# Show an SDE's solution has positive probability to visit every set in the state space

Let $$(\Omega, \mathcal{F},\mathbb{P})$$ be a filtered probability space, let $$b:[0,T]\times \mathbb{R}^n\to \mathbb{R}^n$$ be a continuous function and Lipschitz continuous in the space variable. For each $$x\in \mathbb{R}^n$$, consider the following SDE: $$X_t=x+\int_0^t b(s,X_s)ds+W_t, \quad \forall t\in [0,T].$$ I was wondering whether it is true that, for any $$t\in (0,T]$$ and Borel set $$B\in \mathbb{R}^n$$ with positive Lebesgue measure, we have that $$\mathbb{P}(X_t\in B)>0$$.

I am not sure how to prove the statement rigorously for general Lipschitz continuous drifts. Intuitively I feel it is correct, since for any initial point $$x$$, $$W_t$$ has a positive probability to reach any point within any $$t>0$$.

If $$b$$ is affine in the space variable, then the SDE can be solved explicitly and the solution is Gaussian, which implies the statement. If $$b$$ is bounded, then I can apply the Girsanov Theorem to construct an equivalent probability measure $$Q$$, such that under $$\mathbb{Q}$$, $$X_t$$ is a Brownian motion. Then the equivalence of measures implies the desired statement for the measure $$\mathbb{P}$$. However, for general Lipschitz drift, the Novikov condition $$E^P[\exp(\frac{1}{2}\int_0^T b^2(s,X_s)\,ds)]<\infty$$ may not be satisfied, which prevents me to conclude the result by using the Girsanov Theorem.

A similar question has been asked here, but the answer suggests to apply the Girsanov Theorem.

This is true by Girsanov's theorem, under much more general conditions than boundedness of $$b$$. (Novikov's condition is sufficient but far from necessary.) For instance, if $$b$$ has linear growth in the spatial variable uniformly in time, in the sense that $$\sup_{t,x}|b(t,x)|/(1+|x|) < \infty$$, then Girsanov's theorem can be justified by an argument due to Beneš, given in Corollary 3.5.16 of the Karatzas-Shreve textbook.
More generally, suppose merely that $$b$$ is measurable and that your SDE admits a weak solution with $$\int_0^T|b(t,X_t)|dt < \infty$$ almost surely. Then the law of $$(X_t)_{t \in [0,T]}$$ on path space is equivalent to the law of $$(x+W_t)_{t \in [0,T]}$$ if and only if $$\mathbb{P}\left( \int_0^T |b(t,X_t)|^2dt < \infty \right) = \mathbb{P}\left( \int_0^T |b(t,x+W_t)|^2dt < \infty \right) = 1.$$ See Theorem 7.7 of the Liptser-Shiryaev textbook.