Let $X,Y$ be random variables on a probability space $(\Omega,\mathcal{F},P)$. Show that if $P(|X-Y|>(1/n))=0$ $\forall n\in \Bbb N$,then $X=Y$ almost surely.
I've tried to solve this in the following way:
Taking the limit as $n\to\infty$ on both sides of $$P(|X-Y|>(1/n))=0,$$ you get $$\lim_{n\to\infty}P(|X-Y|>(1/n))=0.$$ Since $|X-Y|>(1/n)$ is a monotone sequence, the continuity property of the probability measure $P$ yields that $$\begin{align} \lim_{n\to\infty}P(|X-Y|>(1/n))&=P(\lim_{n\to\infty}(|X-Y|>(1/n)))\\ &=P((|X-Y|>0)). \end{align}$$ So, we have that $$P((|X-Y|>0))=\lim_{n\to\infty}P(|X-Y|>(1/n))=0,$$ which is our desired result.
Is this correct?
This is correct, there are just some minor formal issues: The monotone sequence you are referring to is the sequence of sets $\{|X-Y|>(1/n)\}\in\mathcal F$, which your notation does not quite reflect. Also, instead of $$P(\lim_{n\to\infty}(|X-Y|>(1/n))),$$ it would be more formally correct (or at least more common notation) to write $$P\left(\bigcup_{n=1}^\infty\{|X-Y|>(1/n)\}\right).$$ Other than that, your solution is fine.