How to express random inequality events in terms of measurable events?

Question

Suppose that $(\Omega,\mathcal{F},P)$ is a probability space, and $X,Y:(\Omega,\mathcal{F}) \rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$ are real random variables. I want to justify why an event like $\{X \leq Y\}$ is $\mathcal{F}$-measurable. My thought is to justify something like $$\{X \leq Y\}=^?\bigcup_{q \in \mathbb{Q}} \{X \leq q\} \cap \{q \leq Y\},$$ where $\{X \leq q\} \cap \{q \leq Y\}$ is clearly $\mathcal{F}$-measurable. I highly suspect I want to use the rationals to approximate the reals in the way above, but I'm not sure how to justify this. Any help would be appreciated!

Answer 1

What you wrote is not quite correct; any outcome where $X=Y=\pi$ is not included in the RHS. Instead, you can prove the complement of $\{X\le Y\}$ is measurable via $$\{X>Y\}=\bigcup_{q\in \mathbb Q} \{X>q\}\cap \{q>Y\}.$$ Proof: Clearly the existence of a $q$ so $X>q$ and $q>Y$ implies $X>Y$. Conversely, if $X>Y$, then the density of $\mathbb Q$ in $\mathbb R$ implies there is a rational $q$ in the interval $(Y,X)$.