Let $P$ and $Q$ be probability measures on $(\Omega, \mathcal{F})$ and suppose $Q \ll P$.
Question. If $(f_n)$ is a nonnegative uniformly integrable sequence with respect to $P$, then is it also uniformly integrable with respect to $Q$?
To establish the affirmative, I need to show that for all $\epsilon > 0$, there exists $K > 0$ such that for all $n$, $$\int_{f_n > K} f_n dQ < \epsilon.$$
Since $Q \ll P$, there exists a $P$-integrable function $g = dQ/dP$ such that $$\int_{f_n > K} f_n dQ = \int_{f_n > K} f_n g dP.$$
So I need to show that the uniform integrability of $(f_n)$ with respect to $P$ implies the uniform integrability of $(f_ng)$.
I'm stuck here. My guess is that the implication just stated does not hold, but I can't think of a counterexample.
We are trying to find all the functions $g$ such that if $\left(f_n\right)_{n\geqslant 1}$ is a uniformly integrable sequence, then so is $\left(gf_n\right)_{n\geqslant 1}$.
Assume that $g$ is not bounded. Then for each integer $n$, the set $A_n:=\left\{g\geqslant n\right\}$ has a positive probability, denoted $p_n$. Define $f:=\sum_{n=1}^{+\infty } c_n\mathbf 1_{A_n}$, where $c_n $is such that $c_n n p_n=1/n$. Then $f$ is integrable while $gf\geqslant \sum_{n=1}^{+\infty } c_nn\mathbf 1_{A_n}$ is not. Therefore, choosing $f_n=f$, we can see that $g$ do not satisfy " $\left(gf_n\right)_{n\geqslant 1}$ is uniformly integrable".
If $g$ is bounded, then $\left(gf_n\right)_{n\geqslant 1}$ is uniformly integrable.
Therefore, the answer to the initial question is yes if the density is essentially bounded, and maybe no if the density if not essentially bounded.