Is integral of adapted separable process adapted?

Question

Assume $f(t,\omega)$ is (i)separable, (ii) measurable as function from $((0,T)×\Omega)$ into $R$ and (iii) is adapted to the filtration $F_t, 0<t<T$ Also $\int_0^Tf^2(s)ds<\infty$ almost sure.
Is integral $\int_0^tf(s,\omega)ds$ measurable with respect to $F_t$?
Definition: A stochastic process $(X(t), t\in I)$ is called separable if there exists a countable sequence $(t_j)$ that is a dense subset of $I$ and a subset N of $\Omega$ with $P(N)=0$ such that, if $\omega \notin N$, then

$$\{X(t, w) \in F \text{ for all } t\in J\} = \{X(t_j, w) \in F \text{ for all } t_j\in J\}$$ for any open subset J of I and for any closed subset F of R
In the book they wrote:
Since the integrand is a separable process that is $F_t$ measurable, the integral is also $F_t$ measurable.
It was a part of the theorem.

Answer 1

Partial answer

The question may be whether the process $f$ is progressively measurable under the assumptions. If yes, we can conclude as follows.

Fix $t \ge 0$. For each $\omega$, $f(\cdot,\omega)$ is square-integrable hence integrable on $[0,t]$

For every $t$, the map $(s,\omega) \mapsto f(s,\omega)$ is measurable on $[0,t] \times \Omega$.

So $\omega \mapsto \int_0^t f(s,\omega)$ is measurable. This result is contained in Fubini theorem for integrable functions in both variables. Here, one may apply this result first with $f_n = f 1_{[|f| \le n}$ and apply Lebesgue dominated convergence theorem to get the pointwise convergence of the integrals of $f_n$ to the integral of $f$ and deduce that $\int_0^t f(s,\omega)$ depends measurably on $\omega$.

Therefore, the process $\int_0^\cdot f(s,\cdot)$ is adapted. And since it is continuous, it is progressively measurable.

Answer 2

In the book they likely mean up to modification eg. in the Karatzas-Shreve book (pg 133) and the M.M.Rao book (3.3 section 8. Theorem.), they talk about progressively measurable modifications.

In terms of counterexamples minus separabilility-assumption, this was answered here Is the integral of an adapted, measurable process adapted?. So the question is whether that example from G.Letta is also separable. I use the content from the answer in More natural example of measurable but not progressive process

First, as mentioned here there is also an analytic notion of separability:

$f\colon A\to B$ is separable, if there exists a countable dense subset $S\subset A$ with the property: for any closed $F\subset B$ and any open $I\subset A$, if $f(t)\in F$ for all $t\in I\cap S$ , then $f(t)\in F$ for all $t\in I$.

By $\mathbb{Q}$-separable, we mean separable with respect to $\mathbb{Q}$.

Take $\Omega$ to be the space of $\mathbb{Q}$-separable lower semi-continuous functions $\omega:[0,\infty)\to\Bbb R$, such that $\int_0^t|\omega(s)|^{2}\,ds<\infty$ for each $t>0$. Let $H_s(\omega):=\omega(s)$ be the coordinate process on $\Omega$, and $(\mathcal F_t)_{t\ge0}$ the filtration generated by $(H_t)_{t\ge 0}$. So we get $\mathcal F_{t}$-adapted.

Separable process

This follows by definition because for $H$ to be separable, it means we take a generic $\omega$ and then indeed $H(\cdot,\omega)=\omega(\cdot)$ is separable wrt to $\mathbb{Q}$.

Measurable Let $\mathcal J$ denote the collection of compact intervals $J\subset[0,\infty)$ and for $J\in\mathcal J$ define $H_J(\omega):=\inf\{H_s(\omega): s\in J\}$. Finally, define $\mathcal G:=\sigma\{H_J: J\in\mathcal J\}$. Then for $c\in\Bbb R$, because $s\mapsto H_s(\omega)$ is lower semi-continuous, $$ \{(s,\omega)\in[0,\infty)\times\Omega: H_s(\omega)>c\}=\cup_{J\in\mathcal J_0}\left(J\times\{H_J>c\}\right), $$ where $\mathcal J_0$ is the subcollection of $\mathcal J$ comprising those intervals with rational endpoints. It follows that $H$ is $\mathcal B[0,\infty)\otimes\mathcal G$ measurable, and clearly $\mathcal F_t\subset\mathcal G$ for each $t>0$.

Not adapted integral

Consider now, for $t>0$ fixed, the integral $X_t:=\int_0^t H_s\,ds$. Supposed that $X_t$ was $\mathcal F_t$-measurable. In this case there would be a countable set $S\subset(0,t]$ with $X_t$ measurable over $\sigma\{H_s:s\in S\}$. In fact, we can take this set to be a subset of the rationals as constructed here.

If $\omega$ and $\omega'$ were distinct elements of $\Omega$ with $\omega(s)=\omega'(s)$ for all $s\in S$, then we would clearly have $X_t(\omega)=X_t(\omega')$. The following choice of $\omega$ and $\omega'$ leads to a contradiction. Take $\omega(s)=1$ for all $s>0$ and $\omega'(s):=1_{U}$ where $$U:=\bigcup_{n=1}^\infty(q_n-2^{-n}\varepsilon,q_n+2^{-n}\varepsilon)$$

for rational $0<\epsilon<1$ (for another choice of $U$ see here). These have different integrals for all large enough $t>0$.

Tricky part: I believe the second function is $\mathbb{Q}$-separable. Lets start with $F=\{1\}$. We cannot have both $\omega'(t)=1$ on $I\cap \mathbb{Q}$ and $\omega'(s)=0$ for some irrational $s\in I$. Due to the density of the rationals, that would mean $s=q_n+2^{-n}\varepsilon$, but here $\varepsilon$ was picked to be rational, so this cannot happen. If now $F=\{0\}$, then we again get a contradiction to having $\omega'(s)=1$ for some irrational $s\in I$ by the density of rationals and the openeness of the intervals.

Any feedback is welcome.