1. ## Cauchy Criterion

Theorem. If $f$ is Riemann integrable then its Riemann sums satisfy the Cauchy Criterion for the existence of the integral.

Cauchy Criterion: For every $\epsilon >0$, there exists $\delta >0$ such that if $P$ is a partition of $[a,b]$, with $||P|| < \delta$ and $\mathcal{R}_{1}(f,P)$ and $\mathcal{R}_{2}(f,P)$ are Riemann sums of $f$ on $P$, then $|\mathcal{R}_{1}(f,P)-\mathcal{R}_{2}(f,P)| < \epsilon$.

Proof. Suppose $f$ is Riemann integrable on $[a,b]$. Then for all $\epsilon>0$, there exists $\delta >0$ such that $|\mathcal{R}(f,P)-I| < \epsilon$ whenever $\mathcal{R}(f,P)$ is a Riemann sum for $f$ corresponding to a partition of $[a,b]$. Thus $|\mathcal{R}_{1}(f,P)-I| < \epsilon/2$ and $|\mathcal{R}_{2}(f,P)-I| < \epsilon/2$. Hence $|\mathcal{R}_{1}(f,P)- \mathcal{R}_{2}(f,P)| < \epsilon$. $\diamond$

Correct?

Also could we replace $\mathcal{R}_{1}(f,P)$ and $\mathcal{R}_{2}(f,P)$ with $\mathcal{U}(f,P)$ and $\mathcal{L}(f,P)$. And so the inequality would then be: $|\mathcal{U}(f,P)-\mathcal{L}(f,P)| < \epsilon$? Because these are not necessarily Riemann sums.

2. Basically $\epsilon/2$ was arbitrary.