Proof of Null Sets of a Product Measure, and Exchanging the Factors

Proof of (i). Let $E\in\mathcal{F}\otimes\mathcal{G}$. By Existence and Uniqueness of the Product Measure (iii), the function $x\mapsto\nu(E_{x})$ lies in $\mathcal{M}^{+}(X,\mathcal{F})$ and

$$(\mu\otimes\nu)(E)=\int_{X}\nu(E_{x})\,d\mu(x).$$

By The Lebesgue Integral and Null Sets (iv), this integral is $0$ if and only if $\nu(E_{x})=0$ for $\mu$-almost every $x\in X$. Likewise, by the second formula of Existence and Uniqueness of the Product Measure (iii), the function $y\mapsto\mu(E^{y})$ lies in $\mathcal{M}^{+}(Y,\mathcal{G})$ and

$$(\mu\otimes\nu)(E)=\int_{Y}\mu(E^{y})\,d\nu(y),$$

and by The Lebesgue Integral and Null Sets (iv) on $(Y,\mathcal{G},\nu)$ this integral is $0$ if and only if $\mu(E^{y})=0$ for $\nu$-almost every $y\in Y$.

Proof of (ii). As $s$ is $(\mathcal{G}\otimes\mathcal{F})/(\mathcal{F}\otimes\mathcal{G})$-measurable, $s^{-1}(E)\in\mathcal{G}\otimes\mathcal{F}$ for every $E\in\mathcal{F}\otimes\mathcal{G}$, and the image measure $\rho=(\nu\otimes\mu)\circ s^{-1}$ of $\nu\otimes\mu$ under $s$, the function $E\mapsto(\nu\otimes\mu)(s^{-1}(E))$ on $\mathcal{F}\otimes\mathcal{G}$, is a measure on $(X\times Y,\mathcal{F}\otimes\mathcal{G})$ (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (i)). For $A\in\mathcal{F}$ and $B\in\mathcal{G}$ we have $s^{-1}(A\times B)=B\times A$ (Rectangles, Sections and the Swap: Set Identities (iv)), so by the rectangle property of the product measure $\nu\otimes\mu$ (Product Measure (i))

$$\rho(A\times B)=(\nu\otimes\mu)(B\times A)=\nu(B)\,\mu(A)=\mu(A)\,\nu(B).$$

Thus $\rho$ is a measure on $(X\times Y,\mathcal{F}\otimes\mathcal{G})$ with the rectangle property of Existence and Uniqueness of the Product Measure (i), and the uniqueness there gives $\rho=\mu\otimes\nu$.