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

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.

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.