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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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    @Nate-Eldredge: Thank you for the correction2012-03-16

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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.

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    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.2012-03-16