Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that the function $ \phi_\mathbb Q : [0,1] \to \{0,1\}$ defined by $$\phi_\mathbb{Q}(w)=\begin{cases} 1 &\text{if } w\in \mathbb Q, \\ 0 & \text{if } w\notin \mathbb Q \end{cases}$$ is Lebesgue integrable and not Riemann integrable.

I proved that the function is Lebesgue integrable as follows:

The function is positive so we just have to prove that $\int_0^1 \phi_\mathbb Q(x) dx <+ \infty$, and it goes like this :

$\int_0^1 \phi_\mathbb Q(x) dx \le\int_0^1 1\ dx = 1 < \infty$

so $\phi_\mathbb{Q}$ is Lebesgue integrable.

However I couldn't find a proof for being not Riemann integrable.

share|cite|improve this question
What is the your definition of the Riemann integrability? If it is through Riemann sums, show that for any partition there is one zero sum and one which equals $1$ – Ilya Jan 19 '13 at 21:05
we studied one definition and it is Rienman sums , i dont know how to connect it with this function – Lofaif Jan 19 '13 at 21:08
@Lofaif: Lebesgue integrable functions must also be measurable; i.e., you have to know that the integral makes sense before claiming that it is finite. So for completeness on the Lebesgue part you could give a reason $f$ is measurable, or show by other means why the Lebesgue integral actually exists. – Jonas Meyer Jan 19 '13 at 21:14
up vote 1 down vote accepted

First of all, w.r.t. Lebesgue integrability. The function is simple: $\phi_\Bbb Q(x) = 1_{\Bbb Q}(x)$ and thus $$ \int_{[0,1]}\phi_\Bbb Q(x)\lambda(\mathrm dx) = \lambda([0,1]\cap\Bbb Q) = 0 $$ just by the definition of the Lebesgue integral for simple functions.

Let us now consider the Riemann integral. The function $f:[0,1]\to\Bbb R$ is Riemann integrable if exists the following limit: $\lim_{\mu(T)\to 0}S_T(f)$ where $$ T = \{0 = t_0<\xi_1<t_1<\dots<t_{n-1}<\xi_n<t_n =1\} $$ is a pointed partition of $[0,1]$, $\mu(T) = \max_k|t_{k} - t_{k-1}|$ and $$ S_T(f) = \sum_{k=1}^nf(\xi_k)(t_k - t_{k-1}) $$ is a Riemann sum. To show that the limit does not exist for $f = \phi_\Bbb Q$ it is sufficient to to show that for any small $\delta$ there are two partitions $T',T''$ such that $\mu(T')\leq\delta$ and $\mu(T'')\leq\delta$ but $0=S_{T'}(f)\neq S_{T''}(f) = 1$. That's easy - we take both partitions to be uniform, but we choose $\xi'_i$ in the first case to be irrational numbers and $\xi''_i$ to be rational.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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