Showing that the integral $\omega \mapsto\int_{\tau_{n-1}}^{\tau_n} f(s,\omega)ds$ is $\mathscr{F}_{\tau_n}$ measurable for stopping times $\tau_n$

Question

Let $f(t,\omega)$ be a progressively measurable process in $L^2(\lambda_T \otimes P)$ and a filtration $\mathscr{F}_t$ be given.

I know that in this case, for each $t_n \in [0,T]$, we have that $$\omega \mapsto\int_{t_{n-1}}^{t_n} f(s,\omega)ds$$ is $\mathscr{F}_{t_n}$-measurable.

I know that this follows from Fubini's theorem.

However, more generally, let $\tau_j$ be a sequence of bounded $\mathscr{F}_t$-stopping times. In this case, how can we show that $\xi_j(\omega):= \int_{\tau_{j-1}(\omega)}^{\tau_j(\omega)} f(s,\omega)ds$ is $\mathscr{F}_{\tau_j}$-measurable if $f$ is progressively measurable?

To be more specific about the context, I have come through this question from the following excerpt from René Schilling's Brownian Motion. Below, $\mu_T(\omega,ds)=dA_s(\omega)$ is the measure on $[-,T]$ induced by the increasing continuous function $s \mapsto _s(\omega),$ i.e. $\mu_T(\omega,[s,t))=A_t(\omega)-A_s(\omega)$ for all $s \le t \le T$.

Answer 1

Since $f(t,\omega)$ is a progressively measurable process, then $ Y(t)=\int_0^t f(s,\omega)\,ds$ is a progressively measurable process too and \begin{gather*} \int_0^{\tau}f(s,\omega)\,ds=Y(\tau)\in \mathscr{F}_{\tau}\\ \int_{\tau_{n-1}}^{\tau_n}f(s,\omega)\,ds=Y(\tau_n)-Y(\tau_{n-1}) \in\mathscr{F}_{\tau_n} \end{gather*}