$mu$ pure point measure if and only if $mu(A)=sum_{xin A} mu(left{xright})$











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$mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
$mu(A)=sum_{xin A} mu(left{xright})$



I have this



If $mu(A)=0$ is hold.



If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...






real-analysis functional-analysis measure-theory







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    up vote
    0
    down vote

    favorite












    $mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
    $mu(A)=sum_{xin A} mu(left{xright})$



    I have this



    If $mu(A)=0$ is hold.



    If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
    Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...






    real-analysis functional-analysis measure-theory







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
      $mu(A)=sum_{xin A} mu(left{xright})$



      I have this



      If $mu(A)=0$ is hold.



      If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
      Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...






      real-analysis functional-analysis measure-theory







      share|cite|improve this question













      $mu$ is a pure point measure if and only if for every $Ain mathcal{B}(X)$, $mathcal{B}(X)$ sigma-algebra Baire. $X$ compact hausdorff
      $mu(A)=sum_{xin A} mu(left{xright})$



      I have this



      If $mu(A)=0$ is hold.



      If $mu(A)>0$, then exists $xin A: mu(left{xright})>0$
      Then $left{x:mu(A)>0right}=bigcup_{xin A} left{x:mu(left{xright})>0right}$ and I do not know how to continue ...






      real-analysis functional-analysis measure-theory




      real-analysis functional-analysis measure-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 17 at 3:10









      eraldcoil

      27619




      27619






















          1 Answer
          1






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          I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.






          share|cite|improve this answer

















          • 1




            Presumably it would be sufficient for $mu$ to be decomposable.
            – copper.hat
            Nov 19 at 0:27










          • thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
            – eraldcoil
            Nov 19 at 2:00












          • How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
            – eraldcoil
            Nov 19 at 2:22






          • 1




            @eraldcoil Yes, that is correct.
            – Kavi Rama Murthy
            Nov 19 at 5:09











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          1 Answer
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          1 Answer
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          active

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          active

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          up vote
          2
          down vote



          accepted










          I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.






          share|cite|improve this answer

















          • 1




            Presumably it would be sufficient for $mu$ to be decomposable.
            – copper.hat
            Nov 19 at 0:27










          • thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
            – eraldcoil
            Nov 19 at 2:00












          • How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
            – eraldcoil
            Nov 19 at 2:22






          • 1




            @eraldcoil Yes, that is correct.
            – Kavi Rama Murthy
            Nov 19 at 5:09















          up vote
          2
          down vote



          accepted










          I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.






          share|cite|improve this answer

















          • 1




            Presumably it would be sufficient for $mu$ to be decomposable.
            – copper.hat
            Nov 19 at 0:27










          • thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
            – eraldcoil
            Nov 19 at 2:00












          • How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
            – eraldcoil
            Nov 19 at 2:22






          • 1




            @eraldcoil Yes, that is correct.
            – Kavi Rama Murthy
            Nov 19 at 5:09













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.






          share|cite|improve this answer












          I think $mu$ is supposed to be a finite measure. ''If" part: ${x:mu{x}>frac 1 n}$ is finite for each $n$ and hence ${x:mu{x}>0}$ is an at most countable set , say ${x_1,x_2,...}$. Verify now that $mu =sum_n mu{x_n}delta_{x_n}$. (You need the fact that $mu{x}=0$ if $x notin {x_1,x_2,...}$). "Only if" part is obvious.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 17 at 5:07









          Kavi Rama Murthy

          43.9k31852




          43.9k31852








          • 1




            Presumably it would be sufficient for $mu$ to be decomposable.
            – copper.hat
            Nov 19 at 0:27










          • thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
            – eraldcoil
            Nov 19 at 2:00












          • How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
            – eraldcoil
            Nov 19 at 2:22






          • 1




            @eraldcoil Yes, that is correct.
            – Kavi Rama Murthy
            Nov 19 at 5:09














          • 1




            Presumably it would be sufficient for $mu$ to be decomposable.
            – copper.hat
            Nov 19 at 0:27










          • thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
            – eraldcoil
            Nov 19 at 2:00












          • How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
            – eraldcoil
            Nov 19 at 2:22






          • 1




            @eraldcoil Yes, that is correct.
            – Kavi Rama Murthy
            Nov 19 at 5:09








          1




          1




          Presumably it would be sufficient for $mu$ to be decomposable.
          – copper.hat
          Nov 19 at 0:27




          Presumably it would be sufficient for $mu$ to be decomposable.
          – copper.hat
          Nov 19 at 0:27












          thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
          – eraldcoil
          Nov 19 at 2:00






          thanks. I have this. Let $B=left{xin A: mu(left{xright})>0right}.$ In math.stackexchange.com/questions/3001899/… $left{x:mu(left{xright})>0right}$ is countable. Now, $mu(A)=mu(Asetminus B)+mu(B)=mu(Asetminus B)+sum_{xin B} mu(left{xright})$, and $Asetminus B=left{xin A:mu(left{xright})=0right}$ How prove $mu(Asetminus B)=0$?
          – eraldcoil
          Nov 19 at 2:00














          How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
          – eraldcoil
          Nov 19 at 2:22




          How prove $mu(Asetminus B)=0$? I have this. Sup. $mu(Asetminus B)>0$ then exists $xin Asetminus B$ such that $mu(left{xright})>0$ a contradiction. It is correct?
          – eraldcoil
          Nov 19 at 2:22




          1




          1




          @eraldcoil Yes, that is correct.
          – Kavi Rama Murthy
          Nov 19 at 5:09




          @eraldcoil Yes, that is correct.
          – Kavi Rama Murthy
          Nov 19 at 5:09


















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