Let $X$ be a compact metric space, $T:X\rightarrow X$ a homeomorphism and $\mu$ be a $T$-invariant probability measure on $X$ such that the set of points with dense orbit in $supp(\mu)$ has full measure. Is the following statement true?


if $\mu$ is the only measure which satisfies that condition, then $\mu$ is ergodic.



A question about ergodicity
Let $Y$ be the support of $\mu$ and suppose that $\mu$ is not ergodic. Then there exists a $T$-invariant measurable set $A\subset Y$ such that $0<\mu(A)<1$. The measures $\mu\_1$, $\mu\_2$ defined by $$\mu\_1(B):=\mu(B \cap A)/\mu(A),$$ $$\mu\_2(B):=\mu(B \setminus A)/\mu(X \setminus A)$$ are distinct and invariant, and each gives full measure to the set of points whose orbit is dense in $Y$, contradicting the uniqueness of $\mu$. (I haven't checked the details but I think that the supports of $\mu\_1$ and $\mu\_2$ should also be $Y$, since they each give positive measure to the set of all points with orbit dense in $Y$, and this set is invariant.)


