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I' m a worker and I'm self studying stochastic calculus. I cannot find a proof of the statement in the title; everybody seems to take it as trivial. Consider two measurable spaces $(\Omega, \mathcal{F})$ and $(E, \mathcal{E})$, where $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$ and $\mathcal{E}$ is a $\sigma$-algebra on $E$. Now, if we take a discrete time process: $X=\{X_n \}_{n \in \mathbb{Z}^+ }$; I wish to show that the map $$ (n,\omega)\mapsto X_n(\omega) $$ is measurable with respect to the product $\sigma$-algebra $\mathcal{T}\otimes \mathcal{F}$, where $\mathcal{T}$ is a $\sigma$-algebra on $\mathbb{Z}^+$ that we take as the power set of $\mathbb{Z^+}$.

By definition of stochastic process the map $$ \omega\mapsto X_n(\omega) $$ is always $\mathcal{F}$-measurable, and I know that the map: $$ n \mapsto X_n(\omega) $$ is $\mathcal{T}$-measurable since $\mathcal{T}$ contains all the subset of $\mathbb{Z}^+$. From this I cannot succeed in prooving that for each $A \in \mathcal{E}$: $$ \{ (n,\omega) : X_n(\omega)\in A \} \in \mathcal{T}\otimes \mathcal{F}. $$ For the very same reason I cannot prove that and adapted stochastic process in discrete time is always progressively measurable.

Can you give me an hint or a reference?

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Fix $A \in \mathcal{E}$. We have

$$\{(n,\omega); X_n(\omega) \in A\} = \bigcup_{n \in \mathbb{N}} \big( \{n\} \times \{\omega \in \Omega; X_n(\omega) \in A\} \big).$$

Since $$\underbrace{ \{n\}}_{\in \mathcal{T}} \times \underbrace{\{\omega \in \Omega; X_n(\omega) \in A\}}_{\in \mathcal{F}} \in \mathcal{T} \otimes \mathcal{F}$$

for each $n \in \mathbb{N}$, it follows that

$$\{(n,\omega); X_n(\omega) \in A\}$$

is a countable union of $\mathcal{T} \otimes \mathcal{F}$-measurable sets and, hence, $$\{(n,\omega); X_n(\omega) \in A\} \in \mathcal{T} \otimes \mathcal{F}.$$

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  • $\begingroup$ This answer should start, fix $A \in \mathcal{E}$ given state space $(E, \mathcal{E})$ $\endgroup$
    – clay
    May 6, 2022 at 17:27

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