These definitions come from Karatzas and Shreve, Brownian Motion and Stochastic Calculus. We may take for granted that $U_F(\alpha,\beta; X(\omega))$, the number of upcrossings over $[\alpha,\beta]$ of the stochastic process $\{X_t \in \mathbb{R}, t \geq 0\}$ on the finite set $F$, is measurable as a function of $\omega$.
Then if $I\subseteq [0,\infty)$ is not necessarily finite, we define $$ U_I(\alpha,\beta; X(\omega)) := \sup\{U_F(\alpha,\beta; X(\omega)); F \subseteq I, F \text{ is finite} \}. $$ The authors make the comment that
"Being the limit of random variables of the form $U_F(\alpha,\beta; X(\omega))$ with finite $F$, $U_{[\sigma,\tau]}(\alpha,\beta; X(\omega))$ is measurable."
This statement is apparently false as the sequence of finite $F_n$'s one chooses for the limit must depend on $\omega$. However, (deary me I hope!) the conclusion is still true. How can we show that $U_I(\alpha,\beta; X(\omega))$ is measurable?
If $X_t$ is RCLL or if one makes a similar continuity assumption then I think the result follows by restricting to $\mathbb{Q}$, but I do not want to make any extra assumptions on $X$ other than that it is a real valued stochastic process, i.e. $X_t$ is $\mathscr{F}/\mathscr{B}(\mathbb{R})$-measurable for all $t\geq 0$.
Edit: I just noticed that in the theorem from which the above quote was taken the author's did assume $X$ has right-continuous paths. However, I would still like to know whether or not the upcrossings are measurable without this assumption.