Krdytkyu

up vote 17 down vote favorite 13 Let $f:[a,b]to mathbb{R}$ be a bounded function and $A$ be the set of its discontinuities. I am asking for a (direct) proof that if $A$ is countable then $f$ is Riemann integrable in $[a,b]$ that doesn't explicitely, or implicitly, require the notion of sets of measure $0$ ( and of course without the use of the Lebesgue Criterion). One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that $A$ is countable. A proof that can't be trivially altered so that it holds even if $lambda(A)=0$ EDIT: Here is the proof of WimC with all the details: Let $epsilon>0$ and $$D=left{d_1,d_2,...right}subseteq A$$ be the ...