Let $X_1,X_2,X_3,...,X_n$ be a sequence of random variables which are defined on the same sample space $\Omega$.
Show that limits in probability are unique almost surely. That is, if $X_n \rightarrow X$ almost surely and $X_n \rightarrow\bar X$ almost surely then $\Bbb P[X=\bar X]=1.$
I'm not exactly sure how to do this question as the limits are in probability. If anyone could tell me a way to prove it it would be much appreciated.
Convergence in probability defines a topology on the space of random variables over $\Omega$. In particular, this topology is metrizable by the Ky Fan metric,
$$\rho(X,Y) = \inf\left\{\varepsilon>0: \mathbb P(|X-Y|>\varepsilon)\leqslant\varepsilon\right\}. $$
By definition, $X_n$ converges to $X$ in probability iff $\lim_{n\to\infty}\rho(X_n,X)=0$. Since metrizable spaces are Hausdorff, limits are unique. Indeed, if $X_n\stackrel p\to X$ and $X_n\stackrel p\to\bar X$ then for any $\varepsilon>0$ we may choose $N$ such that $\rho(X_n,X)<\frac\varepsilon2$ and $\rho\left(X_n,\bar X\right)<\frac\varepsilon2$ for $n\geqslant N$, whence $$\rho\left(X,\bar X\right)\leqslant \rho\left(X,X_N\right)+\rho\left(X_N,\bar X\right)<\frac\varepsilon2+\frac\varepsilon2=\varepsilon. $$ It follows that $X=\bar X$.