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The following is the Thomae function:

$$ f(x) = \begin{cases} \frac{1}{q} & if \quad x = \frac{p}{q}, p \in \mathbb{Z}, q \in \mathbb{N},\text{ gcd(p, q) = 1 } \\ 0 & if \quad x\in \mathbb{Q}^c \quad or \quad x=0. \end{cases} $$

My professor said that integrals and derivatives do not kill each other for Thomae function because $\int_{0}^{x} f(t)dt = 0$ and $\frac {d}{dx} \int_{0}^{x} f(t)dt = 0.$

My question is:

I tried to calculate $\int_{0}^{x} f(t)dt$ and I got $\frac{x}{q}$ and not $0.$ Could anyone show me the detailed calculation of this please?

José Carlos Santos
  • 440,053

1 Answers1

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For each partition $P$ of $[0,x]$, the lower sum of $f$ with respect to $P$ is $0$. Therefore, if $f$ is integrable, $\int_0^xf(t)\,\mathrm dt=0$.

And $f$ is integrable since it is bounded and the set of points at which it is discontinuous is countable.

José Carlos Santos
  • 440,053
  • What theorem are you using ? –  Jun 16 '20 at 22:05
  • A bounded real function whose domain is an interval $[a,b]$ of $\Bbb R$ is Riemann-integrable if and only if the set$${x\in[a,b]\mid f\text{ is discontinuous at }x}$$has measure $0$. But I can prove that $f$ is integrable without it too. – José Carlos Santos Jun 16 '20 at 22:10
  • But I took this information before studying Lebesgue measure ..... so do you have a theorem without using the measure notion? –  Jun 16 '20 at 22:12
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    You will find here a proof (without Lebesgue measure) of the fact that a bounded real function whose domain is an interval $[a,b]$ of $\Bbb R$ is Riemann-integrable if it has only a countable set of points at which it is discontinuous. Or you can use the fact that, for every $\varepsilon>0$, there are only finitely many numbers $y$ on $[0,x]$ such that $f(y)>\varepsilon$ to prove that $\int_0^xf(t),\mathrm dt=0$. – José Carlos Santos Jun 16 '20 at 22:21