I want to construct a function $f : [0,1] \rightarrow \mathbb{R}$ such that it is continuous almost everywhere, and moreover no continuous function $g$ on $[0,1]$ satisfying $$f = g \ \mbox{a.e.}$$

Since $f$ has a pretty specific property, and $f$ is on $[0,1]$, it reminds me of a weird set like Cantor set. I try to use a Lebesgue function on fat Cantor set, but there is a continuous function equals to it outside Fat Cantor. So does not work.

Does anyone has idea about constructing function $f$ ?

  • 2
    $\begingroup$ @Verbe It's possible. Trivial, in fact: $\chi_{[0,1/2]}$ is continuous almost everywhere. $\endgroup$ – David C. Ullrich Nov 28 '17 at 23:07
  • $\begingroup$ oh, so the indicator on $[0,1/2]$ is not continuous only at $x = 1/2$, and no continuous equals to it a.e. $\endgroup$ – user117375 Nov 28 '17 at 23:10
  • $\begingroup$ okay, I kinda see it now. Maybe I go too complicated. Thank you. $\endgroup$ – user117375 Nov 28 '17 at 23:11

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