 Solution

Use the predictable processThe value at zero is irrelevant to the Brownian integral. The numerator is bounded and ${\sigma }_s\ge \delta >0$, so $\theta$ is bounded and $[L{]}_{\infty }={\int }_0^T{\theta }_s^{2}ds$ has a deterministic bound. Part (a) supplies a uniformly integrable density martingale. Define a measure on the entire given sigma-algebra byIts density is positive and has expectation one, so $\mathbb{Q}$ is a probability measure equivalent to $\mathbb{P}$. The Girsanov theorem makes ${\tilde{B}}_t=B_t-{\int }_0^{t\wedge T}({\nu }_s-{\mu }_s)/{\sigma }_{s}ds$ a Brownian motion under $\mathbb{Q}$. For $t\le T$, substitute $dB=d\tilde{B}+\theta dt$: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.