Last step step in Girsanov's theorem proof

I consider the version of Girsanov's theorem presented in this question.

Let us take the particular case that $$\mathbb{F}$$ is the filtration generated by standard Brownian motion $$(W)_{t\in[0;T]}$$ of $$(\Omega, \mathcal{F}, \mathbb{P})$$. We want to show that $$(\widetilde{W}_t)_{t\in[0;T]}$$ is a standard Brownian motion in $$(\Omega, \mathcal{F}, \widetilde{\mathbb{P}})$$, where $$\frac{d\widetilde{\mathbb{P}}}{d\mathbb{P}}=Z_T=e^{-\int_0^T\Theta_udW_u-\frac{1}{2}\int_0^T\Theta^2_udu}$$ and $$(\Theta_t)_{t\in[0;T]}$$ is some $$\mathbb{F}$$-adapted process

The proof proceeds to show this using the Lévy characterization of Brownian motion:

1. $$(\widetilde{W}_t)_{t\in[0;T]}$$ is a continuous process, starting at 0, with quadratic variation $$t$$ (this is all evident)
2. All that remains to be shown is that $$(\widetilde{W}_t)_{t\in[0;T]}$$ is a $$(\widetilde{\mathbb{P}}, \mathbb{F})$$-martingale.

First we show that $$(Z_t)_{t\in[0;T]}$$ is a $$(\mathbb{P}, \mathbb{F})$$-martingale. But then, in order to prove point (2.) above, the proof tries to show that $$(\widetilde{W}_tZ_t)_{t\in[0;T]}$$ is a $$(\mathbb{P}, \mathbb{F})$$-martingale by explaining that: $$d(\widetilde{W}_tZ_t)=\cdots=(-\widetilde{W}\Theta_t+1)Z_tdW_t.$$ has no drift.

However, for $$(\widetilde{W}_tZ_t)_{t\in[0;T]}$$ to be a $$(\mathbb{P}, \mathbb{F})$$-martingale we also must know that this process is $$\mathbb{F}$$-adapted. Can we prove this fact? This seems contrary to the comment made after Corollary 5.3.2 that states:

the filtration generated by $$\widetilde{W}$$ may be different from the filtration generated by $$W$$. Is this not a contradiction?