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I'm learning about unique ergodicity and how a transformation can have different measures with respect to which it's ergodic. I read that, for example, if $T : [0, 1) \to [0, 1)$ is the doubling map $Tx = 2 x \mod 1$, then there are many Borel probability measures on $I$ which make $T$ ergodic. For example, if I choose any $p \in [0, 1]$, then I can define a Borel probability measure $\mu_p$ on the binary cylinders which amounts essentially to flipping a coin with probability $p$ of landing Heads, where Lebesgue measure is $\mu_{1 / 2}$.

However, I noticed that different values of $p$ gave singular measures, i.e. $p \neq q \Rightarrow \mu_p \perp \mu_q$. My questions is if there exist any Lebesgue-continuous Borel probability measures on $[0, 1)$ which $T$ is ergodic with respect to. Are different ergodic measures necessarily singular? Is there some more general theorem about this? Or am I missing some more obvious Lebesgue-continuous ergodic measure for this transformation?

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After thinking on it, I think that these different ergodic measures will necessarily be singular. Basically, if $\mu, \nu$ are distinct ergodic measures, then there's some Borel $A \subseteq I$ such that $\mu(A) = \nu(A)$. Then we know that \begin{align*} \mu \left( \left\{ x \in I : \lim_{k \to \infty} \frac{1}{k} \sum_{j = 1}^k \chi_A \left( T^{j - 1}(x) \right) = \mu(A) \right\} \right) & = 1, \\ \nu \left( \left\{ x \in I : \lim_{k \to \infty} \frac{1}{k} \sum_{j = 1}^k \chi_A \left( T^{j - 1}(x) \right) = \nu(A) \right\} \right) & = 1 . \end{align*} But these two sets must necessarily be disjoint.

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  • $\begingroup$ mark as accepted by clicking check mark if this is right. if ur unsure whether this is right, let me know and ill read question and ur answer $\endgroup$ – mathworker21 Apr 16 '20 at 0:01

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