Radon-Nikodym derivative and conditional probability

Question

In this paper by Diaconis and Zabell from 1982, Theorem 2.1 and the remark after essentially stated that

Given two probability measures $P$ and $Q$ on the same probability space $\Omega$. If $Q\ll P$ and their RN-derivative $\frac{dQ}{dP} \in L^\infty$ , then $Q$ can be obtained from $P$ by conditioning.

Their proof is very easy and the assumption $\frac{dQ}{dP} \in L^\infty$ is essential for it. However I was wondering if it is possible to relax the assumption $\frac{dQ}{dP} \in L^\infty$ or if this is a necessary condition, is it? I was looking at concrete cases of this problem, for example between $\text{Exp}(a)$ and $\text{Exp}(b)$.

I know very little about the literature on this topic, any suggestion is also appreciated.

Answer 1

Theorem 2.1 in the quoted paper of Diaconis and Zabell actually states that boundedness of the ratios $Q(\omega)/P(\omega)$ is a necessary and sufficient condition for obtaining $Q$ from $P$ by conditioning in the discrete case. This is true in the general case as well.

Indeed, conditioning here means that there are a probability space $(\widetilde \Omega,\widetilde P)$ that covers the original space $(\Omega,P)$ and an event $E\subset \widetilde\Omega$ such that $Q$ is the image of the conditional measure $\widetilde P|E$ under the projection $\widetilde\Omega\to\Omega$. Since $\widetilde P|E \le \widetilde P/\widetilde P(E)$, the same inequality holds for the image measures as well, which means that $dQ/dP\le 1/\widetilde P(E)$.