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.
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.