Proof of Existence and Uniqueness of the Product Measure

Throughout we use the $[0,\infty]$ conventions of Measure, Measure Space, and Probability Measure and, for functions with values in $[0,\infty]$, the notion of measurability and the integral of Lebesgue Integral of a Nonnegative Measurable Function. Let $\mathcal{R}$ denote the family of measurable rectangles $A\times B$ ($A\in\mathcal{F}$, $B\in\mathcal{G}$) of Product Sigma-Algebra; $\mathcal{R}$ is a $\pi$-system in the sense of Dynkin's Pi-Lambda Theorem, since $(A\times B)\cap(A'\times B')=(A\cap A')\times(B\cap B')$, it contains $X\times Y$, and by definition it generates $\mathcal{F}\otimes\mathcal{G}$ (Generated Sigma-Algebra).

Step 0 (disjoint decompositions). By $\sigma$-finiteness write $X=\bigcup_i X_i$ with $\mu(X_i)<\infty$ and $Y=\bigcup_n Y_n$ with $\nu(Y_n)<\infty$ ($i,n$ ranging over the natural numbers). Setting $C_1=X_1$, $C_i=X_i\setminus(X_1\cup\dots\cup X_{i-1})$ and similarly $D_n\subseteq Y_n$, we obtain pairwise disjoint sets $C_i\in\mathcal{F}$, $D_n\in\mathcal{G}$ with $\bigcup_iC_i=X$, $\bigcup_nD_n=Y$, $\mu(C_i)\le\mu(X_i)<\infty$, $\nu(D_n)\le\nu(Y_n)<\infty$ (monotonicity of measures). Write $\nu_n(B)=\nu(B\cap D_n)$, a finite measure on $(Y,\mathcal{G})$.

Step 1 (sections). For $E\subseteq X\times Y$ and $x\in X$ let $E_x=\{y\in Y:(x,y)\in E\}$. The class of $E$ with $E_x\in\mathcal{G}$ for every $x$ is a $\sigma$-algebra — sections commute with complements, $((X\times Y)\setminus E)_x=Y\setminus E_x$, and with countable unions — and it contains $\mathcal{R}$, since $(A\times B)_x$ equals $B$ if $x\in A$ and $\emptyset$ otherwise. Hence it contains $\mathcal{F}\otimes\mathcal{G}$.

Step 2 (measurability of section masses). Let $\nu'$ be any finite measure on $(Y,\mathcal{G})$. We claim that for every $E\in\mathcal{F}\otimes\mathcal{G}$ the real-valued function $x\mapsto\nu'(E_x)$ (bounded by $\nu'(Y)$) is $\mathcal{F}$-measurable. Let $\mathcal{L}$ be the class of such $E$. For a rectangle, $\nu'((A\times B)_x)=\nu'(B)\,\mathbf{1}_A(x)$, a simple function; so $\mathcal{R}\subseteq\mathcal{L}$. $\mathcal{L}$ is a $\lambda$-system: it contains $X\times Y$ (constant function $\nu'(Y)$); for $F\subseteq E$ both in $\mathcal{L}$, $\nu'((E\setminus F)_x)=\nu'(E_x)-\nu'(F_x)$ by additivity and finiteness, and differences of real measurable functions are measurable, since $\{u-v>t\}=\bigcup_{q\in\mathbb{Q}}(\{u>q\}\cap\{v<q-t\})$ by density of the rationals; and for $E_1\subseteq E_2\subseteq\dots$ in $\mathcal{L}$, $\nu'((\bigcup_jE_j)_x)=\lim_j\nu'((E_j)_x)$ by continuity from below (write the increasing union as a disjoint union of successive differences and use countable additivity), and a nondecreasing pointwise limit of measurable functions is measurable because $\{\sup_ju_j>t\}=\bigcup_j\{u_j>t\}$. By Dynkin's Pi-Lambda Theorem, $\mathcal{L}\supseteq\sigma(\mathcal{R})=\mathcal{F}\otimes\mathcal{G}$.

Step 3 (existence). For $E\in\mathcal{F}\otimes\mathcal{G}$ define

the series being the supremum of its partial sums; each integrand is a nonnegative real measurable function by Steps 1 and 2. Then $m(\emptyset)=0$, and $m$ is countably additive: if $E^1,E^2,\dots$ are pairwise disjoint, their sections are pairwise disjoint, so $\nu_n((\bigcup_jE^j)_x)=\sum_j\nu_n(E^j_x)$; interchanging the integral with this series is permitted because the partial sums are nondecreasing, finite sums pull out of the integral by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and Monotone Convergence Theorem applies; finally the resulting double series of nonnegative terms may be summed in either order, since both iterated sums equal the supremum of the sums over finite sets of index pairs (any finite set of pairs is contained in a finite rectangle of indices, and partial sums are monotone in both indices). This gives $m(\bigcup_jE^j)=\sum_jm(E^j)$. On rectangles,

by the integral of simple functions, countable additivity of $\nu$ along the disjoint sets $B\cap D_n$, and the $[0,\infty]$ product conventions of Measure, Measure Space, and Probability Measure. So $m$ is a measure on $\mathcal{F}\otimes\mathcal{G}$ with the required rectangle property. It is $\sigma$-finite: the countably many rectangles $C_i\times D_n$ are pairwise disjoint, cover $X\times Y$, and satisfy $m(C_i\times D_n)=\mu(C_i)\nu(D_n)<\infty$.

Step 4 (uniqueness). Let $m'$ be any measure on $\mathcal{F}\otimes\mathcal{G}$ with $m'(A\times B)=\mu(A)\nu(B)$ for all rectangles. Fix $i$ and $n$, and consider the two finite measures $E\mapsto m(E\cap(C_i\times D_n))$ and $E\mapsto m'(E\cap(C_i\times D_n))$ on $\mathcal{F}\otimes\mathcal{G}$. For a rectangle $E=A\times B$, $E\cap(C_i\times D_n)=(A\cap C_i)\times(B\cap D_n)$ is again a rectangle, on which both assign $\mu(A\cap C_i)\,\nu(B\cap D_n)$; in particular both have total mass $\mu(C_i)\nu(D_n)<\infty$. The class of $E$ on which they agree contains $X\times Y$ and is closed under proper differences (additivity, finiteness) and increasing countable unions (continuity from below), i.e. it is a $\lambda$-system containing $\mathcal{R}$, hence equals $\mathcal{F}\otimes\mathcal{G}$ by Dynkin's Pi-Lambda Theorem. Since the sets $C_i\times D_n$ are pairwise disjoint with union $X\times Y$, countable additivity gives, for every $E\in\mathcal{F}\otimes\mathcal{G}$,

(the double sums again ordered arbitrarily, as in Step 3). Hence the product measure exists and is unique, and $\mu\otimes\nu=m$ is $\sigma$-finite. $\blacksquare$