ergodic measure and absolutely continuous measure i've come across this problem in Petersen's "Ergodic Theory":
Let $(X,\mathcal{B},T,\mu)$ be an ergodic dynamical system. Let $\nu\ll\mu$ be a measure un $(X,\mathcal{B})$ such that $\nu T^{-1}\ll\nu$. Show that $\nu=\nu T^{-1}$ and that $\nu$ is a constant multiple of $\mu$.
I've tried solving this using Radon Nykodim derivatives but i've had no success doing it. I would appreciate any help. Thanks!
 A: *

*Your statement is false. Here is a counterexample: let $([0,1),\mu,T)$  an irrational rotation, i.e., $T(x)=x+\alpha\mod{1}$ for irrational $\alpha$ and $\mu$ is the Lebesgue measure. Let $g$ be non negative such that $g\in L^1([0,1),\mu)$ and $\frac{1}{g}\in L^\infty([0,1),\mu)$. Then, $d\nu:=gd\mu$ satisfies the hypothesis but clearly $\nu$ is not necessarily $T$-invariant.


*It is false because in Petersen's book it is written $\nu\circ T^{-1}\leq\nu$ instead of $\nu T^{-1}\ll\nu$.


*Here is a proof of the original problem when $\nu$ is a probability measure (we can always suppose this when $\nu$ is finite): let $g=\frac{\partial\nu}{\partial\mu}$ and put $A=\{g>1\}$. In one hand, we note
$$\nu(T^{-1}A\backslash A)+\nu(T^{-1}A\cap A)=
\nu(T^{-1})\leq
\nu(A)=\nu(A\backslash T^{-1}A)+\nu(T^{-1}A\cap A)$$
So,
$$\nu(T^{-1}A\backslash A)\leq \nu(A\backslash T^{-1}A)\quad (1)$$
On the other hand, (since $\nu(X)=1$) $\mu(A)<1$. If $\mu(A)>0$, since $\mu$ is ergodic
$$\mu(A\Delta T^{-1}A)=2\mu(A\backslash T^{-1}A) >0\quad (2)$$
Then, $(2)$ in the second inequality and $(1)$ in the third inequality give
$$\mu(T^{-1}A\backslash A)=
\mu(A\backslash T^{-1}A)\leq
\int_{A\backslash T^{-1}A}gd\mu<
\mu(A\backslash T^{-1}A)=\mu(T^{-1}A\backslash A)$$
where the last equality follows from the $T$-invariance of $\mu$ in the same way as $(1)$ follows from $\nu\circ T^{-1}\leq\nu$. This contradiction shows that $\mu(A)=0$ and, since $\nu(X)=1$, we also deduce $\mu(\{g<1\})=0$.
