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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, 2017 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, 2017 at 14:37
  • $\begingroup$ Take union over integer $t$. $\endgroup$
    – zhoraster
    Aug 15, 2017 at 15:31
  • $\begingroup$ I'm not being attentive. These properties are equivalent. $\endgroup$
    – zhoraster
    Aug 15, 2017 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, 2017 at 18:36

1 Answer 1

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The first definition is a measurable stochastic process. The second definition is a progressively measurable stochastic process. All progressively measurable processes are measurable. The converse, that is suggested in the comments is not true. All measurable processes are not progressively measurable.

I'm getting this from the Karatzas & Shreve text which defines a measurable stochastic process in 1.6 and a progressively measurable stochastic process in 1.11. Karatzas & Shreve references Chung & Doob for this naming convention.

I will prove that progressively measurability implies measurability.

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.

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