prove $L(f)leq U(f)$












1












$begingroup$


How exactly would I go about proving the following statement?



Given $f:[a,b]tomathbb{R}$ show that $$L(f)leq U(f)$$ where $$L(f)=sup_{Pinmathscr{P}}L(f,P) text{ and } U(f)=inf_{Pinmathscr{P}}U(f,P)$$ where $P$ is any partition of $[a,b]$ and $mathscr{P}$ is the set of all partitions of $[a,b]$.



Intuitively this makes sense because for any partition $P$ made up of the intervals $I_{1},...,I_{n}$ we know that $$m_{k}=inf_{xin I_{k}}f(x)leq M_{k}=sup_{xin I_{k}}f(x)$$



Would I be along the right lines to consider $P$ being the partition used for $L(f)$ and $Q$ being the partition used for $U(f)$ and then let $R$ be a partition defined as $R=Pcup Q$ so that $R$ is a refinement of both $P$ and $Q$?



Thanks!






integration analysis riemann-integration partitions-for-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you say anything about the relative values of any two given lower and upper partitions?
    $endgroup$
    – AnotherJohnDoe
    Dec 18 '18 at 9:29










  • $begingroup$
    How do you guarantee the existence of $P,Q$?
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:43
















1












$begingroup$


How exactly would I go about proving the following statement?



Given $f:[a,b]tomathbb{R}$ show that $$L(f)leq U(f)$$ where $$L(f)=sup_{Pinmathscr{P}}L(f,P) text{ and } U(f)=inf_{Pinmathscr{P}}U(f,P)$$ where $P$ is any partition of $[a,b]$ and $mathscr{P}$ is the set of all partitions of $[a,b]$.



Intuitively this makes sense because for any partition $P$ made up of the intervals $I_{1},...,I_{n}$ we know that $$m_{k}=inf_{xin I_{k}}f(x)leq M_{k}=sup_{xin I_{k}}f(x)$$



Would I be along the right lines to consider $P$ being the partition used for $L(f)$ and $Q$ being the partition used for $U(f)$ and then let $R$ be a partition defined as $R=Pcup Q$ so that $R$ is a refinement of both $P$ and $Q$?



Thanks!






integration analysis riemann-integration partitions-for-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you say anything about the relative values of any two given lower and upper partitions?
    $endgroup$
    – AnotherJohnDoe
    Dec 18 '18 at 9:29










  • $begingroup$
    How do you guarantee the existence of $P,Q$?
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:43














1












1








1





$begingroup$


How exactly would I go about proving the following statement?



Given $f:[a,b]tomathbb{R}$ show that $$L(f)leq U(f)$$ where $$L(f)=sup_{Pinmathscr{P}}L(f,P) text{ and } U(f)=inf_{Pinmathscr{P}}U(f,P)$$ where $P$ is any partition of $[a,b]$ and $mathscr{P}$ is the set of all partitions of $[a,b]$.



Intuitively this makes sense because for any partition $P$ made up of the intervals $I_{1},...,I_{n}$ we know that $$m_{k}=inf_{xin I_{k}}f(x)leq M_{k}=sup_{xin I_{k}}f(x)$$



Would I be along the right lines to consider $P$ being the partition used for $L(f)$ and $Q$ being the partition used for $U(f)$ and then let $R$ be a partition defined as $R=Pcup Q$ so that $R$ is a refinement of both $P$ and $Q$?



Thanks!






integration analysis riemann-integration partitions-for-integration







share|cite|improve this question











$endgroup$




How exactly would I go about proving the following statement?



Given $f:[a,b]tomathbb{R}$ show that $$L(f)leq U(f)$$ where $$L(f)=sup_{Pinmathscr{P}}L(f,P) text{ and } U(f)=inf_{Pinmathscr{P}}U(f,P)$$ where $P$ is any partition of $[a,b]$ and $mathscr{P}$ is the set of all partitions of $[a,b]$.



Intuitively this makes sense because for any partition $P$ made up of the intervals $I_{1},...,I_{n}$ we know that $$m_{k}=inf_{xin I_{k}}f(x)leq M_{k}=sup_{xin I_{k}}f(x)$$



Would I be along the right lines to consider $P$ being the partition used for $L(f)$ and $Q$ being the partition used for $U(f)$ and then let $R$ be a partition defined as $R=Pcup Q$ so that $R$ is a refinement of both $P$ and $Q$?



Thanks!






integration analysis riemann-integration partitions-for-integration




integration analysis riemann-integration partitions-for-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 9:38









Christian Blatter

173k8113326




173k8113326










asked Dec 18 '18 at 9:26









BigWigBigWig

11310




11310












  • $begingroup$
    Can you say anything about the relative values of any two given lower and upper partitions?
    $endgroup$
    – AnotherJohnDoe
    Dec 18 '18 at 9:29










  • $begingroup$
    How do you guarantee the existence of $P,Q$?
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:43


















  • $begingroup$
    Can you say anything about the relative values of any two given lower and upper partitions?
    $endgroup$
    – AnotherJohnDoe
    Dec 18 '18 at 9:29










  • $begingroup$
    How do you guarantee the existence of $P,Q$?
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:43
















$begingroup$
Can you say anything about the relative values of any two given lower and upper partitions?
$endgroup$
– AnotherJohnDoe
Dec 18 '18 at 9:29




$begingroup$
Can you say anything about the relative values of any two given lower and upper partitions?
$endgroup$
– AnotherJohnDoe
Dec 18 '18 at 9:29












$begingroup$
How do you guarantee the existence of $P,Q$?
$endgroup$
– Shubham Johri
Dec 18 '18 at 9:43




$begingroup$
How do you guarantee the existence of $P,Q$?
$endgroup$
– Shubham Johri
Dec 18 '18 at 9:43










2 Answers
2






active

oldest

votes


















1












$begingroup$

Say $displaystyle L(f)>U(f)=inf_{Pinscr P}U(f,P)$



$L(f)$ is not a lower bound for ${U(f,P):Pinmathscr P}thereforeexists P_1inmathscr P$ such that $U(f)le U(f,P_1)<L(f)$. $U(f,P_1)$ is not an upper bound for ${L(f,P):pinmathscr P}therefore exists P_2inmathscr P$ such that $U(f,P_1)<L(f,P_2)le L(f)$.



This is a contradiction, since $U(f,P_1)ge L(f,P_2) forall P_1,P_2inmathscr P$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $P,Q in mathscr{P}$, then we have



    $L(F,P) le U(f,Q)$.



    This gives



    $L(f) le U(f,Q)$ for all $Q in mathscr{P}$,



    hence $L(f) le U(f).$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Say $displaystyle L(f)>U(f)=inf_{Pinscr P}U(f,P)$



      $L(f)$ is not a lower bound for ${U(f,P):Pinmathscr P}thereforeexists P_1inmathscr P$ such that $U(f)le U(f,P_1)<L(f)$. $U(f,P_1)$ is not an upper bound for ${L(f,P):pinmathscr P}therefore exists P_2inmathscr P$ such that $U(f,P_1)<L(f,P_2)le L(f)$.



      This is a contradiction, since $U(f,P_1)ge L(f,P_2) forall P_1,P_2inmathscr P$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Say $displaystyle L(f)>U(f)=inf_{Pinscr P}U(f,P)$



        $L(f)$ is not a lower bound for ${U(f,P):Pinmathscr P}thereforeexists P_1inmathscr P$ such that $U(f)le U(f,P_1)<L(f)$. $U(f,P_1)$ is not an upper bound for ${L(f,P):pinmathscr P}therefore exists P_2inmathscr P$ such that $U(f,P_1)<L(f,P_2)le L(f)$.



        This is a contradiction, since $U(f,P_1)ge L(f,P_2) forall P_1,P_2inmathscr P$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Say $displaystyle L(f)>U(f)=inf_{Pinscr P}U(f,P)$



          $L(f)$ is not a lower bound for ${U(f,P):Pinmathscr P}thereforeexists P_1inmathscr P$ such that $U(f)le U(f,P_1)<L(f)$. $U(f,P_1)$ is not an upper bound for ${L(f,P):pinmathscr P}therefore exists P_2inmathscr P$ such that $U(f,P_1)<L(f,P_2)le L(f)$.



          This is a contradiction, since $U(f,P_1)ge L(f,P_2) forall P_1,P_2inmathscr P$.






          share|cite|improve this answer









          $endgroup$



          Say $displaystyle L(f)>U(f)=inf_{Pinscr P}U(f,P)$



          $L(f)$ is not a lower bound for ${U(f,P):Pinmathscr P}thereforeexists P_1inmathscr P$ such that $U(f)le U(f,P_1)<L(f)$. $U(f,P_1)$ is not an upper bound for ${L(f,P):pinmathscr P}therefore exists P_2inmathscr P$ such that $U(f,P_1)<L(f,P_2)le L(f)$.



          This is a contradiction, since $U(f,P_1)ge L(f,P_2) forall P_1,P_2inmathscr P$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 9:43









          Shubham JohriShubham Johri

          5,192717




          5,192717























              0












              $begingroup$

              If $P,Q in mathscr{P}$, then we have



              $L(F,P) le U(f,Q)$.



              This gives



              $L(f) le U(f,Q)$ for all $Q in mathscr{P}$,



              hence $L(f) le U(f).$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                If $P,Q in mathscr{P}$, then we have



                $L(F,P) le U(f,Q)$.



                This gives



                $L(f) le U(f,Q)$ for all $Q in mathscr{P}$,



                hence $L(f) le U(f).$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If $P,Q in mathscr{P}$, then we have



                  $L(F,P) le U(f,Q)$.



                  This gives



                  $L(f) le U(f,Q)$ for all $Q in mathscr{P}$,



                  hence $L(f) le U(f).$






                  share|cite|improve this answer









                  $endgroup$



                  If $P,Q in mathscr{P}$, then we have



                  $L(F,P) le U(f,Q)$.



                  This gives



                  $L(f) le U(f,Q)$ for all $Q in mathscr{P}$,



                  hence $L(f) le U(f).$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 9:46









                  FredFred

                  46.7k1848




                  46.7k1848






























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