prove $U(P_n,f)−L(P_n,f)$ = $ frac{f(1)−f(0)}{n}$











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Suppose $ f: [0,1]$ -> $R$ is an increasing function. For each $ n in N$ , consider the uniform partition $Pn := ( x_0, x_1,..., x_n )$ of $[0,1]$ where $x_i = i$ for $0≤i≤n$.



Now, I need to prove



a) $U(P_n,f)−L(P_n,f)$ = $ frac{f(1)−f(0)}{n}$



and



b) show $f in R [0, 1]$



I get some idea on part a: $m_i = frac {x_i}{n} $ and $M_i= {x_i}{n}$ , then I am not sure



but for part b, I have no idea (is it something like $epsilon > 0$ )?






real-analysis proof-verification







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    Suppose $ f: [0,1]$ -> $R$ is an increasing function. For each $ n in N$ , consider the uniform partition $Pn := ( x_0, x_1,..., x_n )$ of $[0,1]$ where $x_i = i$ for $0≤i≤n$.



    Now, I need to prove



    a) $U(P_n,f)−L(P_n,f)$ = $ frac{f(1)−f(0)}{n}$



    and



    b) show $f in R [0, 1]$



    I get some idea on part a: $m_i = frac {x_i}{n} $ and $M_i= {x_i}{n}$ , then I am not sure



    but for part b, I have no idea (is it something like $epsilon > 0$ )?






    real-analysis proof-verification







    share|cite|improve this question
























      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      Suppose $ f: [0,1]$ -> $R$ is an increasing function. For each $ n in N$ , consider the uniform partition $Pn := ( x_0, x_1,..., x_n )$ of $[0,1]$ where $x_i = i$ for $0≤i≤n$.



      Now, I need to prove



      a) $U(P_n,f)−L(P_n,f)$ = $ frac{f(1)−f(0)}{n}$



      and



      b) show $f in R [0, 1]$



      I get some idea on part a: $m_i = frac {x_i}{n} $ and $M_i= {x_i}{n}$ , then I am not sure



      but for part b, I have no idea (is it something like $epsilon > 0$ )?






      real-analysis proof-verification







      share|cite|improve this question













      Suppose $ f: [0,1]$ -> $R$ is an increasing function. For each $ n in N$ , consider the uniform partition $Pn := ( x_0, x_1,..., x_n )$ of $[0,1]$ where $x_i = i$ for $0≤i≤n$.



      Now, I need to prove



      a) $U(P_n,f)−L(P_n,f)$ = $ frac{f(1)−f(0)}{n}$



      and



      b) show $f in R [0, 1]$



      I get some idea on part a: $m_i = frac {x_i}{n} $ and $M_i= {x_i}{n}$ , then I am not sure



      but for part b, I have no idea (is it something like $epsilon > 0$ )?






      real-analysis proof-verification




      real-analysis proof-verification






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      asked Nov 22 at 20:05









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          Abridged solution. If $f$ is increasing on $[a, b]$ then $suplimits_{x in [a, b]} f(x) = f(b)$ and $inflimits_{x in [a, b]} f(x) = f(a).$ Q.E.D.



          Addendum. Part (b) follows instantaneously from part (a) if you know the definition of Riemann integral as the equality if the supremum of lower approximations and infimum of upper approximations.






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            Abridged solution. If $f$ is increasing on $[a, b]$ then $suplimits_{x in [a, b]} f(x) = f(b)$ and $inflimits_{x in [a, b]} f(x) = f(a).$ Q.E.D.



            Addendum. Part (b) follows instantaneously from part (a) if you know the definition of Riemann integral as the equality if the supremum of lower approximations and infimum of upper approximations.






            share|cite|improve this answer



























              up vote
              0
              down vote













              Abridged solution. If $f$ is increasing on $[a, b]$ then $suplimits_{x in [a, b]} f(x) = f(b)$ and $inflimits_{x in [a, b]} f(x) = f(a).$ Q.E.D.



              Addendum. Part (b) follows instantaneously from part (a) if you know the definition of Riemann integral as the equality if the supremum of lower approximations and infimum of upper approximations.






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                Abridged solution. If $f$ is increasing on $[a, b]$ then $suplimits_{x in [a, b]} f(x) = f(b)$ and $inflimits_{x in [a, b]} f(x) = f(a).$ Q.E.D.



                Addendum. Part (b) follows instantaneously from part (a) if you know the definition of Riemann integral as the equality if the supremum of lower approximations and infimum of upper approximations.






                share|cite|improve this answer














                Abridged solution. If $f$ is increasing on $[a, b]$ then $suplimits_{x in [a, b]} f(x) = f(b)$ and $inflimits_{x in [a, b]} f(x) = f(a).$ Q.E.D.



                Addendum. Part (b) follows instantaneously from part (a) if you know the definition of Riemann integral as the equality if the supremum of lower approximations and infimum of upper approximations.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 22 at 20:16

























                answered Nov 22 at 20:08









                Will M.

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