Let $\mu_x$ and $\mu_y$ be measures on $X$ and $Y$ respectively, let $\mu$ be the product measure $\mu_x \otimes \mu_y$ and let $f(x,y)$ be $\mu$ integrable on $A\subset X\times Y$ Then $$ \int_A f(x,y) d\mu = \int_X\left(\int_{A_x} f(x,y) d\mu_y\right) d\mu_x = \int_Y\left(\int_{A_y} f(x,y) d\mu_x\right) d\mu_y$$ where $$A_x = \{y\mid (x,y)\in A\} , A_y = \{x\mid (x,y)\in A\}$$

Proof: Assume for now that $f(x,y)\geq 0$ Consider the set $$U = X\times Y\times\mathbb{R}$$ equipped with the measure $$\mu_u = \mu_x \otimes \mu_y \otimes \mu^1 = \mu \otimes \mu^1 = \mu_x \otimes \lambda$$ where $\mu^1$ is ordinary Lebesgue measure and $\lambda = \mu_y \otimes \mu^1$ Also consider the set $W\subset U$ defined by $$W = \{(x,y,z)\mid (x,y)\in A, 0\leq z\leq f(x,y)\}$$ Then $$\mu_{u}\left(W\right) = \int_A f(x,y) d\mu$$ And $$\mu_{u}\left(W\right) = \int_X \lambda\left(W_x\right) d\mu_x$$ where $$W_x = \{(y,z)\mid (x,y,z)\in W\}$$ However, we also have that $$\lambda\left(W_x\right) = \int_{A_x} f(x,y) d\mu_y$$ Combining the last three equations gives us Fubini's theorem. To remove the restriction that $f(x,y)$ be nonnegative, write $f$ as $$f(x,y) = f^{+}(x,y) - f^{-}(x,y)$$ where $$f^{+}(x,y) = \frac{\vert f(x,y)\vert + f(x,y)}{2}, f^{-}(x,y) = \frac{\vert f(x,y)\vert - f(x,y)}{2}$$ are both nonnegative.