every positive integer $j$.
This leads to $P\left(\left|Z\right|>\epsilon\right)\leq\max\left\{ P\left(\left|Z-Z_{j}\right|>\frac{1}{2}\epsilon\right),P\left(\left|Z_{j}\right|>\frac{1}{2}\epsilon\right)\right\} $
for every $j$ and letting $j\to\infty$ we find that $P\left(\left|Z\right|>\epsilon\right)=0$.
This for every $\epsilon$ so it is justified to conclude that $P\left(Z=0\right)=1$.

If $Z_{j}=X_{n_{j}}-X$ then $Z_{j}\stackrel{a.s}{\to}Y-X$ and also
$Z_{j}\stackrel{P}{\to}0$ so applying the lemma we find that $P(Y-X=0)=1$ or equivalently $X=Y$ a.s.
