A question about Fubini's Theorem
Olivia Zamora 
Fubini's theorem says that if a function is integrable with respect to a product measure $\mu(dx, dy) = \mu_1(dx) \times \mu_2(dy)$, that is $\int_A|f(x, y)|d\mu < \infty $, then the double integral can be computed as two iterated integrals with $A = A_x\times A_y$
$$\int_{A_x}\int_{A_y}f(x, y) d\mu_1 d\mu_2 = \int_A f(x, y) d\mu = \int_{A_y}\int_{A_x} f(x, y)d\mu_2 d\mu_1$$
This is often stated as a method to compute the double integral in question. However, I often find in practice that the result is used when it is known that
$$\int_{A_x}\int_{A_y}|f(x, y)|d\mu_1d\mu_2 < \infty$$
which is used to justify an exchange of the order of integration in the computation of
$$\int_{A_x}\int_{A_y}f(x, y)d\mu_1d\mu_2$$
as
$$\int_{A_y}\int_{A_x}f(x, y)d\mu_2d\mu_1$$
However, I've always been confused by this, as it doesn't seem to be what the theorem is saying in the first place. Is it true that if one iterated integral is absolutely integrable, then it is equal to the second? And if so, why is this implied by the statement of Fubini's theorem?
$\endgroup$ 21 Answer
$\begingroup$Just recall that we have Tonelli's Theorem:\begin{align*} \iint|A(x,y)|d\mu(x)\times d\mu(y)=\int_{A_{y}}\int_{A_{x}}|A(x,y)|d\mu(x)d\mu(y)=\int_{A_{x}}\int_{A_{y}}|A(x,y)|d\mu(y)d\mu(x), \end{align*}where one of which is finite would imply another.
$\endgroup$ 1
