I am trying to follow the proof of the following statement, and I fail to see one step.
Let $T \colon X \to X$ be an invertible and ergodic transformation on a measure space $(X, \mathcal{F}, \mu)$. If $\mu$ is non-atomic, then $T$ is conservative.
Proof. We have seen that there exists a wandering set $W \in \mathcal{W}_T$ (the set of all wandering sets) such that the dissipative part is $\mathcal{D}(T) = \bigcup_{n \in \mathbb{Z}}T^nW$. Therefore we have $T^{-1}\mathcal{D}(T) = \mathcal{D}(T)$, and so by ergodicity the system is either conservative or totally dissipative.
Suppose it was totally dissipative, so the set $W$ from above satisfies $\mu(W) >0$. Since $\mu$ is non-atomic, we find $U \subseteq W$ such that $0<\mu(U) < \mu(W)$. If we set
$$ A = \bigcup_{n \in \mathbb{Z}}T^nU, $$ we have that $T^{-1}A = A, \mu(A) > 0$ and $\mu(A^c) >0$, contradicting ergodicity.
The only thing that is unclear to me is why it follows that $\mu(A^c) > 0$. I know that $\mu(W\setminus U) >0$ and tried to use that, but I can't see how. I think this should not be difficult, does anyone see a simple way to see that $\mu(A^c) > 0$?
This follows from the fact that
$$X\setminus T^nU = \cdots \cup T^{n-1}W\cup T^n(W\setminus U)\cup T^{n+1}W\cup\cdots.$$
Thus the complement of $A$ is the union of the iterates of $W\setminus U$. One can also depict the situation pictorially: $X$ is made up of disjoint iterates of $W$, which is made up of two pieces $U$, $W\setminus U$ of positive measure. The iterates of one piece ($U$) constitute $A$; the complement of $A$ then must be the iterates of the other piece ($W\setminus U$).
For reference purposes this is Prop.1.2.1 from Aaronson's Infinite Ergodic Theory, p.22.