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Give me an example of two Riemann-integrable functions $f,g:[0,1]\to[0,1]$ such that $g\circ f$ isn't integrable! I already know the following example: $$f(x)=\begin{cases} 0, & \text{if $x$ is irrational} \\ 1, & \text{if $x=0$}\\ \frac1q, & \text{if $x$ is rational and $x=\frac pq$ such that $q\in\Bbb N$ and $(p,q)=1$} \\ \end{cases}$$ $$g(x) = \begin{cases} 1, & \text{if $x$ is of the form $\frac 1q$such that $q\in \Bbb N$} \\ 0, & \text{otherwise} \\ \end{cases}$$ now observe that $g\circ f$ is a famous example of non-integrable function!

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    $\begingroup$ As it turns out, this particular $f$ and $g$ are not a great example--when you extend the definition of the integral to include more functions, $g \circ f$ integrates fine. $\endgroup$ – 6005 Mar 7 '14 at 20:30
  • $\begingroup$ Extend to what?? $\endgroup$ – k1.M Mar 7 '14 at 21:12
  • $\begingroup$ Lebesgue integration! The integral of $g \circ f$ turns out to be $0$. $\endgroup$ – 6005 Mar 8 '14 at 9:10
  • $\begingroup$ Can someone explain why f is integrable? $\endgroup$ – Zslice Dec 17 '14 at 10:32
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More simply, sticking with the same $f$ you may consider $$g(x) = \begin{cases} 0, & \text{if $x=0$} \\ 1, & \text{if $x\in ]0,1]$} \\ \end{cases}$$

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  • $\begingroup$ in this case$g\circ f$ is integrable! $\endgroup$ – k1.M Mar 7 '14 at 20:49
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    $\begingroup$ Nope it is not ! I'll provide proof if you're not persuaded. $\endgroup$ – Gabriel Romon Mar 7 '14 at 20:54
  • $\begingroup$ excuse me your answer is true and simpler thanks!! $\endgroup$ – k1.M Mar 7 '14 at 20:57
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    $\begingroup$ You seem to like exclamation points!!!!! $\endgroup$ – recursive recursion Mar 7 '14 at 21:26

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