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Let $f:[0,1]to mathbb{R}$, $f(x)=0$ if $xnotin mathbb{Q}$ and $f(x)=frac{1}{q}$ if $x=frac{p}{q}$, $p,q$ coprime. $p$ integer, $q$ natural.

I want prove that $f$ is Riemann integrable.

I have a doubt.
Let $P=left{frac{1}{n},frac{2}{n},ldots, frac{n-1}{n},1right}$
a partition.

Now, $U(f,P)=frac{1}{n}sum_{i=1}^n M_i=frac{1}{n} sum_{i=1}^nsupleft{f(x): xin [frac{i-1}{n},frac{i}{n}]right}=frac{1}{n} sum_{i=1}^{n} frac{1}{n}=frac{1}{n} nfrac{1}{n}=frac{1}{n}leq epsilon$ with $nto infty$.
Therefore $inf_{P} U(f,P)=0.$

It is correct?...

If $xnotin mathbb{Q}$ and $x_n =frac{p_n }{q_n}to x$ $(p_n ,q_n)in mathbb{Z}timesmathbb{Z}$ then of course we have $q_n to infty$ and therefore $f(x_n) =q_n^{-1}to 0$ thus $f$ is continuous at $x.$ Thus the set of points of discontinuity of $f$ has Lebesgue measure zero and therefore $f$ is Riemann integrable.

No, it is not correct. You are assming that$$supleft{f(x),middle|,xinleft[frac{i-1}n,frac inright]right}=frac1n.$$This is false. Take, for instance, $n=3$ and $i=2$. Then $dfrac1n=dfrac13$, but$$supleft{f(x),middle|,xinleft[frac13,frac23right]right}=fleft(frac12right)=frac12.$$