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Oct 8, 2014 at 6:56 comment added Asaf Now in order to compute $d\omega{1}/d\omega{2}$, you just take a nice set $A$ and its char. func. and integrate, then you use the substitution $x=Ty$ and the relation between $\omega_{1},\omega_{2}$ and the uniqueness of the RN derivative (and probably induction on $n$) to conclude the result.
Oct 8, 2014 at 6:56 comment added Asaf You can substitute, and about the notation, this is a common notation, which should be understood as follows - $\mu(f) = \int D_{n}fd\mu_{n}$, by the RN thm, where $D_{n}$ is the derivative. Take $f$ to be char. func. of a nice set (you can also take a nice smooth approx. if you like), we should show that $D_{n}$ equals to the above mentioned formula. For $\omega_{0},\omega_{1}$, the proof is trivial. For $\omega_{2}$ you use the fact that $\omega_{1} \ll \omega_{2}$ and the RN derivative for this relation, to compute $d\mu/d\omega_{2}$.
Oct 8, 2014 at 6:38 comment added user90803 But you can't simply substitute on the radon-nykodim derivative. Actually you only have that $T\omega_n = p \omega_{n-1}$, we don't really have a direct relation between $\omega_n$ and $\omega_{n-1}$. I also don't understand your notation $\frac{d\omega_{n-1}}{d\omega_n}(A)$, you are evaluating a function in a borel set?
Sep 21, 2014 at 20:41 comment added Asaf you should simply substitute $\omega_n , \omega_{n-1}$ with their respective forms, the $p$-factor is reduced at it appears both in the nominator and in the denominator, and you're left with the derivative evaluated at $TA$ (the inveratibility is not a real issue here).
Sep 21, 2014 at 20:19 comment added user90803 Thanks for your help. I proved that $\omega_n(T^{-1}(A)) = p \omega_{n-1}(A)$ but I don't understand your last calculation which leed to the conclusion.
Sep 21, 2014 at 18:34 history answered Asaf CC BY-SA 3.0