Here is the problem I have. $\nu$ and $\mu$ are two positive $\sigma$-finite measures, such that $\nu \ll \mu$ and $\mu \ll \nu$. Also, the function $h$ is the Radon Nikodym derivative such that $h = \frac{d\nu}{d\mu}$. What is then $\frac{d\mu}{d\nu}$?
I get lost in the notation here a bit but I guess the first R.N. derivative implies that
$\nu(E)=\mu(1_Eh)$
The second one should similarly imply that $\mu(E) = \nu(1_Eh)$
for some function $h$ and a set $E$.
Is there anything more to it?
$\frac {d\mu} {d\nu} =\frac 1 h$. Note that $\nu\{h=0\}=\int_{\{h=0\}} h d\mu =0$ so $\frac 1 h$ is well defined almost everywhere with respect to $\nu$.