# Definition of measurability of a stochastic process

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?

• And your thoughts on this are?
– Did
Aug 15, 2017 at 14:19
• 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. Aug 15, 2017 at 14:37
• Take union over integer $t$. Aug 15, 2017 at 15:31
• I'm not being attentive. These properties are equivalent. Aug 15, 2017 at 16:13
• @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 $t$? Aug 15, 2017 at 18:36

The definition of a measurable process means that for any set in $$\mathcal{E}$$, the pre-image under $$f_\infty$$ is in $$\mathcal{F} \otimes \mathcal{B}_{[0,\infty)}$$.
The definition of a progressively measurable process means that for any set in $$\mathcal{E}$$, and any $$t \ge 0$$, the pre-image under $$f_t$$ is in $$\mathcal{F}_t \otimes \mathcal{B}_{[0,t)}$$.
It is clear that $$\mathcal{F}_t \otimes \mathcal{B}_{[0,t)}$$ is a subset of $$\mathcal{F} \otimes \mathcal{B}_{[0,\infty)}$$, so any pre-image in the former is also in the latter.