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Do you know where to find references on the following questions (I suspect they are well-known)?

1) The product $\sigma$-algebra is defined to be the smallest $\sigma$-algebra containing all the "measurable rectangles." Is it possible to write every element in the product $\sigma$-algebra as a sequence of countable unions and intersections of the measurable rectangles? (I suspect the answer is no, but I can't think of a good reason why.)

(More precisely, given two $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$, is it possible to an arbitrarily element of $\mathcal{A}\times \mathcal{B}$ as $\bigcup_{i_{1}=1}^{\infty}\bigcap_{i_{2}=1}^{\infty}\cdots\bigcap_{i_{n}}^{\infty}A_{i_{1},\ldots,i_{n}}\times B_{i_{1},\ldots,i_{n}}$ for some $n$ and all $A_{i_{1},\ldots,i_{n}}\in\mathcal{A},\ B_{i_{1},\ldots,i_{n}}\in\mathcal{B}$?)

2) Given two independent $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ (with respect to a measure $\mu$), why is not possible describe the measure $\mu$ on $\sigma(\mathcal{A}\cup\mathcal{B})$ as a product measure $\mathcal{A}\times \mathcal{B}$ (where we would define the measure of a rectangle $A\times B$ for $A\in\mathcal{A}$ and $B\in\mathcal{B}$ as $\mu(A\cap B)=\mu(A)\mu(B)$)?

Again, it makes sense that there are lot more ways to construct independent $\sigma$-algebras than taking products, but I feel like I don't understand a lot if I can't distinguish between the two.

DCT
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  • See Asaf's answer here for a "bottom-up" description of the $\sigma$-algebra generated by a set. Question 1) is about the special case of the $\sigma$-algebra generated by the measurable rectangles. – t.b. Jul 29 '12 at 00:22
  • What do you mean by independent $\sigma$-algebras? – leo Jul 29 '12 at 23:40

2 Answers2

2

Consider the Cantor Set. It is compact, and therefore measurable. Let $\mathcal{A}$ denote the half-open (right closed, left open) intervals, along with the null set. Then $\mathcal{A}$ is an algebra of sets. $\sigma(\mathcal{A})$ is the Borel sets.

However, the Cantor set is totally disconnected and uncountable. It is not a countable intersection or union of elements of $\mathcal{A}$. The appearance of $\sigma(\mathcal{A})$ in terms of $\mathcal{A}$ is subtle.

Consider a Cantor Set constructed along the line $y=x$ in $[0,1]^2$.

ncmathsadist
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1

1) Certainly not be a finite sequence. You can think of $\sigma(\mathcal{A})$ as being the closure of $\mathcal{A}$ under taking countable unions and countable intersections. To see this, denote this closure by $C(\mathcal{A})$. This definitely forms a $\sigma$-algebra, and hence by definition $\sigma(\mathcal{A}) \subseteq C(\mathcal{A})$; but any $\sigma$-algebra containing $\mathcal{A}$ must contain all sets obtained by countable unions and intersections of elements of $\mathcal{A}$, and so $C(\mathcal{A}) \subseteq \sigma(\mathcal{A})$.

2) What makes you think that $\sigma(\mathcal{A} \cup \mathcal{B})$ is generated by rectangles?

Clive Newstead
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  • You should also mention complements in 1) (unless you assume more on $\mathcal{A}$). – t.b. Jul 29 '12 at 00:38