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Product Sigma Algebra

  • Thread starter Aerostd
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Homework Statement



I should mention beforehand that I do not come from a math background so I may ask some trivial questions.

I am reading the book "Real Analysis" by Folland for a course I am taking and am attempting to understand a definition of product sigma algebra. It is stated in the book that if we have an indexed collection of non-empty sets, [tex] \{ X_{\alpha} \}_{\alpha \in A}[/tex], and we have

[tex] X = \prod_{x \in \alpha}^{} X_{\alpha} [/tex]

and

[tex] \pi_{\alpha} : X \to X_{\alpha} [/tex]

the coordinate maps. If [tex] M_{\alpha} [/tex] is a [tex] \sigma [/tex]-algebra on [tex] X_{\alpha} [/tex] for each [tex] \alpha [/tex], the product [tex] \sigma[/tex]-algebra on [tex] X [/tex] is the [tex] \sigma [/tex]-algebra generated by

{[tex] \pi_{\alpha}^{-1}(E_{\alpha}) : E_{\alpha} \in M_{\alpha}, \alpha \in A [/tex]}.


Now the first time I read this definition(by first time I mean the thousandth time), I thought that one has to just pick one [tex] E_{\alpha} [/tex] from some [tex] M_{\alpha}[/tex] and take it's inverse map(pre image?) to get some collection of sets. Then generating a sigma algebra from that collection would yield the product sigma algebra. However, I was told that we have to take the inverse map of every [tex] E_{\alpha} [/tex] inside every [tex] M_{\alpha} [/tex]. I don't know how this is implied by the definition stated in the book but it makes more sense. Maybe I am not familiar with the notation because I would have expected some "for all" symbols in the definition somewhere.

Secondly, I wanted to think about what [tex] \pi_{\alpha}^{-1}(E_{\alpha}) [/tex] would look like just for a single [tex] E_{\alpha} [/tex].


The Attempt at a Solution




I tried to take an example like this, Consider:

[tex] \{ R_{1}, R_{2} \}[/tex], the real numbers. Then [tex] X = R_{1} \times R_{2} [/tex]. Suppose I have sigma algebras [tex] M_{1}[/tex] and [tex] M_{2} [/tex]. Now, What is [tex] \pi_{\alpha}^{-1} (E_{\alpha})[/tex] for some element in say [tex] M_{1} [/tex]?

Is it [tex] E_{\alpha} \times R_{2} [/tex]?

I don't know. Even this is just a guess. I'm hoping someone can give me some hints on understanding this since I am not able to easily find references on product sigma algebras.
 

Answers and Replies

  • #2
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Maybe I am not familiar with the notation because I would have expected some "for all" symbols in the definition somewhere.
Well, the "for all" symbol is implied by the notation. Let P be a certain property (I'll give examples later), then

[tex]\{x\in A~\vert~P(x)\}[/tex]

means that you take all elements in A which satisfy P. An example:

[tex]\{n^2~\vert~n\in \mathbb{N}\}[/tex]

By definition this means that you take all the squares in [tex]\mathbb{N}[/tex]. Thus, you take all the elements in [tex]\mathbb{N}[/tex], and you square it. Another example:

[tex]\{\{x\}~\vert~x\in \mathbb{R}\}[/tex]

means that you take all the [tex]x\in \mathbb{R}[/tex], and you take the singleton set of it.

So, your defintion:

[tex]\{\pi^{-1}(E_\alpha)~\vert~E_\alpha\in M_\alpha, \alpha\in A\}[/tex]

means that you take all the [tex]\alpha\in A[/tex] and all the [tex]E_\alpha\in M_\alpha[/tex], and you calculate [tex]\pi^{-1}(E_\alpha)[/tex].


Is it [tex] E_{\alpha} \times R_{2} [/tex]?
This is correct! [tex]\pi^{-1}(E_\alpha)=E_\alpha\times R_2[/tex]...
 
  • #3
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Thanks a lot. I guess now I will move on to the next page in the book. I don't know if I should open a new thread or not but here goes.

This is Proposition 1.3 in Folland. I am interested more in the notation than in the proof given in the book. The proposition is

If [tex] A [/tex] is countable, then the product sigma-algebra is the sigma-algebra generated by [tex] \{ \prod_{\alpha \in A} E_{\alpha} : E_{\alpha} \in M_{\alpha} \} [/tex].

The proof given is this:

If [tex] E_{\alpha} \in M_{\alpha} [/tex], then [tex] \pi_{\alpha}^{-1} (E_{\alpha}) = \prod_{\beta \in A} E_{\beta} [/tex] where [tex] E_{\beta} = X_{\beta} [/tex] for [tex] \beta \notin \alpha [/tex].

On the other hand, [tex] \prod_{\alpha \in A} E_{\alpha} = \cap_{\alpha \in A} \pi_{\alpha}^{-1} (E_{\alpha}) [/tex]

Next he uses a lemma to prove this. I am more interested in the two equalities he uses.

Before anything I will ask the stupid question. Shouldn't one of the elements in [tex] M_{\alpha} [/tex] be the empty set? That is there should be some [tex] E_{j}[/tex] which will just be the empty set. Then the cartesian product of all elements in [tex] M_{\alpha}[/tex] should be the empty set. Where am I goofing up?

Anyway again I consider my example where I have [tex] \{ R_{1}, R_{2} \} [/tex], the reals and then have [tex] M_{1} [/tex] for [tex] R_{1} [/tex] and [tex] M_{2} [/tex] for [tex] R_{2} [/tex].

Now suppose I consider [tex] E_{1} [/tex] in [tex] M_{1} [/tex]. Then the first equality should be [tex] \pi_{1}^{-1} E_{1} = E_{1} \times R_{2} [/tex], which I can understand.

Next, in the second equality, the LHS I already asked about before. The RHS looks like it should be = [tex] E_{1} \times R_{1} \cap E_{2} \times R_{1} \cap E_{3} \times R_{1} \cap ........ [/tex], but I don't see how this condenses to [tex] E_{1} \times E_{2} \times .....[/tex] which I am assuming is the RHS.
 

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