# If $(f_n)$ is uniformly integrable wrt $P$ and $Q \ll P$, then is $(f_n)$ uniformly integrable wrt $Q$?

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.

• It is clearly true if $g\in L^{\infty}(\Omega,P)$, and I suspect that condition is also necessary. – Prahlad Vaidyanathan Dec 11 '16 at 18:09
• @PrahladVaidyanathan Yes, thank you for pointing that out. Could you point me to an example where $g$ is not essentially bounded? That would help me get a better sense of what adding this condition amounts to. – grndl Dec 11 '16 at 18:20
• Any unbounded function would work. Say, $\Omega = [0,1], P =$ the Lebesgue measure, $g(x) = 1/x$ and $$Q(E) := \int_E g(x)dP(x)$$Here, $f_n := \chi_{[0,1/n]}$ works as a counterexample to what you are looking for. – Prahlad Vaidyanathan Dec 11 '16 at 19:03
• @PrahladVaidyanathan Thank you, that's helpful. – grndl Dec 11 '16 at 22:14

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.