Show that a certain set is measurable

These measurability results have to be checked carefully, as intuition is a pretty poor guide. Ultimately, the set we are interested in must be described in terms of countable unions and intersections of sets known to lie in ${\cal F}^X_{t_0}$.

Let's start by noting that $f$ is continuous on $[0,t_0)$ if and only if $f$ is continuous on $[0,t_0-\varepsilon]$ for all small $\varepsilon>0$. More formally, we can write $$\{\omega: t\mapsto X_t(\omega)\mbox{ continuous on }[0,t_0)\} =\bigcap_{n=\lceil 1/t_0\rceil}^\infty \{\omega: t\mapsto X_t(\omega)\mbox{ continuous on }[0,t_0-1/n]\}.$$

Hence it suffices to replace $[0,t_0)$ with a closed interval $[0,t^*]$ with $t^*<t_0$. The advantage is that $f$ is continuous on $[0,t^*]$ if and only if $f$ is uniformly continuous. Also, a right continuous function $f$ is uniformly continuous on $[0,t^*]$ if and only if $f$ is uniformly continuous on $\mathbb{Q}\cap[0,t^*]$. That's because a uniformly continuous $f$ on $\mathbb{Q}\cap[0,t^*]$ has a unique continuous extension to $[0,t^*]$, and the right continuity guarantees that we recover the original function.

This means we only need to check countably many values of the function. More formally, we can write the set $$\{\omega: t\mapsto X_t(\omega)\text{ is continuous on }[0,t^*]\}$$ as $$\bigcap_{k\geq 1}\bigcup_{\ell\geq 1}\bigcap_{|s-r|<1/\ell} \{\omega: |X_r(\omega)-X_s(\omega)|<1/k\},$$ where the inner intersection is over the countable set $s,r\in\mathbb{Q}\cap[0,t^*]$. This gives the desired result, since $\{\omega: |X_r(\omega)-X_s(\omega)|<1/k\}\in{\cal F}^X_{t_0}$ for all choices of $r,s,k$.