For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique

If $E=A\times B$ is a measurable rectangle, then $(A\times B{)}_x=\{\begin{array}{ll}B,& x\in A,\\ \mathrm{\varnothing },& x\mathrm{\notin }A,\end{array}$ so $(\mu \times \nu )(A\times B)={\int }_X\nu ((A\times B{)}_x)d\mu ={\int }_X\nu (B){\mathbf{1}}_A(x)d\mu =\mu (A)\nu (B)\mathrm{.}$ This is the rectangle formula.

Let $E_1,E_2,\dots$ be pairwise disjoint measurable subsets of $X\times Y$, and put $F_N:={\bigcup }_{k\le N}{E}_k$. Then $(F_N{)}_x={\bigcup }_{k\le N}(E_k{)}_x$ is a disjoint union, so $\nu ((F_N{)}_x)={\sum }_{k\le N}\nu ((E_k{)}_x)$ for every $x$. Therefore $(\mu \times \nu )(F_N)={\int }_X{\sum }_{k\le N}\nu ((E_k{)}_x)d\mu \mathrm{.}$ Since $F_N\uparrow {\bigcup }_{k\ge 1}{E}_k$, [L2] gives $(\mu \times \nu )\left({\bigcup }_{k\ge 1}{E}_k\right)={\sum }_{k\ge 1}(\mu \times \nu )(E_k)\mathrm{.}$ Thus $\mu \times \nu$ is a measure.

By [A2] and step 1.1, each rectangle $X_n\times Y_n$ has finite product measure $(\mu \times \nu )(X_n\times Y_n)=\mu (X_n)\nu (Y_n)<\mathrm{\infty },$ and ${\bigcup }_n(X_n\times Y_n)=X\times Y\mathrm{.}$ Hence $\mu \times \nu$ is sigma-finite.

Let $\rho$ be another measure on $\mathcal{A}\otimes \mathcal{B}$ with the same rectangle formula. Step 1.1 shows that $\rho$ and $\mu \times \nu$ agree on the generating pi-system of measurable rectangles, and step 2.1 gives the required sigma-finite exhaustion. Therefore [L3] implies $\rho =\mu \times \nu$ on all of $\mathcal{A}\otimes \mathcal{B}$. This proves existence, the rectangle formula, sigma-finiteness, and uniqueness.