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Let $f(x)=x$ if $x\in\mathbf{Q}$ and $f(x)=1-x$ if $x \in\mathbf{R}/\mathbf{Q}$. Show that $f$ is Riemann integrable on $[0,1]$.

I know that there are a few posts on this function, but I didn't see the same approach as mine. So, I would like to know if my prove holds, please and have a feedback.

My attempt is to pass by Darboux upper and lower sums.

To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$.

First, consider the interval $[0,\frac{1}{2}]$. We know that $\int_{0}^{1}f(x)=\int_{0}^{1/2}f(x)+\int_{1/2}^{1}f(x)$. So, if one of the integrals is undefined, we could conclude that it is not integrable on the whole interval (*I'm not sure that is true. Should I show that the integral doesn't exist for both intervals, or one is sufficient?)

Let $M_i=\sup\{f(x):x\in[x_i,x_{i+1}]\}$ and $m_i=\inf\{f(x):x\in[x_i,x_{i+1}]\}$ (We work on $[0,\frac{1}{2}]$ now). We remark by density of $\mathbf{Q}$ and $\mathbf{R}/\mathbf{Q}$, that $M_i=(1-x_i)$ and $m_i=x_i$. Thus,

$\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)=\sum_{i=0}^{n}(1-x_i)(x_{i+1}-x_i)-\sum_{i=0}^{n}x_i(x_{i+1}-x_i)$.

But, $\overline{S}_{\sigma}(f)\ge\int_{0}^{1/2}(1-x)\ dx$ and $\underline{S}_{\sigma}(f)\le \int_{0}^{1/2}x\ dx$ whatever the partition. So, $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)\ge \int_{0}^{1/2}(1-x)\ dx - \int_{0}^{1/2}x\ dx=\frac{1}{4}$.

Thus, there is no partition on the interval $[0,\frac{1}{2}]$ such that $\overline{S}_{\sigma}(f)-\underline{S}_{\sigma}(f)<\epsilon$ for $0<\epsilon<\frac{1}{4}$ and we conclude that $f$ is not integrable on $[0,1/2]$ and so on $[0,1]$ as well.

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    $\begingroup$ The proof is good. $\endgroup$
    – RRL
    Feb 26, 2021 at 21:31
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    $\begingroup$ You seem to have learnt the technique from the answer by @RRL to one of your previous questions. +1 and I feel great about it. $\endgroup$
    – Paramanand Singh
    Feb 27, 2021 at 2:17
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    $\begingroup$ Another note: why are your questions worded "show that something is integrable". They should have been "check whether something is integrable or not". $\endgroup$
    – Paramanand Singh
    Feb 27, 2021 at 2:19
  • $\begingroup$ @Paramanand Singh Thank you for advice, I will do it next time! $\endgroup$
    – Daniil
    Feb 27, 2021 at 7:45

1 Answer 1

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So, if one of the integrals is undefined, we could conclude that it is not integrable on the whole interval (*I'm not sure that is true. Should I show that the integral doesn't exist for both intervals, or one is sufficient?)

If $f$ is Riemann integrable on $[a,b]$, then for any $\epsilon> 0$ there is a partition $P$ of $[a,b]$ such that $U(P,f) - L(P,f) < \epsilon$.

Let $P'$ be the refinement of $P$ obtained by adding the point $c \in (a,b)$ if not already there. Otherwise let $P' = P$. Let $P'' $ be the partition of $[a,c]$ induced by $P'$. It follows that

$$U(P'',f) - L(P'',f) \leqslant U(P',f) - L(P',f) \leqslant U(P,f) - L(P,f) < \epsilon,$$

and by the Riemann criterion $f$ must be Riemann integrable on $[a,c]$.

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  • $\begingroup$ Thank you very much) $\endgroup$
    – Daniil
    Feb 27, 2021 at 7:44

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