Definition 15.8.1 (Product Measure). Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be measure spaces, then there exists a measure $\mu \otimes \nu: \cm \times \cn \to [0, \infty]$ such that:
For each $E \in \cm$ and $F \in \cn$, $\mu \otimes \nu(E \times F) = \mu(E)\nu(F)$.
For any measure $\lambda: \cm \otimes \cn \to [0, \infty]$, $\lambda \le \mu$. For any $A \in \cm \otimes \cn$ with $\mu(A) < \infty$, $\lambda(A) = \mu(A)$. In particular, if $\mu$ is $\sigma$-finite, then $\lambda = \mu$.
The measure $\mu \otimes \nu$ is the product of $\mu$ and $\nu$.
Proof. Let
then $\mathcal{R}$ is an elementary family by Proposition 14.4.3. Let
then $\alg$ is a ring over $X \times Y$. For each pairwise disjoint collection $\seqf{E_j \times F_j}\subset \mathcal{R}$, let
then $\mu \otimes \nu$ is well-defined and finitely additive on $\alg$.
Let $A \times B \in \mathcal{A}$ and $\seq{A_n \times B_n}\subset \mathcal{R}$ such that $A \times B = \bigsqcup_{n \in \natp}A_{n} \times B_{n}$, then for any $x \in X$ and $y \in Y$,
By the Monotone Convergence Theorem, for any $y \in Y$,
By the Monotone Convergence Theorem again,
Therefore $\mu \otimes \nu$ is a premeasure on $\alg$. By Carathéodory’s Extension Theorem, there exists a measure $\mu \otimes \nu$ satisfying (1) and (U).$\square$