Ergodicity in Transformation Implied by Ergodicity in Induced Transformation For a finite, recurrant, invertible, measure preserving dynamical system with transformation $T$, I can show that if $T$ is ergodic, then the induced transformation for any positive-measure set is ergodic. How can I prove the converse - that is, if there exists a positive-measure set for which the induced transformation is ergodic, then $T$ is ergodic?
 A: I was searching for a similar statement and I found this. I don't know if it is OK to provide an answer to such an old thread. 
Let $Y$ be a subset of $X$ such that 
$$ \bigcup_{n\geq 1} T^{-n}Y=X$$
Call $\varphi$ the return time function of $Y$ and $T_Y$ the first return map of $Y$. There are two main ingredients in the proof: for every measurable set $A$, we have
$$T_Y^{-1}(Y\cap A)=\bigcup_{n\geq 1}Y\cap \{\varphi=n\}\cap T^{-n}A.$$
This is essentially a decomposition of $Y$ into disjoint pieces which land to $A$ for the first time at $n$.
The second key idea is that if you have a $T$-invariant set $A$, it induces a $T_Y$-invariant set $A\cap Y$:
$$T_Y^{-1}(Y\cap A)=\bigcup_{n\geq 1}Y\cap \{\varphi=n\}\cap T^{-n}A=\bigcup_{n\geq 1}Y\cap \{\varphi=n\}\cap A=A\cap Y.$$
Since $T_Y$ is ergodic, $Y\cap A$ or $Y\cap A^c$ have zero measure. In the first case, by invariance $T^{-n}Y\cap A$ has zero measure, and by the hypothesis on $Y$ we have
$$
\mu(A)=\mu\left(\bigcup_{n\geq 1}T^{-n}Y\cap A\right)\leq 0
$$
The second case is analogous. The proof works in the infinite measure as well.
A: I didn't check the details but I think the following should work:
first, you have to assume that integral the first return maps is finite. Then, the suspension of the induced transformation is, up to a positive constant, the original measure preserving transformation. If you believe this, then the result you want follows form Lemma 9.24 in the book
http://books.google.ch/books/about/Ergodic_Theory.html?id=PiDET2fS7H4C&redir_esc=y
