How to resolve the issue of two sequences converging to zero for $n, m \to \infty$?

Question

My question is motivated by the following exercise in probability theory:

Let $X_n \to X$ in probability and $X_n \geq Y$ a.s. Show that $X \geq Y$ a.s.

I noticed that for all $n, m \in \mathbb N$: $$P(X < Y) \leq P\left(|X_n - X| > \frac{1}{m}\right) + P\left(X_n < Y+\frac{1}{m}\right)$$

so for the two terms on the RHS holds: $$\forall m \in \mathbb N: \ P\left(|X_n - X| > \frac{1}{m}\right) \to 0 \text{ as } n\to \infty$$ $$\forall n \in \mathbb N: \ P\left(X_n<Y+\frac{1}{m}\right)\to 0 \text{ as } m\to \infty$$

Is there a way to finish the proof from this point?

Note that I'm not asking for a proof of the statement (I've seen another one where the issue doesn't come up).

Answer 1

Just make a small change: Instead of $\mathbb{P}(X<Y)$ consider $\mathbb{P}(X+2/m \leq Y)$. Then, by a very similar argumentation, we find

$$\begin{align*} \mathbb{P}\left(X+2/m \leq Y \right) &= \mathbb{P}\left(X+2/m \leq Y, |X_n-X| \leq \frac{1}{m} \right) + \mathbb{P}\left( X+2/m \leq Y, |X_n-X| > \frac{1}{m} \right) \\ &\leq \underbrace{\mathbb{P}(X_n < Y) }_{0}+ \mathbb{P}(|X_n-X| > \frac{1}{m}). \end{align*}$$

Now let $n \to \infty$ and then $m \to \infty$ to conclude that

$$\mathbb{P}(X<Y) = \lim_{m \to \infty} \mathbb{P}(X+2/m \leq Y) = 0.$$