When is a Lebesgue integrable function a Riemann integrable function?












3












$begingroup$


When is a Lebesgue integrable function a Riemann integrable function ?



And if we have $fin mathcal{L}^1([0,1],lambda)$, does it implies that $f$ is Riemann integrable, and why ?






integration lebesgue-integral riemann-integration







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    When is a Lebesgue integrable function a Riemann integrable function ?



    And if we have $fin mathcal{L}^1([0,1],lambda)$, does it implies that $f$ is Riemann integrable, and why ?






    integration lebesgue-integral riemann-integration







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      When is a Lebesgue integrable function a Riemann integrable function ?



      And if we have $fin mathcal{L}^1([0,1],lambda)$, does it implies that $f$ is Riemann integrable, and why ?






      integration lebesgue-integral riemann-integration







      share|cite|improve this question











      $endgroup$




      When is a Lebesgue integrable function a Riemann integrable function ?



      And if we have $fin mathcal{L}^1([0,1],lambda)$, does it implies that $f$ is Riemann integrable, and why ?






      integration lebesgue-integral riemann-integration




      integration lebesgue-integral riemann-integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 18:54









      Bernard

      119k740113




      119k740113










      asked Jan 15 at 18:42









      Anas BOUALIIAnas BOUALII

      1417




      1417






















          2 Answers
          2






          active

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          10












          $begingroup$

          No, it doesn't.



          It was proven by Lebesgue that a bounded function $f$ is Riemann integrable precisely when the set of points where $f$ is discontinuous has measure zero.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
            $endgroup$
            – Anas BOUALII
            Jan 15 at 18:46












          • $begingroup$
            @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
            $endgroup$
            – Noah Schweber
            Jan 15 at 18:47












          • $begingroup$
            But does every Lebesgue integrable function is a.e continious and bounded ?
            $endgroup$
            – Anas BOUALII
            Jan 15 at 18:49






          • 2




            $begingroup$
            No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
            $endgroup$
            – Martin Argerami
            Jan 15 at 18:52










          • $begingroup$
            @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
            $endgroup$
            – Martin Argerami
            Jan 15 at 19:05



















          5












          $begingroup$

          This is the Riemann-Lebesgue Theorem, which says that $f$ is Riemann integrable if and only if the set of discontinuity points is of measure zero.



          Note that $1_{mathbb{Q}}$ (the indicator of the rationals) is Lebesgue-integrable but not Riemann integrable. On $[0,1]$ it is almost everywhere bounded but not continuous. Here, the set of discontinuity points is of measure $1$. So we cannot say anything about continuity and Lebesgue integrability (except that every continuous function on a set of finite measure is Lebesgue-integrable).






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            No, it doesn't.



            It was proven by Lebesgue that a bounded function $f$ is Riemann integrable precisely when the set of points where $f$ is discontinuous has measure zero.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:46












            • $begingroup$
              @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
              $endgroup$
              – Noah Schweber
              Jan 15 at 18:47












            • $begingroup$
              But does every Lebesgue integrable function is a.e continious and bounded ?
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:49






            • 2




              $begingroup$
              No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
              $endgroup$
              – Martin Argerami
              Jan 15 at 18:52










            • $begingroup$
              @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
              $endgroup$
              – Martin Argerami
              Jan 15 at 19:05
















            10












            $begingroup$

            No, it doesn't.



            It was proven by Lebesgue that a bounded function $f$ is Riemann integrable precisely when the set of points where $f$ is discontinuous has measure zero.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:46












            • $begingroup$
              @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
              $endgroup$
              – Noah Schweber
              Jan 15 at 18:47












            • $begingroup$
              But does every Lebesgue integrable function is a.e continious and bounded ?
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:49






            • 2




              $begingroup$
              No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
              $endgroup$
              – Martin Argerami
              Jan 15 at 18:52










            • $begingroup$
              @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
              $endgroup$
              – Martin Argerami
              Jan 15 at 19:05














            10












            10








            10





            $begingroup$

            No, it doesn't.



            It was proven by Lebesgue that a bounded function $f$ is Riemann integrable precisely when the set of points where $f$ is discontinuous has measure zero.






            share|cite|improve this answer











            $endgroup$



            No, it doesn't.



            It was proven by Lebesgue that a bounded function $f$ is Riemann integrable precisely when the set of points where $f$ is discontinuous has measure zero.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 15 at 19:06

























            answered Jan 15 at 18:44









            Martin ArgeramiMartin Argerami

            126k1182180




            126k1182180












            • $begingroup$
              So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:46












            • $begingroup$
              @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
              $endgroup$
              – Noah Schweber
              Jan 15 at 18:47












            • $begingroup$
              But does every Lebesgue integrable function is a.e continious and bounded ?
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:49






            • 2




              $begingroup$
              No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
              $endgroup$
              – Martin Argerami
              Jan 15 at 18:52










            • $begingroup$
              @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
              $endgroup$
              – Martin Argerami
              Jan 15 at 19:05


















            • $begingroup$
              So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:46












            • $begingroup$
              @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
              $endgroup$
              – Noah Schweber
              Jan 15 at 18:47












            • $begingroup$
              But does every Lebesgue integrable function is a.e continious and bounded ?
              $endgroup$
              – Anas BOUALII
              Jan 15 at 18:49






            • 2




              $begingroup$
              No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
              $endgroup$
              – Martin Argerami
              Jan 15 at 18:52










            • $begingroup$
              @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
              $endgroup$
              – Martin Argerami
              Jan 15 at 19:05
















            $begingroup$
            So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
            $endgroup$
            – Anas BOUALII
            Jan 15 at 18:46






            $begingroup$
            So in what case we can go from lebesgue inegral to reimann integral ? When do we have $int_{a}^{b}f(x)dx=int_{[a,b]}fdlambda$ for a Lebesgue integrable function $f$
            $endgroup$
            – Anas BOUALII
            Jan 15 at 18:46














            $begingroup$
            @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
            $endgroup$
            – Noah Schweber
            Jan 15 at 18:47






            $begingroup$
            @AnasBOUALII Precisely when - as this answer says - the set of discontinuities of $f$ has measure zero. Is there something unsatisfying about this?
            $endgroup$
            – Noah Schweber
            Jan 15 at 18:47














            $begingroup$
            But does every Lebesgue integrable function is a.e continious and bounded ?
            $endgroup$
            – Anas BOUALII
            Jan 15 at 18:49




            $begingroup$
            But does every Lebesgue integrable function is a.e continious and bounded ?
            $endgroup$
            – Anas BOUALII
            Jan 15 at 18:49




            2




            2




            $begingroup$
            No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
            $endgroup$
            – Martin Argerami
            Jan 15 at 18:52




            $begingroup$
            No, of course not. In your setup, the function $1_{mathbb Q}$, the characteristic of the rationals, is everywhere discontinuous. And $f(x)=tfrac1{sqrt x},1_{mathbb Q}$ is unbounded, everywhere discontinuous, and still in $L^1[0,1]$.
            $endgroup$
            – Martin Argerami
            Jan 15 at 18:52












            $begingroup$
            @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
            $endgroup$
            – Martin Argerami
            Jan 15 at 19:05




            $begingroup$
            @bof: yes. My example was about a Lebesgue measurable function which is not Riemann integrable.
            $endgroup$
            – Martin Argerami
            Jan 15 at 19:05











            5












            $begingroup$

            This is the Riemann-Lebesgue Theorem, which says that $f$ is Riemann integrable if and only if the set of discontinuity points is of measure zero.



            Note that $1_{mathbb{Q}}$ (the indicator of the rationals) is Lebesgue-integrable but not Riemann integrable. On $[0,1]$ it is almost everywhere bounded but not continuous. Here, the set of discontinuity points is of measure $1$. So we cannot say anything about continuity and Lebesgue integrability (except that every continuous function on a set of finite measure is Lebesgue-integrable).






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              This is the Riemann-Lebesgue Theorem, which says that $f$ is Riemann integrable if and only if the set of discontinuity points is of measure zero.



              Note that $1_{mathbb{Q}}$ (the indicator of the rationals) is Lebesgue-integrable but not Riemann integrable. On $[0,1]$ it is almost everywhere bounded but not continuous. Here, the set of discontinuity points is of measure $1$. So we cannot say anything about continuity and Lebesgue integrability (except that every continuous function on a set of finite measure is Lebesgue-integrable).






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                This is the Riemann-Lebesgue Theorem, which says that $f$ is Riemann integrable if and only if the set of discontinuity points is of measure zero.



                Note that $1_{mathbb{Q}}$ (the indicator of the rationals) is Lebesgue-integrable but not Riemann integrable. On $[0,1]$ it is almost everywhere bounded but not continuous. Here, the set of discontinuity points is of measure $1$. So we cannot say anything about continuity and Lebesgue integrability (except that every continuous function on a set of finite measure is Lebesgue-integrable).






                share|cite|improve this answer









                $endgroup$



                This is the Riemann-Lebesgue Theorem, which says that $f$ is Riemann integrable if and only if the set of discontinuity points is of measure zero.



                Note that $1_{mathbb{Q}}$ (the indicator of the rationals) is Lebesgue-integrable but not Riemann integrable. On $[0,1]$ it is almost everywhere bounded but not continuous. Here, the set of discontinuity points is of measure $1$. So we cannot say anything about continuity and Lebesgue integrability (except that every continuous function on a set of finite measure is Lebesgue-integrable).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 18:56









                OldGodzillaOldGodzilla

                53925




                53925






























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