Integrability of of $f$ over an Infinite Partition












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$begingroup$


Let $a_n$ be a strictly monotonic sequence such that $a_n rightarrow b$. Moreover, $a_0 = a$ and $a<b$. If $f[a,b] rightarrow mathbb{R}$ is a bounded function, and $f$ is integrable on $[a_n, a_n+1]$ for all $nge0$, is $f$ integrable on $[a,b]$?



The usual examples are of combining finitely many intervals over which $f$ is Riemann integrable, however, could this be extended to the case above?






real-analysis integration riemann-integration







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$endgroup$












  • $begingroup$
    Do you intend $b > a$ and for $(a_n)_n$ to be (strictly) monotonically increasing?
    $endgroup$
    – Eric Towers
    Dec 4 '18 at 3:43










  • $begingroup$
    Yes for both assumptions
    $endgroup$
    – Questions
    Dec 4 '18 at 3:51
















1












$begingroup$


Let $a_n$ be a strictly monotonic sequence such that $a_n rightarrow b$. Moreover, $a_0 = a$ and $a<b$. If $f[a,b] rightarrow mathbb{R}$ is a bounded function, and $f$ is integrable on $[a_n, a_n+1]$ for all $nge0$, is $f$ integrable on $[a,b]$?



The usual examples are of combining finitely many intervals over which $f$ is Riemann integrable, however, could this be extended to the case above?






real-analysis integration riemann-integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you intend $b > a$ and for $(a_n)_n$ to be (strictly) monotonically increasing?
    $endgroup$
    – Eric Towers
    Dec 4 '18 at 3:43










  • $begingroup$
    Yes for both assumptions
    $endgroup$
    – Questions
    Dec 4 '18 at 3:51














1












1








1


1



$begingroup$


Let $a_n$ be a strictly monotonic sequence such that $a_n rightarrow b$. Moreover, $a_0 = a$ and $a<b$. If $f[a,b] rightarrow mathbb{R}$ is a bounded function, and $f$ is integrable on $[a_n, a_n+1]$ for all $nge0$, is $f$ integrable on $[a,b]$?



The usual examples are of combining finitely many intervals over which $f$ is Riemann integrable, however, could this be extended to the case above?






real-analysis integration riemann-integration







share|cite|improve this question











$endgroup$




Let $a_n$ be a strictly monotonic sequence such that $a_n rightarrow b$. Moreover, $a_0 = a$ and $a<b$. If $f[a,b] rightarrow mathbb{R}$ is a bounded function, and $f$ is integrable on $[a_n, a_n+1]$ for all $nge0$, is $f$ integrable on $[a,b]$?



The usual examples are of combining finitely many intervals over which $f$ is Riemann integrable, however, could this be extended to the case above?






real-analysis integration riemann-integration




real-analysis integration riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 3:54







Questions

















asked Dec 4 '18 at 3:23









QuestionsQuestions

83




83












  • $begingroup$
    Do you intend $b > a$ and for $(a_n)_n$ to be (strictly) monotonically increasing?
    $endgroup$
    – Eric Towers
    Dec 4 '18 at 3:43










  • $begingroup$
    Yes for both assumptions
    $endgroup$
    – Questions
    Dec 4 '18 at 3:51


















  • $begingroup$
    Do you intend $b > a$ and for $(a_n)_n$ to be (strictly) monotonically increasing?
    $endgroup$
    – Eric Towers
    Dec 4 '18 at 3:43










  • $begingroup$
    Yes for both assumptions
    $endgroup$
    – Questions
    Dec 4 '18 at 3:51
















$begingroup$
Do you intend $b > a$ and for $(a_n)_n$ to be (strictly) monotonically increasing?
$endgroup$
– Eric Towers
Dec 4 '18 at 3:43




$begingroup$
Do you intend $b > a$ and for $(a_n)_n$ to be (strictly) monotonically increasing?
$endgroup$
– Eric Towers
Dec 4 '18 at 3:43












$begingroup$
Yes for both assumptions
$endgroup$
– Questions
Dec 4 '18 at 3:51




$begingroup$
Yes for both assumptions
$endgroup$
– Questions
Dec 4 '18 at 3:51










2 Answers
2






active

oldest

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1












$begingroup$

You can show that $int_a^bf=lim_{xto b}int_a^x f$. In which case you can potentially extend the definition assuming that for $n$ large you can make $int_{a_n}^xf$ as small as you wish where $x$ is arbitrary in $(a_n,a_{n+1}].$ For then you can show that $lim_{xto b}int_a^xf=lim_{ktoinfty}int_{a}^{a_k}f=lim_{ktoinfty}sum_{n=0}^kint_{a_n}^{a_{n+1}}f$.



If you wished to use Riemann sums instead of just summing the integrals then for all $n$ you could choose partition $P_n$ of $[a_n,a_{n+1}]$ such that $U(f,P_n)-int_{a_n}^{a_{n+1}}f<epsilon/2^{n+1}$ where $epsilon $ is arbitrary positive.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    In each $[a_n,a_{n+1}]$, $f$ is continuous a.e. : the (Lebesgue) measure of points of discontinuity in that interval is zero. (This is Lebesgue's criterion for Riemann integrability -- $f$ bounded on $[a,b]$ is Riemann integrable on that interval if and only if it is continuous a.e. on that interval.) The union of the points of discontinuity over the intervals in your partition is a countable union of measure zero sets so has measure zero (via subadditivity of measure). So $f$ is continuous a.e. on $[a,b]$ and therefore is Riemann integrable on $[a,b]$.






    share|cite|improve this answer









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      2 Answers
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      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      1












      $begingroup$

      You can show that $int_a^bf=lim_{xto b}int_a^x f$. In which case you can potentially extend the definition assuming that for $n$ large you can make $int_{a_n}^xf$ as small as you wish where $x$ is arbitrary in $(a_n,a_{n+1}].$ For then you can show that $lim_{xto b}int_a^xf=lim_{ktoinfty}int_{a}^{a_k}f=lim_{ktoinfty}sum_{n=0}^kint_{a_n}^{a_{n+1}}f$.



      If you wished to use Riemann sums instead of just summing the integrals then for all $n$ you could choose partition $P_n$ of $[a_n,a_{n+1}]$ such that $U(f,P_n)-int_{a_n}^{a_{n+1}}f<epsilon/2^{n+1}$ where $epsilon $ is arbitrary positive.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You can show that $int_a^bf=lim_{xto b}int_a^x f$. In which case you can potentially extend the definition assuming that for $n$ large you can make $int_{a_n}^xf$ as small as you wish where $x$ is arbitrary in $(a_n,a_{n+1}].$ For then you can show that $lim_{xto b}int_a^xf=lim_{ktoinfty}int_{a}^{a_k}f=lim_{ktoinfty}sum_{n=0}^kint_{a_n}^{a_{n+1}}f$.



        If you wished to use Riemann sums instead of just summing the integrals then for all $n$ you could choose partition $P_n$ of $[a_n,a_{n+1}]$ such that $U(f,P_n)-int_{a_n}^{a_{n+1}}f<epsilon/2^{n+1}$ where $epsilon $ is arbitrary positive.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You can show that $int_a^bf=lim_{xto b}int_a^x f$. In which case you can potentially extend the definition assuming that for $n$ large you can make $int_{a_n}^xf$ as small as you wish where $x$ is arbitrary in $(a_n,a_{n+1}].$ For then you can show that $lim_{xto b}int_a^xf=lim_{ktoinfty}int_{a}^{a_k}f=lim_{ktoinfty}sum_{n=0}^kint_{a_n}^{a_{n+1}}f$.



          If you wished to use Riemann sums instead of just summing the integrals then for all $n$ you could choose partition $P_n$ of $[a_n,a_{n+1}]$ such that $U(f,P_n)-int_{a_n}^{a_{n+1}}f<epsilon/2^{n+1}$ where $epsilon $ is arbitrary positive.






          share|cite|improve this answer









          $endgroup$



          You can show that $int_a^bf=lim_{xto b}int_a^x f$. In which case you can potentially extend the definition assuming that for $n$ large you can make $int_{a_n}^xf$ as small as you wish where $x$ is arbitrary in $(a_n,a_{n+1}].$ For then you can show that $lim_{xto b}int_a^xf=lim_{ktoinfty}int_{a}^{a_k}f=lim_{ktoinfty}sum_{n=0}^kint_{a_n}^{a_{n+1}}f$.



          If you wished to use Riemann sums instead of just summing the integrals then for all $n$ you could choose partition $P_n$ of $[a_n,a_{n+1}]$ such that $U(f,P_n)-int_{a_n}^{a_{n+1}}f<epsilon/2^{n+1}$ where $epsilon $ is arbitrary positive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 3:54









          MelodyMelody

          81012




          81012























              1












              $begingroup$

              In each $[a_n,a_{n+1}]$, $f$ is continuous a.e. : the (Lebesgue) measure of points of discontinuity in that interval is zero. (This is Lebesgue's criterion for Riemann integrability -- $f$ bounded on $[a,b]$ is Riemann integrable on that interval if and only if it is continuous a.e. on that interval.) The union of the points of discontinuity over the intervals in your partition is a countable union of measure zero sets so has measure zero (via subadditivity of measure). So $f$ is continuous a.e. on $[a,b]$ and therefore is Riemann integrable on $[a,b]$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                In each $[a_n,a_{n+1}]$, $f$ is continuous a.e. : the (Lebesgue) measure of points of discontinuity in that interval is zero. (This is Lebesgue's criterion for Riemann integrability -- $f$ bounded on $[a,b]$ is Riemann integrable on that interval if and only if it is continuous a.e. on that interval.) The union of the points of discontinuity over the intervals in your partition is a countable union of measure zero sets so has measure zero (via subadditivity of measure). So $f$ is continuous a.e. on $[a,b]$ and therefore is Riemann integrable on $[a,b]$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In each $[a_n,a_{n+1}]$, $f$ is continuous a.e. : the (Lebesgue) measure of points of discontinuity in that interval is zero. (This is Lebesgue's criterion for Riemann integrability -- $f$ bounded on $[a,b]$ is Riemann integrable on that interval if and only if it is continuous a.e. on that interval.) The union of the points of discontinuity over the intervals in your partition is a countable union of measure zero sets so has measure zero (via subadditivity of measure). So $f$ is continuous a.e. on $[a,b]$ and therefore is Riemann integrable on $[a,b]$.






                  share|cite|improve this answer









                  $endgroup$



                  In each $[a_n,a_{n+1}]$, $f$ is continuous a.e. : the (Lebesgue) measure of points of discontinuity in that interval is zero. (This is Lebesgue's criterion for Riemann integrability -- $f$ bounded on $[a,b]$ is Riemann integrable on that interval if and only if it is continuous a.e. on that interval.) The union of the points of discontinuity over the intervals in your partition is a countable union of measure zero sets so has measure zero (via subadditivity of measure). So $f$ is continuous a.e. on $[a,b]$ and therefore is Riemann integrable on $[a,b]$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 4:09









                  Eric TowersEric Towers

                  32.7k22370




                  32.7k22370






























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