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?

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.

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.