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Let $X:=(X_t)_{t\geq 0}$ be a process on a filtered probability space $(\Omega, \mathcal{F}, \Bbb{F}, \Bbb{P})$ where $\Bbb{F} :=(\mathcal{F}_t)_{t\geq 0}$ is a filtration. Now what does it mean for $X$ to be measurable?

I know that measurability tells me that the preimage of sets in the sigma algebra are again in the sigma algebra. But the problem is that I don't see from where to where $X$ goes and which sigma algebras the domain and codomain has.

Can someone help me further?

In a remark on the internet I have seen that $X$ should be $\mathcal{F}\otimes \mathcal{B}([0,\infty))$ measurable. But I don't see why.

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  • $\begingroup$ As far as I remember the $\mathcal{F}\otimes \mathcal{B}([0,\infty))$-measurability of $X$ (that maps $\Omega\times [0,\infty)$ to $\mathbb R$ (typically) or $\mathbb R^d$) is only a mild and technical condition. Much more important than that is the ${\cal F}_t$-measurability of every $X_t$ because that reflects the gain in information. $\endgroup$ Commented Mar 8, 2023 at 14:11

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All rely about what is $X$. You have that $$ X:\Omega \times [0,\infty )\to \mathbb{R} $$ where $\Omega $ is a probability space and $[0,\infty )$ have the standard topology. Then $X$ is measurable if and only if $X^{-1}([a,b])$ belongs to the product $\sigma $-algebra for every $a,b\in \mathbb{Q}$ (where the $\sigma $-algebra of $[0,\infty )$ is assumed to be the Borel $\sigma $-algebra).

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