I have $\sigma$ finite measures $\lambda$ and $\mu$ on the same space $(X, \Sigma)$, and $\nu = \lambda + \mu$. An exercise I have is to show that $\lambda \ll \mu$ if and only if $$\nu(\{d\mu/d\nu(x) = 0\}) = 0$$
To show $\rightarrow$ direction, I think
$$\nu(\{d\mu/d\nu(x) = 0\}) = \lambda(\{d\mu/d\nu(x) = 0\}) + \mu(\{d\mu/d\nu(x) = 0\})$$
and second term is 0 because
$$ \mu(\{d\mu/d\nu(x) = 0\}) = \int_{\{d\mu/d\nu(x) = 0\}}(d\mu/d\nu)d\nu $$
so we are taking it over the space where it is 0. The first term I thought I could use Chain rule like
$$ \lambda(\{d\mu/d\nu(x) = 0\}) = \int_{\{d\mu/d\nu(x) = 0\}} (d\lambda/d\nu)d\nu = \int_{\{d\mu/d\nu(x) = 0\}} (d\lambda/d\mu)(d\mu/d\nu)d\nu $$
.. but my question is - is this a good idea? If it is okay, then how do I make it formal that this term is $0$? Does it simplify to
$$ \int_{\{d\mu/d\nu(x) = 0\}} (d\lambda/d\mu)(d\mu/d\nu)d\nu = (d\lambda/d\mu)(0) $$
or am I not understanding right? I mostly just don't know if pulling things out like this is allowed. Also, ideas for $\leftarrow$ way would be helpful.
Thanks
Where did you use the fact that $\lambda << \mu$?. In the second line of your proof you already know that second term is 0. The first term is 0 because of the hypothesis that $\lambda << \mu$? (Just the definition of absolute continuity, right?). For the reverse implication let $\mu =\mu_1+\mu_2$ be the Lebesgue decomposition of $\mu$ with respect to $\nu$. To show that $\lambda << \mu$ it is enough to show that $\lambda << \mu_1$ (because $\mu_1 \leq \mu$). If $\mu_1 (E)=0$ then $\int_E \frac {d\mu_1} {d\nu} d\nu=0$. But we are assuming that $\frac {d\mu_1} {d\nu} >0$ almost everywhere with respect to $\nu$. Hence $\nu (E)=0$. But $\lambda \leq \nu$ so we get $\lambda (E)=0$. This completes the proof.