Continuity on Measures I was presented on an exercise with the following statement. Given a measurable sapace $(\Omega,\mathcal{F})$ and two finite positive measures $\mu,v$. It is defined: $\int_E (1-f)d\mu=\int_E f dv$  Based on this information is it possible to say $\mu$ is absolutely continuos to $v$, given the fact that what I understand by absolutely continuous measures are the measure in which $\mu(A)=0$ and $v(A)=0$ according to Radon-Nykodim theorem. Can we say based on $\int_E (1-f)d\mu=\int_E f dv$ that $\mu$ and $v$ are absolutely continuous?
 A: To begin, let's get clear about the definition of absolute continuity. If $\mu$ and $\nu$ are measures on $(\Omega, \mathcal{F})$, then we say that $\mu$ is absolutely continuous with respect to $\nu$ and write $\mu \ll \nu$ if $\nu(E) = 0$ implies $\mu(E) = 0$ for all $E \in \mathcal{F}$.
Now, suppose that $(\Omega, \mathcal{F})$ is the Lebesgue measurable space on $[0,1]$. Let $\nu = 0$ and $\mu$ be Lebesgue measure. Then, it is not the case that $\mu \ll \nu$. For example, $\nu([0,1])=0 \ $ but $\mu([0,1])=1$.
But for $f=1$, your equality
$$\int_E 1-f d\mu = \int_E f d\nu$$
holds for every Lebesgue measurable $E$.
If we swap the definitions of $\nu$ and $\mu$ and let $f=0$, then we have your equality again but not $\nu \ll \mu$.
So from $\int_E 1-f d\mu = \int_E f d\nu \ $ for some $f$, we cannot conclude that $\mu \ll \nu$ nor $\nu \ll \mu$.
On the other hand, if $\int_E 1-f d\mu = \int_E f d\nu \ $ for all $f$, then if $\nu(E)=0$, then $\int_E f d\nu = 0$, hence $\int_E 1-f d\mu = 0 \ $ for all $f$. Take $f=0$, then $\mu(E)=0$. Hence, $\mu \ll \nu$. Similarly, we can show $\nu \ll \mu$ by considering $f=1$.
Does this answer your question? To be honest, I'm having some trouble understanding it.
