Can ergodicity be checked by restricting attention to continuous maps?
$begingroup$
The following is well-known:
Theorem. Let $(X, mathcal F, mu)$ be a probability space and $T:Xto X$ be a measure preserving map. Then the following are equivalent:
$bullet$ $T$ is $mu$-ergodic.
$bullet$ If $fin L^2(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
$bullet$ If $fin L^infty(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
I am wondering if we can say more if $X$ has extra structure. In particular, is the following true:
Let $X$ be a compact metric space and $mathcal B$ be its Borel $sigma$-algebra.
Let $mu$ be a Borel probability measure on $X$ and $T:Xto X$ be measure preserving.
Then are the following two statements equivalent:
1) $T$ is $mu$-ergodic.
2) If a continuous map $f:Xto mathbb R$ satisfies that $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
Perhaps one may need to assume that $T$ is continuous and that $mu(U)>0$ for each open set $Usubseteq X$.
measure-theory ergodic-theory
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
$endgroup$
add a comment |
$begingroup$
The following is well-known:
Theorem. Let $(X, mathcal F, mu)$ be a probability space and $T:Xto X$ be a measure preserving map. Then the following are equivalent:
$bullet$ $T$ is $mu$-ergodic.
$bullet$ If $fin L^2(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
$bullet$ If $fin L^infty(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
I am wondering if we can say more if $X$ has extra structure. In particular, is the following true:
Let $X$ be a compact metric space and $mathcal B$ be its Borel $sigma$-algebra.
Let $mu$ be a Borel probability measure on $X$ and $T:Xto X$ be measure preserving.
Then are the following two statements equivalent:
1) $T$ is $mu$-ergodic.
2) If a continuous map $f:Xto mathbb R$ satisfies that $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
Perhaps one may need to assume that $T$ is continuous and that $mu(U)>0$ for each open set $Usubseteq X$.
measure-theory ergodic-theory
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
$endgroup$
add a comment |
$begingroup$
The following is well-known:
Theorem. Let $(X, mathcal F, mu)$ be a probability space and $T:Xto X$ be a measure preserving map. Then the following are equivalent:
$bullet$ $T$ is $mu$-ergodic.
$bullet$ If $fin L^2(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
$bullet$ If $fin L^infty(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
I am wondering if we can say more if $X$ has extra structure. In particular, is the following true:
Let $X$ be a compact metric space and $mathcal B$ be its Borel $sigma$-algebra.
Let $mu$ be a Borel probability measure on $X$ and $T:Xto X$ be measure preserving.
Then are the following two statements equivalent:
1) $T$ is $mu$-ergodic.
2) If a continuous map $f:Xto mathbb R$ satisfies that $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
Perhaps one may need to assume that $T$ is continuous and that $mu(U)>0$ for each open set $Usubseteq X$.
measure-theory ergodic-theory
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
$endgroup$
The following is well-known:
Theorem. Let $(X, mathcal F, mu)$ be a probability space and $T:Xto X$ be a measure preserving map. Then the following are equivalent:
$bullet$ $T$ is $mu$-ergodic.
$bullet$ If $fin L^2(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
$bullet$ If $fin L^infty(X, mu)$ satisfies $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
I am wondering if we can say more if $X$ has extra structure. In particular, is the following true:
Let $X$ be a compact metric space and $mathcal B$ be its Borel $sigma$-algebra.
Let $mu$ be a Borel probability measure on $X$ and $T:Xto X$ be measure preserving.
Then are the following two statements equivalent:
1) $T$ is $mu$-ergodic.
2) If a continuous map $f:Xto mathbb R$ satisfies that $fcirc T=f$ $mu$-a.e., then $f$ is constant $mu$-a.e.
Perhaps one may need to assume that $T$ is continuous and that $mu(U)>0$ for each open set $Usubseteq X$.
measure-theory ergodic-theory
measure-theory ergodic-theory
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
asked Dec 19 '18 at 13:43
caffeinemachinecaffeinemachine
6,66121455
6,66121455
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The following answer has a slight flaw, but it gets the point across. I'll mention and correct the flaw after the answer (editing the answer to make it correct will distract from the main idea).
Consider $X = [-1,1] subseteq mathbb{R}$ and $mu = frac{1}{2}cdottext{Lebesgue}$. Let $T(x) = 1-x$ for $0 le x le 1$ and $Tx = -1-x$ for $-1 le x le 0$. Clearly $T$ is not ergodic. However, there is no continuous $f$ with $f circ T = f$ $mu$-a.e. but with $f$ not constant $mu$-a.e.. Suppose otherwise.
Since $T|_{[-1,0]}$ and $T|_{[0,1]}$ are ergodic and $f|_{[-1,0]}, f|_{[0,1]}$ are $T$-invariant and in $L^1$, $f|_{[-1,0]}$ and $f|_{[0,1]}$ must be $mu$-a.e. constant, which, by continuity, means they are (everywhere) constant. Since $f$ is continuous, these constants must match, which means $f$ is everywhere constant, a contradiction.
The flaw is that $T$ is not well-defined, since $0$ gets mapped to two distinct points. To fix this, have $X = [-1,1] cup {-30}$ and have $T$ be $T(x) = 1-x$ for $0 le x le 1$, $T(x) = -1-x$ for $-1 < x < 0$, $T(-30) = -1, T(-1) = -30$. Then proceed basically the same as before.
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
$endgroup$
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The following answer has a slight flaw, but it gets the point across. I'll mention and correct the flaw after the answer (editing the answer to make it correct will distract from the main idea).
Consider $X = [-1,1] subseteq mathbb{R}$ and $mu = frac{1}{2}cdottext{Lebesgue}$. Let $T(x) = 1-x$ for $0 le x le 1$ and $Tx = -1-x$ for $-1 le x le 0$. Clearly $T$ is not ergodic. However, there is no continuous $f$ with $f circ T = f$ $mu$-a.e. but with $f$ not constant $mu$-a.e.. Suppose otherwise.
Since $T|_{[-1,0]}$ and $T|_{[0,1]}$ are ergodic and $f|_{[-1,0]}, f|_{[0,1]}$ are $T$-invariant and in $L^1$, $f|_{[-1,0]}$ and $f|_{[0,1]}$ must be $mu$-a.e. constant, which, by continuity, means they are (everywhere) constant. Since $f$ is continuous, these constants must match, which means $f$ is everywhere constant, a contradiction.
The flaw is that $T$ is not well-defined, since $0$ gets mapped to two distinct points. To fix this, have $X = [-1,1] cup {-30}$ and have $T$ be $T(x) = 1-x$ for $0 le x le 1$, $T(x) = -1-x$ for $-1 < x < 0$, $T(-30) = -1, T(-1) = -30$. Then proceed basically the same as before.
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
$endgroup$
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
add a comment |
$begingroup$
The following answer has a slight flaw, but it gets the point across. I'll mention and correct the flaw after the answer (editing the answer to make it correct will distract from the main idea).
Consider $X = [-1,1] subseteq mathbb{R}$ and $mu = frac{1}{2}cdottext{Lebesgue}$. Let $T(x) = 1-x$ for $0 le x le 1$ and $Tx = -1-x$ for $-1 le x le 0$. Clearly $T$ is not ergodic. However, there is no continuous $f$ with $f circ T = f$ $mu$-a.e. but with $f$ not constant $mu$-a.e.. Suppose otherwise.
Since $T|_{[-1,0]}$ and $T|_{[0,1]}$ are ergodic and $f|_{[-1,0]}, f|_{[0,1]}$ are $T$-invariant and in $L^1$, $f|_{[-1,0]}$ and $f|_{[0,1]}$ must be $mu$-a.e. constant, which, by continuity, means they are (everywhere) constant. Since $f$ is continuous, these constants must match, which means $f$ is everywhere constant, a contradiction.
The flaw is that $T$ is not well-defined, since $0$ gets mapped to two distinct points. To fix this, have $X = [-1,1] cup {-30}$ and have $T$ be $T(x) = 1-x$ for $0 le x le 1$, $T(x) = -1-x$ for $-1 < x < 0$, $T(-30) = -1, T(-1) = -30$. Then proceed basically the same as before.
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
$endgroup$
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
add a comment |
$begingroup$
The following answer has a slight flaw, but it gets the point across. I'll mention and correct the flaw after the answer (editing the answer to make it correct will distract from the main idea).
Consider $X = [-1,1] subseteq mathbb{R}$ and $mu = frac{1}{2}cdottext{Lebesgue}$. Let $T(x) = 1-x$ for $0 le x le 1$ and $Tx = -1-x$ for $-1 le x le 0$. Clearly $T$ is not ergodic. However, there is no continuous $f$ with $f circ T = f$ $mu$-a.e. but with $f$ not constant $mu$-a.e.. Suppose otherwise.
Since $T|_{[-1,0]}$ and $T|_{[0,1]}$ are ergodic and $f|_{[-1,0]}, f|_{[0,1]}$ are $T$-invariant and in $L^1$, $f|_{[-1,0]}$ and $f|_{[0,1]}$ must be $mu$-a.e. constant, which, by continuity, means they are (everywhere) constant. Since $f$ is continuous, these constants must match, which means $f$ is everywhere constant, a contradiction.
The flaw is that $T$ is not well-defined, since $0$ gets mapped to two distinct points. To fix this, have $X = [-1,1] cup {-30}$ and have $T$ be $T(x) = 1-x$ for $0 le x le 1$, $T(x) = -1-x$ for $-1 < x < 0$, $T(-30) = -1, T(-1) = -30$. Then proceed basically the same as before.
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
$endgroup$
The following answer has a slight flaw, but it gets the point across. I'll mention and correct the flaw after the answer (editing the answer to make it correct will distract from the main idea).
Consider $X = [-1,1] subseteq mathbb{R}$ and $mu = frac{1}{2}cdottext{Lebesgue}$. Let $T(x) = 1-x$ for $0 le x le 1$ and $Tx = -1-x$ for $-1 le x le 0$. Clearly $T$ is not ergodic. However, there is no continuous $f$ with $f circ T = f$ $mu$-a.e. but with $f$ not constant $mu$-a.e.. Suppose otherwise.
Since $T|_{[-1,0]}$ and $T|_{[0,1]}$ are ergodic and $f|_{[-1,0]}, f|_{[0,1]}$ are $T$-invariant and in $L^1$, $f|_{[-1,0]}$ and $f|_{[0,1]}$ must be $mu$-a.e. constant, which, by continuity, means they are (everywhere) constant. Since $f$ is continuous, these constants must match, which means $f$ is everywhere constant, a contradiction.
The flaw is that $T$ is not well-defined, since $0$ gets mapped to two distinct points. To fix this, have $X = [-1,1] cup {-30}$ and have $T$ be $T(x) = 1-x$ for $0 le x le 1$, $T(x) = -1-x$ for $-1 < x < 0$, $T(-30) = -1, T(-1) = -30$. Then proceed basically the same as before.
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
edited Dec 20 '18 at 22:31
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
answered Dec 20 '18 at 22:22
mathworker21mathworker21
9,4561928
9,4561928
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
add a comment |
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
I do not follow how $T|_{[0, 1]}$ is ergodic. I interpret this to mean that the map $Ito I$ taking $x$ to $1-x$ is ergodic (w.r.t the Lebesgue measure). But the set $[0, varepsilon]cup [1-varepsilon, 1]$ is preserved under the inverse image.
$endgroup$
– caffeinemachine
Dec 21 '18 at 9:08
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
@caffeinemachine ah, you're absolutely right, my apologies. However, can't we just choose any ergodic map as $T|_{[0,1]}$? And similarly with $T|_{[-1,0]}$, and then just have $T$ be those two patched together.
$endgroup$
– mathworker21
Dec 21 '18 at 9:34
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
I think you are right. We could define $T:[-1, 1]cup{-30}to [-1, 1]cup{-30}$ as $T(x)=2xpmod{1}$ for $xneq 0, -30$ and $T(0)=-30$, $T(-30)=0$. Then $T$ is not ergodic but any $T$-invariant mao is constant a.e.
$endgroup$
– caffeinemachine
Dec 21 '18 at 12:16
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
$begingroup$
@caffeinemachine I don't see how your $T$ is defined (you're doing mod 1). My point was that we can just take any arbitrary $T_1,T_2$ s.t. $T_1 : [0,1] to [0,1]$ is ergodic and $T_2: {-30}cup[-1,0) to {-30}cup[-1,0)$ are ergodic.
$endgroup$
– mathworker21
Dec 21 '18 at 13:56
add a comment |
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