I want to show the following statement:
If $X_n \rightarrow Y_1$ in probability and $X_n \rightarrow Y_2$ in probability also. Then $Y_1 = Y_2$ almost surely.
Here is my approach:
I want to show $Y_1 = Y_2$ almost surely; so it is equivalent to show $$P(\{X_1 \neq X_2\}) =0$$
Hence, first observe the event $$\{\omega:Y_1(\omega) \neq Y_2(\omega)\} = \bigcup_{n=1}^\infty\{\omega: |Y_1(\omega) - Y_2(\omega)| > \frac{1}{n}\}$$ Now, take probability on above equivalent events and since the event $\{|Y_1 - Y_2| > \frac{1}{n}\}$ is nonincreasing, we have $$P\{\omega:Y_1(\omega) \neq Y_2(\omega)\} = \lim_{n \rightarrow \infty}P\{|Y_1 - Y_2| > \frac{1}{n}\}$$ Then I stuck to conclude further to get the Right-Hand-Side limit becomes $0$. Any thought is appreciated.
Thank you.
