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I had an exam today and one of my question was: Give a function $f$ that is Riemann integrable but not Lebesgue integrable

How is it possible ? I always thought that Riemann $\implies $ Lebesgue, isn't it ?

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    $\begingroup$ Was the question "Prove or disprove"? $\endgroup$ – Joe Johnson 126 Dec 28 '15 at 15:10
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    $\begingroup$ I assume here the question includes "Improper (Riemann) integrals"? $\endgroup$ – Clement C. Dec 28 '15 at 15:11
  • $\begingroup$ no, just to give an exemple of a function that is Riemann integrable but not Lebesgue integrable. I'm a little bit suprised of such a question since I always thought that a Riemann integrable function was Lebesgue integrable. $\endgroup$ – user301068 Dec 28 '15 at 15:11
  • $\begingroup$ @ClementC.: I don't know, but probably. Is it that important ? $\endgroup$ – user301068 Dec 28 '15 at 15:12
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    $\begingroup$ No. Lebesgue integrability of $f$ requires the integrability of $|f|$ as well. Consider integrating $\frac{\sin x}x$ on $(0,\infty)$. $\endgroup$ – MPW Dec 28 '15 at 15:13
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You may like to visit http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ for an excellent introduction to the different integrals. I am reproducing an image from that website for your quick reference.

Relationship between various integrals

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A function $f$ is Lebesgue integrable if and only if $f^+=0\vee f$ and $f^-=0\wedge f$ are integrable. So if $f\geq 0$ is Riemann integrable then it will be also Lebesgue integrable. But if $f$ has not a constant sign, then $f$ can be Riemann integrable but not Lebesgue integrable as $f(x)=\frac{\sin(x)}{x}$ on $[0,\infty [$. And this is because $\int f^+=\int f^-=+\infty$. Therefore $\int f=\int f^+-\int f^-$ is not defined.

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  • $\begingroup$ How can you show that $\int f^+ = +\infty$? $\endgroup$ – Alfredo Lozano Apr 15 '17 at 18:38

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