Cauchy Criterion

**manjohn12** · March 15th 2009, 10:24 PM

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.

**manjohn12** · March 17th 2009, 12:25 PM

Basically $\epsilon/2$ was arbitrary.

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