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's 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 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.