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!
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 $$\nu(T^{-1}A\backslash A)+\nu(T^{-1}A\cap A)= \nu(T^{-1}A)\leq \nu(A)=\nu(A\backslash T^{-1}A)+\nu(T^{-1}A\cap A)$$ So, $$\nu(T^{-1}A\backslash A) \leq \nu(A\backslash T^{-1}A)\quad (1)$$ In the same way, the $T$-invariance of $\mu$ yields $$\mu(T^{-1}A\backslash A) = \mu(A\backslash T^{-1}A) \quad (2)$$ On the other hand, (since $\nu(X)=1$) $\mu(A)<1$. If $\mu(A)>0$, since $\mu$ is ergodic $$\mu(A\Delta T^{-1}A)=2\mu(A\backslash T^{-1}A) >0\quad (3)$$ Then, $(2)$ and the second inequality and $(3)$ give \begin{eqnarray*} \nu(T^{-1}A\backslash A) &=& \int_{T^{-1}A\backslash A} gd\mu \\ &\ge& \mu(T^{-1}A\backslash A) \\ &=& \mu(A\backslash T^{-1}A) \\ &>& \int_{A\backslash T^{-1}A}gd\mu \\ &=& \nu(A\backslash T^{-1}A). \end{eqnarray*} which contradicts $(1)$. Hence $g \ge 1$ $\mu$-almost surely, so $\nu \ge \mu$. Since $\mu$ and $\nu$ are probability measures, we derive $\mu=\nu$ and $g=1$ $\mu$-almost surely.
