Let $f$ be the Cantor function and $\lambda$ the Lebesgue measure in $[0,1]$ define $\nu(A)=\int_{A} f d\lambda$ for all $A \subset [0,1]$ is true that $\nu << \lambda$ or $\lambda << \nu $? If it's true compute the respective Radom-Nikodym derivative.
If $\lambda (A)=0$ how $f$ is positive measurable then i know that $\int_{A} f d \lambda=0$ and these implies that $\nu(A)=0$ hence $\nu << \lambda$ and these implies that exists $\frac{d \nu}{d \lambda}=g$ $\Rightarrow \nu(A)=\int_{A} g d \lambda \Rightarrow \int_{A} f d \lambda = \int_{A} g d \lambda$ but the Radom Nikodym derivative is unique $\lambda$-a.e and therefore $f=g$, $\lambda$-a.e is these correct?
I think to $\lambda$ is not absolute continuous w.r.t $\nu$ but I not sure how counterexample works
Any hint or help I will be very grateful
If $f$ is any non-negqative measurable function on a measure space $(X,\mathcal S, \mu)$ then $\nu (A)=\int_A fd\mu$ defines a measure which is absolutely continuous w.r.t. $\mu$. [$\int_A fd\mu=0$ whenever $\mu(A)=0$ by the definition of integral w.r.t. $\mu$]. The Radon Nikodym derivative is $f$ itself.