Do a.s. right-continuous paths imply product measurability

Question

Let $( X_t )_{ t \geq 0}$ be an $\mathbb{R}^d$-valued stochastic process on the probability space $(\Omega, \mathcal{F}, P)$ which has right-continuous sample paths. The latter means that the map $t \mapsto X_t (\omega)$ is right-continuous for all $\omega \in \Omega$. Using this fact I know how to show that $(t, \omega) \mapsto X_t (\omega)$ is $\mathcal{B}[0, \infty) \otimes \mathcal{F} - \mathcal{B}(\mathbb{R}^d)$-measurable.

But what if instead of the right-continuity of the sample paths, we require a.s. right-continuity, i.e. assume that there is some $\Omega_0 \in \mathcal{F}$ with $P(\Omega_0) = 1$ such that $t \mapsto X_t ( \omega )$ is right-continuous for all $\omega \in \Omega_0$. Does it then follow that $(t, \omega) \mapsto X_t (\omega)$ is $\mathcal{B}[0, \infty) \otimes \mathcal{F} - \mathcal{B}(\mathbb{R}^d)$-measurable? Since I think of the concept of measurability to be related to the measurable space $(\Omega, \mathcal{F})$, and not necesarrily to the measure $P$ on it, it is not clear how to work with this.

Answer 1

Surely not. Let $X_t(\omega)=f(t)I_A(\omega)$ where $f$ is completely arbitrary and $P(A)=0$. Then $P(A^{c})=1$ and $X_t \equiv 0$ for $\omega \in A^{c}$. But $\{(t,\omega): X_t(\omega) >0\}=\{t: f(t) >0\} \times A$. If this set is product measurable then $\{t: f(t) >0\}$ would have to be a Borel set which need not be.