Problem 3. Page 240. Barry Simon. (Associated Lebesgue-Stieltjes measure )












2












$begingroup$


In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
($M$ is a algebra).



Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.



In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$



Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
For any Baire set $A.$



I have this:
$mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$



Why this sum is $0$ when $1/2not in A?$






real-analysis measure-theory lebesgue-measure stieltjes-integral







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
    ($M$ is a algebra).



    Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
    Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.



    In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$



    Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
    For any Baire set $A.$



    I have this:
    $mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$



    Why this sum is $0$ when $1/2not in A?$






    real-analysis measure-theory lebesgue-measure stieltjes-integral







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
      ($M$ is a algebra).



      Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
      Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.



      In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$



      Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
      For any Baire set $A.$



      I have this:
      $mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$



      Why this sum is $0$ when $1/2not in A?$






      real-analysis measure-theory lebesgue-measure stieltjes-integral







      share|cite|improve this question









      $endgroup$




      In $[0,1]$ let $M$ consists of all finite unions of sets of the form $[c,d)$ where $c,d$ are rational, or $[c,1]$ where $c$ is rational.
      ($M$ is a algebra).



      Given a monotone function $bar{alpha}$, if $([c_j,d_j))_{j=1}^{l}$ is a disjoint finite family of half-open intervals.
      Let $alpha(bigcup_{j=1}^{l} [c_j,d_j) )=sum_{j=1}^{l} (bar{alpha}(d_j)-bar{alpha}(c_j))$ finitely additive.



      In $M$, let $bar{alpha}(x)=0$ if $x<1/2$ and $bar{alpha}(x)=1$ if $xgeq 1/2.$



      Prove that the associated Lebesgue-Stieltjes measure $mu$ has $mu(A)=0$ if $1/2notin A$ and $mu(A)=1$ if $1/2in A$.
      For any Baire set $A.$



      I have this:
      $mu(A)=int_{X} mathcal{X}_{A}d(bar{alpha})=lim_n sum_{j=1}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1})-bar{alpha}(x_{j}))$



      Why this sum is $0$ when $1/2not in A?$






      real-analysis measure-theory lebesgue-measure stieltjes-integral




      real-analysis measure-theory lebesgue-measure stieltjes-integral






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 '18 at 20:57









      eraldcoileraldcoil

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      504211






















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          $begingroup$

          In the partition $bigcup_{j=1}^{n-1}[x_j,x_{j+1})$ of $[0,1]$, there is a specific $i$ with $frac{1}{2}in[x_i,x_{i+1})$. Consider $frac{1}{2}notin A$. Then there is a $Ninmathbb{N}$, such that for all $nge N:$ $[x_i,x_{i+1})subseteq [0,1]backslash A$. This implies that $chi_A(x_i)=0$, such that



          $$mu(A)=lim_{nrightarrowinfty} sum_{j=1\jneq i}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-))$$



          Then it follows for each $jin{1,ldots,n-1}$ with $ jneq i$ that either $x_j,x_{j+1}lefrac{1}{2}$ or $x_j,x_{j+1}>frac{1}{2}$, such that



          $$bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-)=0,$$



          such that $mu(A)=0.$ Note, that you have to consider the left sided limits $bar{alpha}(x_{j+1}-)$ and $bar{alpha}(x_{j}-)$ as in the definition of Stieltjes integral.






          share|cite|improve this answer











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            $begingroup$

            In the partition $bigcup_{j=1}^{n-1}[x_j,x_{j+1})$ of $[0,1]$, there is a specific $i$ with $frac{1}{2}in[x_i,x_{i+1})$. Consider $frac{1}{2}notin A$. Then there is a $Ninmathbb{N}$, such that for all $nge N:$ $[x_i,x_{i+1})subseteq [0,1]backslash A$. This implies that $chi_A(x_i)=0$, such that



            $$mu(A)=lim_{nrightarrowinfty} sum_{j=1\jneq i}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-))$$



            Then it follows for each $jin{1,ldots,n-1}$ with $ jneq i$ that either $x_j,x_{j+1}lefrac{1}{2}$ or $x_j,x_{j+1}>frac{1}{2}$, such that



            $$bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-)=0,$$



            such that $mu(A)=0.$ Note, that you have to consider the left sided limits $bar{alpha}(x_{j+1}-)$ and $bar{alpha}(x_{j}-)$ as in the definition of Stieltjes integral.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              In the partition $bigcup_{j=1}^{n-1}[x_j,x_{j+1})$ of $[0,1]$, there is a specific $i$ with $frac{1}{2}in[x_i,x_{i+1})$. Consider $frac{1}{2}notin A$. Then there is a $Ninmathbb{N}$, such that for all $nge N:$ $[x_i,x_{i+1})subseteq [0,1]backslash A$. This implies that $chi_A(x_i)=0$, such that



              $$mu(A)=lim_{nrightarrowinfty} sum_{j=1\jneq i}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-))$$



              Then it follows for each $jin{1,ldots,n-1}$ with $ jneq i$ that either $x_j,x_{j+1}lefrac{1}{2}$ or $x_j,x_{j+1}>frac{1}{2}$, such that



              $$bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-)=0,$$



              such that $mu(A)=0.$ Note, that you have to consider the left sided limits $bar{alpha}(x_{j+1}-)$ and $bar{alpha}(x_{j}-)$ as in the definition of Stieltjes integral.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                In the partition $bigcup_{j=1}^{n-1}[x_j,x_{j+1})$ of $[0,1]$, there is a specific $i$ with $frac{1}{2}in[x_i,x_{i+1})$. Consider $frac{1}{2}notin A$. Then there is a $Ninmathbb{N}$, such that for all $nge N:$ $[x_i,x_{i+1})subseteq [0,1]backslash A$. This implies that $chi_A(x_i)=0$, such that



                $$mu(A)=lim_{nrightarrowinfty} sum_{j=1\jneq i}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-))$$



                Then it follows for each $jin{1,ldots,n-1}$ with $ jneq i$ that either $x_j,x_{j+1}lefrac{1}{2}$ or $x_j,x_{j+1}>frac{1}{2}$, such that



                $$bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-)=0,$$



                such that $mu(A)=0.$ Note, that you have to consider the left sided limits $bar{alpha}(x_{j+1}-)$ and $bar{alpha}(x_{j}-)$ as in the definition of Stieltjes integral.






                share|cite|improve this answer











                $endgroup$



                In the partition $bigcup_{j=1}^{n-1}[x_j,x_{j+1})$ of $[0,1]$, there is a specific $i$ with $frac{1}{2}in[x_i,x_{i+1})$. Consider $frac{1}{2}notin A$. Then there is a $Ninmathbb{N}$, such that for all $nge N:$ $[x_i,x_{i+1})subseteq [0,1]backslash A$. This implies that $chi_A(x_i)=0$, such that



                $$mu(A)=lim_{nrightarrowinfty} sum_{j=1\jneq i}^{n-1} mathcal{X}_{A}(x_j)(bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-))$$



                Then it follows for each $jin{1,ldots,n-1}$ with $ jneq i$ that either $x_j,x_{j+1}lefrac{1}{2}$ or $x_j,x_{j+1}>frac{1}{2}$, such that



                $$bar{alpha}(x_{j+1}-)-bar{alpha}(x_{j}-)=0,$$



                such that $mu(A)=0.$ Note, that you have to consider the left sided limits $bar{alpha}(x_{j+1}-)$ and $bar{alpha}(x_{j}-)$ as in the definition of Stieltjes integral.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 13 at 0:11

























                answered Jan 13 at 0:01









                user408858user408858

                479213




                479213






























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