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Ergodic transformation on a atomless measure space

Ergodicity and Appropiate Partition of the SpaceExistence of ergodic joiningMeasure preserving ergodic map commutes with complementation?Approximate eigenvalues of an ergodic invertible transformationBirkhoff ergodic theoremRokhlin lemma need not hold for arbitrary sequenceErgodicity of a measurable transformation on $mathbbT^2$Showing measure is invariant/ergodic for a skew product.If $T$ has continuous spectrum and $(f, 1)=0$ then $mu_f$ has no atoms.ergodic system has dense orbits a.e., continuity of T

5

1

$begingroup$

I am currently reading Kakutani–Rokhlin lemma and faced a problem which is given below :---

Let $(X,mathscr B,mu,T)$ be an invertible measure preserving system such that $mu(x)=0,forall xin X$. Suppose $T$ is ergodic we have to show, $$mubigg(bigcup_kin Bbb Z,knot=0xin X:T^k(x)=xbigg)=0.$$

I don't how do I start with this problem. Any help will be appreciated.

measure-theory ergodic-theory

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edited Mar 12 at 10:47

Mathlover

asked Mar 12 at 10:42

MathloverMathlover

1697

$endgroup$

add a comment | 

5

1

$begingroup$

I am currently reading Kakutani–Rokhlin lemma and faced a problem which is given below :---

Let $(X,mathscr B,mu,T)$ be an invertible measure preserving system such that $mu(x)=0,forall xin X$. Suppose $T$ is ergodic we have to show, $$mubigg(bigcup_kin Bbb Z,knot=0xin X:T^k(x)=xbigg)=0.$$

I don't how do I start with this problem. Any help will be appreciated.

measure-theory ergodic-theory

share|cite|improve this question

edited Mar 12 at 10:47

Mathlover

asked Mar 12 at 10:42

MathloverMathlover

1697

$endgroup$

add a comment | 

$begingroup$

I am currently reading Kakutani–Rokhlin lemma and faced a problem which is given below :---

Let $(X,mathscr B,mu,T)$ be an invertible measure preserving system such that $mu(x)=0,forall xin X$. Suppose $T$ is ergodic we have to show, $$mubigg(bigcup_kin Bbb Z,knot=0xin X:T^k(x)=xbigg)=0.$$

I don't how do I start with this problem. Any help will be appreciated.

measure-theory ergodic-theory

share|cite|improve this question

edited Mar 12 at 10:47

Mathlover

asked Mar 12 at 10:42

MathloverMathlover

1697

$endgroup$

share|cite|improve this question

edited Mar 12 at 10:47

Mathlover

asked Mar 12 at 10:42

MathloverMathlover

1697

share|cite|improve this question

edited Mar 12 at 10:47

Mathlover

asked Mar 12 at 10:42

MathloverMathlover

1697

asked Mar 12 at 10:42

MathloverMathlover

1697

asked Mar 12 at 10:42

MathloverMathlover

1697

1 Answer
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$begingroup$

I believe the correct condition instead of "$forall xin X: mu(x)=0$" is that $mu$ is atom-free.

You want to show that for every invertible measurable map $T$ on a measurable space $(X,mathscrB)$, the set of $T$-periodic points has measure zero with respect to every atom-free $T$-ergodic measure $mu$.

If this is not the case, there must be an integer $n>0$ such that the set $P_n$ of points having period $n$ has positive measure. By ergodicity, the measure of $P_n$ must in fact be $1$. Since $mu$ is atom-free, there must be a measurable set $Asubsetneq P_n$ such that $0<mu(A)<1/n$. The set $E:=Acup T^-1(A)cupcdotscup T^-(n-1)(A)$ will then be an invariant set with $0<mu(E)<1$, contradicting ergodicity.
Q.E.D.

If you only require that there are no singleton atoms, then the conclusion does not hold. For example, take $X:=[0,1]$ with $mathscrB$ being the $sigma$-algebra generated by singletons, and for each $AinmathscrB$, let
beginalign
mu(A) &:=
begincases
0 & textif $A$ is countable, \
1 & textif $Xsetminus A$ is countable.
endcases
endalign
The identity map $T:xmapsto x$ is measurable, measure-preserving and ergodic. It has no singleton atoms, but it is not non-atomic because the complement of any countable set is an atom. On the other hand, every point is periodic under $T$ with period $1$.

share|cite|improve this answer

answered yesterday

BlackbirdBlackbird

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1

$begingroup$

I believe the correct condition instead of "$forall xin X: mu(x)=0$" is that $mu$ is atom-free.

You want to show that for every invertible measurable map $T$ on a measurable space $(X,mathscrB)$, the set of $T$-periodic points has measure zero with respect to every atom-free $T$-ergodic measure $mu$.

If this is not the case, there must be an integer $n>0$ such that the set $P_n$ of points having period $n$ has positive measure. By ergodicity, the measure of $P_n$ must in fact be $1$. Since $mu$ is atom-free, there must be a measurable set $Asubsetneq P_n$ such that $0<mu(A)<1/n$. The set $E:=Acup T^-1(A)cupcdotscup T^-(n-1)(A)$ will then be an invariant set with $0<mu(E)<1$, contradicting ergodicity.
Q.E.D.

If you only require that there are no singleton atoms, then the conclusion does not hold. For example, take $X:=[0,1]$ with $mathscrB$ being the $sigma$-algebra generated by singletons, and for each $AinmathscrB$, let
beginalign
mu(A) &:=
begincases
0 & textif $A$ is countable, \
1 & textif $Xsetminus A$ is countable.
endcases
endalign
The identity map $T:xmapsto x$ is measurable, measure-preserving and ergodic. It has no singleton atoms, but it is not non-atomic because the complement of any countable set is an atom. On the other hand, every point is periodic under $T$ with period $1$.

share|cite|improve this answer

answered yesterday

BlackbirdBlackbird

1,475711

$endgroup$

add a comment | 

1

$begingroup$

I believe the correct condition instead of "$forall xin X: mu(x)=0$" is that $mu$ is atom-free.

You want to show that for every invertible measurable map $T$ on a measurable space $(X,mathscrB)$, the set of $T$-periodic points has measure zero with respect to every atom-free $T$-ergodic measure $mu$.

If this is not the case, there must be an integer $n>0$ such that the set $P_n$ of points having period $n$ has positive measure. By ergodicity, the measure of $P_n$ must in fact be $1$. Since $mu$ is atom-free, there must be a measurable set $Asubsetneq P_n$ such that $0<mu(A)<1/n$. The set $E:=Acup T^-1(A)cupcdotscup T^-(n-1)(A)$ will then be an invariant set with $0<mu(E)<1$, contradicting ergodicity.
Q.E.D.

If you only require that there are no singleton atoms, then the conclusion does not hold. For example, take $X:=[0,1]$ with $mathscrB$ being the $sigma$-algebra generated by singletons, and for each $AinmathscrB$, let
beginalign
mu(A) &:=
begincases
0 & textif $A$ is countable, \
1 & textif $Xsetminus A$ is countable.
endcases
endalign
The identity map $T:xmapsto x$ is measurable, measure-preserving and ergodic. It has no singleton atoms, but it is not non-atomic because the complement of any countable set is an atom. On the other hand, every point is periodic under $T$ with period $1$.

share|cite|improve this answer

answered yesterday

BlackbirdBlackbird

1,475711

$endgroup$

add a comment | 

$begingroup$

I believe the correct condition instead of "$forall xin X: mu(x)=0$" is that $mu$ is atom-free.

You want to show that for every invertible measurable map $T$ on a measurable space $(X,mathscrB)$, the set of $T$-periodic points has measure zero with respect to every atom-free $T$-ergodic measure $mu$.

If this is not the case, there must be an integer $n>0$ such that the set $P_n$ of points having period $n$ has positive measure. By ergodicity, the measure of $P_n$ must in fact be $1$. Since $mu$ is atom-free, there must be a measurable set $Asubsetneq P_n$ such that $0<mu(A)<1/n$. The set $E:=Acup T^-1(A)cupcdotscup T^-(n-1)(A)$ will then be an invariant set with $0<mu(E)<1$, contradicting ergodicity.
Q.E.D.

If you only require that there are no singleton atoms, then the conclusion does not hold. For example, take $X:=[0,1]$ with $mathscrB$ being the $sigma$-algebra generated by singletons, and for each $AinmathscrB$, let
beginalign
mu(A) &:=
begincases
0 & textif $A$ is countable, \
1 & textif $Xsetminus A$ is countable.
endcases
endalign
The identity map $T:xmapsto x$ is measurable, measure-preserving and ergodic. It has no singleton atoms, but it is not non-atomic because the complement of any countable set is an atom. On the other hand, every point is periodic under $T$ with period $1$.

share|cite|improve this answer

answered yesterday

BlackbirdBlackbird

1,475711

$endgroup$

share|cite|improve this answer

answered yesterday

BlackbirdBlackbird

1,475711

answered yesterday

BlackbirdBlackbird

1,475711

answered yesterday

BlackbirdBlackbird

1,475711