I'm reading section 7.1 The Product Measure Theorem from this lecture note.
Problem 7.1.1. Let $(X, \mathcal{A})$ and $(Y, \mathcal{B})$ be measurable spaces. Is there a natural way to define a measure on the space $X \times Y$ which reflects the structure of the original measure space?
Definition 7.1.2. Let $(X, \mathcal{A})$ and $(Y, \mathcal{B})$ be measurable spaces. A measurable rectangle is a set of the form $A \times B$, where $A \in \mathcal{A}$ and $B \in \mathcal{B}$. Let $Z=X \times Y$ and
$$
\mathcal{Z}_0=\left\{\coprod_{i=1}^n A_i \times B_i \mid A_i \in \mathcal{A}, B_i \in \mathcal{B}\right\}.
$$
Lemma 7.1.3. $\mathcal{Z}_0$ is an algebra in $\mathcal{P}(Z)$.
My understanding It seems from this answer that $\mathcal{Z}_0$ is not generally an algebra and thus Lemma 7.1.3. is not correct. That's why the construction of product measure is usually not by Hahn–Kolmogorov theorem.
Could you confirm if my understanding is fine?
Update Below is a direct construction of product measure I have found.
Theorem 7.9 (Product measure) Let $\left(\Omega_1, \mathcal{F}_1, \mu_1\right)$ and $\left(\Omega_2, \mathcal{F}_2, \mu_2\right)$ be two measurable spaces where $\mu_1$ and $\mu_2$ are $\sigma$-finite measures. There exists a unique measure $\mu$ on $\left(\Omega_1 \times \Omega_2, \mathcal{F}_1 \otimes \mathcal{F}_2\right)$ that satisfies $\mu\left(A_1 \times\right.$ $\left.A_2\right)=\mu_1\left(A_1\right) \mu_2\left(A_2\right)$ for all $A_1 \in \mathcal{F}_1$ and $A_2 \in \mathcal{F}_2$. This measure is called the product measure, written as $\mu=\mu_1 \times \mu_2$.
Proof
- Uniqueness:
First we show that any such measure must be $\sigma$-finite. Since $\mu_1$ and $\mu_2$ are $\sigma$-finite there exist $\left\{A_n\right\}_{n=1}^{\infty} \in \mathcal{F}_1$ and $\left\{B_n\right\}_{n=1}^{\infty} \in F_2$ such that $\bigcup_{n=1}^{\infty} A_n=\Omega_1, \bigcup_{n=1}^{\infty} B_n=\Omega_2, \mu_1\left(A_n\right)$ and $\mu_2\left(B_n\right)$ are finite for all $n$. Consider $\bigcup_{(i, j) \in \mathbb{N}^2} A_i \times B_j$. For and $\left(\omega_1, \omega_2\right) \in \Omega_1 \times \Omega_2$ there exists $i, j$ such that $\omega_1 \in A_i$ and $\omega_2 \in B_j$, which means $\left(\omega_1, \omega_2\right) \in A_i \times b_j$. Hence $\bigcup_{(i, j) \in \mathbb{N}^2} A_i \times B_j=\Omega_1 \times \Omega_2$. For any $(i, j) \in \mathbb{N}^2$ we have $\mu\left(A_i \times B_j\right)=$ $\mu_1\left(A_i\right) \mu_2\left(B_j\right)<\infty$. Since $\mathbb{N}^2$ is a countable set we can conclude that $\mu$ is $\sigma$-finite.
Suppose there are two measures $\mu$ and $\mu^{\prime}$ satisfying the condition in the theorem. Recall that the collection of measurable rectangles $\left\{A_1 \times A_2: A_1 \in \mathcal{F}_1, A_2 \in \mathcal{F}_2\right\}$ is a $\pi$-system. $\mu$ and $\mu^{\prime}$ are both $\sigma$-finite and agree on this $\pi$-system. By Uniqueness theorem they agree on the generated $\sigma$-field $\mathcal{F}_1 \otimes \mathcal{F}_2$, i.e., $\mu=\mu^{\prime}$, which means such measure must be unique.
- Existence:
For any $B \in \mathcal{F}_1 \otimes \mathcal{F}_2$ let $\mu(B)=\int_{\Omega_1} \mu_2\left(B_{\omega_1}\right) d \mu_1\left(\omega_1\right)$ where $B_{\omega_1}=\left\{\omega_2 \in \Omega_2:\left(\omega_1, \omega_2\right) \in \Omega_1 \times \Omega_2\right\}$ as introduced previously. Then $\mu$ is a measure.
For any $A_1 \in \mathcal{F}_1, A_2 \in \mathcal{F}_2$,
$$
\mu\left(A_1 \times A_2\right)=\int_{\Omega_1} \mu_2\left(\left(A_1 \times A_2\right)_{\omega_1}\right) d \mu_1\left(\omega_1\right)=\int_{\Omega_1} \mathbb{1}_{A_1} \mu_2\left(A_2\right) d \mu_1\left(\omega_1\right)=\mu_2\left(A_2\right) \int_{\Omega_1} \mathbb{1}_{A_1} d \mu_1\left(\omega_1\right)=\mu_1\left(A_1\right) \mu_2\left(A_2\right)
$$
Hence such measure exists.
Best Answer
@Snoop It seems from Theorem 7.1.4 that $\coprod_{i=1}^n A_i \times B_i$ in Definition 7.1.2. means $\bigcup_{i=1}^n A_i \times B_i$. Then my above understanding is not correct.
Theorem 7.1.4 [Product Measure Theorem]. Let $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, \lambda)$ be measure spaces. Then there exists a measure $\pi$ on $(X \times Y, \mathcal{A} \times \mathcal{B})$ such that $\pi(A \times B)=\mu(A) \lambda(B)$. Moreover, if $\mu$ and $\lambda$ are $\sigma$-finite, then $\pi$ is unique and $\sigma$-finite. In the case where $\mu$ and $\lambda$ are $\sigma$-finite, we denote the uniquely obtained measure by $$ \pi=\mu \times \lambda $$ and call the measure the product of $\mu$ and $\lambda$.
Proof. Suppose that $A \times B$ can be written as $\sum_{i=1}^{\infty} A_i \times B_i$, where each of the measurable rectangles $A_i \times B_i$ are disjoint. Then $$ \chi_A(x) \chi_B(y)=\chi_{A \times B}(x, y)=\sum_{i=1}^{\infty} \chi_{A_i}(x) \chi_{B_i}(y) $$ for all $x \in X$ and $y \in Y$. Fix $x$ and integrate with respect to $\lambda$. By MCT, $$ \begin{aligned} \int_Y \chi_{A \times B}(x, y) d \lambda(y) & =\int_Y \sum_{i=1}^{\infty} \chi_{A_i}(x) \chi_{B_i}(y) d \lambda(y) \\ \int_Y \chi_A(x) \chi_B(y) d \lambda(y) & =\sum_{i=1}^{\infty} \chi_{A_i}(x) \int_Y \chi_{B_i}(y) d \lambda(y) \\ \chi_A(x) \lambda(B) & =\sum_{i=1}^{\infty} \chi_{A_i}(x) \lambda\left(B_i\right) \end{aligned} $$
Further integrating with respect to $\mu$ yields (again by MCT) $$ \mu(A) \lambda(B)=\sum_{i=1}^{\infty} \mu\left(A_i\right) \lambda\left(B_i\right) \quad(*). $$
Define $\pi_0$ on $\mathcal{Z}_0$ by $\pi_0\left(\cup_{i=1}^n A_i \times B_i\right)=\sum_{i=1}^n \mu\left(A_i\right) \lambda\left(B_i\right)$. Then $\pi_0$ is a measure on $\mathcal{Z}$ by $(*)$ (the only nontrivial issue to check was countable additivity). Caratheodory's Extension Theorem gives us a measure $\pi$ defined on at least $\mathcal{Z}$ that extends $\pi_0$. If $\mu$ and $\lambda$ are $\sigma$-finite, then $\pi_0$ is $\sigma$-finite, so Hahn's Extension Theorem tells us that $\pi$ is unique.