Are $sigma$-finite measures agreeing on a generating set equal?











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Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?



Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.










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  • Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
    – mathworker21
    Dec 1 at 13:58










  • Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
    – Bar Alon
    Dec 1 at 14:23















up vote
0
down vote

favorite
1












Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?



Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.










share|cite|improve this question
























  • Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
    – mathworker21
    Dec 1 at 13:58










  • Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
    – Bar Alon
    Dec 1 at 14:23













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0
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Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?



Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.










share|cite|improve this question















Suppose I have two measures $m,n$ and that they are both $sigma$-finite and agree on some semi-ring $R$ generating their $sigma$-algebras. Must these two measures then agree on the entire $sigma$-algebra?



Of course, the Caratheodory extension theorem states that if the two pre-measures given by the restriction of $m,n$ to $R$ are themselves $sigma$-finite, then their extension to the $sigma$-algebra is unique, but I suspect there might be a case where the restriction of a $sigma$-finite measure to the semi-ring might not remain $sigma$-finite.







real-analysis measure-theory






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edited 2 days ago

























asked Dec 1 at 13:53









Bar Alon

467114




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  • Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
    – mathworker21
    Dec 1 at 13:58










  • Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
    – Bar Alon
    Dec 1 at 14:23


















  • Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
    – mathworker21
    Dec 1 at 13:58










  • Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
    – Bar Alon
    Dec 1 at 14:23
















Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
– mathworker21
Dec 1 at 13:58




Let $mathcal{B}$ be the $sigma$-algebra. Then ${A in mathcal{B} : m(A) = n(A)}$ is a sigma-algebra containing $R$ and is thus $mathcal{B}$.
– mathworker21
Dec 1 at 13:58












Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
– Bar Alon
Dec 1 at 14:23




Don't we run into a problem because the measures may not agree on complements of sets of infinite measure?
– Bar Alon
Dec 1 at 14:23










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It turns out that the answer is no.



Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
$$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.



On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.



Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.






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    1 Answer
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    up vote
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    down vote



    accepted










    It turns out that the answer is no.



    Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
    $$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
    Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.



    On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.



    Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      It turns out that the answer is no.



      Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
      $$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
      Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.



      On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.



      Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        It turns out that the answer is no.



        Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
        $$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
        Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.



        On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.



        Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.






        share|cite|improve this answer












        It turns out that the answer is no.



        Consider the two measures $m,n$ defined on the borel $sigma$-algebra $mathcal{B}(mathbb{R})$ as follows:
        $$m(A)=|Acapmathbb{Q}|,quad n(A)=|Acap(mathbb{Q}cup{sqrt{2}})|$$
        Since $mathbb{Q}$ is countable, they are both $sigma$-finite. Additionally, they agree on the semi-ring $R$ of all half-open intervals (for any $a,bin mathbb{R}$: $m([a,b))=infty=n([a,b))$), and $R$ indeed generates $mathcal{B}(mathbb{R})$.



        On the other hand $m({sqrt{2}})not =n({sqrt{2}})$ which is clearly a borel set.



        Note that in the above example $m, n$ are indeed no longer $sigma$-finite when restricted to $R$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Bar Alon

        467114




        467114






























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