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.

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

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.

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    $\begingroup$ 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

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