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Let $f: [0,1] \to \mathbb{R}$ be defined by $f(x)=0$ if $x$ is irrational and $f(x) = x+1$ if $x$ is rational. I want to prove using an argument by upper and lower sums that $f$ is not Riemann integrable. I can easily see that $L(f) = 0$, but I'm not sure how to find $U(f)$.

Any hint is appreciated.

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    $\begingroup$ $U(f) \ge 1$ because every interval contains a rational number. $\endgroup$
    – Martin R
    Nov 27, 2020 at 13:46
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    $\begingroup$ You can say $U(f) \geq 1 $ $\endgroup$ Nov 27, 2020 at 13:47
  • $\begingroup$ That's true, thanks! $\endgroup$ Nov 27, 2020 at 13:51

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If you split the interval $[0,1]$ into $n$ (equal) subintervals, the maximum value of $f(x)$ on the $k$-th interval is $f\bigl(\frac kn\bigr)=\frac kn + 1$, so $$U_n(f)=\frac1n\sum_{k=1}^n\Bigl(\frac kn+1\Bigr)=\frac1n\biggl(\frac{1+2+\dots+n}n +n\biggr).$$ Can you take it from there?

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  • $\begingroup$ I found that $U_{n}(f) = \frac{3n+1}{2n}$, so its infimum is $3/2$ and then $U(f) \leq 3/2$, but I got stuck there since I can't conclude that $U(f) \not = 0$ from that inequality. $\endgroup$ Nov 27, 2020 at 14:09
  • $\begingroup$ You mean $U(f)\ge 3/2$ since it is the infimum. $\endgroup$
    – Bernard
    Nov 27, 2020 at 14:28
  • $\begingroup$ Right, I was confused. Thanks $\endgroup$ Nov 27, 2020 at 15:07

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