Let $(\Omega, \mathcal{F})$ be a measurable space. Let $P,Q$ be two probability measures on $(\Omega, \mathcal{F})$ s.t. $Q<<P$, i.e. $Q$ is absolutely continuous w.r.t. $P$. Furthermore, let $X,Y$ be two (real-valued) random variables. If $X$ and $Y$ are independent w.r.t. $P$, are they also independent w.r.t $Q$?
Going through the definition of independence via independence of the $\sigma$-algebras $\sigma(X)$ and $\sigma(Y)$, I am getting stuck at the point where I pass from $P$ to $Q$ via the Radon-Nykodym derivative, since it isn't independent of $X$ and $Y$. My suspicion is, that the claim is false but I cannot come up with an immediate counterexample.
Let $\Omega = \{(1, 1), (1, 0), (0, 1), (0, 0)\}$, and let $X(\omega) = \omega_1$ and $Y(\omega)=\omega_2$. If $P$ assigns probability $1/4$ to each atom, then $X$ and $Y$ are independent. Any other probability measure $Q$ on these four atoms will satisfy $Q \ll P$.