Generator for Product Sigma-Algebra

Theorem

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

Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively.

Then $\GG_1 \times \GG_2$ is a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.

Proof

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We begin by stating that this result is in fact incorrect.

For a simple counter-example, consider $X = Y = \set {1, 2}$ both equipped with the power set sigma algebra, that is, $\Sigma_1 = \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 2} }$.

Then clearly, we can take $\GG_1 = \GG_2 = \set 1$.

This article, or a section of it, needs explaining. In particular: How is $\set 1$ a generator for $\set {1, 2}$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

However, note that $\GG_1 \times \GG_2 = \set {\tuple {1, 1} }$ and hence the sigma algebra generated by this set will have exactly four elements.

Clearly, $\Sigma_1 \times \Sigma_2$ contains more than four elements and hence the product sigma algebra, i.e., the sigma algebra generated by the product $\Sigma_1\times\Sigma_2$ will contain more than four elements.

We also state and prove a minor modification of the above stated result in this light.

Theorem

Let $X, Y, \Sigma_1, \Sigma_2, \GG_1, \GG_2$ be as before and let $X \in \GG_1$ and $Y \in \GG_2$.

Then $\Sigma_1 \otimes \Sigma_2$ is generated by $\GG_1 \times \GG_2$.

Proof:

We begin by defining $\Sigma$ as the sigma algebra generated by $G_1 \times G_2$.

It is clear that $\Sigma \subset \Sigma_1 \otimes \Sigma_2$.

For the reverse inclusion, set:

$M_1 := \set {A \subset X: A \times Y \in \Sigma}$

Clearly, $\GG_1 \subset M_1$.

Therefore, $M_1$ is non-empty.

We claim that $M_1$ is a sigma algebra.

Note that $X \in M_1$ obviously.

Also, $A \in M_1$ suffices $\paren {X \setminus A} \times Y = \paren {X \times Y} \setminus \paren {A \times Y} \in \Sigma$, proving closure of $M_1$ under complements.

Also, if $A_n \in M_1$ for $n \in \N$, then:

$\ds \paren {\bigcup_{n \mathop = 1}^\infty A_n} \times Y = \bigcup_{n \mathop = 1}^\infty \paren {A_n \times Y} \in M_1$

proving closure under countable unions and establishing our claim.

Now since $M_1$ is a sigma algebra containing $\GG_1$, $M_1 \supset \Sigma_1$.

Similarly, setting $M_2 := \set {B \subset Y: X \times B \in \Sigma}$ gives us that $M_2 \supset \Sigma_2$.

Now, if $C \in \Sigma_1$ and $D \in \Sigma_2$, then clearly $C \in M_1$ and $D \in M_2$ giving $C \times Y, X \times D \in \Sigma$

Hence:

$C \times D = \paren {C \times Y} \cap \paren {X \times D} \in \Sigma$

proving:

$\Sigma_1 \times \Sigma_2 \subset \Sigma$

and hence giving:

$\Sigma_1 \otimes \Sigma_2 \subset \Sigma$

establishing the result.

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $13.3$