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Math Help - Cauchy Criterion

  1. #1
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    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.
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  2. #2
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    Basically  \epsilon/2 was arbitrary.
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