How can I decide whether this function is integrable?

Question

A function $f$ is defined on $[0,1]$ such that for every

irrational number in $[0,1]$, $f(x)=1$ and
rational number in $[0,1]$, $f(x) = \dfrac{a-2}{a}$, where $a$ is the smallest natural number for which $ax$ is an integer.

Does $\int_0^1 f(x)dx$ exist? If yes, what's the value?

I have a hunch that it is not integrable, but somehow unable to proceed.

integrability defined in term of lower and upper sums, i.e. infimums and supremums — ali mansoor, Jan 18 '21 at 19:15
Hint: If you change $f$ at finitely many places, its supremum of lower sums and its infimum of upper sums do not change. — Hagen von Eitzen, Jan 18 '21 at 19:15
Compare your function with the Thomae's function and see whether you can modify a proof of its integrability (you can find those on MSE or somewhere else) — bjorn93, Jan 18 '21 at 20:21
I agree with bjorn93. If you understand the proof that Thomae's function is integrable, then you can imitate it in your case. (And conversely.) — GEdgar, Jan 18 '21 at 21:38

macton · Answer 1 · 2021-01-20T07:44:36.560

0

Consider $g(x) = 1-f(x)$, so it suffices to solve $$ \int_{[0,1]}f(x) dx = 1 - \int_{[0,1]} g(x) dx $$ where $g(x)$ is almost the Thomae's function as comments mentioned, and you can refer to here for a proof, you just need to modify a little.

edited Jan 20 '21 at 07:44

answered Jan 19 '21 at 09:35

macton

679

shouldn't the 1st statement be "the upper sum of $f$ is 1? – ali mansoor Jan 20 '21 at 07:04
well I am dumb, sorry :( nevertheless the second one should be correct. – macton Jan 20 '21 at 07:45

How can I decide whether this function is integrable?

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