A bounded function with one point of discontinuity is Riemann integrable












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The lemma states: Assume $f$ is bounded on $[a,b]$ and continuous on $[a,b]$ except at $a$. Then $f$ is integrable on $[a,b]$.



I am really stuck on this one. I would really appreciate some help. Thanks!










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    0












    $begingroup$


    The lemma states: Assume $f$ is bounded on $[a,b]$ and continuous on $[a,b]$ except at $a$. Then $f$ is integrable on $[a,b]$.



    I am really stuck on this one. I would really appreciate some help. Thanks!










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      The lemma states: Assume $f$ is bounded on $[a,b]$ and continuous on $[a,b]$ except at $a$. Then $f$ is integrable on $[a,b]$.



      I am really stuck on this one. I would really appreciate some help. Thanks!










      share|cite|improve this question











      $endgroup$




      The lemma states: Assume $f$ is bounded on $[a,b]$ and continuous on $[a,b]$ except at $a$. Then $f$ is integrable on $[a,b]$.



      I am really stuck on this one. I would really appreciate some help. Thanks!







      real-analysis integration






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      edited Jul 30 '14 at 7:49









      Jyrki Lahtonen

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      asked Dec 8 '12 at 15:55









      user42864user42864

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          Let $epsilon > 0$ be given. Now, our problem is at $a$, so we could like to isolate $a$. Choose $x_0 = a$ and $x_1$ close enough to $a$ so that $$(max_{[x_0,x_1]}f(x)-min_{[x_0,x_1]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ I'll let you fill in the details of why this is possible, but you must use boundedness of $f$. Now, $f$ is continuous on the interval $[x_1,b]$ hence integrable, and so we can find a partition ${y_i}$ of that interval so that $$sum_{i=1}^n (max_{[y_i,y_{i+1}]}f(x)-min_{[y_i,y_{i+1}]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ How can we combine this partition with $x_0,x_1$?






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            $begingroup$

            Let $epsilon > 0$ be given. Now, our problem is at $a$, so we could like to isolate $a$. Choose $x_0 = a$ and $x_1$ close enough to $a$ so that $$(max_{[x_0,x_1]}f(x)-min_{[x_0,x_1]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ I'll let you fill in the details of why this is possible, but you must use boundedness of $f$. Now, $f$ is continuous on the interval $[x_1,b]$ hence integrable, and so we can find a partition ${y_i}$ of that interval so that $$sum_{i=1}^n (max_{[y_i,y_{i+1}]}f(x)-min_{[y_i,y_{i+1}]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ How can we combine this partition with $x_0,x_1$?






            share|cite|improve this answer











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              1












              $begingroup$

              Let $epsilon > 0$ be given. Now, our problem is at $a$, so we could like to isolate $a$. Choose $x_0 = a$ and $x_1$ close enough to $a$ so that $$(max_{[x_0,x_1]}f(x)-min_{[x_0,x_1]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ I'll let you fill in the details of why this is possible, but you must use boundedness of $f$. Now, $f$ is continuous on the interval $[x_1,b]$ hence integrable, and so we can find a partition ${y_i}$ of that interval so that $$sum_{i=1}^n (max_{[y_i,y_{i+1}]}f(x)-min_{[y_i,y_{i+1}]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ How can we combine this partition with $x_0,x_1$?






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $epsilon > 0$ be given. Now, our problem is at $a$, so we could like to isolate $a$. Choose $x_0 = a$ and $x_1$ close enough to $a$ so that $$(max_{[x_0,x_1]}f(x)-min_{[x_0,x_1]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ I'll let you fill in the details of why this is possible, but you must use boundedness of $f$. Now, $f$ is continuous on the interval $[x_1,b]$ hence integrable, and so we can find a partition ${y_i}$ of that interval so that $$sum_{i=1}^n (max_{[y_i,y_{i+1}]}f(x)-min_{[y_i,y_{i+1}]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ How can we combine this partition with $x_0,x_1$?






                share|cite|improve this answer











                $endgroup$



                Let $epsilon > 0$ be given. Now, our problem is at $a$, so we could like to isolate $a$. Choose $x_0 = a$ and $x_1$ close enough to $a$ so that $$(max_{[x_0,x_1]}f(x)-min_{[x_0,x_1]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ I'll let you fill in the details of why this is possible, but you must use boundedness of $f$. Now, $f$ is continuous on the interval $[x_1,b]$ hence integrable, and so we can find a partition ${y_i}$ of that interval so that $$sum_{i=1}^n (max_{[y_i,y_{i+1}]}f(x)-min_{[y_i,y_{i+1}]}f(x))(x_1 - x_0) < frac{epsilon}{2}.$$ How can we combine this partition with $x_0,x_1$?







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Apr 8 '18 at 22:27









                Paolo Leonetti

                11.5k21550




                11.5k21550










                answered Dec 8 '12 at 17:00









                Zach L.Zach L.

                5,2501131




                5,2501131






























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