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I've seen proofs stating Dirichlet is not Riemann Integrable, and the intuition for that was that the function was discontinuous in every point. My doubt is why the Thomae's function doesn't act in a similar way? I would really like to know the intuition that could guide through the usual proof of this fact.

Any help would be appreciated.

An old man in the sea.
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3 Answers3

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Thomae's function is defined by $f(x):={1\over q}$ when $x={p\over q}$ in lowest terms, and $f(x):=0$ otherwise, i.e., if $x$ is irrational.

Claim: This $f$ is Riemann integrable over $[0,1]$, and $\int_0^1 f(x)\>dx=0$.

Proof: Let an $\epsilon>0$ be given. There is an $n\in{\mathbb N}$ with ${1\over n}<\epsilon$. The points $x\in[0,1]$ with $f(x)\geq\epsilon$ have denominator $<n$. There are at most $$2+\sum_{k=2}^{n-1}(k-1)\leq {n^2\over2}$$ such "bad" points. Consider now the partition ${\cal P}_N$ of $[0,1]$ into $N>{n^2\over\epsilon}$ equal parts of width $h:={1\over N}$. At most $n^2$ of the subintervals can contain a "bad" point $x$, and in any case $f(x)\leq1$ there. There are less than $N$ "good" intervals, and $f(x)<\epsilon$ for all $x$ in a "good" interval. It follows that $$0=L(f,{\cal P}_N)<U(f,{\cal P}_N)<n^2\cdot1\cdot\>h +N\cdot\epsilon\cdot h={n^2\over N}+\epsilon<2\epsilon\ .\tag{1}$$

As $\epsilon>0$ was arbitrary this proves the claim; and $(1)$ then immediately implies $\int_0^1 f(x)\>dx=0$, as well.

Christian Blatter
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  • Is it important to choose the width of the partition so small that at most one rational number is in a part of the interval? How did you come up with the choice $N \gt \frac{n^2}{\varepsilon}$? Does this choice satisfy this criterium? – mdcq Dec 30 '17 at 23:51
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    @philmcole: We only have to deal with the "bad" rational numbers. There are $\leq {n^2\over2}$ of them. For all other rational numbers we have $f(x)<\epsilon$ anyway. (That's the point of Thomae's function.) – Christian Blatter Dec 31 '17 at 10:14
  • Thanks! But I still don't understand why you choose $N \gt \frac{n^2}{\varepsilon}$? Wouldn't simply $N \gt \frac{n^2}{2}$ suffice? – mdcq Dec 31 '17 at 10:49
  • @philmcole: In $(1)$ it is used that $N>n^2/\epsilon$. – Christian Blatter Dec 31 '17 at 11:00
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Proof of continuity of Thomae Function at irrationals.

This can be helpful.

And anyway the idea is that the Thomae's function is continuous in all irrational numbers, but discontinuous in rational. Then, we have that the measure of rational numbers is $0$, and our function is bounded on $[0,1]$, hence it is Riemann integrable.

Community
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Mesmerized student
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A bounded function $f$ is said to be Riemann integrable if and only if $f$ is continuous almost everywhere, which is the same thing as saying that $f$ is Riemann integrable if and only if $f$ is continuous everywhere except for a null set.

The Thomae's function is not continuous in the rationals. But the rationals are a null set.

On the other hand the Dirichlet function is not continuous in the irrationals and the irrationals are not a null set, therefore it is not continuous almost everywhere, so it is not Riemann integrable.

José Carlos Santos
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PedroGonçalves
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    Your starting line is written in the form of a definition whereas in reality is a deep and difficult theorem. It is best not to use measure theory and instead apply the criterion of upper lower Darboux sums to check integrability of Dirichlet and Thomae functions. – Paramanand Singh Mar 11 '18 at 17:21