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This is a continuation of my previous question and inspired by Arturo Margidin's reply.

Suppose there are a collection of measurable spaces $(X_i, \mathbb{S}_i), i \in I$. Let $\mathbf{X}=\prod_{i\in I}X_i$, and let $\mathbb{S}=\prod_{i\in I} \mathbb{S}_i$ be the product $\sigma$-algebra.

  1. Suppose $A_{i_0}\subseteq X_{i_0}$ is such that $A_{i_0}\times\prod_{j\neq i_0}X_j\in\mathbb{S}$. Does it follow that $A_{i_0}\in\mathbb{S}_{i_0}$?

  2. More generally, suppose we have a family of subsets, $A_i\subseteq X_i$, and we know that $A_j\in\mathbb{S}_j$ for all $j\neq i_0$ and that $\prod_{i\in I}A_i\in\mathbb{S}$. Does it follow that $A_{i_0}\in \mathbb{S}_{i_0}$?

  3. Basically, I would like to reconstruct each $\mathbb{S}_i$ from $\prod_{i \in I} \mathbb{S}_i$. If you have other approaches, please don't hesitate to reply. In particular, Part 1 and Part 2 are attempts based on projection of measurable rectangles in the product $\sigma$-algebra. I was wondering if it is possible to go from sectioning a measurable set in the product $\sigma$-algebra?

Thanks and regards!

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Is your index set $I$ finite, countably many, or arbitrary? – Morning Feb 22 '11 at 13:55
@Tim: I think your use of quantifiers was obscuring, rather than clarifying, your questions 1 and 2; in 1, it seemed at first as if you were selecting $A_i$ for all $X_i$ and asking about all of them simultaneously. I've reworded them, hopefully to make them clearer. – Arturo Magidin Feb 22 '11 at 14:23
@Morning: $I$ is arbitrary. But if your treatment is different for different cardinality of $I$, I also would like to hear. – Tim Feb 22 '11 at 14:26
@Arturo: Thanks! In my original wording, for example for Part 1, I intended to say $i_0$ as used in your wording is arbitrary in $I$. – Tim Feb 22 '11 at 14:27
@Tim: This is immediate from the phrasing. – Arturo Magidin Feb 22 '11 at 14:44
up vote 2 down vote accepted

Consider $I$ to be countable as suggested by Arturo Magidin. We show part (2). Define $$ {\cal G} = \{\pi_i^{-1}(A):A\in \mathbb S_i, i\in I\}. $$ The product $\sigma$-field $\mathbb S$ is generated by $\cal G$. Fix $i_0$, and pick arbitrary $x_i\in A_i$ for every $i\in I, i\neq i_0$. (Note that one needs to assume $\mathbb S_i\ni A_i\neq \emptyset$, otherwise the statement is obviously not correct.)

Define a mapping $$ T_{i_0} : X_{i_0}\ni x_{i_0} \mapsto \{x_i\}_{t\in I}\in {\bf X}. $$ Observe that if $T_{i_0}$ is $\mathbb S_{i_0}/\mathbb S$ measurable, then $\prod_{i\in I}A_i\in\mathbb S$ implies that $T_{i_0}^{-1}(\prod_{i\in I}A_i) = A_{i_0}\in\mathbb S_{i_0}$, and we are done.

It remains to show that $T_{i_0}$ is measurable and it suffices to show that $T_{i_0}^{-1}(G)\in\mathbb S_{i_0}$ for all $G\in \cal G$. This is true since for all $i\in I, A\in \mathbb S_i$, $T_{i_0}^{-1}(\pi_i^{-1}(A))$ equals (1) $A$ if $i=i_0$, (2) $X_{i_0}$ if $i\neq i_0, x_i\in A$, and (3) $\emptyset$ if $i\neq i_0, x_i\notin A$.

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it seems good, and if it is, then it answers (1) by taking $A_j = X_j$ with $j\neq i$. – Arturo Magidin Feb 22 '11 at 19:25
@Arturo Right, (1) is a special case of (2). Thanks. – Morning Feb 22 '11 at 22:54

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