The definition of integrability, as I understand it, is that $\forall \epsilon > 0$, there exists a partition $P$ such that the difference between lower and upper Darboux sums of $f$ with respect to this parition is within $\epsilon$.
With this said, many proofs involve exploiting the fact that we could write $\forall \epsilon > 0$, there is a partition $P$ such that the upper and lower sums are within $\frac{\epsilon}{2}$. Taking the sum eventually yields a result that is within $\epsilon$.
My question is: how can we reason that the difference is $\frac{\epsilon}{2}$? Is the argument that, since $\epsilon$ is any real number greater than $0$, surely dividing it by $2$ would yield another number greater than $0$, which is effectively our new $\epsilon$? I don't know if I'm quite following such logic, because we nonetheless end up with $\epsilon$ in the end. It seems like the $\frac{\epsilon}{2}$ is a finer point than $\epsilon$, though our lower bound at $0$ probably implies we can find an infinite number of irrational numbers $> 0$ such that this is true.