Let $(X,\mathcal{B})$ be a standard Borel space and $T:X\to X$ a Borel automorphism. Suppose that $\mu$ is a probability measure on $X$ which is:
- Continuous: i.e. $\mu(\{x\})=0$ for all $x\in X$.
- Quasi invariant: $\forall A\in\mathcal{B} (\mu(A)=0\iff \mu(TA)=0$.
- Ergodic: $\forall A\in\mathcal{B} (TA=A\implies \mu(A)=0\vee \mu(A)=1)$.
Now let $W\in\mathcal{B}$ be wandering, i.e. $T^n W\cap T^m W=\emptyset$ if $n\neq m$. Must a wandering set have $0$ measure? It is a well known fact that this holds if $\mu$ is $T$-invariant, but what happens under these conditions?
If we assume $\mu(W)>0$ for some wandering set $W$, then, letting $A:=\bigcup_{n\in\mathbb{Z}}T^n W$, then $\mu(A)\ge \mu(W)>0$ and $TA=A$, so $\mu(A)=1$. We also have $\mu(A)=\sum_{n\in\mathbb{Z}}\mu(T^nW)$, and each term of the series must be positive, by quasi-invariance, but that's about it.
EDIT: I think the following works, if someone could verify it I'd be very grateful.
Let $W$ be a wandering set, and suppose $\mu(W)>0$. Since $\mu$ is continuous and $(X,\mathcal{B})$ is standard, $\mu$ is non atomic, so there exists $W_1,W_2\in\mathcal{B}$ such that $W=W_1\cup W_2$, $W_1\cap W_2=\emptyset$ and $\mu(W_1)\mu(W_2)>0$. Now let $B=\bigcup_{g\in G}gW_1$. We have $\mu(B)\ge \mu(W_1)>0$. Since $\mu$ is ergodic, it follows that $\mu(B)=1$. Since $W$ is wandering, we have that $W_2$ and $B$ are disjoint, which leads to a contradiction since $\mu(W_2)>0$. The fact that $\mu$ is quasi invariant was seemingly never used, which seems strange since it is mentioned in this paper by Shelah and Weiss (beginning of page 2).