1
$\begingroup$

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:
$\bar{I}(f,P)=\sum_{j=0}^{n}(a_{j+1}-a_j)\sup\left \{f(x)|a_j\leq x\leq a_{j+1} \right \}$
$\underline{I}(f,P)=\sum_{j=0}^{n}(a_{j+1}-a_j)\inf\left \{f(x)|a_j\leq x\leq a_{j+1} \right \}$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

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

$\underline{\int_{0}^{t}}fdx=-t^2/2$

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 $\lim_{n \to \infty}\sum_{j=1}^n (t/n) \frac{t(j+1)}{n}=t^2/2$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.