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Let $f:[0,1]\to \mathbb{R}$ $$ f(x)=\begin{cases}x & x\ \mbox{is rational} \\ -x & x\ \mbox{is irrational} \end{cases} $$ Prove that the function $f$ is not integrable.

Use darboux sums and upper and lower integrals in your proof.

I think that the reason it is not integrable is because the lower and upper integral are not equal. I think this is true because the limit of the lower darboux sum does not equal the limit of the upper darboux sum. I think this is true because the limits do not exist, but I do not know how to prove that in this case.

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    $\begingroup$ Well, as $\mu([0,1]\cap\mathbb Q)=0$, the integral is $-1/2$... Please specify your integral. $\endgroup$ – user2345215 Nov 16 '14 at 21:20
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    $\begingroup$ I think she is talking about the Riemann integral. HINT: Look at Riemann sums and their connection to the integral. $\endgroup$ – GenericNickname Nov 16 '14 at 21:22
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    $\begingroup$ The first parts of what you think are correct. But the limits of the upper and the lower Darboux sums exist. They are just not equal. For an interval $[a,b] \subset [0,1]$, what are $\inf \{ f(x) : x \in [a,b]\}$ and $\sup \{ f(x) : x\in [a,b]\}$? $\endgroup$ – Daniel Fischer Nov 16 '14 at 21:59
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That is not the Dirichlet function. Anyway, let's work with the function you gave. If you take a partition $P$ of $[0,1]$ notice that the upper sum of $f$ over that partition is greater than 1 and the lower sum of $f$ over that partition is minor than -1. Consequently $$U(f,P)-L(f,P)>2$$ i.e, the Cauchy criteria cannot be applied, so the function is not integrable.

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