# Product Sigma Algebra

## 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, $$\{ X_{\alpha} \}_{\alpha \in A}$$, and we have

$$X = \prod_{x \in \alpha}^{} X_{\alpha}$$

and

$$\pi_{\alpha} : X \to X_{\alpha}$$

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

{$$\pi_{\alpha}^{-1}(E_{\alpha}) : E_{\alpha} \in M_{\alpha}, \alpha \in A$$}.

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 $$E_{\alpha}$$ from some $$M_{\alpha}$$ 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 $$E_{\alpha}$$ inside every $$M_{\alpha}$$. 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 $$\pi_{\alpha}^{-1}(E_{\alpha})$$ would look like just for a single $$E_{\alpha}$$.

## The Attempt at a Solution

I tried to take an example like this, Consider:

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

Is it $$E_{\alpha} \times R_{2}$$?

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.

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

$$\{x\in A~\vert~P(x)\}$$

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

$$\{n^2~\vert~n\in \mathbb{N}\}$$

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

$$\{\{x\}~\vert~x\in \mathbb{R}\}$$

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

$$\{\pi^{-1}(E_\alpha)~\vert~E_\alpha\in M_\alpha, \alpha\in A\}$$

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

Is it $$E_{\alpha} \times R_{2}$$?

This is correct! $$\pi^{-1}(E_\alpha)=E_\alpha\times R_2$$...

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 $$A$$ is countable, then the product sigma-algebra is the sigma-algebra generated by $$\{ \prod_{\alpha \in A} E_{\alpha} : E_{\alpha} \in M_{\alpha} \}$$.

The proof given is this:

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

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

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 $$M_{\alpha}$$ be the empty set? That is there should be some $$E_{j}$$ which will just be the empty set. Then the cartesian product of all elements in $$M_{\alpha}$$ should be the empty set. Where am I goofing up?

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

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

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