Proof of $\sigma(A^n)\supset\sigma\bigg(\sigma (A)^n\bigg)$?

Let ${\mathcal B}_n$ be the Borel $\sigma$-algebra on ${\mathbb R}^n$. Then it's not hard to show that $${\mathcal B}_n=\sigma(A^n)$$ where $$A=\{(-\infty,a]: a\in{\mathbb Q}\}.$$ Let ${\mathcal B}={\mathcal B}_1$. I want to show that ${\mathcal B}_n=\sigma({\mathcal B}^n)$. In other words $$\sigma(A^n)=\sigma\bigg(\sigma (A)^n\bigg).$$ It's not hard to see that $$\sigma(A^n)\subset\sigma\bigg(\sigma (A)^n\bigg)$$ since L.H. is the smallest $\sigma$-algebra containing $A^n$ while R.H. is a $\sigma$-algebra containing $A^n$. How can I go on to show $$\sigma(A^n)\supset\sigma\bigg(\sigma (A)^n\bigg)?$$

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You should write $\mathcal A:=\{(-\infty,a],a\in\Bbb Q\}$. – Davide Giraudo Oct 15 '12 at 19:51
@DavideGiraudo:Edited accordingly, thanks. – Jack Oct 15 '12 at 20:35
You didn't explain the notation $A^n$. Also $\mathcal B^n$. – GEdgar Oct 15 '12 at 21:30

It suffices to show that $\sigma(A)^n \subset \sigma(A^n)$. Let $S \in \sigma(A)^n$, say $$S = A_1 \times A_2 \times \cdots \times A_n$$ where $A_i \in \sigma(A)$. We want to show $S \in \sigma(A^n)$. We will show that the set $$S_1 = A_1 \times \mathbb{R} \times \cdots \times \mathbb{R}$$ is in $\sigma(A^n)$. Because $A_1 \in \sigma(A)$ we have $$S_1 \in \sigma(A \times \{\mathbb{R}\} \times \cdots \{\mathbb{R}\}) \subset \sigma(A^n).$$ The sets $S_2,\ldots,S_n$ defined analogously for the other coordinates are in $\sigma(A^n)$ by the same argument, so $S = S_1 \cap \cdots \cap S_n \in \sigma(A^n)$.

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Let $\mathcal{B}_m=\big\{\prod_{i\neq m}\mathbb{R}\times B: B\in\sigma({A})\big\}$. Show that $\mathcal{B}_m$ is a $\sigma$-algebra and $f_m:\sigma(A)\to\mathcal{B}_m$ given by $f_m(B)=\prod_{i\neq m}\mathbb{R}\times B$ an isomorphism of measurable spaces. This shows that $\mathcal{B}_m$ is the smallest $\sigma$-algebra containing all sets of the form $\prod_{i\neq m}\mathbb{R}\times A$ and hence $\mathcal{B}_m\subseteq \sigma(A^n)$. Now every element in $\sigma(A)^n$ is the intersection of $n$ elements $B_1,\ldots,B_n$ with $B_m\in\mathcal{B}_m$. Since $\sigma$-algebras are closed under finite intersections, $\sigma(A)^n\subseteq\sigma(A^n)$.

So $\sigma(A^n)$ is a $\sigma$-algebra containing $\sigma(A)^n$ and since $\sigma\big(\sigma(A)^n\big)$ is the smallest such $\sigma$-algebra, $$\sigma\big(\sigma(A)^n\big)\subseteq\sigma(A^n).$$

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This is a result from a more general proposition (Prop.1.5 Real Analysis by Folland):

Suppose $X_1,\cdots,X_n$ are metric spaces and let $X=\prod_{1}^n X_j$, equipped with the product metric. Then $\bigotimes_1^nB_{X_j}\subset B_X$. If the $X_j$'s are separable, then $\bigotimes_1^nB_{X_j}=B_X$. ($B_Y$ denotes the Borel sigma algebra for the metric space $Y$.)

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