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It is true that, under some conditions, given a measure-preserving transformation $T$, we can always construct a $T$-invariant probability. I am wondering whether we can do a converse. See Parry's Topics in ergodic theory p14

Given a probability space $(X,\mathcal{B},\mu)$, can we always find a measure-preserving transformation $T:X \rightarrow X$ such that $\mu$ is $T$-invariant, except the identity?

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    $\begingroup$ The identity map. $\endgroup$ – Jon Bannon Mar 27 '11 at 20:07
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    $\begingroup$ You are right, I should exclude the identity. $\endgroup$ – Po C. Mar 27 '11 at 20:12
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For a counterexample to Jon's implicit rewording of your question, consider a purely atomic probability space in which the atoms all have different measure.

For a positive result, consider any purely non atomic probability space. It is measure isomorphic to $[0,1]$ with Lebesgue measure. I guess for non separable purely non atomic probability spaces you just need to apply Maharam's classification theorem.

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  • $\begingroup$ @Jon: In the category of measurable spaces and equivalence classes of non-singular maps every non-complete measurable space is equivalent to its completion. $\endgroup$ – Dmitri Pavlov Mar 27 '11 at 22:02
  • $\begingroup$ Thanks, Dmitri. I was being silly... clearly, the right notion here is 'isomorphism mod 0'. $\endgroup$ – Jon Bannon Mar 27 '11 at 22:51

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