I have a question regarding finding the Radon Nikodym derivative of a measure w.r.t. another measure constructed from the first one. The situation is the following:
Take a measure space $(\Omega, \mathcal{A}, P)$, with $P$ being a probability measure. Now construct a second measure from $P$ via $P \mapsto Q := f(P)$, with a $\mathcal{A}$-measurable function $f$ being chosen in a way such that the resulting $Q$ again is a probability measure over $(\Omega, \mathcal{A})$. Furthermore, $f$ is required to conserve the kernel of $P$, i.e. $\{A \subset \mathcal{A}: P(A) = 0\} = \{A \subset \mathcal{A}: Q(A) = 0\}$.
(Take for instance $Q(A) = \sqrt{P(A)} \; \forall A \in \mathcal{A}$, for the monotonicity with which the square root is increasing amounts for the $\sigma$-additivity, and $Q(\Omega) = 1$ as well as $Q(\emptyset) = 0$). Hence these two measures are $\sigma$-finite by definition, and clearly equivalent.
My question is now, is there a way to determine the Radon Nikodym derivative $g = \frac{dQ}{dP}$, or $\tilde{g} = \frac{dP}{dQ}$ ? I don't know exactly but I guess that $g = f'$ is too easy and would not work ?
A first idea was that one could write $g$ as a decomposition of simple functions but it is not quite clear how to explicitly expand the simple functions into characteristic functions since I do not know which sets from $\mathcal{A}$ to choose.
Thanks a lot in advance!