limit in probability is almost surely unique? I read this proposition in a book, which was not proved.
And I cannot verify it myself.
Could anyone help me out here?
If $$X_{n}\rightarrow X$$ in probability and $$X_{n}\rightarrow Y$$ almost surely,
then $$P(X=Y)=1.$$
An alternative version is that the p-limit of a sequence is almost surely unique.
Thanks for your time.
Cheers.
 A: Back to basics: Assume that $X_n\to X$ in probability and that $X_n\to Y$ in probability. Then, for every positive $x$, $P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x)$ converges to zero since both terms do. Now, $$[|X-Y|\geqslant 2x]\subseteq[|X_n-X|\geqslant x]\cup[|X_n-Y|\geqslant x],$$ hence, for every $n$, 
$$
P(|X-Y|\geqslant2x)\leqslant P(|X_n-X|\geqslant x)+P(|X_n-Y|\geqslant x).
$$ 
Considering the limit of the RHS when $n\to+\infty$, this proves that $P(|X-Y|\geqslant2x)=0$. This holds for every positive $x$ and $$[X\ne Y]=\bigcup_{k\geqslant1}[|X-Y|\geqslant k^{-1}],$$ hence $P(X\ne Y)=0$. This means that $X=Y$ almost surely.
Note: The hypothesis that $X_n\to X$ in probability and $X_n\to Y$ in probability, which we used above, is weaker than the hypothesis that $X_n\to X$ in probability and $X_n\to Y$ almost surely.
A: For the sake of having an answer:
We know the following fact:

If $X_n \to Y$ in probability then there is a subsequence $X_{n_k} \to Y$ almost surely.

So take such a subsequence. As $X_{n} \to X$ a.s. we also have $X_{n_k} \to X$ a.s. and thus $X = Y$ a.s. because the almost sure limit of a sequence is unique a.e. (if it exists).
This is just fleshing out your last comment a bit more formally.
A: See, for example, p. 150 in the book An Intermediate Course in Probability by Allan Gut‏ (in particular, the proof of Theorem 2.1(ii) on p. 151).
