I have come across a statement I am not sure how to prove.
Assume you have a filtered probability space $(\Omega, \mathcal{F},\{\mathcal{F}_t\},P)$. A stochastic process $X_t$ is progressively measurable if for every $t\ge 0$, the function $f(\omega,s) $from $\Omega\times[0,t]$ to $\mathbb{R}$ given by $f(\omega,s)=X_s(\omega)$, is $\mathcal{F}_t\otimes\mathcal{B}([0,t])$ measurable.
Prog is the smallest sigma-algebra on $\Omega\times[0,\infty)$ such that all progressively measurable processes are Prog-measurable.
Obviously we have that every progressively measurable process is Prog-measurable. However, assume you have a function $g: \Omega\times[0,\infty)\rightarrow \mathbb{R}$, and assume that it is Prog-measurable, then it is stated that the stochastic process $Y_t$ defined by $Y_t(\omega)=g(\omega,t)$ is progressively measurable.
Do you know how to show this?
Attempt:
My only idea is using something like the Doob-Dynkin lemma. The Doob-Dynkin lemma states that if a function $f$ is $\sigma(h)$ measurable then there exists a borel function k, such that $f=k(h).$ The problem I have is that Prog is not generated by a single progressively measurable process it is generated by an infinite amount, so I don't think I can use the Doob-Dynkin lemma directly? But if I can prove that if $g$ is Prog-measurable, then it is some kind of limit or some kind of Borel function of the progressively measurable processes, then it will also be progressively measurable. Is there a way to do this?
Best Answer
This is a consequence of the functional form of the monotone class theorem as found, for example, at https://almostsuremath.com/2019/10/27/the-functional-monotone-class-theorem/ (see Theorem 1 there).
The key observation is that Prog is generated by an algebra of bounded process that contains the constant processes and is stable under bounded pointwise convergence.