# Cauchy Criterion

• Mar 15th 2009, 10:24 PM
manjohn12
Cauchy Criterion
Theorem. If $\displaystyle f$ is Riemann integrable then its Riemann sums satisfy the Cauchy Criterion for the existence of the integral.

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

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

Correct?

Also could we replace $\displaystyle \mathcal{R}_{1}(f,P)$ and $\displaystyle \mathcal{R}_{2}(f,P)$ with $\displaystyle \mathcal{U}(f,P)$ and $\displaystyle \mathcal{L}(f,P)$. And so the inequality would then be: $\displaystyle |\mathcal{U}(f,P)-\mathcal{L}(f,P)| < \epsilon$? Because these are not necessarily Riemann sums.
• Mar 17th 2009, 12:25 PM
manjohn12
Basically $\displaystyle \epsilon/2$ was arbitrary.