Prove that $underline{int_a^b} f(x)dx=sup_{xi}int_a^b xi(x)dx$, $xi$ a step-function.












2












$begingroup$



Let $f:[a,b]tomathbb R$ be a bounded nonnegative function. Prove that
$$
underline{int_a^b} f(x)dx=sup_{xi}int_a^b xi(x)dx,
$$
where $xi$ belongs to the set of step-functions such that $xi(x)leq f(x)$ for all $xin[a,b]$.



Show that an analogous result holds if we take $xi$ continuous or $xi$ integrable (keeping the hypothesis $xi(x)leq f(x)$ for all $xin[a,b]$).




Let $X$ be the set of step-functions $xi:[a,b]tomathbb R$ such that $xi(x)leq f(x)$ for all $xin[a,b]$. Take $P={a=t_0,t_1,ldots,t_{n-1},t_n=b}$ be a partition of $[a,b]$. So
$$
underline{int_a^b} f(x)dx = sup_{P} s(f;P),
$$
where $s(f;P)=sum_{i=1}^n m_i(f)(t_i-t_{i-1})$ and $m_i(f)=underset{xin[t_{i-1},t_i]}{inf} f(x)$.



Let $xiin X$. From $xi(x)leq f(x)$ for all $xin[a,b]$, we get
$$
int_a^b xi(x)dx=underline{int_a^b} xi(x)dxleq underline{int_a^b} f(x)dx Longrightarrow underset{xi}{sup} int_a^b xi(x)dx leq underline{int_a^b} f(x)dx.
$$
Define $xi_1:[a,b]tomathbb R$ by $xi_1(x)=m_i(f)$ for all $xin[t_{i-1},t_i]$. Hence,
$$
int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
$$



Thus the supremum is attained. Therefore,
$$
underset{xi}{sup} int_a^b xi(x)dx = underline{int_a^b} f(x)dx.
$$



Is my proof correct? If not, why?






real-analysis proof-verification riemann-integration







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $f:[a,b]tomathbb R$ be a bounded nonnegative function. Prove that
    $$
    underline{int_a^b} f(x)dx=sup_{xi}int_a^b xi(x)dx,
    $$
    where $xi$ belongs to the set of step-functions such that $xi(x)leq f(x)$ for all $xin[a,b]$.



    Show that an analogous result holds if we take $xi$ continuous or $xi$ integrable (keeping the hypothesis $xi(x)leq f(x)$ for all $xin[a,b]$).




    Let $X$ be the set of step-functions $xi:[a,b]tomathbb R$ such that $xi(x)leq f(x)$ for all $xin[a,b]$. Take $P={a=t_0,t_1,ldots,t_{n-1},t_n=b}$ be a partition of $[a,b]$. So
    $$
    underline{int_a^b} f(x)dx = sup_{P} s(f;P),
    $$
    where $s(f;P)=sum_{i=1}^n m_i(f)(t_i-t_{i-1})$ and $m_i(f)=underset{xin[t_{i-1},t_i]}{inf} f(x)$.



    Let $xiin X$. From $xi(x)leq f(x)$ for all $xin[a,b]$, we get
    $$
    int_a^b xi(x)dx=underline{int_a^b} xi(x)dxleq underline{int_a^b} f(x)dx Longrightarrow underset{xi}{sup} int_a^b xi(x)dx leq underline{int_a^b} f(x)dx.
    $$
    Define $xi_1:[a,b]tomathbb R$ by $xi_1(x)=m_i(f)$ for all $xin[t_{i-1},t_i]$. Hence,
    $$
    int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
    $$



    Thus the supremum is attained. Therefore,
    $$
    underset{xi}{sup} int_a^b xi(x)dx = underline{int_a^b} f(x)dx.
    $$



    Is my proof correct? If not, why?






    real-analysis proof-verification riemann-integration







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Let $f:[a,b]tomathbb R$ be a bounded nonnegative function. Prove that
      $$
      underline{int_a^b} f(x)dx=sup_{xi}int_a^b xi(x)dx,
      $$
      where $xi$ belongs to the set of step-functions such that $xi(x)leq f(x)$ for all $xin[a,b]$.



      Show that an analogous result holds if we take $xi$ continuous or $xi$ integrable (keeping the hypothesis $xi(x)leq f(x)$ for all $xin[a,b]$).




      Let $X$ be the set of step-functions $xi:[a,b]tomathbb R$ such that $xi(x)leq f(x)$ for all $xin[a,b]$. Take $P={a=t_0,t_1,ldots,t_{n-1},t_n=b}$ be a partition of $[a,b]$. So
      $$
      underline{int_a^b} f(x)dx = sup_{P} s(f;P),
      $$
      where $s(f;P)=sum_{i=1}^n m_i(f)(t_i-t_{i-1})$ and $m_i(f)=underset{xin[t_{i-1},t_i]}{inf} f(x)$.



      Let $xiin X$. From $xi(x)leq f(x)$ for all $xin[a,b]$, we get
      $$
      int_a^b xi(x)dx=underline{int_a^b} xi(x)dxleq underline{int_a^b} f(x)dx Longrightarrow underset{xi}{sup} int_a^b xi(x)dx leq underline{int_a^b} f(x)dx.
      $$
      Define $xi_1:[a,b]tomathbb R$ by $xi_1(x)=m_i(f)$ for all $xin[t_{i-1},t_i]$. Hence,
      $$
      int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
      $$



      Thus the supremum is attained. Therefore,
      $$
      underset{xi}{sup} int_a^b xi(x)dx = underline{int_a^b} f(x)dx.
      $$



      Is my proof correct? If not, why?






      real-analysis proof-verification riemann-integration







      share|cite|improve this question











      $endgroup$





      Let $f:[a,b]tomathbb R$ be a bounded nonnegative function. Prove that
      $$
      underline{int_a^b} f(x)dx=sup_{xi}int_a^b xi(x)dx,
      $$
      where $xi$ belongs to the set of step-functions such that $xi(x)leq f(x)$ for all $xin[a,b]$.



      Show that an analogous result holds if we take $xi$ continuous or $xi$ integrable (keeping the hypothesis $xi(x)leq f(x)$ for all $xin[a,b]$).




      Let $X$ be the set of step-functions $xi:[a,b]tomathbb R$ such that $xi(x)leq f(x)$ for all $xin[a,b]$. Take $P={a=t_0,t_1,ldots,t_{n-1},t_n=b}$ be a partition of $[a,b]$. So
      $$
      underline{int_a^b} f(x)dx = sup_{P} s(f;P),
      $$
      where $s(f;P)=sum_{i=1}^n m_i(f)(t_i-t_{i-1})$ and $m_i(f)=underset{xin[t_{i-1},t_i]}{inf} f(x)$.



      Let $xiin X$. From $xi(x)leq f(x)$ for all $xin[a,b]$, we get
      $$
      int_a^b xi(x)dx=underline{int_a^b} xi(x)dxleq underline{int_a^b} f(x)dx Longrightarrow underset{xi}{sup} int_a^b xi(x)dx leq underline{int_a^b} f(x)dx.
      $$
      Define $xi_1:[a,b]tomathbb R$ by $xi_1(x)=m_i(f)$ for all $xin[t_{i-1},t_i]$. Hence,
      $$
      int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
      $$



      Thus the supremum is attained. Therefore,
      $$
      underset{xi}{sup} int_a^b xi(x)dx = underline{int_a^b} f(x)dx.
      $$



      Is my proof correct? If not, why?






      real-analysis proof-verification riemann-integration




      real-analysis proof-verification riemann-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 31 at 17:23







      Ali Khan

















      asked Nov 3 '18 at 2:51









      Ali KhanAli Khan

      17312




      17312






















          1 Answer
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          $begingroup$

          You are close to being correct.



          The function $xi_1$ you introduce belongs to $X_1$, the collection of step functions dominated by $f$ that correspond to lower Riemann sums. You are correct in writing



          $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) $$



          but not correct in claiming attainment of the supremum by a step fuction through the subsequent implication



          $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
          $$



          The equality $int_a^b xi_1(x)dx = s(f;P)$ for a specific partition does not imply $int_a^b xi_1(x)dx = sup_{P} s(f;P)$.



          What is true, however, is that, since there is a one-to-one correspondence between functions $xi_1 in X_1$ and lower sums,



          $$sup_{xi_1 in X_1}int_a^b xi_1(x)dx = sup_{P} s(f;P) = underline{int_a^b} f(x)dx$$



          Since $X_1 subset X$ it follows that



          $$left{int_a^b xi_1(x) , dx ,,|,, xi_1 in X_1right} subset left{int_a^b xi(x) , dx ,,|,, xi in Xright},$$



          and



          $$tag{*}underline{int_a^b} f(x),dx = sup_{xi_1 in X_1} int_a^b xi_1(x) , dx leqslant sup_{xi in X} int_a^b xi(x) , dx leqslant underline{int_a^b} f(x),dx $$



          Recall that you correctly proved the far right inequality in (*), and we have as desired,



          $$underline{int_a^b} f(x),dx = sup_{xi in X} int_a^b xi(x) , dx $$






          share|cite|improve this answer









          $endgroup$














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            active

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            active

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            2












            $begingroup$

            You are close to being correct.



            The function $xi_1$ you introduce belongs to $X_1$, the collection of step functions dominated by $f$ that correspond to lower Riemann sums. You are correct in writing



            $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) $$



            but not correct in claiming attainment of the supremum by a step fuction through the subsequent implication



            $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
            $$



            The equality $int_a^b xi_1(x)dx = s(f;P)$ for a specific partition does not imply $int_a^b xi_1(x)dx = sup_{P} s(f;P)$.



            What is true, however, is that, since there is a one-to-one correspondence between functions $xi_1 in X_1$ and lower sums,



            $$sup_{xi_1 in X_1}int_a^b xi_1(x)dx = sup_{P} s(f;P) = underline{int_a^b} f(x)dx$$



            Since $X_1 subset X$ it follows that



            $$left{int_a^b xi_1(x) , dx ,,|,, xi_1 in X_1right} subset left{int_a^b xi(x) , dx ,,|,, xi in Xright},$$



            and



            $$tag{*}underline{int_a^b} f(x),dx = sup_{xi_1 in X_1} int_a^b xi_1(x) , dx leqslant sup_{xi in X} int_a^b xi(x) , dx leqslant underline{int_a^b} f(x),dx $$



            Recall that you correctly proved the far right inequality in (*), and we have as desired,



            $$underline{int_a^b} f(x),dx = sup_{xi in X} int_a^b xi(x) , dx $$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              You are close to being correct.



              The function $xi_1$ you introduce belongs to $X_1$, the collection of step functions dominated by $f$ that correspond to lower Riemann sums. You are correct in writing



              $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) $$



              but not correct in claiming attainment of the supremum by a step fuction through the subsequent implication



              $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
              $$



              The equality $int_a^b xi_1(x)dx = s(f;P)$ for a specific partition does not imply $int_a^b xi_1(x)dx = sup_{P} s(f;P)$.



              What is true, however, is that, since there is a one-to-one correspondence between functions $xi_1 in X_1$ and lower sums,



              $$sup_{xi_1 in X_1}int_a^b xi_1(x)dx = sup_{P} s(f;P) = underline{int_a^b} f(x)dx$$



              Since $X_1 subset X$ it follows that



              $$left{int_a^b xi_1(x) , dx ,,|,, xi_1 in X_1right} subset left{int_a^b xi(x) , dx ,,|,, xi in Xright},$$



              and



              $$tag{*}underline{int_a^b} f(x),dx = sup_{xi_1 in X_1} int_a^b xi_1(x) , dx leqslant sup_{xi in X} int_a^b xi(x) , dx leqslant underline{int_a^b} f(x),dx $$



              Recall that you correctly proved the far right inequality in (*), and we have as desired,



              $$underline{int_a^b} f(x),dx = sup_{xi in X} int_a^b xi(x) , dx $$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                You are close to being correct.



                The function $xi_1$ you introduce belongs to $X_1$, the collection of step functions dominated by $f$ that correspond to lower Riemann sums. You are correct in writing



                $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) $$



                but not correct in claiming attainment of the supremum by a step fuction through the subsequent implication



                $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
                $$



                The equality $int_a^b xi_1(x)dx = s(f;P)$ for a specific partition does not imply $int_a^b xi_1(x)dx = sup_{P} s(f;P)$.



                What is true, however, is that, since there is a one-to-one correspondence between functions $xi_1 in X_1$ and lower sums,



                $$sup_{xi_1 in X_1}int_a^b xi_1(x)dx = sup_{P} s(f;P) = underline{int_a^b} f(x)dx$$



                Since $X_1 subset X$ it follows that



                $$left{int_a^b xi_1(x) , dx ,,|,, xi_1 in X_1right} subset left{int_a^b xi(x) , dx ,,|,, xi in Xright},$$



                and



                $$tag{*}underline{int_a^b} f(x),dx = sup_{xi_1 in X_1} int_a^b xi_1(x) , dx leqslant sup_{xi in X} int_a^b xi(x) , dx leqslant underline{int_a^b} f(x),dx $$



                Recall that you correctly proved the far right inequality in (*), and we have as desired,



                $$underline{int_a^b} f(x),dx = sup_{xi in X} int_a^b xi(x) , dx $$






                share|cite|improve this answer









                $endgroup$



                You are close to being correct.



                The function $xi_1$ you introduce belongs to $X_1$, the collection of step functions dominated by $f$ that correspond to lower Riemann sums. You are correct in writing



                $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) $$



                but not correct in claiming attainment of the supremum by a step fuction through the subsequent implication



                $$int_a^b xi_1(x)dx = sum_{i=1}^n m_i(f)(t_i-t_{i-1}) = s(f;P) Longrightarrow int_a^b xi_1(x)dx = sup_{P} s(f;P)
                $$



                The equality $int_a^b xi_1(x)dx = s(f;P)$ for a specific partition does not imply $int_a^b xi_1(x)dx = sup_{P} s(f;P)$.



                What is true, however, is that, since there is a one-to-one correspondence between functions $xi_1 in X_1$ and lower sums,



                $$sup_{xi_1 in X_1}int_a^b xi_1(x)dx = sup_{P} s(f;P) = underline{int_a^b} f(x)dx$$



                Since $X_1 subset X$ it follows that



                $$left{int_a^b xi_1(x) , dx ,,|,, xi_1 in X_1right} subset left{int_a^b xi(x) , dx ,,|,, xi in Xright},$$



                and



                $$tag{*}underline{int_a^b} f(x),dx = sup_{xi_1 in X_1} int_a^b xi_1(x) , dx leqslant sup_{xi in X} int_a^b xi(x) , dx leqslant underline{int_a^b} f(x),dx $$



                Recall that you correctly proved the far right inequality in (*), and we have as desired,



                $$underline{int_a^b} f(x),dx = sup_{xi in X} int_a^b xi(x) , dx $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 3 '18 at 5:01









                RRLRRL

                53.5k52574




                53.5k52574






























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