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Integrability of multiple integrals

  1. Jan 27, 2008 #1
    Q1) Let f(x,y)=3, if x E Q
    f(x,y)=2y, if y E QC

    Show that
    1 3
    ∫ ∫ f(x,y)dydx exists
    0 0
    but the function f is not (Riemann) integrable over the rectangle [0,1]x[0,3]


    I proved that the iterated integral exists and equal 9, but I am completely stuck with the second part (i.e. to prove that the function f is not integrable over the rectangle [0,1]x[0,3] ), how can I do it? Can someone please help me?

    Thanks!
     
  2. jcsd
  3. Jan 28, 2008 #2
    May someone be nice enough to answer this theoretical question?
     
  4. Jan 28, 2008 #3

    HallsofIvy

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    It would help if you would ask a plausible question. "f(x,y)=3, if x E Q
    f(x,y)=2y, if y E QC" doesn't make sense. What is f(2, [itex]\pi[/itex) where x is rational and q is irrational? What is f(x,y) if x is not rational? What is f(x,y) if y is not irrational?
     
  5. Jan 29, 2008 #4
    Oh sorry, it should be y E Q instead of x E Q.
    i.e.
    Let f(x,y)=3, if y E Q
    f(x,y)=2y, if y E QC

    and I am having trouble with proving that f is not (Riemann) integrable over the rectangle [0,1]x[0,3]
     
  6. Jan 29, 2008 #5

    HallsofIvy

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    Okay so that is really the same as f(y)= 3 if y[itex]\in[/itex]Q, f(y)= 2y if y[itex]\in[/itex]QC. What theorems do you have to work with? Since that function is discontinuous at every value of y except 3/2, It's set of discontinuities is not a null set and so it is not Riemann integrable.
     
  7. Jan 29, 2008 #6
    Theorem:
    f is Riemann integrable iff for all epsilon>0, there exists a partition P s.t. upper Riemann sum - lower Riemann sum < epsilon

    How can I prove using this theorem?
     
  8. Jan 29, 2008 #7

    HallsofIvy

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    Now matter how small the cells in your partition are, they all include points with rational y coordinate and points with irrational y coordinates. As long as [itex]y\le 1[/itex], 2y< 3 so the highest value of the function in that cell is 3. Since that is true in any cell, the "upper sum" for any partition is 3. Also, the smallest value of the function in any cell is 2y* where y* is the y at which the function has its smallest value in that cell. The "lower sum" then is 1 (the integral of 2y over the entire interval) for any partition. Taking the limit as the partition size goes to 0, we still get 3 as the limit of the upper sums and 1 as the limit of the lower sums. Since those two limits are not the same, the function is not Riemann integrable on the interval.
     
  9. Jan 30, 2008 #8
    But for y=sqrt3=1.71, which is in the retangle [0,1]x[0x3]
    f(x,y)=2(sqrt3)=3.42 > 3
    So the max value is not always 3, right?
     
  10. Jan 30, 2008 #9

    HallsofIvy

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    Oh, blast! I was looking at [itex]0\le x\le 1[/itex], [itex]0\le y\le 1[/itex]!
     
  11. Jan 30, 2008 #10
    The question says "prove the function f is not (Riemann) integrable over the rectangle [0,1]x[0,3]"
     
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