Riemann-Darboux Integrability of Subinterval












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I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.



Theorem: Suppose $f$ is Riemann-Darboux integrable on $[a,b]$. Let $cin(a,b)$. Then, $f$ is Riemann-Darboux integrable on the intervals $[a,c]$ and $[c,b]$.



Attempted Proof: Since $f$ is Riemann-Darboux integrable on $[a,b]$, for arbitrary $epsilon>0$, there exists a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<epsilon$.



Let $n_p,n_{p*}$, and $n_{p'}$ be the number of partition parts in $P$, $P^*$, and $P'$, respectively. Also, let $m_i=inf_{xin[x_{i-1},x_i]}f(x)$ and $M_i=sup_{xin[x_{i-1},x_i]}f(x)$.



Consider the partition of $[a,c]$ given by $P^*=Pcap[a,c]$. Then, $$U(f,P^*)-L(f,P^*)=sum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[a,c]$.



Next, consider the partition $[a,b]$ given by $P'=Pcap[c,b]$. Then,



$$U(f,P')-L(f,P')=sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[c,b]$. $square$



Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.






real-analysis integration







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    0














    I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.



    Theorem: Suppose $f$ is Riemann-Darboux integrable on $[a,b]$. Let $cin(a,b)$. Then, $f$ is Riemann-Darboux integrable on the intervals $[a,c]$ and $[c,b]$.



    Attempted Proof: Since $f$ is Riemann-Darboux integrable on $[a,b]$, for arbitrary $epsilon>0$, there exists a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<epsilon$.



    Let $n_p,n_{p*}$, and $n_{p'}$ be the number of partition parts in $P$, $P^*$, and $P'$, respectively. Also, let $m_i=inf_{xin[x_{i-1},x_i]}f(x)$ and $M_i=sup_{xin[x_{i-1},x_i]}f(x)$.



    Consider the partition of $[a,c]$ given by $P^*=Pcap[a,c]$. Then, $$U(f,P^*)-L(f,P^*)=sum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[a,c]$.



    Next, consider the partition $[a,b]$ given by $P'=Pcap[c,b]$. Then,



    $$U(f,P')-L(f,P')=sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[c,b]$. $square$



    Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.






    real-analysis integration







    share|cite|improve this question

























      0












      0








      0







      I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.



      Theorem: Suppose $f$ is Riemann-Darboux integrable on $[a,b]$. Let $cin(a,b)$. Then, $f$ is Riemann-Darboux integrable on the intervals $[a,c]$ and $[c,b]$.



      Attempted Proof: Since $f$ is Riemann-Darboux integrable on $[a,b]$, for arbitrary $epsilon>0$, there exists a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<epsilon$.



      Let $n_p,n_{p*}$, and $n_{p'}$ be the number of partition parts in $P$, $P^*$, and $P'$, respectively. Also, let $m_i=inf_{xin[x_{i-1},x_i]}f(x)$ and $M_i=sup_{xin[x_{i-1},x_i]}f(x)$.



      Consider the partition of $[a,c]$ given by $P^*=Pcap[a,c]$. Then, $$U(f,P^*)-L(f,P^*)=sum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[a,c]$.



      Next, consider the partition $[a,b]$ given by $P'=Pcap[c,b]$. Then,



      $$U(f,P')-L(f,P')=sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[c,b]$. $square$



      Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.






      real-analysis integration







      share|cite|improve this question













      I'm studying Riemann-Darboux integration. I'm trying to prove the following rather intuitive notion for integrals. Please let me know if you find any errors in this proof, as I'm self-studying this topic.



      Theorem: Suppose $f$ is Riemann-Darboux integrable on $[a,b]$. Let $cin(a,b)$. Then, $f$ is Riemann-Darboux integrable on the intervals $[a,c]$ and $[c,b]$.



      Attempted Proof: Since $f$ is Riemann-Darboux integrable on $[a,b]$, for arbitrary $epsilon>0$, there exists a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)<epsilon$.



      Let $n_p,n_{p*}$, and $n_{p'}$ be the number of partition parts in $P$, $P^*$, and $P'$, respectively. Also, let $m_i=inf_{xin[x_{i-1},x_i]}f(x)$ and $M_i=sup_{xin[x_{i-1},x_i]}f(x)$.



      Consider the partition of $[a,c]$ given by $P^*=Pcap[a,c]$. Then, $$U(f,P^*)-L(f,P^*)=sum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[a,c]$.



      Next, consider the partition $[a,b]$ given by $P'=Pcap[c,b]$. Then,



      $$U(f,P')-L(f,P')=sum_{i=n_{p*}+1}^{n_{p'}}(M_i-m_i)Delta x_ilesum_{i=1}^{n_{p*}}(M_i-m_i)Delta x_i+ sum_{n_{p*}+1}^{n_p}(M_i-m_i)Delta x_i=sum_{i=1}^{n_p}(M_i-m_i)Delta x_i=U(f,P)-L(f,P)<epsilon$$ Therefore, $f$ is integrable on $[c,b]$. $square$



      Any and all feedback, or alternative proofs are appreciated. I love to see different arguments to expand my skill set.






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      asked Nov 28 '18 at 3:48









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          This is essentially fine, although you are tacitly assuming that $c$ itself is a member of the partition, $P$. This isn't a big deal though - a $P$ such that
          $$U(P,f)-L(P,f)<epsilon$$
          is guaranteed by integrability, and you can always just add $c$ to this partition if it is not already there. Namely, if $P_{c}$ is this new partition, called a refinement of $P$, one must have
          $$U(P_{c},f)-L(P_{c},f)leq U(P,f)-L(P,f)<epsilon$$
          and you can proceed as you have done in your proof.



          An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $[a,b]$, it is bounded and continuous up to a null set on $[a,b]$, and hence, on $[a,c]$ and $[c,b]$ as well.






          share|cite|improve this answer





















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            This is essentially fine, although you are tacitly assuming that $c$ itself is a member of the partition, $P$. This isn't a big deal though - a $P$ such that
            $$U(P,f)-L(P,f)<epsilon$$
            is guaranteed by integrability, and you can always just add $c$ to this partition if it is not already there. Namely, if $P_{c}$ is this new partition, called a refinement of $P$, one must have
            $$U(P_{c},f)-L(P_{c},f)leq U(P,f)-L(P,f)<epsilon$$
            and you can proceed as you have done in your proof.



            An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $[a,b]$, it is bounded and continuous up to a null set on $[a,b]$, and hence, on $[a,c]$ and $[c,b]$ as well.






            share|cite|improve this answer


























              2














              This is essentially fine, although you are tacitly assuming that $c$ itself is a member of the partition, $P$. This isn't a big deal though - a $P$ such that
              $$U(P,f)-L(P,f)<epsilon$$
              is guaranteed by integrability, and you can always just add $c$ to this partition if it is not already there. Namely, if $P_{c}$ is this new partition, called a refinement of $P$, one must have
              $$U(P_{c},f)-L(P_{c},f)leq U(P,f)-L(P,f)<epsilon$$
              and you can proceed as you have done in your proof.



              An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $[a,b]$, it is bounded and continuous up to a null set on $[a,b]$, and hence, on $[a,c]$ and $[c,b]$ as well.






              share|cite|improve this answer
























                2












                2








                2






                This is essentially fine, although you are tacitly assuming that $c$ itself is a member of the partition, $P$. This isn't a big deal though - a $P$ such that
                $$U(P,f)-L(P,f)<epsilon$$
                is guaranteed by integrability, and you can always just add $c$ to this partition if it is not already there. Namely, if $P_{c}$ is this new partition, called a refinement of $P$, one must have
                $$U(P_{c},f)-L(P_{c},f)leq U(P,f)-L(P,f)<epsilon$$
                and you can proceed as you have done in your proof.



                An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $[a,b]$, it is bounded and continuous up to a null set on $[a,b]$, and hence, on $[a,c]$ and $[c,b]$ as well.






                share|cite|improve this answer












                This is essentially fine, although you are tacitly assuming that $c$ itself is a member of the partition, $P$. This isn't a big deal though - a $P$ such that
                $$U(P,f)-L(P,f)<epsilon$$
                is guaranteed by integrability, and you can always just add $c$ to this partition if it is not already there. Namely, if $P_{c}$ is this new partition, called a refinement of $P$, one must have
                $$U(P_{c},f)-L(P_{c},f)leq U(P,f)-L(P,f)<epsilon$$
                and you can proceed as you have done in your proof.



                An alternative approach is given by Lebesgue's Criterion for Riemann Integrability, which states that Riemann/Darboux Integrability is equivalent to boundedness plus continuity up to a null set (see Wikipedia for a good explanation of null sets). Since your functions is integrable on $[a,b]$, it is bounded and continuous up to a null set on $[a,b]$, and hence, on $[a,c]$ and $[c,b]$ as well.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 6:46









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