Here is the statement I'd like to prove: Let $u(x):\mathbb{R}^d\to\mathbb{R}^d$ such that $\sup_x|u(x)|\leq K$. Define $X_t$ by $dX_t=u(X_t)dt+\sigma B_t$ for $\sigma>0$ and $X_0=x$. For any open set $A\subset\mathbb{R}^d$, assume $P(B_t\in A)>0$ holds, show that $P(X_t\in A)>0$ holds.
I first show that $\operatorname{E}[f(x+\sigma B_t)]=\operatorname{E}[\Lambda_t f(X_t)]$ holds for some process $\Lambda_t$ and some $f:\mathbb{R}^d\to\mathbb{R}$. I take $\Lambda_t=\exp\left(\int_0^t\frac{u(X_s)}{\sigma}dB_s-\frac{1}{2}\int_0^t \frac{|u(X_s)|^2}{\sigma^2}ds\right)$ (the exponential martingale) and $f(x)=\textbf{1}_A(x)$. By the 1st Girsanov theorem, $$\begin{aligned}0&<P\left(B_t\in\frac{A-x}{\sigma}\right)\\&=P(\sigma B_t\in A-x)\\&=\operatorname{E}[\textbf{1}_A(x+\sigma B_t)]\\&=\operatorname{E}[\Lambda_t\textbf{1}_A(X_t)]\\&\leq(\operatorname{E}[\Lambda_t^2])^{1/2}\operatorname{E}[\textbf{1}_A(X_t)]^{1/2},\end{aligned}$$ Then, I have to show $\operatorname{E}[\Lambda_t^2]<\infty$ to conclude. But I have trouble reaching this conclusion. I'm trying to use the assumption on boundedness of solution. I also know that $\Lambda_t$ satisfies the SDE $d\Lambda_t=\Lambda_t\sigma_tdB_t$, where $\sigma_t$ is some bounded adaptive stochastic process.I'd like to know how I can proceed. I also find a similar question on this here: https://mathoverflow.net/questions/267663/moment-bounds-on-exponential-martingale, but it seems to overwhelmed to my case.