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 $0$"
measure-theory operator-theory ergodic-theory
asked Sep 9 at 10:29
sleeve chen
2,83331647
add a comment |Â
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 $0$"
measure-theory operator-theory ergodic-theory
asked Sep 9 at 10:29
sleeve chen
2,83331647
add a comment |Â
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 $0$"
measure-theory operator-theory ergodic-theory
asked Sep 9 at 10:29
sleeve chen
2,83331647
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
measure-theory operator-theory ergodic-theory
asked Sep 9 at 10:29
sleeve chen
2,83331647
asked Sep 9 at 10:29
sleeve chen
2,83331647
asked Sep 9 at 10:29
sleeve chen
2,83331647
asked Sep 9 at 10:29
sleeve chen
2,83331647
asked Sep 9 at 10:29
sleeve chen
2,83331647
2,83331647
add a comment |Â
add a comment |Â
1 Answer
1
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votes
up vote
2
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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.
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
add a comment |Â
1 Answer
1
active
oldest
votes
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.
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
add a comment |Â
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.
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
add a comment |Â
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.
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
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.
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
answered Sep 9 at 11:16
Davide Giraudo
122k16147250
122k16147250
add a comment |Â
add a comment |Â
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