The problem is to show that if I have $T: (X, \mu) \rightarrow (X, \mu)$ that preserves $\mu$ and is ergodic then the number of preimages of a point is well-defined and is constant $\mu$-ae.
I thought to consider the family of sets $E_n = \{y \in X: \#T^{-1}(y) = n\}$. This family is disjoint and her union is X. If I could prove that $T^{-1}(E_n) = E_n$ for all $n \geq 0$, then, using the ergodicity, I know that $\mu(E_n) = 0$ or $1$, but I couldn't prove that.
Any suggestions?
Maybe, my way is incorrect.
Besides, I need to show that if the number of preimages is greater than $1$, then, entropy is positive.
It's not true. Consider a space $X$ consisting of three points $0,1,2$ with $$ T(0) = 1,\ T(1) = 2, \ T(2) = 1 $$ $$ \mu(\{0\}) = 0, \ \mu(\{1\}) = \mu(\{2\}) = 1/2 $$ The transformation $T$ is measure-preserving and ergodic (in fact the only $T$-invariant sets are $\emptyset$ and $X$). But the number of preimages is $2$ for the point $1$ and $1$ for the point $2$, so $\mu(E_1) = \mu(E_2) = 1/2$.
More generally, given any ergodic measure-preserving transformation $T$ on a probability measure space $(X, \Sigma, \mu)$ and $A \subset X$ with $0 < \mu(A) < 1$, we can modify this by taking an additional copy $\tilde{A}$ of $A$, have our new transformation $\tilde{T}$ map each $\tilde{a} \in \tilde{A}$ to the corresponding $a \in A$ and otherwise agree with $T$, and take the new measure $\tilde{\mu}$ to be $0$ on $\tilde{A}$ and the same as $\mu$ on $X$. Then $\tilde{T}$ is ergodic for $\tilde{\mu}$, and each member of $A$ has one more pre-image under $\tilde{T}$ than it had under $A$, but each member of $X \backslash A$ has the same number of pre-images as it had before.