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?