 Finite-horizon drift replacement by a change of measure

For $dX=\mu dt+\sigma dB$ on $[0,T]$, take $\theta =(\nu -\mu )/\sigma$ and use density $\mathcal{E}(\int \theta dB{)}_T$. If $\mu ,\nu$ are bounded and $\sigma$ is continuous and bounded below by a positive constant, the exponential has bounded bracket and defines an equivalent measure. Under it, $\tilde{B}=B-\int \theta dt$ is Brownian and $dX=\nu dt+\sigma d\tilde{B}$. Set $\theta =0$ after $T$ to define the density on the entire original sigma-algebra and the new Brownian motion on the whole time axis.