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Homework Help: Reimann Integral definition confusion.

  1. May 22, 2012 #1
    My lecture notes say:

    Let f:[a,b]->R be bounded.
    F is said to rienmann integrable if:

    [tex] L=\int_{a}^{b} f(x)=U [/tex]

    where :




    but everywhere else(internet) there's a definition with epsilon.
    I have the epsilon stuff later under "riemann criterion" so was wondering if the above definition is okay or I copied something wrong
  2. jcsd
  3. May 22, 2012 #2
  4. May 22, 2012 #3
    Never heard of a darboux integral will this definition suffice if asked for a riemann integral?
  5. May 22, 2012 #4


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    Actually, this is an equality in the limit, i.e., you need to be able to make the difference

    L-U as small as possible, i.e., show that there are partitions such that, for all e>0,

    L-U<e .

    Or, if you really want to go over the top, use net-convergence on the partitions.
  6. May 22, 2012 #5


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    Yes, the "epsilon stuff" is hidden in the sup and inf- they are limits. If they are equal, then f is "integrable" and their common value is the integral of f.

    Here is an example in which it does NOT work and so f is not integrable: let f(x)= 0 if x is rational and 1 if x is irrational. Now, divide the interval from a to b into n sub-intervals. Inside any such sub-interval, there exist both rational and irrational numbers so f(x) has values of both 0 and 1 inside each interval. That means that each [itex]L(f, P)[/itex], the sum of the minimum height and base of each interval, is 0 and each [itex]U(f, P)[/itex], the sum of the maximum height and base of each interval is 1. The "sup" is 1 and the "inf" is 0. They are not the same so f is not integrable.
  7. May 22, 2012 #6
    Darboux sum is important here.

    We define upper Darboux sum and lower Darboux sum, and say that function is Riemann-integrable iff [itex]\displaystyle\sup_{P}L_{f,P} = \displaystyle\inf_{P}U_{f,P}[/itex] iff [itex]\forall \varepsilon>0 ~\forall P ~ \exists \delta>0~ \mbox{if} ~\lambda(P)<\delta ~\mbox{then} ~|U_{f,P}-L_{f,P}|<\varepsilon[/itex], where [itex]\lambda =
    \displaystyle\max_{1\le k \le n} |x_{k}-x_{k-1}|[/itex]

    We supposed that [itex]f[/itex] is bounded.
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