It is actually exercise 8.8 in Oksendal SDE book. Let $B_t$ be a Brownian motion and $Z_t$ be a 1-dimensional Ito process of the form $$dZ_t = u(t,\omega) dt + dB_t.$$ Let $\mathcal{Z}_t$ be the $\sigma$-algebra generated by $\{Z_s : s \leq t\}$. Let $$ dN_t = \Big(u(t,\omega) - E[u(t,\omega)|\mathcal{Z}_t]\Big)dt + dB_t.$$ I want to prove that $N_t$ is a Brownian motion.
Let $\mathcal{N}_t$ be the $\sigma$-algebra generated by $\{N_s : s \leq t\}$. I think that it is enough to prove that $E\Big[u(t,\omega) - E[u(t,\omega)|\mathcal{Z}_t] \hspace{2pt} \Big| \hspace{2pt} \mathcal{N}_t\Big] = 0$. It is sufficient to prove that $\mathcal{N}_t \subset \mathcal{Z}_t$, since $$ E\Big[u(t,\omega) - E[u(t,\omega)|\mathcal{Z}_t] \hspace{2pt} \Big| \hspace{2pt} \mathcal{N}_t\Big] = E\Big[u(t,\omega) \hspace{2pt} \Big| \hspace{2pt} \mathcal{N}_t\Big] - E\Big[E[u(t,\omega)|\mathcal{Z}_t] \hspace{2pt} \Big| \hspace{2pt} \mathcal{N}_t\Big]=0.$$ We have $dN_t = dZ_t - E[u(t,\omega)|\mathcal{Z}_t]dt$ and it is equivalent to $$ N_t = N_0+ Z_t -Z_0 + g(t), $$ for some $g(t)$. But I do not know to prove the subset relation due to the presence of $N_0$ and $Z_0$.