= Solution
Use the predictable process
$$
\theta_s=\frac{\nu_s-\mu_s}{\sigma_s}\mathbf1_{[0,T]}(s),\qquad L_t=\int_0^t\theta_s\,dB_s.
$$
The value at zero is irrelevant to the Brownian integral. The numerator is bounded and $\sigma_s\geq\delta>0$, so $\theta$ is bounded and $[L]_\infty=\int_0^T\theta_s^2ds$ has a deterministic bound. Part (a) supplies a <uniformly integrable> density <martingale>. Define a measure on the entire given sigma-algebra by
$$
\boxed{\mathbb Q(A)=\mathbb E_{\mathbb P}\left[\mathbf1_A\exp\left(\int_0^T\frac{\nu_s-\mu_s}{\sigma_s}\,dB_s-\frac12\int_0^T\left(\frac{\nu_s-\mu_s}{\sigma_s}\right)^2ds\right)\right],\quad A\in\mathcal F.}
$$
Its density is positive and has <expectation> one, so $\mathbb Q$ is a probability measure equivalent to $\mathbb P$. The <Girsanov theorem> makes $\widetilde B_t=B_t-\int_0^{t\wedge T}(\nu_s-\mu_s)/\sigma_s\,ds$ a <Brownian motion> under $\mathbb Q$. For $t\leq T$, substitute $dB=d\widetilde B+\theta\,dt$:
$$
\boxed{X_t=\int_0^t\nu_s\,ds+\int_0^t\sigma_s\,d\widetilde B_s.}
$$
Continuity of $\sigma$ gives pathwise local square integrability, even though it is not assumed bounded. Equivalence preserves this property. This is <finite-horizon drift replacement by a change of measure>.
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