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Thread: Riemann Sum

  1. #1
    Dec 2008

    Riemann Sum


    I am drawing a blank here. I know it has to do with the fact that the rationals are dense in the reals, but somehow it is not coming together for me. Any idea?

    Let f : [ 0,1] \times [0,1] \rightarrow \mathbb{R} be defined by f(x, y)=\begin{Bmatrix} 1 & x\text{ rational} \\ 2y & x \text{ irrational.} \end{Bmatrix}
    Show that the iterated integral \int_0^1 [\int_0^1 f(x,y)dy]dx exists but that f is not integrable.

    I am supposed to use the following to prove the part where it is not integrable:

    Suppose f: R \rightarrow \mathbb{R} is bounded form the sum S_n=\sum_{jk=0}^{n-1} f(C_{jk})\triangle x \triangle y = \sum_{jk=0}^{n-1} f(C_{jk}) \triangle A.
    If the sequence (S_n)_n converges to a limit S as n \rightarrow \infty and if the limit S is the same for any choice of sample points C_{jk} in R_{jk}, then we say that f is integrable over R.

    In other words, I can't use the uppersum not equal to the lower sum strategy since it is not cover in class which I am taking now.

    Any help is appreciated. Thanks for your time!
    Last edited by chabmgph; Dec 13th 2008 at 07:54 PM.
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