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Can one use these facts to show that the Thomae function (aka the popcorn function) $$ t(x) = \begin{cases} 0 & \text{if}~ x~\text{is irrational}\\ 1/n & \text{if}~x=m/n,~\text{where}~m,n\in \mathbb N,n\gt 0,\gcd(m,n) =1 \end{cases} $$

is continuous at every irrational points and discontinuous at all rational points in $\mathbb R$ ?

Facts:
(i) the set of all irrational numbers in $\mathbb R$ is not an $F_{\sigma}$ set.
(ii) the set of points of continuity of any function $f$ is a $G_{\delta}$ set.
(iii) there is no function that is discontinuous at all irrational points and continuous at all rational points .

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No. You'll have to show that $t$ is discontinuous at every rational and continuous at every irrational directly. The only thing that the facts will tell you is that $t$ is not continuous precisely on the rationals. –  David Mitra Apr 28 '12 at 19:01
    
@DavidMitra: oh okay. Thanks. –  Linda Apr 28 '12 at 19:41
    
@DavidMitra: Can I use the fact that $t(x)$ continuous at every irrational point and discontinuous at every rational point to show that $t(x)$ is Riemann integrable? –  Linda Apr 28 '12 at 20:09
    
If you can appeal to the theorem that a function is Riemann integrable iff it is bounded and the set of points where it is discontinuous has measure zero, then yes. But, you could prove the result directly. See here for example. –  David Mitra Apr 28 '12 at 20:49
    
@DavidMitra: Since $\mathbb Q$ is an $F_\sigma$ and the set of discontinuities of any function is an $F_\sigma$ set, can't I say that $t(x)$ is discontinuous at the rationals? –  Linda May 6 '12 at 4:09

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