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 $\mathbb{P}(X\_t\in B)>0$.




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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](https://mathoverflow.net/questions/267477/probability-that-sde-visits-any-point), but the answer suggests to apply the Girsanov Theorem.



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


