Showing there is a unique invariant measure on the unit circumference












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












$begingroup$



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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021863%2fshowing-there-is-a-unique-invariant-measure-on-the-unit-circumference%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021863%2fshowing-there-is-a-unique-invariant-measure-on-the-unit-circumference%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Verónica Boquete

Image
Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more