Product Sigma-Algebra Generated by Projections

Theorem
Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.

Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X \times Y$.

Let $\pr_1: X \times Y \to X$ and $\pr_2: X \times Y \to Y$ be the first and second projections, respectively.

Then:


 * $\Sigma_1 \otimes \Sigma_2 = \map \sigma {\pr_1, \pr_2}$

where $\sigma$ denotes generated $\sigma$-algebra.

Proof
Let $R = \{A \times B: A \in \Sigma_1, B \in \Sigma_2\}$ is a generator for $ \Sigma_1 \otimes \Sigma_2$ be the Definition:Product Sigma-Algebra.

Let's notice that $\pr_1^{-1}(A) = A \times Y \in R$ y $\pr_2^{-1}(B) = X \times B \in R$ for any $A \in \Sigma_1$ and $B \in \Sigma_2$.

Let $G = \{\pr_i^{-1}(C_i): C_i \in \Sigma_i, i = 1, 2\}$. Thus, $G \subseteq R$.

However, for any $A \times B \in R$:


 * $A \times B = \pr_1^{-1}(A) \cap \pr_2^{-1}(B) \in \sigma(G)$

So we have, $R \subseteq \sigma(G)$.

But, $G \subseteq R$ and $R \subseteq \sigma(G)$, we use Generated Sigma-Algebra Preserves Subset and ($\sigma(G) = \sigma(\sigma(G))$). Thus, $\sigma(R) = \sigma(G)$.

It follows from the definition:


 * $\Sigma_1 \otimes \Sigma_2 = \sigma(R) = \sigma(G) = \sigma (\pr_1, \pr_2)$