About definition of Ergodic theorem

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Let $(X,Sigma, mu)$ be a probability space, and $T:Xrightarrow X$ be a measure-preserving transformation. We say $mu$ is ergodic with respect to $T$ if




for every $Ein Sigma $ with $T^-1(E) = E$, either $mu(E)=1$ or $0$.




I have a very fundamental problem about this definition:



the result "$mu(E)=1$ or $0$" does not have any information about $T$ or does not say anything about $T$. So is the condition "$T^-1(E)=E, forall Ein Sigma$" necessary?



$T^-1(E)=E$ should depend on the choice of $T$; I pick a particular $T$ such that $T^-1(E)=E$. However, $T$ does not show up in "$mu(E)=1$ or $0$"










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    Let $(X,Sigma, mu)$ be a probability space, and $T:Xrightarrow X$ be a measure-preserving transformation. We say $mu$ is ergodic with respect to $T$ if




    for every $Ein Sigma $ with $T^-1(E) = E$, either $mu(E)=1$ or $0$.




    I have a very fundamental problem about this definition:



    the result "$mu(E)=1$ or $0$" does not have any information about $T$ or does not say anything about $T$. So is the condition "$T^-1(E)=E, forall Ein Sigma$" necessary?



    $T^-1(E)=E$ should depend on the choice of $T$; I pick a particular $T$ such that $T^-1(E)=E$. However, $T$ does not show up in "$mu(E)=1$ or $0$"










    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $(X,Sigma, mu)$ be a probability space, and $T:Xrightarrow X$ be a measure-preserving transformation. We say $mu$ is ergodic with respect to $T$ if




      for every $Ein Sigma $ with $T^-1(E) = E$, either $mu(E)=1$ or $0$.




      I have a very fundamental problem about this definition:



      the result "$mu(E)=1$ or $0$" does not have any information about $T$ or does not say anything about $T$. So is the condition "$T^-1(E)=E, forall Ein Sigma$" necessary?



      $T^-1(E)=E$ should depend on the choice of $T$; I pick a particular $T$ such that $T^-1(E)=E$. However, $T$ does not show up in "$mu(E)=1$ or $0$"










      share|cite|improve this question













      Let $(X,Sigma, mu)$ be a probability space, and $T:Xrightarrow X$ be a measure-preserving transformation. We say $mu$ is ergodic with respect to $T$ if




      for every $Ein Sigma $ with $T^-1(E) = E$, either $mu(E)=1$ or $0$.




      I have a very fundamental problem about this definition:



      the result "$mu(E)=1$ or $0$" does not have any information about $T$ or does not say anything about $T$. So is the condition "$T^-1(E)=E, forall Ein Sigma$" necessary?



      $T^-1(E)=E$ should depend on the choice of $T$; I pick a particular $T$ such that $T^-1(E)=E$. However, $T$ does not show up in "$mu(E)=1$ or $0$"







      measure-theory operator-theory ergodic-theory






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      asked Sep 9 at 10:29









      sleeve chen

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          For any $Ein Sigma$, let $P(E)$ be the implication "$left(T^-1E=Eright)Rightarrow left(mu(E)in 0,1right)$". Ergodicity of $mu$ with respect to $T$ means that fro all $EinSigma$, proposition $P(E)$ is true. Since the first part of the implication involves $T$, the definition of ergodicity also implies $T$.



          Note also that "$T^-1(E)=E, forall Ein Sigma$" does not hold in general. For example, if $Sigma$ contains the singletons, this would force $T$to be the identity.






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            1 Answer
            1






            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            For any $Ein Sigma$, let $P(E)$ be the implication "$left(T^-1E=Eright)Rightarrow left(mu(E)in 0,1right)$". Ergodicity of $mu$ with respect to $T$ means that fro all $EinSigma$, proposition $P(E)$ is true. Since the first part of the implication involves $T$, the definition of ergodicity also implies $T$.



            Note also that "$T^-1(E)=E, forall Ein Sigma$" does not hold in general. For example, if $Sigma$ contains the singletons, this would force $T$to be the identity.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              For any $Ein Sigma$, let $P(E)$ be the implication "$left(T^-1E=Eright)Rightarrow left(mu(E)in 0,1right)$". Ergodicity of $mu$ with respect to $T$ means that fro all $EinSigma$, proposition $P(E)$ is true. Since the first part of the implication involves $T$, the definition of ergodicity also implies $T$.



              Note also that "$T^-1(E)=E, forall Ein Sigma$" does not hold in general. For example, if $Sigma$ contains the singletons, this would force $T$to be the identity.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                For any $Ein Sigma$, let $P(E)$ be the implication "$left(T^-1E=Eright)Rightarrow left(mu(E)in 0,1right)$". Ergodicity of $mu$ with respect to $T$ means that fro all $EinSigma$, proposition $P(E)$ is true. Since the first part of the implication involves $T$, the definition of ergodicity also implies $T$.



                Note also that "$T^-1(E)=E, forall Ein Sigma$" does not hold in general. For example, if $Sigma$ contains the singletons, this would force $T$to be the identity.






                share|cite|improve this answer












                For any $Ein Sigma$, let $P(E)$ be the implication "$left(T^-1E=Eright)Rightarrow left(mu(E)in 0,1right)$". Ergodicity of $mu$ with respect to $T$ means that fro all $EinSigma$, proposition $P(E)$ is true. Since the first part of the implication involves $T$, the definition of ergodicity also implies $T$.



                Note also that "$T^-1(E)=E, forall Ein Sigma$" does not hold in general. For example, if $Sigma$ contains the singletons, this would force $T$to be the identity.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 9 at 11:16









                Davide Giraudo

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