In this [paper](https://statweb.stanford.edu/%7Ecgates/PERSI/papers/zabell82.pdf) 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.



Radon-Nikodym derivative and conditional probability
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 $dQ/dP\le 1/\widetilde P(E)$.


