Setup of the question:
Assume you have a Brownian motion $B_t$ on a probability space $(\Omega,\mathcal{F},P)$. Let $\mathcal{F}_t$ be the natural filtration of the Brownian motion, and let $\bar{\mathcal{F}}_t$ be the natural filtration where we have also added the events with probability zero. Fix $T>0$. Let $\phi$ be a progressively measurable process such that $P(\int_0^T\phi_t^2dt<\infty)=1.$
Define $Z_t:$
$$Z_t\doteq\exp\left(\int_o^t\phi_sdB_s-\frac{1}{2}\int_0^2\phi_s^2ds\right).$$
Assume that $z_t$ is a martingale on $[0,T]$ with respect to $\bar{\mathcal{F}}_t$ .
Define the equivalent probability $Q$ on $(\Omega,\mathcal{F})$ such that $Z_T$ is the Radon-Nikodym derivative of $Q$ with respect to $P$.
Then Girsanovs theorem tells us that $\tilde{B}_t\doteq B_t-\int_0^t\phi_sds$ is a $\bar{\mathcal{F}}_t$ Brownian motion with respect to $Q$ on $[0,T]$.
Question
By Girsanovs theorem we obviously have that $\tilde{B}_t$ is $\bar{\mathcal{F}}_t$ measurable. But I am wondering if $\tilde{B}_t$ generates $\bar{\mathcal{F}}_t, t \le T$? Define $\bar{\mathcal{G}}_t$ as the filtration generated by $\tilde{B}_t$ which also contain the $Q$ zero events. Since $P$ and $Q$ are equivalent it also contains the $P$ zero events. Is $\bar{\mathcal{G}}_t$ strictly smaller than $\bar{\mathcal{F}}_t$, or are they equal?
Attempt
If I can show that $B_t$ is either a limit of $\tilde{B}_t$ or a Borel-function of $\tilde{B}_t$ the result will follow. The problem is $\phi_s$. I know that $\tilde{B}_t=B_t-\int_0^t\phi_sds$ is $\bar{\mathcal{G}}_t$-measurable. But do I know if $\int_0^t\phi_sds$ is $\bar{\mathcal{G}}_t$-measurable?, if yes, we would be done. Any idea on how to finish it?