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Exodd
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automorphisms of a measurable space can be approximated by continuous automorphismsmeasure preserving maps?

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Exodd
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Suppose first that $D=[0,1]$ is equipped with the usual Lebesgue measure, and that $\varphi$ is a measure-preserving transformation $\varphi:D\to D$ that is bijective and whose inverse is also measure preserving (called automorphism).

Is it true that there existexists a sequence of continuous automorphisms measure-preserving transformations $\varphi_n:D\to D$ converging in measure to $\varphi$?

Suppose first that $D=[0,1]$ is equipped with the usual Lebesgue measure, and that $\varphi$ is a measure-preserving transformation $\varphi:D\to D$ that is bijective and whose inverse is also measure preserving (called automorphism).

Is it true that there exist a sequence of continuous automorphisms $\varphi_n:D\to D$ converging in measure to $\varphi$?

Suppose first that $D=[0,1]$ is equipped with the usual Lebesgue measure, and that $\varphi$ is a measure-preserving transformation $\varphi:D\to D$ that is bijective and whose inverse is also measure preserving (called automorphism).

Is it true that there exists a sequence of continuous measure-preserving transformations $\varphi_n:D\to D$ converging in measure to $\varphi$?

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Exodd
Exodd
  • 201
  • 1
  • 5

automorphisms of a measurable space can be approximated by continuous automorphisms?

Suppose first that $D=[0,1]$ is equipped with the usual Lebesgue measure, and that $\varphi$ is a measure-preserving transformation $\varphi:D\to D$ that is bijective and whose inverse is also measure preserving (called automorphism).

Is it true that there exist a sequence of continuous automorphisms $\varphi_n:D\to D$ converging in measure to $\varphi$?