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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. –  Arturo Magidin Feb 21 '11 at 5:51
    
@Arturo: you are right on both. –  Tim Feb 21 '11 at 5:53
    
@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"). –  Arturo Magidin Feb 21 '11 at 6:03
    
@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. –  Tim Feb 21 '11 at 6:09
    
@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). –  Arturo Magidin Feb 21 '11 at 6:11

1 Answer 1

up vote 5 down vote accepted

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.

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