# product $\sigma$-algebra on uncountable index set

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.

• You might be interested in this: math.stackexchange.com/q/23030/177453 – Dasherman Apr 27 '20 at 13:01
• @Dasherman thanks! i think this is what i thought about written down in a much better way – GhostAmarth Apr 27 '20 at 13:10

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