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

Question

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.

Answer 1

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.