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David Berman
David Berman
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Hi everyone! I am currently studying the basic theory of measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost...

Question: Suppose $T$ is an invertible measure-preserving map of standard probability measure space $(X,\mu)$. Suppose that $TA=A$ for all measurable subsets $A\subset{}X$, where the equality is up to sets of measure $0$. Prove that the set of those $x$ where $Tx\neq{}x$ has measure $0$.

Hi everyone! I am currently studying the basic theory of measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost...

Question: Suppose $T$ is an invertible measure-preserving map of probability measure space $(X,\mu)$. Suppose that $TA=A$ for all measurable subsets $A\subset{}X$, where the equality is up to sets of measure $0$. Prove that the set of those $x$ where $Tx\neq{}x$ has measure $0$.

Hi everyone! I am currently studying the basic theory of measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost...

Question: Suppose $T$ is an invertible measure-preserving map of standard probability measure space $(X,\mu)$. Suppose that $TA=A$ for all measurable subsets $A\subset{}X$, where the equality is up to sets of measure $0$. Prove that the set of those $x$ where $Tx\neq{}x$ has measure $0$.

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David Berman
David Berman
  • 23
  • 4
Source Link
David Berman
David Berman
  • 23
  • 4

trivial map on $\sigma-$algebra $\mod{}0$ is trivial

Hi everyone! I am currently studying the basic theory of measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost...

Question: Suppose $T$ is an invertible measure-preserving map of probability measure space $(X,\mu)$. Suppose that $TA=A$ for all measurable subsets $A\subset{}X$, where the equality is up to sets of measure $0$. Prove that the set of those $x$ where $Tx\neq{}x$ has measure $0$.