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Homework Help: Prove that a double integral exists

  1. Jul 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f(x,y) = 1 if x = 1/3 and y is rational, and let f(x,y) = 0 otherwise. Show that the double integral of f over the region Q = [0,1]x[0,1] in R2 exists (SSQ f dA exists) yet the integral from 0 to 1 of f(1/3, y) does not exist.

    (sorry for the weird way of writing, I'm not sure how to do integral signs)

    2. Relevant equations

    3. The attempt at a solution
    For the second part - a function is integrable if its set of discontinuities is finite or has zero content. However, irrational numbers are infinitely dense in any interval so the set of discontinuities are infinite and they cnanot be covered by a finite amoutn of intervals with arbitrarily small lengths, so the function f(1/3, y) is not integrable

    But the second part is kind of counter-intuitive to me. No matter how i partition that square, as long as the partitions have real lengths, then the lower riemann summ will always be 0 (since every square that contains a point where x= 1/3 and y is rational, will also contain points where x is not 1/3 and/or y is not rational) and the upper riemann summ will always be non-zero (since, any partitioning into squares of real lengths, will result in at least some squares containing points where x = 1/3 and y is rational). So, the lower integral and upper integral will not agree (one is zero, one non-zero), so how can the double integal over the region exist?
  2. jcsd
  3. Jul 7, 2009 #2


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    Not every 'square' (really the partition can consist of rectangles as well) contains a point where x=1/3. How about [1/2,1]x[1/2,1]? How about [0,1/4]x[0,1]? Etc etc. If you draw small enough rectangles, very few of them will overlap x=1/3 and they can have very small total area.
  4. Jul 7, 2009 #3
    Ok, I think I know what you're saying
    The integral exists if the difference between the upper and lower riemann sums can be made arbitrarily small (There exist partitions p of Q such that Spf - spf < epsilon for any epsilon >0). I know spf is always 0, and, while Spf will not be zero for any partition into rectangles of real lengths, it can be made to approach zero by refining the partitions and making the total area of all the rectangless that contain x = 1/3 approach zero (since the area containing points where f=1 is reduced, the upper riemann sum is also reduced).

    So, for any epsilon we choose, there exists a particion p of Q such that Spf - spf < e

    is this what you meant?
  5. Jul 7, 2009 #4


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    That's exactly what I mean. If all of the rectangles have side length less than e/2, then all of the rectangles that touch x=1/3 will lie between x=1/3-e/2 and x=1/3+e/2. Hence the total area is less than e.
  6. Jul 7, 2009 #5
    It makes a lot more sense now.. :) Gotta watch out for my faulty intuition haha. Thank you so much for your help!
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