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Integration, Zero Content

  1. Jan 1, 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 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!
     
  2. jcsd
  3. Jan 2, 2008 #2
    The dirichlet function d(y) = 1 y in Q, d(y) = 0 y not in Q is not Riemann integrable.
    So for fixed x = 0, f(x,y) = d(y) is not integrable with respect to y.

    But [tex]\int_0^1 f(x,y) dx = 0[/tex], (regardless if f(0,y) = 1 or 0) so you can integrate the first expression.
     
  4. Jan 3, 2008 #3
    I haven't learnt about the dirichlet function, so I don't quite get your points. Are there any other ways of explaining? Anyone want to give it another try?

    Thanks!
     
  5. Jan 3, 2008 #4

    EnumaElish

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    Homework Helper

    Please do not cross post.
     
  6. Jan 21, 2008 #5
    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, am I 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

    I am not too sure about this, could someone please give an affirmative answer?
     
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