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Let $I$ be an index set and $(\Omega_i, \mathcal{A}_i)$ measurable spaces for every $i \in I$. Then the product $\sigma-$algebra is defined by $$\bigotimes_{i \in I} \mathcal{A}_i := \sigma(E)$$ with $$ E:= \{pr_i^{-1}(A_i) : i \in I,\ A_i \in \mathcal{A}_i\}$$

So far I've only seen the product $\sigma$-algebra defined for $|I| = 2$: $$ \mathcal{A}_1 \otimes \mathcal{A}_2 := \sigma(A_1 \times A_2 : A_1 \in \mathcal{A}_1, A_2 \in \mathcal{A}_2) $$

I see how these two definition coincide, since $(A_1 \times \Omega_2) \cap (\Omega_1 \times A_2) = A_1 \times A_2$. Similarly you can define the product $\sigma$-algebra for countabe $I$ as

$$ \bigotimes_{i \in I} \mathcal{A}_i = \sigma\left(\left\{\prod_{i \in I} A_i : A_i \in \mathcal{A}_i \right\}\right) $$

since $$\prod_{i \in I} A_i = \bigcap_{i \in I} \Big(A_i \times \prod_{j \in I\setminus\{i\}} \Omega_j \Big) \in \sigma(E)$$ as a countable intersection. (those cartesian products above are supposed to be in the order given by $I$)

My question is: Does this also work for uncountable $I$?

We can't directly argue $\bigcap_{i \in I} \Big(A_i \times \prod_{j \in I\setminus\{i\}} \Omega_j \Big) \in \sigma(E)$ since this intersection is not countable, so I would think they are indeed different but i can't find a counterexample.


EDIT:

Maybe one could use a cardinality argument? Something like: For a set $\prod_{i \in I} A_i \in \sigma(E)$ there are either only countably many $A_i = \Omega_i$ or countably many $A_i \neq \Omega_i$. But the other $\sigma$-algebra would also allow mixed cases where there are both uncountably many $A_i \neq \Omega_i$ and uncountably many $A_i = \Omega_i$. This isn‘t really written formally down though.

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    $\begingroup$ You might be interested in this: math.stackexchange.com/q/23030/177453 $\endgroup$ – Dasherman Apr 27 '20 at 13:01
  • $\begingroup$ @Dasherman thanks! i think this is what i thought about written down in a much better way $\endgroup$ – GhostAmarth Apr 27 '20 at 13:10
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I think the main idea in my edit above gives a good approach:

Let $B = pr_i^{-1}(A_i) = A_i \times \prod_{j \in I\setminus\{i\}} \Omega_j \in E$. Then $B^c = (\Omega_i \setminus A_i) \times \prod_{j \in I\setminus\{i\}} \emptyset$.

In both sets there are only countably many factors that are $\neq \Omega_i$ and $\neq \emptyset$ respectively. Since countable unions of countable sets are countable, countable unions/intersections of $B_k \in E$ have only countably many factors $\neq \Omega_i$ or $\neq \emptyset$. The same applies if you take the complement of such a union/intersection. This means

$$\sigma(E) \subseteq \left\{\prod_{i \in J} A_i \times \prod_{i \in I \setminus J} \Omega_i: A_i \in \mathcal{A}_i,\ J \subseteq I, J \ \text{countable}\right\} \cup \left\{ \prod_{i \in J} A_i \times \prod_{i \in I \setminus J} \emptyset: A_i \in \mathcal{A}_i,\ J \subseteq I, J \ \text{countable} \right\}$$

and therefore $\sigma(E) \neq \sigma\left( \left\{ \prod_{i \in I}A_i : A_i \in \mathcal{A}_i \right\} \right)$.

If there's something wrong please let me know. I scribbled it down so this probably isn't 100% correct.

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