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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.

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1 Answer 1

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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.)

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