Basically was arbitrary.
Theorem. If is Riemann integrable then its Riemann sums satisfy the Cauchy Criterion for the existence of the integral.
Cauchy Criterion: For every , there exists such that if is a partition of , with and and are Riemann sums of on , then .
Proof. Suppose is Riemann integrable on . Then for all , there exists such that whenever is a Riemann sum for corresponding to a partition of . Thus and . Hence .
Also could we replace and with and . And so the inequality would then be: ? Because these are not necessarily Riemann sums.