Background
The Radon-Nikodym Theorem tells us that for two sigma finite measures $\mu$ and $\nu$, that $\nu$ is absolutely continuous with respect to $\mu$ if and only if there exists a function $f$ such that it is the Radon-Nikodym derivative $\frac{d \nu}{d \mu}.$ In my course on Measure Theory (and the corresponding textbook “Measure Theory” by Donald Cohn), there is not much exploration into the Radon-Nikodym derivative itself and so I am left wondering more about when it exists and the implications of this.
Question
Assuming that the two measures $\mu$ and $\nu$ are both absolutely continuous between one another and are sigma finite on the measurable space $(\Omega, \mathscr{F})$, then by the Radon-Nikodym Theorem, both $\frac{d \nu}{d \mu}$ and $\frac{d \mu}{d \nu}$ are guaranteed to exist.
However, it isn't clear to me exactly what the relationship between the two derivatives will be (if there exits one). And if there is an easy way to calculate one from the other (which can sometimes be done with derivatives in other areas of mathematics).
I would be grateful for any guidance here. I assume (if there exits a connection between the two) that this is just a standard result, however, I haven't been able to find anything on it online or in the textbooks that I have access to and so any help or references here would be appreciated.
Best Answer
First we show that $\frac{d\mu}{d\nu}\neq 0$, $\mu$-surely, \begin{align*} \int \mathbf 1\left( \frac{d\mu}{d\nu} = 0 \right) d\mu &=\int \mathbf 1\left( \frac{d\mu}{d\nu} = 0 \right)\cdot \frac{d\mu}{d\nu} d\nu=0 \end{align*}
One can show that $\frac{d \nu}{d \mu} = \frac{1}{\frac{d \mu}{d \nu}}$ $\mu$-surely (or equivalently $\nu$-surely), indeed for any measurable $A$, \begin{align*} \int_A \frac{1}{\frac{d \mu}{d \nu}} d\mu&=\int_A \frac{1}{\frac{d \mu}{d \nu}} \cdot \frac{d\mu}{d\nu} d\nu\\ &=\int_A d\nu\\ &=\nu(A) \end{align*} and the Radon-Nikodym derivative is (almost surely) unique.