Stieltjes integral of proportion function












1












$begingroup$


Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:



$F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$



Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:



(1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?



(2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
$mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?



(3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?



(4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?



I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.






measure-theory stieltjes-integral







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:



    $F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$



    Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:



    (1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?



    (2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
    $mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?



    (3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?



    (4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?



    I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.






    measure-theory stieltjes-integral







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:



      $F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$



      Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:



      (1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?



      (2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
      $mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?



      (3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?



      (4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?



      I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.






      measure-theory stieltjes-integral







      share|cite|improve this question











      $endgroup$




      Given a finite set of real numbers $x_1leq x_2leq ... leq x_n$, define the function:



      $F(x):= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)$



      Considering it's induced Stieltjes measure $mu_F$, I would like to verify (or eliminate) a few properties:



      (1) Is $mu_Fbig( (-infty,x_1) big)=0$ and $mu_Fbig( (x_n,infty) big)=0$?



      (2) Is $mu_F$ supported on ${ x_k }_{k=1}^n$, i.e
      $mu_F(mathbb{R})= underset{k=1}{overset{n}{sum}} mu_F({ x_k })$ ?



      (3) Is $mu_F big( [x_k,x_{k+1}] big)= frac{1}{n} $ for all $kin [n-1]$?



      (4) Is $int_{-infty}^tf(x)dF(x)= frac{1}{n} underset{k=1}{overset{n}{sum}} chi_{(-infty, x_k]}(x)cdot f(x_k)$?



      I am trying to understand integration against this Stieljes measure and I am pretty weak on the subject of Stieltjes integral, so I would appreciate pointers on how to relate to this.






      measure-theory stieltjes-integral




      measure-theory stieltjes-integral






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Oct 24 '18 at 15:43







      Keen-ameteur

















      asked Oct 23 '18 at 14:48









      Keen-ameteurKeen-ameteur

      1,367316




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          1 Answer
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          $begingroup$

          You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:



          begin{align}
          mu_F([c,d))&=F(d-)-F(c-)\
          mu_F((c,d])&=F(d+)-F(c+)\
          mu_F((c,d))&=F(d-)-F(c+)\
          mu_F([c,d])&=F(d+)-F(c-)
          end{align}



          (1) is true. You should try to solve this on your own.



          (2) is also true, actually it holds



          $$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$



          (3) Is not true, since



          $$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$



          (4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that



          $$int_{(-infty, t]}f(x)dF(x)$$
          and
          $$int_{(-infty, t)}f(x)dF(x)$$
          can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)



          Please consider checking my answer. Thanks! :-)






          share|cite|improve this answer











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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes









            1












            $begingroup$

            You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:



            begin{align}
            mu_F([c,d))&=F(d-)-F(c-)\
            mu_F((c,d])&=F(d+)-F(c+)\
            mu_F((c,d))&=F(d-)-F(c+)\
            mu_F([c,d])&=F(d+)-F(c-)
            end{align}



            (1) is true. You should try to solve this on your own.



            (2) is also true, actually it holds



            $$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$



            (3) Is not true, since



            $$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$



            (4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that



            $$int_{(-infty, t]}f(x)dF(x)$$
            and
            $$int_{(-infty, t)}f(x)dF(x)$$
            can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)



            Please consider checking my answer. Thanks! :-)






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:



              begin{align}
              mu_F([c,d))&=F(d-)-F(c-)\
              mu_F((c,d])&=F(d+)-F(c+)\
              mu_F((c,d))&=F(d-)-F(c+)\
              mu_F([c,d])&=F(d+)-F(c-)
              end{align}



              (1) is true. You should try to solve this on your own.



              (2) is also true, actually it holds



              $$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$



              (3) Is not true, since



              $$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$



              (4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that



              $$int_{(-infty, t]}f(x)dF(x)$$
              and
              $$int_{(-infty, t)}f(x)dF(x)$$
              can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)



              Please consider checking my answer. Thanks! :-)






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:



                begin{align}
                mu_F([c,d))&=F(d-)-F(c-)\
                mu_F((c,d])&=F(d+)-F(c+)\
                mu_F((c,d))&=F(d-)-F(c+)\
                mu_F([c,d])&=F(d+)-F(c-)
                end{align}



                (1) is true. You should try to solve this on your own.



                (2) is also true, actually it holds



                $$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$



                (3) Is not true, since



                $$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$



                (4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that



                $$int_{(-infty, t]}f(x)dF(x)$$
                and
                $$int_{(-infty, t)}f(x)dF(x)$$
                can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)



                Please consider checking my answer. Thanks! :-)






                share|cite|improve this answer











                $endgroup$



                You have $mathcal{B}(mathbb{R})= sigma({[c,d):c,dinmathbb{R}, c<d})$. You should make yourself clear, that it holds:



                begin{align}
                mu_F([c,d))&=F(d-)-F(c-)\
                mu_F((c,d])&=F(d+)-F(c+)\
                mu_F((c,d))&=F(d-)-F(c+)\
                mu_F([c,d])&=F(d+)-F(c-)
                end{align}



                (1) is true. You should try to solve this on your own.



                (2) is also true, actually it holds



                $$mu_F=-frac{1}{n}sum_{i=1}^ndelta_{x_k}$$



                (3) Is not true, since



                $$mu_F([x_k,x_{k+1}])=F(x_{k+1}+)-F(x_k-)=frac{1}{n}big[(n-(k+1))-(n-(k-1))]=-frac{2}{n}$$



                (4) Does not make sense. $x$ has to be integrated, so the right hand side should not depend on $x$. Just try to solve the integral, since you know the measure from (2). Note, that



                $$int_{(-infty, t]}f(x)dF(x)$$
                and
                $$int_{(-infty, t)}f(x)dF(x)$$
                can have different values. (What happens, if $t=x_k$ for some $kin{1,ldots, n}$?)



                Please consider checking my answer. Thanks! :-)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 14 at 5:19

























                answered Jan 14 at 4:42









                user408858user408858

                482213




                482213






























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