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Set of continuity points of a real function

I want to prove that there does not exist a function on [0,1] which is continuous on rational number,and it is not continuous on irrational number. Someone suggest that if this function exists,it is Riemann integrable. Then use Riemann Lebesgue Lemma, it is incorrect. But I don't know how to prove it is integrable.

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  • It seems like you could show that if a function is continuous at a point $x$ then it must be continuous in some interval $(x-\epsilon, x+\epsilon)$. Your result would then follow. – MJD Mar 16 '12 at 02:00
  • @MarkDominus any more details – 89085731 Mar 16 '12 at 02:01
  • http://math.stackexchange.com/questions/67620/set-of-continuity-points-of-a-real-function –  Mar 16 '12 at 02:02
  • I have voted to close as dupe of the question Byron linked to. – Aryabhata Mar 16 '12 at 02:07
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    Not quite a duplicate: this one can be answered without the Baire Category Theorem. – Robert Israel Mar 16 '12 at 02:09
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    @MarkDominus: This doesn't work. It's possible for a function to be continuous at exactly one point. For example, $f(x) = x 1_\mathbb{Q}(x)$ is continuous at $x=0$ but at no other point. – Nate Eldredge Mar 16 '12 at 02:13
  • @RobertIsrael: Even without question 67620, I believe it is a dupe of an earlier question. Of course, I haven't been able to find it yet. – Aryabhata Mar 16 '12 at 02:38
  • @Nate-Eldredge: Thank you for the correction – MJD Mar 16 '12 at 02:51

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Suppose function $f$ is continuous at all rationals. Let $r_n$ be an enumeration of the rationals in $[0,1]$. Construct a sequence of nested closed intervals $I_n = [a_n, b_n]$ with $I_0 = [0,1]$ as follows. Given $I_n$, take a rational number $t_{n+1} \in (a_n, b_n)$ that is not $r_{n+1}$, and let $I_{n+1} = [t_{n+1}-\delta, t_{n+1}+\delta]$ where $0 < \delta < \min(|r_{n+1} - t_{n+1}|, |a_n - t_{n+1}|, |b_n - t_{n+1}|)$ and $\delta$ is small enough that $|f(x) - f(t_{n+1})| < 1/(n+1)$ for all $x \in I_{n+1}$. The intersection of these intervals contains no rationals, but does contain a point where $f$ is continuous.

Robert Israel
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  • Your Proof is brilliant. But as I say in the problem,I want to prove this function is integrable,can you help me that?thanks. – 89085731 Mar 16 '12 at 02:54
  • No. A function that is discontinuous at all irrationals can't be Riemann integrable, by the Lebesgue criterion for Riemann integrability. On the other hand, a function that is continuous at all rationals can fail to be Riemann integrable. – Robert Israel Mar 16 '12 at 15:14