Let $\mu_1,\mu_2$ be sigma finite measures. Prove that $\mu_1\ll \mu_2$ and $\mu_2\ll \mu_1$ $\iff \exists f$, where $f$ is strictly positive such that $\nu(A)=\int_A fd\mu$ for all measurable $A$.
It is the strictly positive nature of $f$ which I am having trouble with. Let $\mu_1\ll \mu_2$ and $\mu_2\ll \mu_1$. I know that there exist $f_1,f_2\geq 0$ s.t. $\nu(A)=\int_A f_1d\mu$ and $\mu(A)=\int_A f_2d\nu$ but am not sure how that would lead me to a strictly positive density function. Any suggestions?
By Radon Nikodym there is a nonnegative function $f$ so that $\nu(A)=\int_A f d\mu$ for all measurable $A$. Consider the measurable set $S=\{x:f(x)=0\}$. Then $\nu(S)=\int_S f(x) d\mu=0$. Therefore $\mu(S)=0$.
Define $$\tilde f(x)=\begin{cases}f(x)&\text{ if } x\not \in S\\ 24 &\text{ if }x\in S.\end{cases}$$
Then for any measurable $A$ we have that $$\nu(A)=\int_A f(x)d\mu=\int_A \tilde f(x)d\mu,$$
because changing a function on a set of measure zero doesn't change integral.