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.