Several other questions have noted the fact that the standard definition of conditional probability, $$P(A\mid B) = \frac{P(A \cap B)}{P(B)},$$ is undefined when $P(B) = 0$. Wikipedia avoids dividing by $0$ using any decreasing sequence of events $B_n \rightarrow B$ with $P(B_n) > 0$, and taking $$P(A|B) = \lim_{n \rightarrow \infty} P(A|B_n).$$ However, this methods defies my intuition when $A$ is also a zero-probability event. If $P(A) = 0$, then clearly the above limit equals $0$. Yet it seems that some zero-probability events should have positive probabilities conditional on other zero-probability events.
For instance, let $X \sim \text{Unif}[0, 9]$. Clearly, $\{X \in \mathbb{Z}\}$ and $\{\sqrt{X} \in \mathbb{Z}\}$ are both zero-probability events. Yet it seems obvious that if the realization of $X$ is integral, then $\sqrt{X}$ has a nonzero chance of also being integral. Indeed, based on the fact that $0, 1, 4,$ and $9$ have integral square roots, I'd expect $$P(\sqrt{X} \in \mathbb{Z} \mid X \in \mathbb{Z}) = \frac{4}{10}.$$ Is there a way to formalize this?