Let $T : X\to X$ be a continuous transformation on a compact metric
space $X$ and let $\mu$ be an invariant Borel probability measure for
$T$ (i.e., $\mu(T^{-1}(A)) = \mu(A)$ for all Borel sets $A$), which is
ergodic (that is, if $T^{-1}(A) = A$, then $\mu(A)\in\{0,1\}$). Assume
that $\mu$ has full support, i.e., $\operatorname{supp}(\mu) = X$.
Does it then follow that $\mu$ is the unique ergodic measure?
Well, suppose that there exists another ergodic measure $\nu$ for $T$ on $X$. Then I know that $\mu$ and $\nu$ must be mutually singular. That is, there exist sets $E,F\subset X$ such that $E\cap F = \emptyset$ and $\mu(E) = \nu(F) = 1$. But from there I can't infer a contradiction. And, more, the statement does not hold for arbitrary measures $\mu$ and $\nu$. For example, consider $X = [0,1]$ with $\mu=$ Lebesgue measure and $\nu=\delta_0$. Then $\mu$ and $\nu$ are mutually singular and $\mu$ has full support.
What I can show is that the support of $\nu$ is a $\mu$-null set and has empty interior. But that's all. Or is that maybe all about it, and such ergodic measures can coexist?
Best Answer
Hint:
Both of those measures are invariant and ergodic w/r/t $T:x\mapsto 2x \mod{1}$.