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

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

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

    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 06:54 PM.
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