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I have Thomae's Function defined as follows:

$f(x): [0,1] \to \mathbb R$

$f(x) = q$ if $x$ is rational and $x = p/q$, $0$ otherwise (please note that this is the usual definition of Thomae's function, just with the difference in $q$ and $1/q$

I need to show that $f$ is Lebesgue integrable and compute its integral. I know its value is $0$, however I'm unsure how to start. I think this definition of Thomaes's Function is confusing me, as we are dealing with $q$ and not $1/q$. I found this to be helpful: Integral of Thomae's function, however it involves Thomaes's Function with $f$ taking on the value $1/q$.

Any assistance would be appreciated.

EDIT: I believe this boils down to showing that the function is zero a.e. (and thus the integral is zero), but again, I'm unsure how to begin.

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Tadpole
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    You've said that the function is zero for all values except the rationals, isn't that good enough? – Rolf Hoyer Apr 12 '15 at 21:55
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    Ah, I think I know: You're saying the set with all the rational points is countable, and all countable sets have measure zero... therefore the function is zero a.e. and hence the integral is 0. Correct? – Tadpole Apr 12 '15 at 22:09
  • Yes, the hard thing would be proving Riemann integrability without knowing about measure. – Rolf Hoyer Apr 12 '15 at 22:12
  • Right, because every proof I've seen proving Riemann integrability involves Lebesque integrability criterion. Thanks for the response, it really narrowed down where I needed to look. – Tadpole Apr 12 '15 at 22:15

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