Showing there is a unique invariant measure on the unit circumference












1












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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







share|cite|improve this question











$endgroup$












  • $begingroup$
    very interesting exercise. Can I ask where you found it?
    $endgroup$
    – Masacroso
    Dec 1 '18 at 21:45












  • $begingroup$
    The Lebesgue measure of $Omega$ in the plane is zero.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:02












  • $begingroup$
    You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:22










  • $begingroup$
    @Masacroso It was given by a Professor of mine. I think he invented it.
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:31










  • $begingroup$
    @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 $mu(Omega)=1$
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:33


















1












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    very interesting exercise. Can I ask where you found it?
    $endgroup$
    – Masacroso
    Dec 1 '18 at 21:45












  • $begingroup$
    The Lebesgue measure of $Omega$ in the plane is zero.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:02












  • $begingroup$
    You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:22










  • $begingroup$
    @Masacroso It was given by a Professor of mine. I think he invented it.
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:31










  • $begingroup$
    @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 $mu(Omega)=1$
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:33
















1












1








1





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 21:46









Bernard

119k639112




119k639112










asked Dec 1 '18 at 21:37









Pedro GomesPedro Gomes

1,7532721




1,7532721












  • $begingroup$
    very interesting exercise. Can I ask where you found it?
    $endgroup$
    – Masacroso
    Dec 1 '18 at 21:45












  • $begingroup$
    The Lebesgue measure of $Omega$ in the plane is zero.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:02












  • $begingroup$
    You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:22










  • $begingroup$
    @Masacroso It was given by a Professor of mine. I think he invented it.
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:31










  • $begingroup$
    @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 $mu(Omega)=1$
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:33




















  • $begingroup$
    very interesting exercise. Can I ask where you found it?
    $endgroup$
    – Masacroso
    Dec 1 '18 at 21:45












  • $begingroup$
    The Lebesgue measure of $Omega$ in the plane is zero.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:02












  • $begingroup$
    You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
    $endgroup$
    – copper.hat
    Dec 1 '18 at 22:22










  • $begingroup$
    @Masacroso It was given by a Professor of mine. I think he invented it.
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:31










  • $begingroup$
    @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 $mu(Omega)=1$
    $endgroup$
    – Pedro Gomes
    Dec 1 '18 at 22:33


















$begingroup$
very interesting exercise. Can I ask where you found it?
$endgroup$
– Masacroso
Dec 1 '18 at 21:45






$begingroup$
very interesting exercise. Can I ask where you found it?
$endgroup$
– Masacroso
Dec 1 '18 at 21:45














$begingroup$
The Lebesgue measure of $Omega$ in the plane is zero.
$endgroup$
– copper.hat
Dec 1 '18 at 22:02






$begingroup$
The Lebesgue measure of $Omega$ in the plane is zero.
$endgroup$
– copper.hat
Dec 1 '18 at 22:02














$begingroup$
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
$endgroup$
– copper.hat
Dec 1 '18 at 22:22




$begingroup$
You need to correctly define one invariant measure and show that any invariant probability measure must agree on the 'intervals'.
$endgroup$
– copper.hat
Dec 1 '18 at 22:22












$begingroup$
@Masacroso It was given by a Professor of mine. I think he invented it.
$endgroup$
– Pedro Gomes
Dec 1 '18 at 22:31




$begingroup$
@Masacroso It was given by a Professor of mine. I think he invented it.
$endgroup$
– Pedro Gomes
Dec 1 '18 at 22:31












$begingroup$
@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 $mu(Omega)=1$
$endgroup$
– Pedro Gomes
Dec 1 '18 at 22:33






$begingroup$
@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 $mu(Omega)=1$
$endgroup$
– Pedro Gomes
Dec 1 '18 at 22:33












1 Answer
1






active

oldest

votes


















2












$begingroup$

$Omega={(cos theta, sin theta): 0leq theta<2pi}$, 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 $mu(A(B))=frac{Leb(B)}{2pi}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer! What about uniqueness? Have you got any thoughts?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 11:44












  • $begingroup$
    I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
    $endgroup$
    – John_Wick
    Dec 2 '18 at 14:09










  • $begingroup$
    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)?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 16:32










  • $begingroup$
    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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
    $endgroup$
    – John_Wick
    Dec 2 '18 at 17:16











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

$Omega={(cos theta, sin theta): 0leq theta<2pi}$, 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 $mu(A(B))=frac{Leb(B)}{2pi}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer! What about uniqueness? Have you got any thoughts?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 11:44












  • $begingroup$
    I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
    $endgroup$
    – John_Wick
    Dec 2 '18 at 14:09










  • $begingroup$
    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)?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 16:32










  • $begingroup$
    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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
    $endgroup$
    – John_Wick
    Dec 2 '18 at 17:16
















2












$begingroup$

$Omega={(cos theta, sin theta): 0leq theta<2pi}$, 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 $mu(A(B))=frac{Leb(B)}{2pi}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for your answer! What about uniqueness? Have you got any thoughts?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 11:44












  • $begingroup$
    I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
    $endgroup$
    – John_Wick
    Dec 2 '18 at 14:09










  • $begingroup$
    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)?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 16:32










  • $begingroup$
    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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
    $endgroup$
    – John_Wick
    Dec 2 '18 at 17:16














2












2








2





$begingroup$

$Omega={(cos theta, sin theta): 0leq theta<2pi}$, 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 $mu(A(B))=frac{Leb(B)}{2pi}.$






share|cite|improve this answer









$endgroup$



$Omega={(cos theta, sin theta): 0leq theta<2pi}$, 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 $mu(A(B))=frac{Leb(B)}{2pi}.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 1:56









John_WickJohn_Wick

1,486111




1,486111












  • $begingroup$
    Thanks for your answer! What about uniqueness? Have you got any thoughts?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 11:44












  • $begingroup$
    I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
    $endgroup$
    – John_Wick
    Dec 2 '18 at 14:09










  • $begingroup$
    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)?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 16:32










  • $begingroup$
    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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
    $endgroup$
    – John_Wick
    Dec 2 '18 at 17:16


















  • $begingroup$
    Thanks for your answer! What about uniqueness? Have you got any thoughts?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 11:44












  • $begingroup$
    I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
    $endgroup$
    – John_Wick
    Dec 2 '18 at 14:09










  • $begingroup$
    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)?
    $endgroup$
    – Pedro Gomes
    Dec 2 '18 at 16:32










  • $begingroup$
    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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
    $endgroup$
    – John_Wick
    Dec 2 '18 at 17:16
















$begingroup$
Thanks for your answer! What about uniqueness? Have you got any thoughts?
$endgroup$
– Pedro Gomes
Dec 2 '18 at 11:44






$begingroup$
Thanks for your answer! What about uniqueness? Have you got any thoughts?
$endgroup$
– Pedro Gomes
Dec 2 '18 at 11:44














$begingroup$
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
$endgroup$
– John_Wick
Dec 2 '18 at 14:09




$begingroup$
I gave you the proof of uniqueness. If the measure satisfies those conditions then it has to be of that unique form. Look closely.
$endgroup$
– John_Wick
Dec 2 '18 at 14:09












$begingroup$
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)?
$endgroup$
– Pedro Gomes
Dec 2 '18 at 16:32




$begingroup$
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)?
$endgroup$
– Pedro Gomes
Dec 2 '18 at 16:32












$begingroup$
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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
$endgroup$
– John_Wick
Dec 2 '18 at 17:16




$begingroup$
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 $mu(A(B))=frac{2pi x}{2pi}=frac{leb(B)}{2pi}.$
$endgroup$
– John_Wick
Dec 2 '18 at 17:16


















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