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.
Best Answer
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 $B^c = (\Omega_i \setminus A_i) \times \prod_{j \in I\setminus\{i\}} \Omega_j$.
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
$$E \subseteq \left\{A \times \prod_{i \in I \setminus J} \Omega_i: A\in\mathcal P\left(\prod_{j\in J} A_j\right),\ J \subseteq I, J \ \text{countable}\right\} =: \mathcal S.$$
One can verify that $\mathcal S$ is a $\sigma$-algebra, see https://math.stackexchange.com/a/23073/721316, thus $\sigma(E)\subseteq \mathcal S$. But obviously $\prod_{i\in I} A_i\notin\mathcal S$ if $A_i \in\mathcal A_i\setminus \{\emptyset,\, \Omega_i\}$ for uncountably many $i\in I$.