Suppose $X_n\overset{P}{\rightarrow} X$ and $X_n\overset{P}{\rightarrow} Y$. Show that $P(X=Y)=1$. I'm thinking about trying to prove this result by contradiction. For instance, if $P(Y>X)=p>0$, then at least one of $X_n\overset{P}{\rightarrow} X$ or $X_n\overset{P}{\rightarrow} Y$ has to fail, and the same would go if $P(Y<X)=p>0$. But I do not know how to prove this.
 A: Using the triangle inequality and the fact that
$$\{|X-X_n|+|X_n-Y|>\varepsilon\}\subset\{|X-X_n|>\varepsilon/2\}\cup\{|X_n-Y|>\varepsilon/2\},
$$
we obtain
\begin{align*}
P(|X-Y|>\varepsilon)&=P(|X-X_n+X_n-Y|>\varepsilon)\\
&\le P(|X-X_n|+|X_n-Y|>\varepsilon)\\
&\le P(|X-X_n|>\varepsilon/2)+P(|X_n-Y|>\varepsilon/2).
\end{align*}
If we let $n\to\infty$, we see that
$$
P(|X-Y|>\varepsilon)=0
$$
for each $\varepsilon>0$. It means that $P(X=Y)=1$ since otherwise we would get a contradiction.
A: Let $(\Omega,\mathcal{F},P)$ denote the underlying probability triplet. Then, $X_n\overset{P}{\to}X$ implies $\exists A\in\mathcal{F}$ such that
$$
P(A)=1\quad\text{and}\quad X_{n_j}(\omega)\to X(\omega)\forall \omega\in A
$$
where $\{X_{n_j}\}$ is a subsequence of $\{X_n\}$. Note that $X_n\overset{P}{\to}Y$ implies $X_{n_j}\overset{P}{\to}Y$ so there is a further subsequence $\{X_k^*\}$ and $B\in\mathcal{F}$ such that $\forall\omega\in B$ we have $X_k^*(\omega)\to Y(\omega)$. Let $C=A\cap B$. Then,
$$
P(C)=1-P(C^c)=1-P(A^c\cup B^c)\ge 1-P(A^c)-P(B^c)=1-0-0=1.
$$
So $P(C)=1$ and
$$
\forall\omega: X_k^*(\omega)\to X(\omega), X_k^*(\omega)\to Y(\omega)\implies\omega\in C:X(\omega)=Y(\omega).
$$
A: The event {Y>a>b>X} is contained in the union of {$|X_n-X|>(b-a)/2$} and the event {$|X_n-Y|>(b-a)/2$}. Letting $n \rightarrow \infty$ we see that {Y>a>b>X} has probability 0. Varying a and b ever rationals we see that {Y>X} is a null event. Similarly, {X>Y} is a null event. Hence X=Y almost surely.
