Showing that $f(x)=1$ if $x=frac{1}{n}$, $0$ otherwise on [0,1] is Riemann Integrable












3












$begingroup$


I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:



$$f(x) =
left{
begin{array}{ll}
1 & mbox{if } x = frac{1}{n} \
0 & mbox{otherwise}
end{array}
right.$$



For the upper and lower Riemann sum I am using the following definitions:



$$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$



With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?






riemann-sum







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    3












    $begingroup$


    I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:



    $$f(x) =
    left{
    begin{array}{ll}
    1 & mbox{if } x = frac{1}{n} \
    0 & mbox{otherwise}
    end{array}
    right.$$



    For the upper and lower Riemann sum I am using the following definitions:



    $$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$



    With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?






    riemann-sum







    share|cite|improve this question









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      3












      3








      3


      1



      $begingroup$


      I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:



      $$f(x) =
      left{
      begin{array}{ll}
      1 & mbox{if } x = frac{1}{n} \
      0 & mbox{otherwise}
      end{array}
      right.$$



      For the upper and lower Riemann sum I am using the following definitions:



      $$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$



      With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?






      riemann-sum







      share|cite|improve this question









      $endgroup$




      I have to show that the following function $f:[0,1]rightarrowmathbb{R}$ is Riemann Integrable:



      $$f(x) =
      left{
      begin{array}{ll}
      1 & mbox{if } x = frac{1}{n} \
      0 & mbox{otherwise}
      end{array}
      right.$$



      For the upper and lower Riemann sum I am using the following definitions:



      $$S_{l}(f,V)=sum^{n}_{j=1}inf_{I(j)}(f)(x_j-x_{j-1})$$



      With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V={0,x_1,...,1}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set ${frac{1}{n}:ninmathbb{N}}$. But I can't make the proof concrete. Could anybody help me out?






      riemann-sum




      riemann-sum






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      asked Jun 20 '14 at 13:44









      GehaktmolenGehaktmolen

      17611




      17611






















          2 Answers
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          $begingroup$

          Try the following:



          The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.



          Can you continue from this?






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.



            For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.



            Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.



            Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.



            Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.






            share|cite|improve this answer









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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

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              active

              oldest

              votes









              0












              $begingroup$

              Try the following:



              The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.



              Can you continue from this?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Try the following:



                The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.



                Can you continue from this?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Try the following:



                  The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.



                  Can you continue from this?






                  share|cite|improve this answer









                  $endgroup$



                  Try the following:



                  The set $F={xin [0,1]: f(x)>epsilon }$ is finite for every $epsilon>0$. Then you can form a partition such that if an interval contains some $xin F$ then it have no other. Finally you can choose the partition such that the sum of interval who contains some $xin F$ is $<epsilon$. Separate the interval wich cover $F$ and those which don't.



                  Can you continue from this?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 20 '14 at 14:12









                  YTSYTS

                  2,260827




                  2,260827























                      0












                      $begingroup$

                      $ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.



                      For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.



                      Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.



                      Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.



                      Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.



                        For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.



                        Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.



                        Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.



                        Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.



                          For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.



                          Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.



                          Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.



                          Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.






                          share|cite|improve this answer









                          $endgroup$



                          $ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.



                          For every $epsilon>0$, there exists an integer $N_0$ such that $1/N_0<epsilon$. Let $N=max{N_0,5}$.



                          Partition $[0,1]$ with $V={0,x_1, x_2,ldots,x_{4N-5}}$ where $x_1=frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=frac{1}{N^3}$, and $x_{2k}<frac{1}{2N-k}<x_{2k+1}$.



                          Then, $$S_u(f,V)= 1cdot frac{1}{2N} + frac{1}{N^3} cdot (2N-1)<frac{1}{2N}+frac{2N}{N^3}=frac{1}{2N}+frac{2}{N^2}$$.



                          Since $Nge 5$, $frac{2}{N^2}<frac{1}{2N}$. Therefore, $S_u(f,V)<frac{1}{N}<epsilon$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jun 20 '14 at 14:13









                          BadoeBadoe

                          385111




                          385111






























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