Proof of Cauchy Criterion for Riemann Integrals

[S.N.]

Homework Statement

Some proofs I've looked at vary, but they generally follow the format show here: http://en.wikibooks.org/wiki/Real_Analysis/Riemann_integration#Theorem_.28Cauchy_Criterion.29

This isn't a question about an exercise, but rather a request for a clarification or a way of putting part of the proof in more understandable terms. My issue is the <= direction of the proof. I don't understand much about it. We're picking a $$\delta$$, sure, but then I don't see where the 1/n comes from, or what the business with m<n is, or just about anything in the 'reverse direction' of this proof. Does anyone have maybe a link to a proof with more heuristics, or if you have the time and patience, maybe a more intuitive way of describing that's going on?

Thanks for any help.

 

[lippyka]

Homework Equations The Attempt at a Solution I think the 1/n comes from the fact that if we choose \delta to be smaller than 1/n, then it must also be smaller than 1/m for any m>n. This is because 1/m is larger than 1/n, so \delta will have to be smaller than both. The business with m<n is to show that the sum of the \delta's is still bounded by the upper sum of the partition, which is the same as the supremum of the upper sums, or U(f,P). We do this by picking an arbitrary m and showing that \sum_{k=1}^{n} \delta_k <= U(f,P). To do this, we use the fact that \delta_k < 1/m for all k<=m, and \delta_k < 1/n for all k>m. This implies that \sum_{k=1}^{n} \delta_k < \sum_{k=1}^{m} \frac{1}{m} + \sum_{k=m+1}^{n}\frac{1}{n} = m\cdot\frac{1}{m} + (n-m)\cdot\frac{1}{n}= 1. Since U(f,P) is an upper bound, then \sum_{k=1}^{n} \delta_k <= U(f,P).