# Generator of the $\sigma$-algebra on a product space

Let $\Gamma$ be an arbitrary index set and $(S,\mathcal{S})$ a measure space. I want to study the product space $S^\Gamma$. For that we define

$$X_\gamma : S^\Gamma \to S, \hspace{12pt}\omega=(\omega_\gamma)_{\gamma \in \Gamma} \mapsto X_\gamma(\omega):=\omega_\gamma$$

where $X_\gamma$ is the coordinate map on the $\gamma$-th coordinate. Moreover, we define

$$\mathcal{S}^\Gamma=\sigma(X_\gamma,\gamma \in \Gamma):=\sigma(\{\{X_\gamma \in A_\gamma\};\gamma\in \Gamma,A_\gamma\in \mathcal{S}\})$$

this should be the smallest $\sigma$-algebra, such that all $X_\gamma$'s are measurable. Now my question, why is the following set a generator of this $\sigma$-algebra?

$$M:=\{\omega\in S^\Gamma;\omega_j\in A_j,j\in J\}=\prod_{j\in J}A_j\times \prod_{k\in J^c}S$$

for all $J\subset \Gamma$ which are finite and $A_j\in \mathcal{S}$. Why is this true? It would be appreciated if someone could give me a proof or post a reference.

As I saw the statement, I had to think about the product topology. This can be described in a similar way (using a basis). However, in topology one has often that a property is true just for finite intersections or similar things. In measure theory this is not the case. So I do not see where the finiteness come in. Thank you for your help.

cheers

math

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## 2 Answers

Similar as in product topology we care about finite intersections, in measure theory we care about countable operations. But note one thing: let $\Gamma$ be infinite and $J\subset \Gamma$ be some infinite countable set. Let us numerate: $j_1,j_2,\dots, j_n,\dots \in J$.

If $\mathcal S^\Gamma$ is the product $\sigma$-algebra, and $A_j\in\mathcal S$ then any set of the form $$B_n = \prod\limits_{k=1}^nA_{j_n}\times \prod_{k\in \Gamma\setminus\{j_1,j_2,\dots,j_n\}}S\in \mathcal S^\Gamma$$ for any finite $n$, hence $$\prod\limits_{k\in J}A_{j}\times \prod_{k\in J^c}S = \bigcap\limits_{n=1}^\infty B_n\in \mathcal S^\Gamma.$$

So it means that there is no need to ask the product $\sigma$-algebra to be generated by a basis of products where there are possibly countably many sets which are not equal to $S$.

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@ Ilya: Thank you for your answer: In fact, it's just a simplification? There's no restriction to finite subset, but (of course) it's much easier to handle with, right? With not finite, I always mean countable infinite. –  math Mar 16 '12 at 13:15
@math In fact, it is just more easy to define it through finite sets, isn't it? –  Ilya Mar 16 '12 at 13:17
@ Ilya: I agree. As I said, this is just for simplicity. There would be no need. I was not sure if it would fail, if we would allow countable infinite sets. That's the reason for the question. Of course the definition makes absolutely sense, "keep it as easy as possible." –  math Mar 16 '12 at 13:20

Let $J\subset \Gamma$ be finite and let $A_j\in \mathcal S$ for each $j\in \Gamma$. Then $$\prod_{j\in J} A_j\times \prod_{k\in J^c} S = \bigcap_{j\in J} X_j^{-1}(A_j)$$

So $M \subset \sigma(X_\gamma,\, \gamma\in \Gamma)$ and therefore $\sigma(M)\subset \sigma(X_\gamma, \, \gamma\in \Gamma)$.

On the other hand, each $X_\gamma$ is measurable w.r.t $\sigma(M)$, so we also get the other inclusion $\sigma(M)\supset \sigma(X_\gamma, \, \gamma\in \Gamma)$ and therefore

$$\sigma(M)= \sigma(X_\gamma, \, \gamma\in \Gamma)$$

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@ Sam: Thanks for your quick answer! $X^{-1}_j(A_j)=S\times S\times\dots\times S \times A_j\times S\times\dots$ Where the $A_j$ is at the $j$-th element, right? Why do we look just at finite subsets of $\Gamma$ instead of arbitrary subsets? –  math Mar 16 '12 at 13:01
@math: I don't think arbitrary products of subsets are in the sigma algebra generated by $(X_\gamma)$. One could pass to countable subsets of $\Gamma$, but looking for a generating collection one usually is interested in the "smaller" of two alternatives. –  Sam Mar 16 '12 at 18:22
@ Sam: sorry for not being precise! With arbitrary I meant countable. (Everything else doesn't make sense). Again thanks for your help –  math Mar 16 '12 at 19:17
@math: Ah, ok. Then you are right: We could also look at countably infinite subsets if we wished. –  Sam Mar 16 '12 at 20:09