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Math Help - Riemann Epsilon

  1. #1
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    Riemann Epsilon

    Lemma. Let  [a,b] \subseteq K \subseteq \mathbb{R} , and let  f: \mathbb{K} \to \mathbb{R} be a function. Let  P be a partition of  [a,b] , and let  \epsilon >0 be a real number. Suppose that whenever  \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 . Then these statements follow:

    1.  f is bounded on  [a,b] .

    2.  \mathcal{U}(f,P)- \mathcal{L}(f,P) \leq \epsilon

    3. If  Q is any refinement of  P , and  \mathcal{R}(f,P) and  \mathcal{R}(f,Q) are Riemann sums, then  |\mathcal{R}(f,P)- \mathcal{R}(f,Q)| \leq \epsilon

    Proof. (1) Suppose  f is unbounded. Then there is a sequence  (y_n) in  [a,b] that converges to  y \in [a,b] such that for every  n \in \mathbb{N} ,  |f(y_n)| > n . Then the set of Riemann sums of  f corresponding to  P is an unbounded set of real numbers. (2)  \mathcal{U}(f,P) and  \mathcal{L}(f,P) don't have to be Riemann sums. And since we are taking a "maximum" difference, we can get to  \epsilon . (3) I think that  \mathcal{R}(f,Q) \leq \mathcal{R}(f,P) . And so the inequaliy follows.  \diamond

    Is this correct?

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  2. #2
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    Basically is the fact that  \mathcal{U}(f,P) and  \mathcal{L}(f,P) are not Riemann sums they key idea here?
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