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Math Help - computing iterated integrals

  1. #1
    GTO
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    computing iterated integrals

    Let (X,M,\mu) and (Y,N,\nu) be measure spaces. Let X=Y=[0,1] with M=N the Borel \sigma algebra. Let \mu be Lebesgue measure and \nu be the counting measure. Define the diagonal in Xx Y as D=\{(x,x) : x \in [0,1]\}, and let 1_D be the indicator function.
    Compute the following

    \int_{X \times Y} 1_D d(\mu \times \nu), \int_X \int_Y 1_D d \nu d\mu, and \int_Y \int_X 1_D d \mu d\nu

    this is really hard. please help me
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    Quote Originally Posted by GTO View Post
    Let (X,M,\mu) and (Y,N,\nu) be measure spaces. Let X=Y=[0,1] with M=N the Borel \sigma algebra. Let \mu be Lebesgue measure and \nu be the counting measure. Define the diagonal in Xx Y as D=\{(x,x) : x \in [0,1]\}, and let 1_D be the indicator function.
    Compute the following

    \int_{X \times Y} 1_D d(\mu \times \nu), \int_X \int_Y 1_D d \nu d\mu, and \int_Y \int_X 1_D d \mu d\nu

    this is really hard. please help me
    Start by visualising it geometrically. The function 1_D(x,y) on the unit square is 1 on the diagonal y=x and 0 everywhere else. Now think about integration with respect to the two measures. When you integrate a function f(x,y) with respect to \mu you are taking the Lebesgue integral of f as a function of x, keeping y constant. In particular, \int_X1_D(x,y)\,d\mu(x) = 0 (because 1_D(x,y) = 0 for almost all x). But when you integrate a function f(x,y) with respect to \nu you are taking the integral of f (as a function of y) with respect to counting measure, keeping x constant. In particular, \int_Y1_D(x,y)\,d\nu(y) = 1 (because 1_D(x,y) = 1 when y=x, and 1_D(x,y) = 0 at all other values of y).

    Therefore \int_X \int_Y 1_D(x,y)\, d \nu(y) d\mu(x) = \int_X1\,d\mu(x) = 1, but \int_Y \int_X 1_D(x,y)\, d \mu(x) d\nu(y) = \int_Y0\,d\nu(y) = 0.

    Now what about the integral with respect to the product measure, \int_{X \times Y} 1_D\, d(\mu \times \nu) ? If this were finite then Fubini's theorem would say that it must be equal to both of the above repeated integrals. But those two integrals are different, so the conclusion must be that \int_{X \times Y} 1_D\, d(\mu \times \nu) is not finite.
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