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Why is the Dirichlet function,

$$f(x) = \begin{cases}x : x \in \mathbb{Q}, \\ 0 : x \notin \mathbb{Q}, \end{cases}$$

not Riemann integrable on any interval $[a, b]$?

user2850514
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2 Answers2

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Because $\mathbb Q$ is dense and discrete in $\mathbb R$, so $$\sup_{x\in[\xi,\eta]} f(x) = \eta \neq 0 = \inf_{x\in[\xi,\eta]} f(x)$$ Thus, $U(f;[a,b],Z) > 0=L(f;[a,b],Z)$ for any partiton $Z$

[I have edited to reflect $f(x) = x\chi_{\mathbb Q}(x)$, the previous version was designed for $f(x) = \chi_{\mathbb Q}(x)$]

AlexR
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    Thanks for the response, unfortunately I don't recognise anything in the answer! Is there any way you could explain it using Riemann sums? – user2850514 Feb 05 '14 at 22:51
  • @user2850514 $U$ denotes the upper riemann sum and $L$ the lower. They are defined by $$U(f;[a,b],Z) = \sum_{I\in Z} |I| \sup_{x\in I} f(x)$$ and $$L(f;[a,b],Z) = \sum_{I\in Z} |I| \inf_{x\in I} f(x)$$ where $I$ are the intervals in $Z$, $[a,b] = \bigcup_{I\in Z} I$ and $|I|$ is the interval length. – AlexR Feb 05 '14 at 22:54
  • Does this help or will you provide your notational conventions so I can adapt the answer to it? – AlexR Feb 05 '14 at 22:55
  • I am not familiar with sup and inf, more familiar with the terminology Steven has used. – user2850514 Feb 05 '14 at 22:57
  • @user2850514 Unfortunately, the definition of a Riemann integral heavily relies on $\inf$ and $\sup$. You could substitute them for $\min$ and $\max$ if the interval endpoints are chosen to be rational. – AlexR Feb 05 '14 at 22:58
  • Can you explain what they are? I may know of them in a different form – user2850514 Feb 05 '14 at 22:59
  • $\inf$ is the "biggest lower bound". That means it coincides with the minimum $\min$ if there is a point wich attains it. It's somewhat a generalisation of $\min$, since $$\inf [0,1] = \min [0,1] = 0$$ but $\min (0,1]$ doesn't exist and $\inf (0,1] = 0$. The analogous is for $\sup$ and $\max$. See here for details – AlexR Feb 05 '14 at 23:02
  • @AlexR you are actually using the, conceptually cleaner, Darboux integral rather than the Riemann integral. The two notions of integrability are equivalent. – Steven Gubkin Feb 05 '14 at 23:03
  • @StevenGubkin Thanks for the hint. It was introduced to me as the Riemann integral in my first calculus lessons (it's been a while though). It should be noted that the two arising definitions of integrability are equivalent. – AlexR Feb 05 '14 at 23:04
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Say $a,b$ are positive. No matter what partition of the interval you choose, you can always pick a rational representative. The Riemann sum is then the same as the riemann sum for $f(x) = x$, so you get $(b-a)/2$ However, if you were to always pick an irrational representative of each subinterval, you would get $0$. These answers are different, so the limit of the sequence of Riemann sums depends on the choice of the representatives. So it is not Riemann integrable.

Steven Gubkin
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  • I have attempted to find the Riemann sum of both, although am stuggling to work with the summations for rational $x$. – user2850514 Feb 05 '14 at 22:58
  • You do not need to. If you choose all your $x_i$ rational, then the sum you get is exactly a riemann sum for the function $f(x)=x$, which you should already know how to integrate. – Steven Gubkin Feb 05 '14 at 23:00
  • Ah yes that makes sense. Following on from that, what if we have an interval [-a, a], an integral of such would give 0, so do we modify and consider above and below the curve? – user2850514 Feb 05 '14 at 23:07
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    In that case, you can choose rational representatives in $[-a,0)$ and irrational representatives in $[0,a]$; or vice versa. The two choices give different answers, as before. – TonyK Feb 05 '14 at 23:56