# 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!

1. 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.
2. It is false because in Petersen's book it is written $$\nu\circ T^{-1}\leq\nu$$ instead of $$\nu T^{-1}\ll\nu$$.
3. 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.
• How do we get $\mu(T^{-1}A\backslash A)\leq \int_{T^{-1}A\backslash A}gd\mu$ in the last sequence of inequalities? On $T^{-1}A\backslash A$ we have $g\leq 1$. Shouldn't it be the reverse inequality ? Commented Jul 11, 2021 at 20:10
• @Noobmathematician Being $\mu$ a $T$-invariant measure, we have $\mu(T^{-1}A\setminus A) = \mu(T^{-1}A) - \mu(A\cap T^{-1}A) = \mu(A) - \mu(A\cap T^{-1}A) = \mu(A\setminus T^{-1}A)$. Thus, $$\mu(T^{-1}A\setminus A) = \mu(A\setminus T^{-1}A) \leq \int_{A\setminus T^{-1}A}g\ d\mu$$ since $g > 1$ on $A\setminus T^{-1}A$. Commented Jul 15, 2021 at 13:19