Let $\mathcal{F}$ denote the collection of all sets $S \subset X \times Y$ for which the mapping
$$
x \mapsto \chi_{S}(x, y)
$$
is $\mu$-integrable for each $y \in Y$ and the mapping
$$
y \mapsto \int_{X} \chi_{S}(x, y) d \mu(x)
$$
is $\nu$-integrable. For $S \in \mathcal{F}$ we write
$$
\rho(S) \equiv \int_{Y}\left[\int_{X} \chi_{S}(x, y) d \mu(x)\right] d \nu(y)
$$
Note $\mathcal{P}{0} \subset \mathcal{F}$ and $$ \rho(A \times B)=\mu(A) \nu(B)\left(A \times B \in \mathcal{P}{0}\right)
$$
If $A_{1} \times B_{1}, A_{2} \times B_{2} \in P_{0}$, then
$$
\left(A_{1} \times B_{1}\right) \cap\left(A_{2} \times B_{2}\right)=\left(A_{1} \cap A_{2}\right) \times\left(B_{1} \cap B_{2}\right) \in P_{0}
$$
and
$$
\left(A_{1} \times B_{1}\right)-\left(A_{2} \times B_{2}\right)=\left(\left(A_{1}-A_{2}\right) \times B_{1}\right) \cup\left(\left(A_{1} \cap A_{2}\right) \times\left(B_{1}-B_{2}\right)\right)
$$
MATH112 COURSE NOTES ：
$$
(\mu \times \nu)(R-S)=0
$$
hence
$$
\rho(R-S)=0
$$
Thus
$$
\mu{x \mid(x, y) \in S}=\mu{x \mid(x, y) \in R}
$$
for $v$ a.e. $y \in Y$, and
$$
(\mu \times \nu)(S)=\rho(R)=\int \mu{x \mid(x, y) \in S} d \nu(y) .
$$