TITLE: Infinite product of measurable spaces
QUESTION [41 upvotes]: Suppose there is a family (can be
infinite) of measurable spaces. What
are the usual ways to define a sigma
algebra on their Cartesian product?

There is one way in the context of
defining product measure on
planetmath. Let $(E_i, B_i)$ be
measurable spaces, where $i \in I$
is an index set, possibly infinite.
We define their product as follows:

let $E= \prod_{i \in I} E_i$  , the    Cartesian product of $E_i$,

let $B=\sigma((B_i)i \in I)$ , the    smallest sigma algebra
containing    subsets of E of the
form $\prod_{i    \in I}B_i$  where
$B_i=E_i$ for all    but a finite
number of $i \in I$ .


I was wondering why it is required
that "$B_i=E_i$ for all but a finite
number of $i \in I$"?


Thanks and regards!

ADDED:
I was wondering if the product sigma algebra defined in 2 is the smallest sigma algebra such that any tuple composed of one measurable set from each individual sigma algebra is measurable?

REPLY [3 votes]: Details of $\textbf{lemma}$ mencionated by Michael Greinecker : Let
$L=\{E \subseteq X; E\in \sigma(C)\,\mbox{for some} \, C\subseteq Y,\,\mbox{with C countable}\}$ 
See that the following hold:
$\bullet$ X, $\emptyset \in L$ (Obvious); 
$\bullet$ If, $E\in L$, then $E^{c}\in\sigma(C_E)$ for $C_E\subseteq Y$ (countable) such that $E\in\sigma(C_E)$;
$\bullet$ if $\{E_n\}_{n\in\mathbb{N}} \subseteq L$, we can take $\displaystyle C=\bigcup_{n=1}^{\infty}C_{E_n}$ with $C_{E_n}$ in the same sense of above item. Then C is countable and $\displaystyle\bigcup_{n=1}^{\infty}E_n \in \sigma(C)$.
This proof that L is an $\sigma$-algebra. Finnaly observes that
$\bullet$ $Y\subseteq L$ (Using $\sigma$-algebra generating by each element of Y);
$\bullet$ For all countable $C\subseteq Y$ see that $\sigma(C)\subseteq \sigma(Y)$, and this implies that $L\subseteq \sigma(Y)$.
Because $\sigma(Y)$ is the smallest $\sigma$ -algebra containg Y, we have $L=\sigma(Y)$.
This proof that each event just depend of a countable many sets of generators. 
$\textbf{More about}:$
In the case of product, by the observation of Michael you have that the product measure is generated by the sets $\prod_{n\in C}Y_n \times \prod_{n\in I-C}X_n$ with $C\subseteq I$ countable, this implies that for every $E\subseteq X$ exist an at most countable $C\subseteq I$ and $E_C$ in product sigma algebra of $(X_i, B_i)_{i\in C}$ satisfying $$E=\pi_{C}^{-1}(E_C)\hspace{3cm}(1)$$ for $\pi_C : \prod_{i\in I} X_{i} \to  \prod_{i\in C} X_{i}$ the natural projection. This results says that, the product sigma algebra is not generated by the sets $\prod_{i\in I} Y_{i}$ with $Y_i \in B_i$ for all $i\in I(uncontable)$, indeed if every $Y_i\neq X_i$, for uncontable many of index $i\in I$, not exist an countable $C\subset I$ and $E_C$ such that (1) holds.