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Integration, Zero Content

  1. Jan 2, 2008 #1
    1) Let f(x,y)=1 for x=0, y E Q
    f(x,y)=0 otherwise
    on R=[0,1] x [0,1]

    Then
    1 1
    ∫ ∫ f(x,y) dxdy = 0 exists
    0 0
    but
    1 1
    ∫ ∫ f(x,y) dydx does not exist
    0 0



    I don't understand why the first iterated Riemann integral exists, but the second interated integral does not exist, can someone please explain (perhaps in terms of the concept of zero content) ?

    Thank you!
     
    Last edited: Jan 3, 2008
  2. jcsd
  3. Jan 2, 2008 #2
    None of these integral exixts. Simply because the function [tex]f(x,y)[/tex] is not integrable since [tex]y \in \mathbb{Q}[/tex]
     
  4. Jan 2, 2008 #3
    I could be way off, but I'll try.

    In the first case, you're integrating over x first, so y is fixed. For x in [0, 1], f(x,y) is 0 except at x = 0. So you're integrating a function that's zero except on a set of measure zero, so the first part of the iterated integral is 0. Then for the second part, you're just integrating 0, so the whole integral is 0.

    In the second case, you're integrating over y first, so x is fixed. For x fixed and not equal to 0, you get the zero function, but for x = 0, you're integrating the characteristic function of the rationals, which isn't integrable.

    That could be way off though. I need to work on this particular area more.
     
  5. Jan 2, 2008 #4

    Dick

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    Riemann integrable, right? Just think of upper and lower sums for the dy integration (at x=0). The dx integration is zero. The set in R^2 has zero content. It's just the iterated integral that has problems. In the Lebesgue sense, both integrals exist and are zero.
     
    Last edited: Jan 3, 2008
  6. Jan 3, 2008 #5
    Yes, I am using the definition of Riemann integral.

    But I don't understand your point. Which set in R^2 that you are referring to has zero content? And are you commenting on the first (dxdy) or the second (dydx) iterated integral?

    Thanks!
     
  7. Jan 3, 2008 #6

    Dick

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    The set I'm referring to is the set where the function is nonzero. But doing this via iterated integrals doesn't have much to do with that. And I'm talking about the second iterated integral. The integral of f(0,y)dy does not exist (since it's 1 for y rational and 0 for y irrational - the upper sums are all 1 and the lower sums are all zero). The integral of f(x,y)dy does exist for all x not equal to zero (since the function is identically 0).
     
  8. Jan 9, 2008 #7
    I know that S={(0,y)|0<y<1} is where f is discontinuous and S has zero content in R2.

    To claim that
    1
    ∫ f(x,y) dx = 0
    0
    which of the following theorem are you actually using? (I am confused whether this is a 1-dimensional or a 2-dimensional integral)

    Theorem: If f is bounded on [a,b] and the set of pionts in [a,b] at which f is discontinuous has zero content, then f is integrable on [a,b].

    Theorem: Suppose f is a bounded function on the rectangle R. If the set of points in R at which f is discontinuous has zero content, then f is integrable on R.

    Could somebody help? Thanks!!
     
  9. Jan 9, 2008 #8
    Theorem: If f is bounded on [a,b] and the set of pionts in [a,b] at which f is discontinuous has zero content, then f is integrable on [a,b].

    That one. The whole point of the exercise, as far as I can tell, is that although the integral of f(x,y) over the rectangle exists (and is 0), the iterated integrals don't agree with that integral.
     
  10. Jan 9, 2008 #9
    For each fixed y, f(x,y) is zero except at a single point where it is (possibly) 1, so the integral
    1
    ∫ f(x,y) dx should be zero, I think.
    0

    But there are "infinitely many" fixed y's in the interval [0,1], then how can
    1
    ∫ f(x,y) dx = 0 ?
    0

    I am still very confused. Can someone please explain in greater detail? I really want to understand it. I appreciate for your help!
     
  11. Jan 11, 2008 #10
    Can someone kindly help?
     
  12. Jan 11, 2008 #11

    Dick

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    I think we've been over this before. You just repeated the same statement with two different answers. What do you really think? What is the integral of f(0,y)dy? Really, I want an answer. Would this answer have any impact on one of the forms of the iterated integral? This is open to some debate. But I would like your opinion.
     
  13. Jan 14, 2008 #12
    I believe that
    1
    ∫ f(x,y) dx = 0
    0
    We treat y as a constant here, but the thing that I am in trouble with is that there are inifinte number of possible y's in [0,1], would this affect the answer?



    Also, if
    1
    I f(0,y) dy
    0
    does not exist (I believe so), why does it imply that
    1
    I f(x,y) dy
    0
    does not exist as well?


    Thanks!
     
  14. Jan 14, 2008 #13

    Dick

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    For x not equal zero the dy integral exists and is zero. Let F(x) be the integral of f(x,y)dy. Then F(0) is undefined and F(x)=0 for x not equal to zero. The iterated integral is integral F(x)dx. Is an integral undefined if the function is undefined at a single point? Technically, I suppose so, since it's hard to define an upper or lower sum that includes that point, but you'll have to check the fine print in your definition. That's all that's wrong with the iterated integral.
     
  15. Jan 17, 2008 #14
    1
    ∫ f(x,y) dy = F(x) exists for x not =0
    0

    1
    ∫ f(0,y) dy = F(0) does not exist
    0

    Theorem: If f is bounded on [a,b] and continuous at all except finitely many points in [a,b], then f is (Riemann) integrable on [a,b].

    In our case, x=0 is just a single point (finitely many points!!)
    So actually
    1 1
    ∫ ∫ f(x,y) dydx
    0 0
    1
    =∫ F(x) dx exists as well and equal zero, right?
    0


    If so, then my claim in the first post that
    1 1
    ∫ ∫ f(x,y) dydx does not exist must be wrong
    0 0
     
  16. Jan 17, 2008 #15

    Dick

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    That's exactly what I asked you. Generally the non-existence of a function at a single point doesn't mean the integral is undefined. That's why I said you have to check the fine print in the integral definition. Judging from the theorem you quoted, both iterated integrals are defined and equal to zero.
     
  17. Jan 18, 2008 #16
    I know this is off topic

    but Dick, theres is a topic in the announcement forum in which you are heavilly complimented for your help with homework. You should check it out.
     
  18. Jan 23, 2008 #17
    Actually, would
    1
    ∫ F(x) dx be an improper intergral?
    0
    (improper since the function F(x) is undefined at 0 and only at 0, and 0 is part of our integration range)
    Now, is an improper integral a Riemann integral?

    Also, I asked my tutors and people from other forums and I received answers at two opposite ends (I am very confused, too) about the existence of the following integral in the Riemann sense,
    1 1
    ∫ ∫ f(x,y) dydx
    0 0

    Would anyone be able to provide an affirmative answer? More discussions are welcomed...

    Thanks again!
     
  19. Jan 23, 2008 #18

    Dick

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    You're welcome, but this whole thread is getting overworn. In any realistic sense, as opposed to technically legalistic sense, both iterated integrals exist and are equal to zero.
     
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