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Definition: A stochastic process $(X_t)_{t \in [0,\infty)}$ on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t \in [0,\infty)},\mathbb{P})$ with values in a measurable space $(E,\mathcal{E})$ is called measurable if the map $$ f_{\infty} \colon \Omega \times [0,\infty) \rightarrow E \colon (\omega,s) \mapsto X_s(\omega) $$ is $\mathcal{F} \otimes \mathcal{B}_{[0,\infty)}$-measurable.

Sometimes it is also required that the map $$ f_t \colon \Omega \times [0,t) \rightarrow E \colon (\omega,s) \mapsto X_s(\omega) $$ is $\mathcal{F} \otimes \mathcal{B}_{[0,t)}$-measurable for every $t \in [0,\infty)$.

Now my question is, are those two definitions equivalent? If not, is there one which implies the other?

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  • $\begingroup$ And your thoughts on this are? $\endgroup$ – Did Aug 15 '17 at 14:19
  • $\begingroup$ Assume that $f_{\infty}$ is measurable. Then for $B \in \mathcal{E}$ and all $t \in [0,\infty)$ $$ f_{t}^{-1}(B) = \underbrace{f_{\infty}^{-1}(B)}_{\in \mathcal{F} \otimes \mathcal{B}_{[0,\infty)}} \cap \underbrace{(\Omega \times [0,t))}_{\in \mathcal{F} \otimes \mathcal{B}_{[0,t)} } \in \mathcal{F} \otimes \mathcal{B}_{[0,t)}. $$ Hence the first definition implies the second. $\endgroup$ – user146309 Aug 15 '17 at 14:37
  • $\begingroup$ Take union over integer $t$. $\endgroup$ – zhoraster Aug 15 '17 at 15:31
  • $\begingroup$ I'm not being attentive. These properties are equivalent. $\endgroup$ – zhoraster Aug 15 '17 at 16:13
  • $\begingroup$ @zhoraster Is my proof of the first implication in my comment above correct? For the second implication, why can we take the union over all integer $t$? $\endgroup$ – user146309 Aug 15 '17 at 18:36

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