Below is the Exercise 32.6 ( billingsley's probability and measure) from the Section of Radon- Nikodym Theorem.
I have done the question (a) but I am unable to do the other two parts (b) and (c), even though I might have the solution I do not understand it.
For (b) note that if $\,\mathrm d \nu =h\,\mathrm d \mu $ and $\,\mathrm d \mu =g \,\mathrm d \nu $ for some $h\in L^1(\mu)$ and some $g\in L^1(\nu)$ then $\,\mathrm d \nu =hg\,\mathrm d \nu $, hence $hg=1$ $\nu $-almost everywhere, what imply that $h=1/g\ $ $\nu$-almost everywhere, what is equivalent to the expression that you want to prove.