2
$\begingroup$

Is possible to construct a bounded non Riemann integrable real function such that the set of discontinuity points has empty interior? I know that if the set of discontinuity points is a null set then the answer is no. Every null set has empty interior but the converse is not true.

$\endgroup$
  • $\begingroup$ Do you mean for $f:\Bbb R\to\Bbb R$? $\endgroup$ – Elliot G May 19 '16 at 1:38
1
$\begingroup$

Certainly. Say $K$ is a "fat Cantor set". So in particular $K$ is compact, has empty interior and positive Lebesgue measure.

Now define, for example, $$ f(x) = \begin{cases} 0,&(x \in K), \\\sin(1 / (1+d(x,K))),&(x\notin K). \end{cases}$$Then $f$ is discontinuous precisely on $K$ (and hence $f$ is not Riemann integrable, since $K$ is not a null set).


Note This is fairly stupid; as zhw points out we could simply let $f=\chi_K$ (that is, $f(t)=1$ for $t\in K$, $0$ for $t\notin K$). Replacing my stupidity with his non-stupidity seems like cheating, so I'll leave the above as it is.

$\endgroup$
  • 1
    $\begingroup$ Why not define $f =\chi_K?$ $\endgroup$ – zhw. May 19 '16 at 1:44
  • $\begingroup$ @zhw. Good question. No good answer available, sorry. $\endgroup$ – David C. Ullrich May 19 '16 at 2:36
  • $\begingroup$ lol, my only complaint is I'm zhw., not zhu. $\endgroup$ – zhw. May 19 '16 at 3:27
  • $\begingroup$ @zhw. Sorry.... $\endgroup$ – David C. Ullrich May 19 '16 at 4:24

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.