I´m trying to find a solution for the Excercise 6.2.1 from the book "Foundations of Ergodic Theory", written for Marcelo Viana and Oliviera.
The Problem:
Show that if $u$ is a measurable solution of the cohomological equation (6.2.2)$u(x+\alpha)-u(x)=\phi(x)$, then $h :\mathbb{T}^2→\mathbb{T}^2$,$h(x,y)=(x,y+u(x))$ is an ergodic equivalence between $(f_{0},m)$ and $(f ,m)$, that is, $h$ is an invertible measurable transformation that preserves the measure m and conjugates the two maps $f$ and $f_{0}$. Deduce that $(f ,m) $ cannot be ergodic.
My Aproach
The inverse of $h$ obviuosly is the function
$$h^{-1}(x,y)=(x,y-u(x))$$
hence $h$ is invertible. The function $h$ is a conjugation: this is equivalent to the fact that $u$ is a solution of the cohomological equation.
Now, what I'm still can not prove is that $h$ preserves the mesure m. I know from Mañé's books that must be prove using the Fubbini's Theorem: some like the following:
For $E\subset \mathbb{T}^2$:
$$m(h^{-1}(E))=\int_{\mathbb{T}^2}1_{h^{-1}(A)}dm$$ $$=\int_{\mathbb{T}^2}1_{A}(h(x,y))dm$$ $$=\int_{\mathbb{T}^2}1_{A}((x,y+u(x)))dm$$ $$=\int_{\mathbb{S}^1}\left[\int_{\mathbb{S}^1}1_{A}((x,y+u(x)))d\nu(y)\right]d\mu(x)$$ ......... $$=\int_{\mathbb{S}^1}\left[\int_{\mathbb{S}^1}1_{A}((x,y))d\mu(x)\right]d\nu(y)$$ $$=m(E)$$
My question is: How can I (fill the ditails and) complete the proof using Fubbini's Theorem?