This is part of an exercise from Carother's Real Analysis:

Let $G$ be open such that $\mathbb{Q}\cap[0,1] \subset G$ and $m(G)<\frac{1}{2}$. Show that $\chi_G\notin\mathcal{R}[0,1]$.


  • $\chi_G(x)=\begin{cases}1 \text{ if } x\in G\\ 0 \text{ if } x\notin G\end{cases}$

  • $\mathcal{R}[0,1]$ is the space of Riemann Integrable functions on the interval $[0,1]$

  • $m(\cdot)$ denotes the Lebesgue measure


  • I want to show this by using Lebesgue's Criterion for Riemann Integration- namely that a function is Riemann Integrable iff the measure of the set of discontinuities of the function has measure zero.

  • I also know that the measure of $\mathbb{Q}$ is zero, so since $m(\mathbb{Q})=0$, then $m(\mathbb{Q\cap [0,1]})=0$. (because $\mathbb{Q}\cap[0,1]\subset\mathbb{Q}\implies m(\mathbb{Q}\cap[0,1])\leq m(\mathbb{Q})=0$, and measure is nonnegative.) Well, this means that $m(G)=0$, so I don't see where I'm supposed to use the information that $m(G)<\frac{1}{2}$ from the assumption in the question.

  • I also know that $\chi_{\mathbb{Q}}$ and $\chi_{\mathbb{R}\setminus\mathbb{Q}}$ are both not Riemann Integrable.

  • Anyway, I now try to show that the measure of the discontinuities of the characteristic function of $G$ is non-zero.

    Let $D(f)$ denote the set of discontinuities of a function $f$, and let $\omega_f(x)$ denote the oscillation of $f$ in the open interval $I$.

\begin{align*} &m(D(f))\neq 0\\ &\iff m\left(\bigcup_{n=1}^{\infty}\left\{x\in [0,1]:\omega_f(x)\geq \frac{1}{n}\right\}\right)\neq 0\\ &\iff m\left(\bigcup_{n=1}^{\infty}\left\{x\in [0,1]:\inf_{x\in [0,1]}\sup_{s,t\in [0,1]}|\chi_G(s)-\chi_G(t)|\geq \frac{1}{n}\right\}\right)\neq 0\\ \end{align*}

(I've been told by members on MSE that the notation that I learned for the oscillation is confusing, so just as a note:

$\omega_f(x) = \inf_{\epsilon>0} \sup_{s,t \in B(x,\epsilon)} |f(t)-f(s)|$

is an alternative definition for the oscillation of $f$ on an open ball $B(x,\epsilon)$.)

So for any $\epsilon$-ball I take centered at a rational $r\cap [0,1]: r\in\mathbb{Q}$, it will always intersect an irrational in $[0,1]$ as well, which is not in $G$, because both $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$. This is important, because this means that $\chi_G$ will have infinitely many discontinuities. I think there are uncountably many discontinuities since the discontinuities would be the irrationals in $[0,1]$.

I'm not sure how to proceed from here.

I'm still fairly new to the notion of measure, so if there is a better way to show this, then please feel free to enlighten me. Thanks.

  • 1
    $\begingroup$ There's either a typo in the book or you copied something wrong, because "$G=\{\mathbb{Q}\cap[0,1]\}$ be open" doesn't make sense. Strictly speaking, $\{\mathbb{Q}\cap[0,1]\}$ is a set containing exactly one element, namely the set of rational numbers in the unit interval. If I overlook the outer pair of curly brackets, then this is the set of rationals in the unit interval, and this set is not open. That said, my guess is that you only need to show various ways of choosing points in the subintervals (forming a Riemann sum) lead to different limiting results. $\endgroup$ – Dave L. Renfro Feb 11 '15 at 21:18
  • $\begingroup$ @DaveL.Renfro, it's possible I misunderstood something. Here is what it says in the book exactly: Let $G$ be an open set containing the rationals in $[0,1]$ with $m(G)<\frac{1}{2}$. Prove that $f=\chi_G$ is not Riemann integrable on $[0,1]$ $\endgroup$ – Sujaan Kunalan Feb 11 '15 at 21:20
  • $\begingroup$ Try Riemann sums when all the "evaluation points" are chosen rational. (Think about why you can do this.) The Riemann sums will all have (total) value $1,$ so as the norm of the partitions approaches zero, you'll get (a limit of) $1.$ Now see if you can somehow, by making use of the fact that $G$ has measure less than $\frac{1}{2}$ (less than $1$ is really enough I would think) to get a sequence of Riemann sums whose values are bounded under $1$ as the norms of the partitions approach zero. But maybe the Lebesgue criteria is easier, I don't know right off ... $\endgroup$ – Dave L. Renfro Feb 11 '15 at 21:28

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