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Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $(X_t)_{t\ge 0}$ be a real-valued almost surely continuous stochastic process on $(\Omega,\mathcal A,\operatorname P)$.

Let $N\subseteq\Omega$ be a $\operatorname P$-null set such that $X(\omega,\;\cdot\;)$ is continuous for all $\omega\in\tilde\Omega:=\Omega\setminus N$. It's well known that a (surely) one-sided continuous stochastic process $(Y_t)_{t\ge 0}$ on $(\Omega,\mathcal A,\operatorname P)$ is product measurable, i.e. $$\Omega\times[0,\infty)\to\mathbb R\;,\;\;\;(\omega,t)\mapsto Y_t(\omega)$$ is $\mathcal A$-$\mathcal B([0,\infty))$-measurable.

Can we show that $X$ is product measurable? Clearly, we can conclude that $$\tilde\Omega\times[0,\infty)\;,\;\;\;(\omega,t)\mapsto X_t(\omega)$$ is $\left.\mathcal A\right|_{\tilde \Omega}$-$\mathcal B([0,\infty))$-measurable, but that's not the product measurability of $X$.

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  • $\begingroup$ $\left.\mathcal A\right|_{\tilde\Omega}$ denotes the trace $\sigma$-algebra of $\tilde\Omega$ in $\mathcal A$. $\endgroup$
    – 0xbadf00d
    Commented Jan 30, 2016 at 16:02

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No. It may happen that for certain $\omega\in N$, $t\mapsto X_t(\omega)$ is non-measurable. In this case $X$ cannot be jointly measurable on $\Omega\times[0,\infty)$.

On the other hand, if you alter $X$ by by setting $\tilde X_t(\omega)=0$ for all $t\ge 0$ and all $\omega\in N$ while setting $\tilde X_t(\omega)=X_t(\omega)$ for $t\ge 0$ and $\omega\in\tilde\Omega$, then $\tilde X$ is jointly measurable and $\Bbb P[\tilde X_t= X_t,\forall t]=1$..

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  • $\begingroup$ I've asked the question cause I've seen people using the product measurability of a Brownian motion. I want to note that this led me to the following question: math.stackexchange.com/questions/1633453/… $\endgroup$
    – 0xbadf00d
    Commented Jan 30, 2016 at 18:47

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