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I was studying the various types of convergence and have studied almost sure convergence.I understand that almost sure convergence means $ Pr (\lim_{n\to \infty} X_n = X) = 1 $.

I came across a statement "If $X_n \rightarrow X $ almost surely and $X_n \rightarrow Y $ almost surely then X=Y almost surely". I was trying to prove the same but was unable to do so. Any suggestions?

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Suppose that $X_n\to X$ almost surely as $n\to\infty$ and $X_n\to Y$ almost surely as $n\to\infty$. Then there exists $\Omega'\subset\Omega$ such that $\Pr(\Omega')=1$ and for each $\omega\in\Omega'$ $$ |X_n(\omega)-X(\omega)|\to0 $$ as $n\to\infty$. Similarly, there exists $\Omega''\subset\Omega$ such that $\Pr(\Omega'')=1$ and for each $\omega\in\Omega''$ $$ |X_n(\omega)-Y(\omega)|\to0 $$ as $n\to\infty$. Hence, we have that $$ |X(\omega)-Y(\omega)| \le|X(\omega)-X_n(\omega)|+|X_n(\omega)-Y(\omega)|\to0 $$ as $n\to\infty$ for each $\omega\in\Omega'\cap\Omega''$ and $X(\omega)=Y(\omega)$ for each $\omega\in\Omega'\cap\Omega''$. But $\Pr(\Omega'\cap\Omega'')=1-\Pr(\Omega'^c\cup\Omega''^c)=1$ since the union of two null events must be also null. So we obtain that $X=Y$ almost surely.

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Let $\Omega_X$ be the set of full measure on which $X_n \rightarrow X$. Define $\Omega_Y$ likewise. Note that $\Omega_0:=\Omega_X \cap \Omega_Y$ has full measure as well. On $\Omega_0$, $X_n \rightarrow X$ AND $X_n \rightarrow Y$. Hence $X=Y$ a.s due to uniqueness of limit.

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