Let $\mu,\ \nu$ be measures, $\mu\ll\nu$ and $\nu\ll \mu$. Show that $\frac{d\mu}{d\nu} = (\frac{d\nu}{d\mu})^{-1}$. Let $\mu,\  \nu$ be measures, $\mu\ll\nu$ and $\nu\ll\mu$. Show that $\frac{d\mu}{d\nu} = (\frac{d\nu}{d\mu})^{-1}$.
There are not any other assumptions, so is it necessary for $\mu, \nu$ to be $\sigma$-finite? Does  expression $\frac{d\mu}{d\nu}$ make sense without making such assumption?
Any hints would be great!
 A: It suffices to check $\nu(A) = \int_A(\frac{d\mu}{d\nu})^{-1}d\mu$
Since $\mu(A) = \int_A\frac{d\mu}{d\nu}d\nu$, approximating $(\frac{d\mu}{d\nu})^{-1}$ by monotonically increasing simple functions $f^i = \sum_{j = 0}^{N^i}a_j^i\chi_{A_j^i}$ from below,
$$\int_Af^id\mu = \sum_{j = 0}^{N^i}a_j^i\mu(A_j^i\cap A) = \sum_{j = 0}^{N^i}a_j^i\int_{A_j^i\cap A}\frac{d\mu}{d\nu}d\nu = \sum_{j = 0}^{N^i}\int_{A_j^i\cap A}a_j^i\frac{d\mu}{d\nu}d\nu = $$
$$\sum_{j = 0}^{N^i}\int_{ A}a_j^i\chi_{A_j^i}\frac{d\mu}{d\nu}d\nu = \int_{ A}\sum_{j = 0}^{N^i}a_j^i\chi_{A_j^i}\frac{d\mu}{d\nu}d\nu = \int_{ A}f^i\frac{d\mu}{d\nu}d\nu$$
now $f^i$ isotone $\implies$ $f^i\frac{d\mu}{d\nu}$ isotone. It is now the job of the Monotone Convergence Theorem to tell us:
$$lim_{i \to \infty}\int_{ A}f^i\frac{d\mu}{d\nu}d\nu = \int_{ A}lim_{i \to \infty}f^i\frac{d\mu}{d\nu}d\nu = \int_{ A}(\frac{d\mu}{d\nu})^{-1}\frac{d\mu}{d\nu}d\nu = \int_{ A}d\nu = \nu(A)$$
I believe Radon Nikodym derivatives are only defined for $\sigma$-finite measures.
