Showing there is a unique invariant measure on the unit circumference
Exercise:Assume that $Omega$ is a circumference of radius $1$ and centred at the origin of $mathbb{R}^2$. Show that there exists a unique measure $mu$ defined on $mathscr{B}_{Omega}$ such that $mu(Omega)=1$ and $mu$ is invariant for all rotations centred at the origin.
I tried to solve the question the following way:
I can define de measure using the Lebesgue measure $mu(A)=frac{lambda(A)}{lambda(Omega)}=frac{lambda(A)}{2pi}$ for $A subseteqOmega$
If $A_j$ is a disjoint sequence of sets so that $bigcup_{jinmathbb{N}}A_j=Omega$
If there were two measures $mu_1$ and $mu_2$ then since by assumption $mu_1(Omega)=1$ and $mu_2(Omega)=1$ then $mu_2(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_2(A_j)=mu_2(Omega)=1=mu_1(Omega)=mu_1(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_1(A_j)$
Since the measure is invariant to the rotations I think it does not matter the measure attributed to each single $A_j$, so what it needs to be assured in order to have uniqueness is that both $mu_1$ and $mu_2$ attribute the same set of values to the different sets. Otherwise if one was the Dirac measure and the other was not the uniqueness would fail.
Question:
How should I solve the problem?
Thanks in advance!
real-analysis measure-theory
edited Dec 1 '18 at 21:46
Bernard
118k639112
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
|
show 2 more comments
Exercise:Assume that $Omega$ is a circumference of radius $1$ and centred at the origin of $mathbb{R}^2$. Show that there exists a unique measure $mu$ defined on $mathscr{B}_{Omega}$ such that $mu(Omega)=1$ and $mu$ is invariant for all rotations centred at the origin.
I tried to solve the question the following way:
I can define de measure using the Lebesgue measure $mu(A)=frac{lambda(A)}{lambda(Omega)}=frac{lambda(A)}{2pi}$ for $A subseteqOmega$
If $A_j$ is a disjoint sequence of sets so that $bigcup_{jinmathbb{N}}A_j=Omega$
If there were two measures $mu_1$ and $mu_2$ then since by assumption $mu_1(Omega)=1$ and $mu_2(Omega)=1$ then $mu_2(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_2(A_j)=mu_2(Omega)=1=mu_1(Omega)=mu_1(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_1(A_j)$
Since the measure is invariant to the rotations I think it does not matter the measure attributed to each single $A_j$, so what it needs to be assured in order to have uniqueness is that both $mu_1$ and $mu_2$ attribute the same set of values to the different sets. Otherwise if one was the Dirac measure and the other was not the uniqueness would fail.
Question:
How should I solve the problem?
Thanks in advance!
real-analysis measure-theory
edited Dec 1 '18 at 21:46
Bernard
118k639112
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
very interesting exercise. Can I ask where you found it?
– Masacroso
Dec 1 '18 at 21:45
The Lebesgue measure of $Omega$ in the plane is zero.
– copper.hat
Dec 1 '18 at 22:02
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
– copper.hat
Dec 1 '18 at 22:22
@Masacroso It was given by a Professor of mine. I think he invented it.
– Pedro Gomes
Dec 1 '18 at 22:31
@copper.hat Could you point me to that proof of the fact an invariant probability measure must agree on the intervals? What I defined is in a certain sense a probability measure since $mu(Omega)=1$
– Pedro Gomes
Dec 1 '18 at 22:33
|
show 2 more comments
Exercise:Assume that $Omega$ is a circumference of radius $1$ and centred at the origin of $mathbb{R}^2$. Show that there exists a unique measure $mu$ defined on $mathscr{B}_{Omega}$ such that $mu(Omega)=1$ and $mu$ is invariant for all rotations centred at the origin.
I tried to solve the question the following way:
I can define de measure using the Lebesgue measure $mu(A)=frac{lambda(A)}{lambda(Omega)}=frac{lambda(A)}{2pi}$ for $A subseteqOmega$
If $A_j$ is a disjoint sequence of sets so that $bigcup_{jinmathbb{N}}A_j=Omega$
If there were two measures $mu_1$ and $mu_2$ then since by assumption $mu_1(Omega)=1$ and $mu_2(Omega)=1$ then $mu_2(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_2(A_j)=mu_2(Omega)=1=mu_1(Omega)=mu_1(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_1(A_j)$
Since the measure is invariant to the rotations I think it does not matter the measure attributed to each single $A_j$, so what it needs to be assured in order to have uniqueness is that both $mu_1$ and $mu_2$ attribute the same set of values to the different sets. Otherwise if one was the Dirac measure and the other was not the uniqueness would fail.
Question:
How should I solve the problem?
Thanks in advance!
real-analysis measure-theory
edited Dec 1 '18 at 21:46
Bernard
118k639112
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
Exercise:Assume that $Omega$ is a circumference of radius $1$ and centred at the origin of $mathbb{R}^2$. Show that there exists a unique measure $mu$ defined on $mathscr{B}_{Omega}$ such that $mu(Omega)=1$ and $mu$ is invariant for all rotations centred at the origin.
I tried to solve the question the following way:
I can define de measure using the Lebesgue measure $mu(A)=frac{lambda(A)}{lambda(Omega)}=frac{lambda(A)}{2pi}$ for $A subseteqOmega$
If $A_j$ is a disjoint sequence of sets so that $bigcup_{jinmathbb{N}}A_j=Omega$
If there were two measures $mu_1$ and $mu_2$ then since by assumption $mu_1(Omega)=1$ and $mu_2(Omega)=1$ then $mu_2(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_2(A_j)=mu_2(Omega)=1=mu_1(Omega)=mu_1(bigcup_{jinmathbb{N}}A_j)=sum_{jinmathbb{N}}mu_1(A_j)$
Since the measure is invariant to the rotations I think it does not matter the measure attributed to each single $A_j$, so what it needs to be assured in order to have uniqueness is that both $mu_1$ and $mu_2$ attribute the same set of values to the different sets. Otherwise if one was the Dirac measure and the other was not the uniqueness would fail.
Question:
How should I solve the problem?
Thanks in advance!
real-analysis measure-theory
real-analysis measure-theory
edited Dec 1 '18 at 21:46
Bernard
118k639112
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
edited Dec 1 '18 at 21:46
Bernard
118k639112
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
edited Dec 1 '18 at 21:46
Bernard
118k639112
edited Dec 1 '18 at 21:46
Bernard
118k639112
edited Dec 1 '18 at 21:46
Bernard
118k639112
118k639112
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
asked Dec 1 '18 at 21:37
Pedro Gomes
1,7012720
1,7012720
very interesting exercise. Can I ask where you found it?
– Masacroso
Dec 1 '18 at 21:45
The Lebesgue measure of $Omega$ in the plane is zero.
– copper.hat
Dec 1 '18 at 22:02
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
– copper.hat
Dec 1 '18 at 22:22
@Masacroso It was given by a Professor of mine. I think he invented it.
– Pedro Gomes
Dec 1 '18 at 22:31
@copper.hat Could you point me to that proof of the fact an invariant probability measure must agree on the intervals? What I defined is in a certain sense a probability measure since $mu(Omega)=1$
– Pedro Gomes
Dec 1 '18 at 22:33
|
show 2 more comments
very interesting exercise. Can I ask where you found it?
– Masacroso
Dec 1 '18 at 21:45
The Lebesgue measure of $Omega$ in the plane is zero.
– copper.hat
Dec 1 '18 at 22:02
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
– copper.hat
Dec 1 '18 at 22:22
@Masacroso It was given by a Professor of mine. I think he invented it.
– Pedro Gomes
Dec 1 '18 at 22:31
@copper.hat Could you point me to that proof of the fact an invariant probability measure must agree on the intervals? What I defined is in a certain sense a probability measure since $mu(Omega)=1$
– Pedro Gomes
Dec 1 '18 at 22:33
very interesting exercise. Can I ask where you found it?
– Masacroso
Dec 1 '18 at 21:45
very interesting exercise. Can I ask where you found it?
– Masacroso
Dec 1 '18 at 21:45
The Lebesgue measure of $Omega$ in the plane is zero.
– copper.hat
Dec 1 '18 at 22:02
The Lebesgue measure of $Omega$ in the plane is zero.
– copper.hat
Dec 1 '18 at 22:02
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
– copper.hat
Dec 1 '18 at 22:22
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
– copper.hat
Dec 1 '18 at 22:22
@Masacroso It was given by a Professor of mine. I think he invented it.
– Pedro Gomes
Dec 1 '18 at 22:31
@Masacroso It was given by a Professor of mine. I think he invented it.
– Pedro Gomes
Dec 1 '18 at 22:31
@copper.hat Could you point me to that proof of the fact an invariant probability measure must agree on the intervals? What I defined is in a certain sense a probability measure since $mu(Omega)=1$
– Pedro Gomes
Dec 1 '18 at 22:33
@copper.hat Could you point me to that proof of the fact an invariant probability measure must agree on the intervals? What I defined is in a certain sense a probability measure since $mu(Omega)=1$
– Pedro Gomes
Dec 1 '18 at 22:33
|
show 2 more comments
1 Answer
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$Omega={(cos theta, sin theta): 0leq theta<2pi}$, and $mathcal{A}={A(B):={(cos theta, sin theta):thetain B}:Binmathcal{B}_{[0,2pi)}}$. For any $2times 2$ orthogonal matrix $P_alpha=left[begin{array}{cc}cos alpha & sin alpha\-sin alpha&cosalphaend{array}right]$ we have $mu circ P_alpha=mu.$ Now $mu(A([0,2pi)))=1$ and hence $1=mu(A([0,2pi)))=sum_{i=1}^n mu(A([2pifrac{i-1}{n},2pifrac{i}{n})))=sum_{i=1}^n mu circ P_{-2pifrac{i-1}{n}} (A([2pifrac{i-1}{n},2pifrac{i}{n})))=n mu(A([0,frac{2pi}{n})))Rightarrow mu(A([0,frac{2pi}{n})))=1/n.$
Now for any interval $Bsubset [0,2pi)$ of length $2pi/n$, we have $mu(A(B))=1/n$ and for any interval $Bsubset [0,2pi)$ of length $2pi m/n$ with $m/n< 1$, we have $mu(A(B))=m/n$. Hence, by continuty of measure for any $0<x<1$ any interval $Bsubset [0,2pi)$ of length $2pi x$ with $xleq 1$, we have $mu(A(B))=x.$ As $mathcal{C}={A(B):Bsubset [0,2pi)text{ is an interval of length }2pi xtext{ with }0leq x< 1}$ is a field which generates $mathcal{A}.$ By extension of measure we get $mu(A(B))=frac{Leb(B)}{2pi}.$
answered Dec 2 '18 at 1:56
John_Wick
1,366111
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
add a comment |
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$Omega={(cos theta, sin theta): 0leq theta<2pi}$, and $mathcal{A}={A(B):={(cos theta, sin theta):thetain B}:Binmathcal{B}_{[0,2pi)}}$. For any $2times 2$ orthogonal matrix $P_alpha=left[begin{array}{cc}cos alpha & sin alpha\-sin alpha&cosalphaend{array}right]$ we have $mu circ P_alpha=mu.$ Now $mu(A([0,2pi)))=1$ and hence $1=mu(A([0,2pi)))=sum_{i=1}^n mu(A([2pifrac{i-1}{n},2pifrac{i}{n})))=sum_{i=1}^n mu circ P_{-2pifrac{i-1}{n}} (A([2pifrac{i-1}{n},2pifrac{i}{n})))=n mu(A([0,frac{2pi}{n})))Rightarrow mu(A([0,frac{2pi}{n})))=1/n.$
Now for any interval $Bsubset [0,2pi)$ of length $2pi/n$, we have $mu(A(B))=1/n$ and for any interval $Bsubset [0,2pi)$ of length $2pi m/n$ with $m/n< 1$, we have $mu(A(B))=m/n$. Hence, by continuty of measure for any $0<x<1$ any interval $Bsubset [0,2pi)$ of length $2pi x$ with $xleq 1$, we have $mu(A(B))=x.$ As $mathcal{C}={A(B):Bsubset [0,2pi)text{ is an interval of length }2pi xtext{ with }0leq x< 1}$ is a field which generates $mathcal{A}.$ By extension of measure we get $mu(A(B))=frac{Leb(B)}{2pi}.$
answered Dec 2 '18 at 1:56
John_Wick
1,366111
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
add a comment |
$Omega={(cos theta, sin theta): 0leq theta<2pi}$, and $mathcal{A}={A(B):={(cos theta, sin theta):thetain B}:Binmathcal{B}_{[0,2pi)}}$. For any $2times 2$ orthogonal matrix $P_alpha=left[begin{array}{cc}cos alpha & sin alpha\-sin alpha&cosalphaend{array}right]$ we have $mu circ P_alpha=mu.$ Now $mu(A([0,2pi)))=1$ and hence $1=mu(A([0,2pi)))=sum_{i=1}^n mu(A([2pifrac{i-1}{n},2pifrac{i}{n})))=sum_{i=1}^n mu circ P_{-2pifrac{i-1}{n}} (A([2pifrac{i-1}{n},2pifrac{i}{n})))=n mu(A([0,frac{2pi}{n})))Rightarrow mu(A([0,frac{2pi}{n})))=1/n.$
Now for any interval $Bsubset [0,2pi)$ of length $2pi/n$, we have $mu(A(B))=1/n$ and for any interval $Bsubset [0,2pi)$ of length $2pi m/n$ with $m/n< 1$, we have $mu(A(B))=m/n$. Hence, by continuty of measure for any $0<x<1$ any interval $Bsubset [0,2pi)$ of length $2pi x$ with $xleq 1$, we have $mu(A(B))=x.$ As $mathcal{C}={A(B):Bsubset [0,2pi)text{ is an interval of length }2pi xtext{ with }0leq x< 1}$ is a field which generates $mathcal{A}.$ By extension of measure we get $mu(A(B))=frac{Leb(B)}{2pi}.$
answered Dec 2 '18 at 1:56
John_Wick
1,366111
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
add a comment |
$Omega={(cos theta, sin theta): 0leq theta<2pi}$, and $mathcal{A}={A(B):={(cos theta, sin theta):thetain B}:Binmathcal{B}_{[0,2pi)}}$. For any $2times 2$ orthogonal matrix $P_alpha=left[begin{array}{cc}cos alpha & sin alpha\-sin alpha&cosalphaend{array}right]$ we have $mu circ P_alpha=mu.$ Now $mu(A([0,2pi)))=1$ and hence $1=mu(A([0,2pi)))=sum_{i=1}^n mu(A([2pifrac{i-1}{n},2pifrac{i}{n})))=sum_{i=1}^n mu circ P_{-2pifrac{i-1}{n}} (A([2pifrac{i-1}{n},2pifrac{i}{n})))=n mu(A([0,frac{2pi}{n})))Rightarrow mu(A([0,frac{2pi}{n})))=1/n.$
Now for any interval $Bsubset [0,2pi)$ of length $2pi/n$, we have $mu(A(B))=1/n$ and for any interval $Bsubset [0,2pi)$ of length $2pi m/n$ with $m/n< 1$, we have $mu(A(B))=m/n$. Hence, by continuty of measure for any $0<x<1$ any interval $Bsubset [0,2pi)$ of length $2pi x$ with $xleq 1$, we have $mu(A(B))=x.$ As $mathcal{C}={A(B):Bsubset [0,2pi)text{ is an interval of length }2pi xtext{ with }0leq x< 1}$ is a field which generates $mathcal{A}.$ By extension of measure we get $mu(A(B))=frac{Leb(B)}{2pi}.$
answered Dec 2 '18 at 1:56
John_Wick
1,366111
$Omega={(cos theta, sin theta): 0leq theta<2pi}$, and $mathcal{A}={A(B):={(cos theta, sin theta):thetain B}:Binmathcal{B}_{[0,2pi)}}$. For any $2times 2$ orthogonal matrix $P_alpha=left[begin{array}{cc}cos alpha & sin alpha\-sin alpha&cosalphaend{array}right]$ we have $mu circ P_alpha=mu.$ Now $mu(A([0,2pi)))=1$ and hence $1=mu(A([0,2pi)))=sum_{i=1}^n mu(A([2pifrac{i-1}{n},2pifrac{i}{n})))=sum_{i=1}^n mu circ P_{-2pifrac{i-1}{n}} (A([2pifrac{i-1}{n},2pifrac{i}{n})))=n mu(A([0,frac{2pi}{n})))Rightarrow mu(A([0,frac{2pi}{n})))=1/n.$
Now for any interval $Bsubset [0,2pi)$ of length $2pi/n$, we have $mu(A(B))=1/n$ and for any interval $Bsubset [0,2pi)$ of length $2pi m/n$ with $m/n< 1$, we have $mu(A(B))=m/n$. Hence, by continuty of measure for any $0<x<1$ any interval $Bsubset [0,2pi)$ of length $2pi x$ with $xleq 1$, we have $mu(A(B))=x.$ As $mathcal{C}={A(B):Bsubset [0,2pi)text{ is an interval of length }2pi xtext{ with }0leq x< 1}$ is a field which generates $mathcal{A}.$ By extension of measure we get $mu(A(B))=frac{Leb(B)}{2pi}.$
answered Dec 2 '18 at 1:56
John_Wick
1,366111
answered Dec 2 '18 at 1:56
John_Wick
1,366111
answered Dec 2 '18 at 1:56
John_Wick
1,366111
answered Dec 2 '18 at 1:56
John_Wick
1,366111
1,366111
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
add a comment |
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
Thanks for your answer! What about uniqueness? Have you got any thoughts?
– Pedro Gomes
Dec 2 '18 at 11:44
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
– John_Wick
Dec 2 '18 at 14:09
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
I am not understanding what you mean by extension? Are you referring to Caratheodory extension theorem? How does that gives you the normalized Lebesgue measure(An extension theorem)?
– Pedro Gomes
Dec 2 '18 at 16:32
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
Look at theorem 3.1 in Billingsley, Probability, and Measure for extension theorem. Also, $mu(A(B))=x$ for an interval $Bsubset [0,2pi)$ of length $2pi x$ implies $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
– John_Wick
Dec 2 '18 at 17:16
add a comment |
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very interesting exercise. Can I ask where you found it?
– Masacroso
Dec 1 '18 at 21:45
The Lebesgue measure of $Omega$ in the plane is zero.
– copper.hat
Dec 1 '18 at 22:02
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
– copper.hat
Dec 1 '18 at 22:22
@Masacroso It was given by a Professor of mine. I think he invented it.
– Pedro Gomes
Dec 1 '18 at 22:31
@copper.hat Could you point me to that proof of the fact an invariant probability measure must agree on the intervals? What I defined is in a certain sense a probability measure since $mu(Omega)=1$
– Pedro Gomes
Dec 1 '18 at 22:33