Fourier uniqueness theorem












1














I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:



Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
$$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.



I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:



Question What is a clean way to prove this?



Any hint/help is highly appreciated.






probability-theory probability-distributions fourier-transform characteristic-functions







share|cite|improve this question



























    1














    I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:



    Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
    $$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
    Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.



    I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:



    Question What is a clean way to prove this?



    Any hint/help is highly appreciated.






    probability-theory probability-distributions fourier-transform characteristic-functions







    share|cite|improve this question

























      1












      1








      1







      I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:



      Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
      $$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
      Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.



      I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:



      Question What is a clean way to prove this?



      Any hint/help is highly appreciated.






      probability-theory probability-distributions fourier-transform characteristic-functions







      share|cite|improve this question













      I am reading Theorem 26.2 in the book "Probability and Measure" (2nd ed.) by Patrick Billingsley. In the proof of that theorem, the author claims that:



      Fact Let $mu$ and $nu$ be probability measures on the Borel subsets of $mathbb{R}$, and set
      $$A=left{;left(a,bright];middle|;mu({a})=mu({b})=nu({a})=nu({b})=0right}.$$
      Then $sigma(A)$ coincides with the Borel sigma-algebra on $mathbb{R}$.



      I have been thinking about this result, but have absolutely no clues on this. It does not seem straightforward. So my question is:



      Question What is a clean way to prove this?



      Any hint/help is highly appreciated.






      probability-theory probability-distributions fourier-transform characteristic-functions




      probability-theory probability-distributions fourier-transform characteristic-functions






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 26 '18 at 6:19









      weirdo

      420210




      420210






















          1 Answer
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          Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.






          share|cite|improve this answer





















          • Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
            – weirdo
            Nov 26 '18 at 7:30






          • 1




            @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
            – Kavi Rama Murthy
            Nov 26 '18 at 7:32












          • I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
            – weirdo
            Nov 26 '18 at 7:59








          • 1




            @weirdo You are right.
            – Kavi Rama Murthy
            Nov 26 '18 at 8:00











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          1 Answer
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          4














          Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.






          share|cite|improve this answer





















          • Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
            – weirdo
            Nov 26 '18 at 7:30






          • 1




            @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
            – Kavi Rama Murthy
            Nov 26 '18 at 7:32












          • I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
            – weirdo
            Nov 26 '18 at 7:59








          • 1




            @weirdo You are right.
            – Kavi Rama Murthy
            Nov 26 '18 at 8:00
















          4














          Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.






          share|cite|improve this answer





















          • Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
            – weirdo
            Nov 26 '18 at 7:30






          • 1




            @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
            – Kavi Rama Murthy
            Nov 26 '18 at 7:32












          • I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
            – weirdo
            Nov 26 '18 at 7:59








          • 1




            @weirdo You are right.
            – Kavi Rama Murthy
            Nov 26 '18 at 8:00














          4












          4








          4






          Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.






          share|cite|improve this answer












          Hint: $mu {x}=nu {x}=0$ for all but countable number of $x$. So there is a countable set ${t_1,t_2,...}$ such that the given class contains all intervals $(a,b]$ whose end points are not in this countable set. Can you express any interval in terms of these using countable unions /intersections? (The complement of any countable set is dense). Will be glad to provide more details if needed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 '18 at 6:28









          Kavi Rama Murthy

          50.7k31854




          50.7k31854












          • Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
            – weirdo
            Nov 26 '18 at 7:30






          • 1




            @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
            – Kavi Rama Murthy
            Nov 26 '18 at 7:32












          • I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
            – weirdo
            Nov 26 '18 at 7:59








          • 1




            @weirdo You are right.
            – Kavi Rama Murthy
            Nov 26 '18 at 8:00


















          • Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
            – weirdo
            Nov 26 '18 at 7:30






          • 1




            @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
            – Kavi Rama Murthy
            Nov 26 '18 at 7:32












          • I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
            – weirdo
            Nov 26 '18 at 7:59








          • 1




            @weirdo You are right.
            – Kavi Rama Murthy
            Nov 26 '18 at 8:00
















          Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
          – weirdo
          Nov 26 '18 at 7:30




          Would you mind explaning why $mu{x}=nu{x}$ for all but countable number of $x$?
          – weirdo
          Nov 26 '18 at 7:30




          1




          1




          @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
          – Kavi Rama Murthy
          Nov 26 '18 at 7:32






          @weirdo There can only be $n$ points $x$ such that $mu {x} >frac 1 n$ because $mu$ is a probability measure. Take union of these finite sets to get a countable set $S$ such that $mu {x}=0$ whenever $x notin S$.
          – Kavi Rama Murthy
          Nov 26 '18 at 7:32














          I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
          – weirdo
          Nov 26 '18 at 7:59






          I just want to make sure I did this correctly. So the end points of intervals in $A$ belong to a dense set. Thus, given $alpha<beta$ in $mathbb{R}$, there exist $alpha_{n} downarrow alpha$ (the sequence belongs to the dense set) and $beta_{n}uparrow beta$ and $alpha_nneqalpha$, $beta_nneqbeta$. So we have $(alpha,beta)=cup (alpha_{n},beta_{n})$. Is this what you meant?
          – weirdo
          Nov 26 '18 at 7:59






          1




          1




          @weirdo You are right.
          – Kavi Rama Murthy
          Nov 26 '18 at 8:00




          @weirdo You are right.
          – Kavi Rama Murthy
          Nov 26 '18 at 8:00


















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