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Proving that a function is Riemann integrable

  1. Feb 6, 2006 #1
    "Let (r_n) be any list of rational numbers in [0,1]. Let (a_n) be a sequence such that 1>a_1>a_2>...>a_n>... converging to 0. Let f be defined such that

    f=0 if x is irrational
    =a_n if x is equal to r_x

    Prove that f is Riemann integrable."

    We are doing integrals from Darboux' approach, so no tagged partitions or whatnot. I have to somehow show that the upper sum L(f,P) and lower sum L(f,P), for some partition P, differ by less than a given e>0.

    L(f,P) = 0 for any partition, so |U(f,P) - L(f,P)|=U(f,P). I have no idea how to proceed, because if p is a partition of n equal parts, [tex]U(f,P) \leq \sum_{i=0}^{n}\frac{1}{n}[/tex] which diverges. I am unable to refine the inequality.
     
  2. jcsd
  3. Feb 6, 2006 #2

    NateTG

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    1.You need to take advantage of the fact that the [itex]a_n[/itex] converge to zero.
    2.You don't know that L(f,P) is going to be 0 since the [itex]a_n[/itex] can be less than zero.

    Hint:
    If you know that for n>N, a_n < 1/k can you produce an upper bound for U(f,P) as the norm of the partition goes to zero?
     
    Last edited: Feb 6, 2006
  4. Feb 6, 2006 #3
    a_n is a strictly decreasing sequence converging to 0, a_n is therefore never less than 0.

    I think I have an idea on how to proceed, regardless.
     
  5. Feb 6, 2006 #4

    NateTG

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    <opens mouth, switches feet>
    Oh, I missed that. It makes things a bit more convenient.
     
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