 Solution

Let ${\mathcal{T}}_n$ be the first exit from $[1/n,1-1/n]$. On that compact interval the drift $v/2$ is bounded. The Girsanov theorem therefore gives, through every fixed time $t$, a probability measure equivalent to the original one under which the stopped process $X_{\cdot \wedge {\mathcal{T}}_n}$ has Brownian increments before ${\mathcal{T}}_n$.

If $A\subset (0,1)$ has Lebesgue measure zero, the normal distribution of Brownian motion givesSince $\{t<\mathcal{T}\}={\bigcup }_n\{t<{\mathcal{T}}_n\}$, countable subadditivity gives $\mathbb{P}(X_t\in A,t<\mathcal{T})=0$. Thus the killed law at time $t$ is absolutely continuous with respect to Lebesgue measure on $(0,1)$.