Will this integral be progressively measurable?

Question

Assume you have a function: $F(t,x,\omega)$: $[0,T]\times E\times \Omega \rightarrow \mathbb{R}$, which is predictable (predictable is explained below). Each of the three spaces can be viewed as 3 measure spaces of their own. $([0,T],\mathcal{B}([0,T]),\mathcal{L})$ , $(E,\mathcal{B}(E),\nu)$, we assume that $\nu$ is $\sigma$-finite, and $(\Omega,\mathcal{F},P)$. We also assume that the probability space has a filtration $\{\mathcal{F}_t\}$. We also assume that $E \subset \mathbb{R}$

That $F$ is predictable means that $F$ as a function of 3 variables is measurable with respect to the $\sigma$-algebra $\mathcal{G}$, which means that if $A\subset \mathcal{G}$, then $A\subset \mathcal{P}([0,T])\times \mathcal{P}(E)\times\mathcal{P}(\Omega)$. And $\mathcal{G}$ is generated by the functions $h(t,x,\omega)\rightarrow \mathbb{R}$ such that:

1. If you keep $t$ fixed, then as a function of $(x,\omega)$ h is $\mathcal{B}(E)\times \mathcal{F}_t$-measurable.
2. If you keep $x$ and $\omega$ fixed, $h$ is left-continuous as a function of $t$.

Now assume that $F$ is predictable and that $P(\int_0^T\int_E F(t,x,\omega)^2\nu(dx)dt<\infty)=1$.

What can we then say about the measurability as a function of $(t,\omega)$ of $K(t,\omega)=\int_E F(t,x,\omega)^2\nu(dx)$?

What I need for an argument I am reading in a book to work(it is a stopping-time argument), is that this $K$ has to be progressively measurable. That is, if we for any $t$, restrict $K(s,\omega)$ from $[0,T]\times \Omega$ to $[0,t]\times\Omega$, then it is $\mathcal{B}([0,t])\times \mathcal{F}_t$-measurable.

I assume that it is very likely that this is the case, but I don't know how to prove it. Do you know if it is the case? How can I prove it?

PS: My goal is to show that $\int_0^t\int_EF(s,x,\omega)^2\nu(dx)ds$ is progressively measurable, is it a way to see directly that this is progressively measurable as a function of $(t,\omega)$?

Answer 1

Let $\mathcal P$ denote the predictable $\sigma$-algebra on $[0,\infty)\times\Omega$. What you have described is a function $F$ for which $(x,(t,\omega))\mapsto F(t,x,\omega)$ is $\mathcal B(E)\otimes\mathcal P$-measurable. Now just apply Fubini/Tonelli theorem to deduce that $(t,\omega)\mapsto\int_E F(t,x,\omega)^2\,\nu(dx)$ is $\mathcal P$-measurable, i.e. predictable.