I have a property that I would like to prove for stochastic integrals, it seems very intuitive that it should hold, but I don't know how to prove it. The property is:
Let $B_t$ be a brownian motion on a filtered probability space $(\Omega,\mathcal{F},\mathcal{F}_t,P)$. Assume that $f$ is a progressively measurable process on the probability space with $P(\int_0^T|f_t|^2dt<\infty)=1.$
Let $I(t)$ be a continuous version of the stochastic integral $\int_0^tfdB_t$ for $t\in[0,T]$. Then there exists a set $A\in \mathcal{F}$ with $P(A)=1$ so that if $\omega^*\in A$ and there exists $t_1(\omega^*),t_2(\omega^*)$ where $f(t,\omega^*)=0$ for all $t \in [t_1(\omega^*),t_2(\omega^*)]$ we have
$$[I_{t_1(\omega^*)}](\omega^*)=[I_{t_2(\omega^*)}](\omega^*).$$
The intuitive reason as to why this should hold is that if we take the stochastic integral of $0$ we get zero. But the problem here is that the stochastic integral is not defined $\omega$-wise but as a limit in probability(you can take subsequences so you get limits a.s.) But in these limits we have integrals of simple functions and you don't have that the simple functions necessarily are zero on $[t_1(\omega^*),t_2(\omega^*)]$.
Do you have any idea on how to solve this?