Mixing
If $muneq nu$ then $frac{1}{2}(mu+nu)$ is not ergodic
$begingroup$
The following problem looks simple, but it seems there is more needed than just using the definition of ergodicity.
Let $(X, mathcal{F}, mu, T)$ and $(X, mathcal{F}, nu, T)$ be two measure preserving dynamical systems. Show
that if $muneq nu$, then the measure $rho := frac{1}{2}(mu+nu)$ is not ergodic for $T$.
These are my ideas:
Suppose $T$ is ergodic. Since $muneq nu$, there is a set $Ain mathcal{F}: mu(A)neq nu(A)$, so wlog we can assume $mu(A)<rho(A)<nu(A)$, whence it follows that $rho(A)notinlbrace 0,1rbrace$ and hence can't be $T$-invariant. It seems like we must use that $T$-invariant sets form a $sigma$-algebra.
Is there anyone with further ideas?
ergodic-theory
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
$endgroup$
add a comment |
$begingroup$
The following problem looks simple, but it seems there is more needed than just using the definition of ergodicity.
Let $(X, mathcal{F}, mu, T)$ and $(X, mathcal{F}, nu, T)$ be two measure preserving dynamical systems. Show
that if $muneq nu$, then the measure $rho := frac{1}{2}(mu+nu)$ is not ergodic for $T$.
These are my ideas:
Suppose $T$ is ergodic. Since $muneq nu$, there is a set $Ain mathcal{F}: mu(A)neq nu(A)$, so wlog we can assume $mu(A)<rho(A)<nu(A)$, whence it follows that $rho(A)notinlbrace 0,1rbrace$ and hence can't be $T$-invariant. It seems like we must use that $T$-invariant sets form a $sigma$-algebra.
Is there anyone with further ideas?
ergodic-theory
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
$endgroup$
$begingroup$
If you're looking for a good book I'd suggest "Ergodic Theory: with a view towards Number Theory" by Einsiedler and Ward. I recall seeing stronger result in there.
$endgroup$
– BigbearZzz
Jan 15 at 20:05
1
$begingroup$
I already have a good book with a.o. Knopp's lemma, mixing theorems and much more ergodic theorems. There is no link, actually it is a syllabus.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 20:32
2
$begingroup$
It can't be an extreme point and so it can't be ergodic.
$endgroup$
– John B
Jan 16 at 14:00
$begingroup$
That is only true for if $T$ is continuous on a compact metric space isn't it?
$endgroup$
– Rocco van Vreumingen
Jan 16 at 17:49
$begingroup$
Well, this theorem is a "only if" form, so one implication may be true without these conditions.
$endgroup$
– Rocco van Vreumingen
Jan 16 at 18:53
add a comment |
$begingroup$
The following problem looks simple, but it seems there is more needed than just using the definition of ergodicity.
Let $(X, mathcal{F}, mu, T)$ and $(X, mathcal{F}, nu, T)$ be two measure preserving dynamical systems. Show
that if $muneq nu$, then the measure $rho := frac{1}{2}(mu+nu)$ is not ergodic for $T$.
These are my ideas:
Suppose $T$ is ergodic. Since $muneq nu$, there is a set $Ain mathcal{F}: mu(A)neq nu(A)$, so wlog we can assume $mu(A)<rho(A)<nu(A)$, whence it follows that $rho(A)notinlbrace 0,1rbrace$ and hence can't be $T$-invariant. It seems like we must use that $T$-invariant sets form a $sigma$-algebra.
Is there anyone with further ideas?
ergodic-theory
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
$endgroup$
The following problem looks simple, but it seems there is more needed than just using the definition of ergodicity.
Let $(X, mathcal{F}, mu, T)$ and $(X, mathcal{F}, nu, T)$ be two measure preserving dynamical systems. Show
that if $muneq nu$, then the measure $rho := frac{1}{2}(mu+nu)$ is not ergodic for $T$.
These are my ideas:
Suppose $T$ is ergodic. Since $muneq nu$, there is a set $Ain mathcal{F}: mu(A)neq nu(A)$, so wlog we can assume $mu(A)<rho(A)<nu(A)$, whence it follows that $rho(A)notinlbrace 0,1rbrace$ and hence can't be $T$-invariant. It seems like we must use that $T$-invariant sets form a $sigma$-algebra.
Is there anyone with further ideas?
ergodic-theory
ergodic-theory
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
asked Jan 15 at 19:48
Rocco van VreumingenRocco van Vreumingen
928
928
$begingroup$
If you're looking for a good book I'd suggest "Ergodic Theory: with a view towards Number Theory" by Einsiedler and Ward. I recall seeing stronger result in there.
$endgroup$
– BigbearZzz
Jan 15 at 20:05
1
$begingroup$
I already have a good book with a.o. Knopp's lemma, mixing theorems and much more ergodic theorems. There is no link, actually it is a syllabus.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 20:32
2
$begingroup$
It can't be an extreme point and so it can't be ergodic.
$endgroup$
– John B
Jan 16 at 14:00
$begingroup$
That is only true for if $T$ is continuous on a compact metric space isn't it?
$endgroup$
– Rocco van Vreumingen
Jan 16 at 17:49
$begingroup$
Well, this theorem is a "only if" form, so one implication may be true without these conditions.
$endgroup$
– Rocco van Vreumingen
Jan 16 at 18:53
add a comment |
$begingroup$
If you're looking for a good book I'd suggest "Ergodic Theory: with a view towards Number Theory" by Einsiedler and Ward. I recall seeing stronger result in there.
$endgroup$
– BigbearZzz
Jan 15 at 20:05
1
$begingroup$
I already have a good book with a.o. Knopp's lemma, mixing theorems and much more ergodic theorems. There is no link, actually it is a syllabus.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 20:32
2
$begingroup$
It can't be an extreme point and so it can't be ergodic.
$endgroup$
– John B
Jan 16 at 14:00
$begingroup$
That is only true for if $T$ is continuous on a compact metric space isn't it?
$endgroup$
– Rocco van Vreumingen
Jan 16 at 17:49
$begingroup$
Well, this theorem is a "only if" form, so one implication may be true without these conditions.
$endgroup$
– Rocco van Vreumingen
Jan 16 at 18:53
$begingroup$
If you're looking for a good book I'd suggest "Ergodic Theory: with a view towards Number Theory" by Einsiedler and Ward. I recall seeing stronger result in there.
$endgroup$
– BigbearZzz
Jan 15 at 20:05
$begingroup$
If you're looking for a good book I'd suggest "Ergodic Theory: with a view towards Number Theory" by Einsiedler and Ward. I recall seeing stronger result in there.
$endgroup$
– BigbearZzz
Jan 15 at 20:05
1
1
$begingroup$
I already have a good book with a.o. Knopp's lemma, mixing theorems and much more ergodic theorems. There is no link, actually it is a syllabus.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 20:32
$begingroup$
I already have a good book with a.o. Knopp's lemma, mixing theorems and much more ergodic theorems. There is no link, actually it is a syllabus.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 20:32
2
2
$begingroup$
It can't be an extreme point and so it can't be ergodic.
$endgroup$
– John B
Jan 16 at 14:00
$begingroup$
It can't be an extreme point and so it can't be ergodic.
$endgroup$
– John B
Jan 16 at 14:00
$begingroup$
That is only true for if $T$ is continuous on a compact metric space isn't it?
$endgroup$
– Rocco van Vreumingen
Jan 16 at 17:49
$begingroup$
That is only true for if $T$ is continuous on a compact metric space isn't it?
$endgroup$
– Rocco van Vreumingen
Jan 16 at 17:49
$begingroup$
Well, this theorem is a "only if" form, so one implication may be true without these conditions.
$endgroup$
– Rocco van Vreumingen
Jan 16 at 18:53
$begingroup$
Well, this theorem is a "only if" form, so one implication may be true without these conditions.
$endgroup$
– Rocco van Vreumingen
Jan 16 at 18:53
add a comment |
1 Answer
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$begingroup$
Below is an argument based on the pointwise ergodic theorem. There is also a direct argument without using the ergodic theorem, which you can find (if I remember correctly) in the book of Walters, but I find the following to be more conceptual.
Suppose that $rho$ is ergodic for $T$. This implies that both $mu$ and $nu$ are ergodic as well. Let $Bsubseteq X$ be a measurable set, and consider its characteristic function $1_B(cdot)$. By the ergodic theorem, there exists a measurable set $Ysubseteq X$ with $rho(Y)=1$ such that, for each $xin Y$, the ergodic average
begin{align}
overline{1_B}(x) &:=
lim_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}1_B(T^k x)
end{align}
exists and satisfies $overline{1_B}(x)=int1_B,mathrm{d}rho=rho(B)$.
Similarly, there exists a measurable set $Y_musubseteq X$ with $mu(Y_mu)=1$ such that $overline{1_B}(x)=int1_B,mathrm{d}mu=mu(B)$ for each $xin Y_mu$.
But $rho(Y)=1$ implies $mu(Y)=1$, thus $mu(Ycap Y_mu)=1$. In particular, $Ycap Y_muneqvarnothing$. Take $xin Ycap Y_mu$. We have $rho(B)=overline{1_B}(x)=mu(B)$. It follows that $rho=mu=nu$.
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
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$begingroup$
Below is an argument based on the pointwise ergodic theorem. There is also a direct argument without using the ergodic theorem, which you can find (if I remember correctly) in the book of Walters, but I find the following to be more conceptual.
Suppose that $rho$ is ergodic for $T$. This implies that both $mu$ and $nu$ are ergodic as well. Let $Bsubseteq X$ be a measurable set, and consider its characteristic function $1_B(cdot)$. By the ergodic theorem, there exists a measurable set $Ysubseteq X$ with $rho(Y)=1$ such that, for each $xin Y$, the ergodic average
begin{align}
overline{1_B}(x) &:=
lim_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}1_B(T^k x)
end{align}
exists and satisfies $overline{1_B}(x)=int1_B,mathrm{d}rho=rho(B)$.
Similarly, there exists a measurable set $Y_musubseteq X$ with $mu(Y_mu)=1$ such that $overline{1_B}(x)=int1_B,mathrm{d}mu=mu(B)$ for each $xin Y_mu$.
But $rho(Y)=1$ implies $mu(Y)=1$, thus $mu(Ycap Y_mu)=1$. In particular, $Ycap Y_muneqvarnothing$. Take $xin Ycap Y_mu$. We have $rho(B)=overline{1_B}(x)=mu(B)$. It follows that $rho=mu=nu$.
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
$endgroup$
add a comment |
$begingroup$
Below is an argument based on the pointwise ergodic theorem. There is also a direct argument without using the ergodic theorem, which you can find (if I remember correctly) in the book of Walters, but I find the following to be more conceptual.
Suppose that $rho$ is ergodic for $T$. This implies that both $mu$ and $nu$ are ergodic as well. Let $Bsubseteq X$ be a measurable set, and consider its characteristic function $1_B(cdot)$. By the ergodic theorem, there exists a measurable set $Ysubseteq X$ with $rho(Y)=1$ such that, for each $xin Y$, the ergodic average
begin{align}
overline{1_B}(x) &:=
lim_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}1_B(T^k x)
end{align}
exists and satisfies $overline{1_B}(x)=int1_B,mathrm{d}rho=rho(B)$.
Similarly, there exists a measurable set $Y_musubseteq X$ with $mu(Y_mu)=1$ such that $overline{1_B}(x)=int1_B,mathrm{d}mu=mu(B)$ for each $xin Y_mu$.
But $rho(Y)=1$ implies $mu(Y)=1$, thus $mu(Ycap Y_mu)=1$. In particular, $Ycap Y_muneqvarnothing$. Take $xin Ycap Y_mu$. We have $rho(B)=overline{1_B}(x)=mu(B)$. It follows that $rho=mu=nu$.
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
$endgroup$
add a comment |
$begingroup$
Below is an argument based on the pointwise ergodic theorem. There is also a direct argument without using the ergodic theorem, which you can find (if I remember correctly) in the book of Walters, but I find the following to be more conceptual.
Suppose that $rho$ is ergodic for $T$. This implies that both $mu$ and $nu$ are ergodic as well. Let $Bsubseteq X$ be a measurable set, and consider its characteristic function $1_B(cdot)$. By the ergodic theorem, there exists a measurable set $Ysubseteq X$ with $rho(Y)=1$ such that, for each $xin Y$, the ergodic average
begin{align}
overline{1_B}(x) &:=
lim_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}1_B(T^k x)
end{align}
exists and satisfies $overline{1_B}(x)=int1_B,mathrm{d}rho=rho(B)$.
Similarly, there exists a measurable set $Y_musubseteq X$ with $mu(Y_mu)=1$ such that $overline{1_B}(x)=int1_B,mathrm{d}mu=mu(B)$ for each $xin Y_mu$.
But $rho(Y)=1$ implies $mu(Y)=1$, thus $mu(Ycap Y_mu)=1$. In particular, $Ycap Y_muneqvarnothing$. Take $xin Ycap Y_mu$. We have $rho(B)=overline{1_B}(x)=mu(B)$. It follows that $rho=mu=nu$.
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
$endgroup$
Below is an argument based on the pointwise ergodic theorem. There is also a direct argument without using the ergodic theorem, which you can find (if I remember correctly) in the book of Walters, but I find the following to be more conceptual.
Suppose that $rho$ is ergodic for $T$. This implies that both $mu$ and $nu$ are ergodic as well. Let $Bsubseteq X$ be a measurable set, and consider its characteristic function $1_B(cdot)$. By the ergodic theorem, there exists a measurable set $Ysubseteq X$ with $rho(Y)=1$ such that, for each $xin Y$, the ergodic average
begin{align}
overline{1_B}(x) &:=
lim_{ntoinfty}frac{1}{n}sum_{k=0}^{n-1}1_B(T^k x)
end{align}
exists and satisfies $overline{1_B}(x)=int1_B,mathrm{d}rho=rho(B)$.
Similarly, there exists a measurable set $Y_musubseteq X$ with $mu(Y_mu)=1$ such that $overline{1_B}(x)=int1_B,mathrm{d}mu=mu(B)$ for each $xin Y_mu$.
But $rho(Y)=1$ implies $mu(Y)=1$, thus $mu(Ycap Y_mu)=1$. In particular, $Ycap Y_muneqvarnothing$. Take $xin Ycap Y_mu$. We have $rho(B)=overline{1_B}(x)=mu(B)$. It follows that $rho=mu=nu$.
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
answered Jan 17 at 4:58
BlackbirdBlackbird
1,465711
1,465711
add a comment |
add a comment |
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$begingroup$
If you're looking for a good book I'd suggest "Ergodic Theory: with a view towards Number Theory" by Einsiedler and Ward. I recall seeing stronger result in there.
$endgroup$
– BigbearZzz
Jan 15 at 20:05
1
$begingroup$
I already have a good book with a.o. Knopp's lemma, mixing theorems and much more ergodic theorems. There is no link, actually it is a syllabus.
$endgroup$
– Rocco van Vreumingen
Jan 15 at 20:32
2
$begingroup$
It can't be an extreme point and so it can't be ergodic.
$endgroup$
– John B
Jan 16 at 14:00
$begingroup$
That is only true for if $T$ is continuous on a compact metric space isn't it?
$endgroup$
– Rocco van Vreumingen
Jan 16 at 17:49
$begingroup$
Well, this theorem is a "only if" form, so one implication may be true without these conditions.
$endgroup$
– Rocco van Vreumingen
Jan 16 at 18:53