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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.2012-04-28
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    @DavidMitra: oh okay. Thanks.2012-04-28
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    @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?2012-04-28
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    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](http://math.stackexchange.com/questions/106488/upper-riemann-sum-of-a-function/106500#106500) for example.2012-04-28
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    @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?2012-05-06
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    Not from just the fact that the set of discontinuities of a function is a $F_\sigma$. There are lots of $F_\sigma$ sets (including the empty set). Nothing in that statement would imply that a function is *actually* discontinuous on a given $F_\sigma$. Even if you've already proven that $t$ is continuous on the irrationals, you can't apriori say anything about the behaviour of $t$ on the rationals. But, of course, it is easy to show though that $t$ is discontinuous at any rational $q$, for $t(q)>0$ and in any nhood of $t$ you can find an irrational $p$ and thus a value $f(p)=0$.2012-05-06
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    @DavidMitra: Thanks. I am asking because, the above facts were asked in the same question that asked to show that $t(x)$ has the stated properties. Showing that $t(x)$ is continuous at the irrationals is bit daunting I think.2012-05-06

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