Let $(B_{t},\mathcal{F}_{t})$ be a standard 1-d Brownian motion on some $(\Omega, \mathcal{F}, \mathbb{P}) \ $, and let's assume $\mathcal{F}_{t}$ is the augmentation of $\mathcal{F}_{t}^{W}$.
Let $T>0$ and define $Z_{T}=\exp\{W_{T}-\frac{1}{2}T\} \ $, then $\frac{dP^{T}}{dP}=Z_{T}$ defines a probability measure $P^{T}$ on $\mathcal{F}_{T}$, which is equivalent (i.e. mutually absolutely continuous) w.r.t. $P$ on $F_{T}$.
If we assume $A \in \mathcal{F}_{T}, \ B \in \mathcal{F}_{T}$ are two independent events under $P$, then do we have that $A$ and $B$ are independent under $P^{T}$?
Intuitively I think the answer should be no, but I can't think of a counterexample. Could anyone help on this? Thanks a lot!