1. ## computing iterated integrals

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

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

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

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

Start by visualising it geometrically. The function $\displaystyle 1_D(x,y)$ 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 $\displaystyle \mu$ you are taking the Lebesgue integral of f as a function of x, keeping y constant. In particular, $\displaystyle \int_X1_D(x,y)\,d\mu(x) = 0$ (because $\displaystyle 1_D(x,y) = 0$ for almost all x). But when you integrate a function f(x,y) with respect to $\displaystyle \nu$ you are taking the integral of f (as a function of y) with respect to counting measure, keeping x constant. In particular, $\displaystyle \int_Y1_D(x,y)\,d\nu(y) = 1$ (because $\displaystyle 1_D(x,y) = 1$ when y=x, and $\displaystyle 1_D(x,y) = 0$ at all other values of y).
Therefore $\displaystyle \int_X \int_Y 1_D(x,y)\, d \nu(y) d\mu(x) = \int_X1\,d\mu(x) = 1$, but $\displaystyle \int_Y \int_X 1_D(x,y)\, d \mu(x) d\nu(y) = \int_Y0\,d\nu(y) = 0$.
Now what about the integral with respect to the product measure, $\displaystyle \int_{X \times Y} 1_D\, d(\mu \times \nu)$ ? 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 $\displaystyle \int_{X \times Y} 1_D\, d(\mu \times \nu)$ is not finite.