I am going through Durett by myself (this isn't homework), and exercize 1.7.1 asks to prove that $$\int_X \int_Y |f(x,y)| \mu_2(dy)\mu_1(dx)$$ $$\implies \int_X\int_Y f(x,y) \mu_2(dy)\mu_1(dx) = \int_{X \times Y} f d (\mu_1 \times \mu_2) = \int_Y \int_X f(x,y) \mu_1(dx)\mu_2(dy)$$
But I don't understand why this needs a proof? Fubini's theorem is stated in the book as: $f \geq 0$ or $\int_{X \times Y} |f| d(\mu_1 \times \mu_2) < \infty$, then $$\int_X\int_Y f(x,y) \mu_2(dy)\mu_1(dx) = \int_{X \times Y} f d (\mu_1 \times \mu_2) = \int_Y \int_X f(x,y) \mu_1(dx)\mu_2(dy)$$
But surely the condition $\int_X \int_Y |f(x,y)| \mu_2(dy)\mu_1(dx) < \infty$ is exactly the same as $\int_{X \times Y} |f| d(\mu_1 \times \mu_2) < \infty$? Isn't this how the product measure is defined? I would really appreciate it if someone could explain the difference, and how to proce the statement in the exercise. Thanks!
Edit: I think I understand now that the problem is we are sort of 'going backwards', i.e., Fubini shows that $\int_{X \times Y} |f| d(\mu_1 \times \mu_2) < \infty $ implies the result, so I'm guessing it wants us to show that $\int_X \int_Y |f(x,y)| \mu_2(dy)\mu_1(dx) \implies \int_{X \times Y} |f| d(\mu_1 \times \mu_2) < \infty$?
But isn't this a straightforward consequence of Fubini, since $|f(x,y)| \geq 0$, so $\int_Y |f(x,y)| \mu_2(dy)\mu_1(dx) = \int_{X \times Y} |f| d(\mu_1 \times \mu_2)$? Or is this all that they want us to say?
What you have written in the edited part is perfect. You are supposed to apply Fubini's Theorem twice, first for 'non-negative' case and then for 'integrable case'.