Prove that the limit of a random variable sequence (converges in probability) is almost surely uniuque.
Let $X, Y$ random variables and let a random variable sequence $(X_n)_n$ such that $X_n$ converges in probability to $X$ and $X_n$ converges in probability to $Y$ (i.e. $X_n \overset{P}{\rightarrow} X, X_n\overset{P}{\rightarrow} Y$). Prove that $\mathbb{P}(X=Y)=1$.
Let $\varepsilon>0$. Let $\omega\in\Omega$ (if there exists such $\omega$) such that $|X^{-1}(\omega)-Y^{-1}(\omega)|>\varepsilon$. Thus, for all $n\in\mathbb{N}$, definitely $|X^{-1}(\omega)-X_n^{-1}(\omega)|>\varepsilon$ or $|X_n^{-1}(\omega)-Y^{-1}(\omega)|>\varepsilon$. Thus, $$ \\ \{|X-Y|>\varepsilon\}\subseteq(\{|X_n-X|>\ {\varepsilon\over 3}\}\cup\{|X_n-Y|>{\varepsilon\over 3}\})\ $$ Thus, $$ \\ \mathbb{P}(|X-Y|>{1\over n})\leq\mathbb{P}(|X_n-X|>{\varepsilon\over 3}\cup|X_n-Y|\leq{\varepsilon\over 3})\ \\ \leq\mathbb{P}(|X_n-X|>{\varepsilon\over 3})+\mathbb{P}(|X_n-Y|>{\varepsilon\over 3})\underset{n\to\infty}{\rightarrow} 0\ $$ Thus, $$ \\ \mathbb{P}(|X-Y|>0)=0\Rightarrow \mathbb{P}(X=Y)=1\ $$