I want to prove that if $X_n \to X$ in probability and $X_n \to Y$ in probability then $X = Y$ almost surely.
By definition, for any $\epsilon > 0$ we have:
$$\lim\limits_{n \to +\infty} \mathbb{P}(|X_n - X| \geq \epsilon) = 0$$ $$\lim\limits_{n \to +\infty} \mathbb{P}(|X_n - Y| \geq \epsilon) = 0$$
I should now prove that $$\mathbb{P}(X \neq Y) = 0$$
Maybe I should show that for any $\epsilon > 0$
$$\mathbb{P}(|X - Y| > \epsilon) = 0$$
which is equivalent to what I want to prove, but isn't helpful, because $|X - Y|$ is not greater $|X_n - X| + |X_n - Y|$ anyway (not vise-versa).
Any ideas, how to proceed?
For each $\epsilon>0$, \begin{align} \mathsf{P}(|X - Y| > \epsilon) &=\mathsf{P}(|X -X_n+X_n- Y| > \epsilon) \\ &\le \mathsf{P}(|X_n-X|+|X_n- Y| > \epsilon) \\ &\le \mathsf{P}(|X_n-X| > \epsilon/2)+\mathsf{P}(|X_n- Y| > \epsilon/2)\to 0 \end{align} as $n\to\infty$, which implies that $\mathsf{P}(|X - Y| > \epsilon)=0$. Thus, $$ \mathsf{P}(X\ne Y)=\mathsf{P}\!\left(\bigcup_{k\ge 1}|X-Y|>k^{-1}\right)=\lim_{k\to\infty}\mathsf{P}(|X-Y|>k^{-1})=0. $$