I am reading a section about Generalized Riemann Integral (Kurzweil-Henstock), and there was a problem on that section to provide an example of a function $f$ on $[0,1]$ that is Generalized Riemann Integrable, but its square $f^2$ is NOT Generalized Riemann Integrable.

I couldn't come up with a single function that does the job. Therefore, I appreciate if anyone can provide me with an example of a function $f$ that satisfies the conditions mentioned above. Thanks!


Consider $f(x)=\frac{1}{\sqrt{x}}$ if $x\ne 0$ and any value if $x=0.$

  • $\begingroup$ I am a bit confused about this example. Why is the function $f^2(x)=\frac{1}{x}$ NOT Generalized Riemann Integrable? What goes wrong? $\endgroup$ – Pat_Ho Jul 7 '14 at 23:09
  • 1
    $\begingroup$ $\int_{\epsilon}^1\frac{1}{x}dx=\ln1-\ln\epsilon=-\ln\epsilon.$ When $\epsilon\to 0$ it is $\ln \epsilon\to-\infty.$ $\endgroup$ – mfl Jul 7 '14 at 23:11
  • $\begingroup$ Sorry to bother. $f(x)=\frac{1}{\sqrt{x}}$ if $x\ne 0$ and $f$(x) = 100 if $x=0$. How to calculate this integral from [0,1]? $\endgroup$ – Bear and bunny Nov 27 '14 at 0:31
  • $\begingroup$ $\int_0^1 f(x)dx=[2\sqrt{x}]_0^1=2.$ The value of the function at a particular point doesn't change value of the integral. $\endgroup$ – mfl Nov 27 '14 at 8:37

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