# Every discrete time stochastic process is (jointly) measurable.

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?

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, $$\{(n,\omega); X_n(\omega) \in A\} \in \mathcal{T} \otimes \mathcal{F}.$$

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