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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?

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    Would you like to show that $v$ is absolutely continuous with respect to $\mu$ given that $$ \int_E (1-f)d\mu = \int_E f dv$$ for any $\mathcal{F}$-measurable set $E$ and any measurable function $f$? Or there exists $f$ such that this equality holds for any measurable set $E$? You need more details, measures are not just absolutely continuous they are absolutely continuous with respect to other measure, e.g. assuming that if $\mu(A) = 0$ then $v(A) = 0$, we have that $v$ is abs. continuous w.r.t. $\mu$ etc.https://en.wikipedia.org/wiki/Absolute_continuity#Absolute_continuity_of_measures. – 2017-02-22
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    Thanks for the reply.If you manipulate$\int_E (1-f)d\mu)=\int_E fdv$ you get $\mu(E)=\int f d(v+\mu)$ so according to Radon-Nykodim theorem $\mu$ must be absolutely continuos to $v+\mu$, so there must be a function$f$ that satisfies the equality. But, my question is: For that to be true should $\mu$ be continuous to $v$ as well? Can you deduce it of $\int_E (1-f)d\mu)=\int_E fdv$? – 2017-02-22

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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.