Prove that $f(x)=\begin{cases}x & \text{ if $x$ is rational } \\ -x & \text{ if $x$ is irrational} \end{cases}$ is not Riemann integrable on [0,1]
I'm trying to workout the upper integral and the lower integral. And to say that they are not equal. But got stuck in between.
Definitions that I have been using:
$\bar{I}(f,P)=\sum_{j=0}^{n}(a_{j+1}-a_j)\sup\left \{f(x)|a_j\leq x\leq a_{j+1} \right \}$
$\underline{I}(f,P)=\sum_{j=0}^{n}(a_{j+1}-a_j)\inf\left \{f(x)|a_j\leq x\leq a_{j+1} \right \}$

$\overline{\int_{0}^{1}}f dx= \underset{\epsilon\rightarrow 0}{\lim}\sup\left \{ \overline{I}(f,P)| P\text{ is a partition on [0,1] with a mesh size}\leq\epsilon \right \}$
$\underline{\int_{0}^{1}}f dx= \underset{\epsilon\rightarrow 0}{\lim}\sup\left \{ \underline{I}(f,P)| P\text{ is a partition on [0,1] with a mesh size}\leq\epsilon \right \}$

  • $\begingroup$ What have you tried? In every interval of size $\epsilon$ there are contained both - irrational and rational numbers. $\endgroup$ – Student7 Dec 30 '18 at 20:49
  • 1
    $\begingroup$ The function is continuous in $0$ only, so the function can't be Riemannintegrable by Lebesgue's integral criterion (if you are familiar with that). $\endgroup$ – user370967 Dec 30 '18 at 20:50
  • $\begingroup$ Almost a duplicate. $\endgroup$ – Dietrich Burde Dec 30 '18 at 21:01

Since $\mathbb{Q}$ and $\mathbb{R} \backslash \mathbb{Q}$ are dense. In each Interval there $[a_j,a_{j+1}]$ $\sup\{f(x):x \in [a_j,a_{j+1}]\}= a_{j+1}$ and $\inf\{f(x):x \in [a_j,a_{j+1}]\}= -a_{j+1}$ Therefore

$\overline{\int_{0}^{t}}f dx=t^2/2$,


Lets assume we want to calculate $\overline{\int_{0}^{t}}f (x) dx$. We take the partition $Z_n=[0,t/n,(2t)/n,...,t]$ note that if $a_j$ denotes the jth term in the partition. $a_j-a_{j+1}=t/n$ and $ t/n\to 0$ for $n\to \infty$ since the supremum of $f$ on each interval is $a_{j+1}=t(j+1)/n$ the integral equals $\lim_{n \to \infty}\sum_{j=1}^n (t/n) \frac{t(j+1)}{n}=t^2/2$

  • $\begingroup$ Could you please let me know your calculation for one of the integrals $\endgroup$ – DD90 Dec 30 '18 at 21:23
  • $\begingroup$ Thank you! it works $\endgroup$ – DD90 Dec 30 '18 at 22:21

The upper integral is $1$ and the lower is $-1$. All you need to observe is that every non-degenerate interval contains a rational number and also an irrational number.

  • $\begingroup$ Thanks but the problem is when we get the suprimum on $(a_j,a_{j+1})$ it gives $a_{j+1}$ but not 1. And also for the infimum it gives $-a_{j+1}$. So could you please explain $\endgroup$ – DD90 Dec 30 '18 at 20:56
  • $\begingroup$ I do not understand your concerns. $\endgroup$ – A. Pongrácz Dec 30 '18 at 21:02
  • $\begingroup$ The issue with my approach is that I can show that the upper integral is less than or equal to 1 but not the equality (And vice versa ) $\endgroup$ – DD90 Dec 30 '18 at 21:08
  • $\begingroup$ That is exactly the point of my answer, please try to understand its details. $\endgroup$ – A. Pongrácz Dec 30 '18 at 21:09

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