Let $f(x)=x$ for $x$ irrational and $f(x)=0$ for $x$ rational. Show that $f$ Darboux integrable (lower and upper) on $[0,1]$ and $$(\underline{D})\int_{0}^{1}f=0,\quad (\overline{D})\int_{0}^{1}f=\frac{1}{2}.$$

  • $\begingroup$ Can you use the fact that Darboux integrability is equivalent to Riemann integrability? $\endgroup$ Dec 22, 2015 at 11:23
  • $\begingroup$ @SirJective yes, i got it, but i still confused with the form of function. I mean how i can distinguish rational and irrational? Those are too many... $\endgroup$
    – naomi72
    Dec 22, 2015 at 11:49
  • $\begingroup$ What is the $D$ with the over and underline supposed to refer to next to the integral? $\endgroup$ Dec 22, 2015 at 22:58
  • $\begingroup$ I assume that the line indicates if it is the upper or lower integral correct? If so, then $f$ is only said to be Darboux integrable if the upper is equal to the lower $\endgroup$ Dec 14, 2019 at 3:53

1 Answer 1


So we know that Darboux integrability is equivalent to Riemann integrability. Let's consider the discontinuities of $f$. Given that $f$ is discontinuous at every rational in the interval, and the rationals are a countable set, we have that $\mathbb{Q} \cap [0,1]$ is countable. Since it is countable, it forms a zero set, and hence the set of discontinuities of $f$ is a zero set.

Now given that the function is defined on $[0,1]$ as $f(x)=x$ if $x$ is irrational, we can see that the function is bounded by $1$. By the Riemann-Lebesgue Theorem we have that a bounded function $f:[a,b]\to \mathbb{R}$ with a set of discontinuities which form a zero set is Riemann integrable. As mentioned in the comments, since it is Riemann integrable it is Darboux integrable.

I'm not sure about your notation for the second half of the question. If you explain it in the comments I will edit this answer to make it complete.

  • 1
    $\begingroup$ I am fairly certain this function is discontinuous everywhere, that is why the upper and lower integrals are different $\endgroup$ Dec 14, 2019 at 3:50

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