Is the image of a metric automorphism under a measure-preserving mapping also a metric automorphism?
$begingroup$
I'm having my first tiny little bits of ergodic theory so please forgive the probable naivete of the question.
So, I'm looking at the first page of chapter 8 of Cornfeld, Fomin, Sinai's "Ergodic Theory".
They state:
Now let us show that an arbitrary automorphism $T'$ of the measure space $(M', mathscr{F}, mu)$ naturally gives rise to stationary random processes.
Consider some countable partition $$xi=(C_1,cdots,C_k),1leq kleqinfty$$
For the state space choose $Y=(1,2,cdots,k)$ and put $M=prod_{n=-infty}^{infty}Y^{(n)}$ ; $Y^{(n)}=Y$.
Consider the map $phi:M'rightarrow M$ defined as follows: the $n$th coordinate of the point $phi x'$ equals $j$ if and only if $(T')^nx'in C_j$. The map $phi$ is measurable. It transforms the measure $mu'$ on $M'$ onto a certain measure $mu$ on $M$: $ mu(A)equivmu'(phi^{-1}A), Ainmathscr A$, while the transformation $T'$ is the shift on $M$. From the fact that $T'$ is an automorphism it follows that the random process $(M,mathscr A, mu)$ is stationary (i.e. its measure is shift-ivariant (e.d.)).
Now, I have problems with the bold text: first, $T'$ does not act on $M$, so I guess they mean: "$phi T'phi^{-1}$ is the shift on $M$"; then, I can't see how the fact that $T'$ preserves the measure $mu'$ of $M'$ should guarantee me that $phi T'phi^{-1}$ preserves the $M$-measure $mu$.
I actually tried to sketch a proof:
$$mu(B) = mu'(phi^{-1}(B))= mu'(T'phi^{-1}(B))$$
And also:
$$mu(phi T'phi^{-1}B) = mu'(phi^{-1}phi T'phi^{-1}B)$$
Now, if I got things right I would want the two to be equal; yet, it seems to me that this would be the case only if $phi$ itself was an isomorphism, so that $phi^{-1}phiequiv id_{M'}$.
What am I missing here?
Thanks in advance!
EDIT: I also found the same statement in Sinai's "Metric entropy of dynamical systems":
Take a finite partition $xi = {C_1 , C_2 , cdots , C_r }$ of $M$ . It generates a stationary random
process of probability theory with values $1, 2, cdots , r$ if one uses the formula:
$w_k(x) = jquad$ if $quad x ∈ T^{−k} C_j , −∞ < k < ∞ .$
...just to be more confident that the statement is, in itself, correct!
measure-theory dynamical-systems ergodic-theory
asked Jan 23 at 19:49
Rocco Rocco
15410
$endgroup$
add a comment |
$begingroup$
I'm having my first tiny little bits of ergodic theory so please forgive the probable naivete of the question.
So, I'm looking at the first page of chapter 8 of Cornfeld, Fomin, Sinai's "Ergodic Theory".
They state:
Now let us show that an arbitrary automorphism $T'$ of the measure space $(M', mathscr{F}, mu)$ naturally gives rise to stationary random processes.
Consider some countable partition $$xi=(C_1,cdots,C_k),1leq kleqinfty$$
For the state space choose $Y=(1,2,cdots,k)$ and put $M=prod_{n=-infty}^{infty}Y^{(n)}$ ; $Y^{(n)}=Y$.
Consider the map $phi:M'rightarrow M$ defined as follows: the $n$th coordinate of the point $phi x'$ equals $j$ if and only if $(T')^nx'in C_j$. The map $phi$ is measurable. It transforms the measure $mu'$ on $M'$ onto a certain measure $mu$ on $M$: $ mu(A)equivmu'(phi^{-1}A), Ainmathscr A$, while the transformation $T'$ is the shift on $M$. From the fact that $T'$ is an automorphism it follows that the random process $(M,mathscr A, mu)$ is stationary (i.e. its measure is shift-ivariant (e.d.)).
Now, I have problems with the bold text: first, $T'$ does not act on $M$, so I guess they mean: "$phi T'phi^{-1}$ is the shift on $M$"; then, I can't see how the fact that $T'$ preserves the measure $mu'$ of $M'$ should guarantee me that $phi T'phi^{-1}$ preserves the $M$-measure $mu$.
I actually tried to sketch a proof:
$$mu(B) = mu'(phi^{-1}(B))= mu'(T'phi^{-1}(B))$$
And also:
$$mu(phi T'phi^{-1}B) = mu'(phi^{-1}phi T'phi^{-1}B)$$
Now, if I got things right I would want the two to be equal; yet, it seems to me that this would be the case only if $phi$ itself was an isomorphism, so that $phi^{-1}phiequiv id_{M'}$.
What am I missing here?
Thanks in advance!
EDIT: I also found the same statement in Sinai's "Metric entropy of dynamical systems":
Take a finite partition $xi = {C_1 , C_2 , cdots , C_r }$ of $M$ . It generates a stationary random
process of probability theory with values $1, 2, cdots , r$ if one uses the formula:
$w_k(x) = jquad$ if $quad x ∈ T^{−k} C_j , −∞ < k < ∞ .$
...just to be more confident that the statement is, in itself, correct!
measure-theory dynamical-systems ergodic-theory
asked Jan 23 at 19:49
Rocco Rocco
15410
$endgroup$
add a comment |
$begingroup$
I'm having my first tiny little bits of ergodic theory so please forgive the probable naivete of the question.
So, I'm looking at the first page of chapter 8 of Cornfeld, Fomin, Sinai's "Ergodic Theory".
They state:
Now let us show that an arbitrary automorphism $T'$ of the measure space $(M', mathscr{F}, mu)$ naturally gives rise to stationary random processes.
Consider some countable partition $$xi=(C_1,cdots,C_k),1leq kleqinfty$$
For the state space choose $Y=(1,2,cdots,k)$ and put $M=prod_{n=-infty}^{infty}Y^{(n)}$ ; $Y^{(n)}=Y$.
Consider the map $phi:M'rightarrow M$ defined as follows: the $n$th coordinate of the point $phi x'$ equals $j$ if and only if $(T')^nx'in C_j$. The map $phi$ is measurable. It transforms the measure $mu'$ on $M'$ onto a certain measure $mu$ on $M$: $ mu(A)equivmu'(phi^{-1}A), Ainmathscr A$, while the transformation $T'$ is the shift on $M$. From the fact that $T'$ is an automorphism it follows that the random process $(M,mathscr A, mu)$ is stationary (i.e. its measure is shift-ivariant (e.d.)).
Now, I have problems with the bold text: first, $T'$ does not act on $M$, so I guess they mean: "$phi T'phi^{-1}$ is the shift on $M$"; then, I can't see how the fact that $T'$ preserves the measure $mu'$ of $M'$ should guarantee me that $phi T'phi^{-1}$ preserves the $M$-measure $mu$.
I actually tried to sketch a proof:
$$mu(B) = mu'(phi^{-1}(B))= mu'(T'phi^{-1}(B))$$
And also:
$$mu(phi T'phi^{-1}B) = mu'(phi^{-1}phi T'phi^{-1}B)$$
Now, if I got things right I would want the two to be equal; yet, it seems to me that this would be the case only if $phi$ itself was an isomorphism, so that $phi^{-1}phiequiv id_{M'}$.
What am I missing here?
Thanks in advance!
EDIT: I also found the same statement in Sinai's "Metric entropy of dynamical systems":
Take a finite partition $xi = {C_1 , C_2 , cdots , C_r }$ of $M$ . It generates a stationary random
process of probability theory with values $1, 2, cdots , r$ if one uses the formula:
$w_k(x) = jquad$ if $quad x ∈ T^{−k} C_j , −∞ < k < ∞ .$
...just to be more confident that the statement is, in itself, correct!
measure-theory dynamical-systems ergodic-theory
asked Jan 23 at 19:49
Rocco Rocco
15410
$endgroup$
I'm having my first tiny little bits of ergodic theory so please forgive the probable naivete of the question.
So, I'm looking at the first page of chapter 8 of Cornfeld, Fomin, Sinai's "Ergodic Theory".
They state:
Now let us show that an arbitrary automorphism $T'$ of the measure space $(M', mathscr{F}, mu)$ naturally gives rise to stationary random processes.
Consider some countable partition $$xi=(C_1,cdots,C_k),1leq kleqinfty$$
For the state space choose $Y=(1,2,cdots,k)$ and put $M=prod_{n=-infty}^{infty}Y^{(n)}$ ; $Y^{(n)}=Y$.
Consider the map $phi:M'rightarrow M$ defined as follows: the $n$th coordinate of the point $phi x'$ equals $j$ if and only if $(T')^nx'in C_j$. The map $phi$ is measurable. It transforms the measure $mu'$ on $M'$ onto a certain measure $mu$ on $M$: $ mu(A)equivmu'(phi^{-1}A), Ainmathscr A$, while the transformation $T'$ is the shift on $M$. From the fact that $T'$ is an automorphism it follows that the random process $(M,mathscr A, mu)$ is stationary (i.e. its measure is shift-ivariant (e.d.)).
Now, I have problems with the bold text: first, $T'$ does not act on $M$, so I guess they mean: "$phi T'phi^{-1}$ is the shift on $M$"; then, I can't see how the fact that $T'$ preserves the measure $mu'$ of $M'$ should guarantee me that $phi T'phi^{-1}$ preserves the $M$-measure $mu$.
I actually tried to sketch a proof:
$$mu(B) = mu'(phi^{-1}(B))= mu'(T'phi^{-1}(B))$$
And also:
$$mu(phi T'phi^{-1}B) = mu'(phi^{-1}phi T'phi^{-1}B)$$
Now, if I got things right I would want the two to be equal; yet, it seems to me that this would be the case only if $phi$ itself was an isomorphism, so that $phi^{-1}phiequiv id_{M'}$.
What am I missing here?
Thanks in advance!
EDIT: I also found the same statement in Sinai's "Metric entropy of dynamical systems":
Take a finite partition $xi = {C_1 , C_2 , cdots , C_r }$ of $M$ . It generates a stationary random
process of probability theory with values $1, 2, cdots , r$ if one uses the formula:
$w_k(x) = jquad$ if $quad x ∈ T^{−k} C_j , −∞ < k < ∞ .$
...just to be more confident that the statement is, in itself, correct!
measure-theory dynamical-systems ergodic-theory
measure-theory dynamical-systems ergodic-theory
asked Jan 23 at 19:49
Rocco Rocco
15410
asked Jan 23 at 19:49
Rocco Rocco
15410
edited Jan 23 at 22:15
Rocco
asked Jan 23 at 19:49
Rocco Rocco
15410
asked Jan 23 at 19:49
Rocco Rocco
15410
asked Jan 23 at 19:49
Rocco Rocco
15410
15410
add a comment |
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