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This is a continuation of my previous question. I admit I still don't fully understand the concept of infinite product of $\sigma$-algebras.

Let $(E_i, \mathbb{B}_i)$ be measurable spaces, where $i \in I$ is an index set, possibly infinite.

I understood the definition of product sigma algebra. However, I am not sure after complement and/or countable union/intersection on those generating subsets, if $\prod_{i \in I} B_i$, $\forall B_i \in \mathbb{B}_i$, is measurable wrt the product sigma algebra $\prod_{i \in I} \mathbb{B}_i$?

I was expecting a "not always" answer to my first question. In this scenario, I was wondering if the product of all proper measurable subsets is always nonmeasurable wrt the product sigma algebra, i.e. is $\prod_{i \in I} B_i$, $\forall B_i \in \mathbb{B}_i$ and $B_i \neq E_i$, always nonmeasurable wrt the product sigma algebra $\prod_{i \in I} \mathbb{B}_i$?

Could you explain why? Thanks and regards!

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    The question is a bit mangled. By "measurable with respect to the product sigma algebra...", you really mean "**lies** in the sigma algebra...", I think. Second, I think you are asking whether a product in which $B_i\neq E_i$ for *infinitely many $i$* can lie in the product or not.2011-02-21
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    @Arturo: you are right on both.2011-02-21
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    @Tim: Maybe you can rewrite it? I honestly see no difference between a "no" answer to your first question, and the second question; it seems that if the answer to your first question is indeed "no", then the second question has necessarily an answer of "no" (otherwise, the answer to your first question would either be "yes", or "sometimes").2011-02-21
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    @Arturo: I was expecting a "not always" answer to my first question according to your previous comment. In this scenario, I asked in the second question, if the product of all proper measurable subsets is always nonmeasurable wrt the product sigma algebra.2011-02-21
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    @Tim: Fine; then please rewrite the question so that it is clear. You can also take care of that out-of-place "measurable" (here you are really just talking about belonging or not belonging to a $\sigma$-algebra).2011-02-21
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    @Arturo: Updated. I was wondering what you mean by "out-of-place 'measurable' (here you are really just talking about belonging or not belonging to a σ-algebra)"?2011-02-21
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    @Tim: Your question is really whether or not a particular kind of set lies or does not lie in the product sigma algebra. Sure, you can think of $(X,\prod \mathbb{B}_i)$ as a measure space, and call the sets in $\prod \mathbb{B}_i$ measurable sets, but I wonder if that's not putting too many layers on what is really a question about whether or not sets are in the $\sigma$-algebra generated by the pullbacks of the $\mathbb{B}_i$ under the canonical projections.2011-02-21
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    @Tim: In the countable case, you'll always get all such sets. For the uncountable case, I think you won't. Try showing that the collection of all $\prod B_i$ with $B_i\neq E_i$ for at most countably many $i$ is a $\sigma$-algebra.2011-02-21
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    @Arturo: Correct me if I am wrong. I think a set together with a sigma algebra on it is called a measurable space, while a measurable space together with a measure on its sigma algebra is called a measure space. So a measurable space does not assume the existence of measure.2011-02-21
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    @Tim: Could you clarify what you mean when you write $\prod_{i \in I} B_i$, $\forall B_i \in \mathbb{B}_i$? The "for all" does not make sense because you don't make an assertion, but just write down a set.2011-02-21
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    @Rasmus: I meant one $B_i$ from each $\mathbb{B}_i$, $i \in I$, and $B_i$ can be any from $\mathbb{B}_i$. My question is if it is true that $\prod_{i \in I} B_i$ is always in the product $\sigma$-algebra.2011-02-21
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    @Tim. Yes, it's called a measure space.2011-02-21
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    @Arturo: pardon me, what is called a measure space? Is my understanding about measurable space and measure space in the one before my last comment correct? i.e. A measurable space is just about sigma algebra, while a measure space is about sigma algebra and measure2011-02-21
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    @Tim: Yes, "Measurable" is set with $\sigma$-algebra; "measure" is measurable with a measure. Sorry if I confused you last night; it was late, and I was overdue for bed.2011-02-21
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    @Arturo: Sorry to hear that. Wish you a great day.2011-02-21

1 Answers 1

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Let $B_i\in\mathbb{B}_i$ be nonempty for each $i\in I$, and consider the set $I_0 = \{i\in I\mid B_i\neq E_i\}$. If $I_0$ is countable, then $\prod B_i\in\prod\mathbb{B}_i$. If the set $I_0$ is uncountable, then $\prod B_i\notin\prod\mathbb{B}_i$. In particular, if $I$ is countable then you get all the "box" sets, and if it is uncountable, you will never have a set $\prod B_i$ where $\emptyset\neq B_i\neq E_i$ for all $i\in I$.

The key is that the product $\sigma$-algebra is the smallest $\sigma$-algebra that makes all the canonical projections measurable; that is, such that for every $i\in I$, every $B_i\mathbb{B}_i$, $\pi_i^{-1}(B_i)\in\prod\mathbb{B}_i$.

Let $X=\prod\limits_{i\in I}E_i$, and $\mathbb{B}=\prod\limits_{i\in I}\mathbb{B}_i$, for simplicity.

First, assume that $I_0$ is countable, and let $B_i\in\mathbb{B}_i$ for each $i$; I claim that $\prod\limits{i\in I}B_i\in\mathbb{B}$.

To see this, for each $i\in I_0$ let $S_i =\pi_i^{-1}(E_i-B_i)= (E_i-B_i)\times\prod\limits_{j\neq i}E_j$. Then $S_i\in\mathbb{B}$, and hence so is $\mathop{\cup}\limits_{i\in I_0}S_i$. Since $I_0$ is countable, this is an element of $\mathbb{B}$, being a countable union of elements of $\mathbb{B}$. An element $(e_i)\in X$ lies in $\cup S_i$ if and only if $e_i\notin B_i$ for some $i\in I_0$. Therefore, $(e_i)\in X-\cup S_i$ if and only if $e_i\in B_i$ for all $i\in I_0$. That is, $X-\cup S_i = \mathop{\prod}\limits_{i\in I_0}B_i\times\mathop{\prod}\limits_{i\notin I_0}E_i = \prod\limits_{i\in I}B_i$. This is the complement of an element of $\mathbb{B}$, hence lies in $\mathbb{B}$.

Note however that $\mathbb{B}$ will usually contain other stuff besides these sets. Even in the case where $I=\{1,2\}$, you usually have more than just the "box" sets in the $\sigma$ algebra. For example, the product of the Borel $\sigma$-algebra on $[0,1]$ with itself contains the set $([\frac{1}{4},\frac{1}{2}]\times[\frac{1}{4},\frac{1}{2}])\cup([\frac{3}{4},1]\times[\frac{3}{4},1])$, which is not of the form $A\times B$ with $A$ and $B$ Borel subsets of $[0,1]$.

Now assume that $I_0$ is uncountable. I will construct a $\sigma$-algebra that necessarily contains $\mathbb{B}$ (but need not be equal to it), and which does not contain any set of the form $\prod\limits{i\in I}B_i$ in which the subset $I_0 = \{i\in I\mid B_i\neq E_i\}$ is uncountable (in particular, when it is equal to all of $I$).

Let $S$ be the $\sigma$-algebra on $\prod\limits_{i\in I}E_i$ that consists of all sets $A$ of the following form: there exists a countable set $J\subseteq I$, and an element $B\in \mathcal{P}\left(\prod\limits_{j\in J}E_i\right)$, such that $A = B\times\prod\limits_{i\notin J}E_i.$ That is, $A$ is an arbitrary thing in the $J$-coordinates, but is all of $E_i$ in the non-$J$ coordinates.

I claim that $S$ is a $\sigma$-algebra on $X$. It contains the empty set by selecting $B=\emptyset$. It is closed under complements, as $X-A = \left((\prod\limits_{j\in J}E_j) - B\right)\times\prod\limits_{i\notin J}E_i$ is of the same form. And it is closed under countable unions: if $A_n$ is a subset of the desired form, with associated countable $J_n\subseteq I$, letting $J=\cup J_n$ we have that $\cup A_n$ is of the form $C\times\prod\limits_{i\notin J}E_i$ with $C$ a subset of $\prod_{j\in J}E_j$. Since each $J_n$ is countable, $J$ is countable, so this set is in $S$. Thus, $S$ is a $\sigma$-algebra on $X$.

Note also that $S$ contains the $\sigma$-algebra $\mathbb{B}$, since the generators of $\mathbb{B}$ all lie in $S$.

However, if $B_i\in\mathbb{B}_i$ for each $i$, and the set $J=\{i\in I\mid B_i\neq E_i\}$ is uncountable, then $\prod B_i\notin S$, and hence does not lie in $\mathbb{B}$ either. Thus, $\mathbb{B}$ does not contain any set which is a product of elements of the different $\mathbb{B}_i$ in which uncountably many components are not the entire space.