I'm learning the measure theory, and a theorem about product $\sigma$-algebra goes like
Let $X_\alpha\neq\varnothing$, $\mathcal M_\alpha$ be $\sigma$-algebra on $X_\alpha$, $\forall\alpha\in\Lambda$. If $\Lambda$ is countable, then
$$\bigotimes_{\alpha\in\Lambda} \mathcal M_\alpha=\sigma\left(\left\{\prod_{\alpha\in\Lambda} E_\alpha:E_\alpha\in\mathcal M_\alpha\right\}\right),$$
where the left symbol $\otimes$ means the product $\sigma$-algebra, i.e., the $\sigma$-algebra generated by $\{\pi_\alpha^{-1}(E_\alpha): E_\alpha\in\mathcal M_\alpha\}$.
I have already know how the theorem is proved, and the condition "countable" is applied when showing that the intersection is closed.
Now I want to give a counterexample, that if the indice set $\Lambda$ is uncountable, the equation doesn't hold.
I'm looking at the case
$X_\alpha=\{0,1\}$, $\mathcal M_\alpha=\{\varnothing, \{0\}, \{1\}, \{0,1\}\}$, and $\Lambda=\mathbb{R}^1$.
And I want to find a set contained in the right-hand $\sigma$-algebra but not in the $\bigotimes_{\alpha\in\mathbb{R}^1} \mathcal M_\alpha$. However, I have totally no idea how to start with. Could anybody help me?
=================================================
I have an idea, for every uncountable set $\tilde{E}$, consider the set $\prod_{\alpha\in\mathbb R^1} \tilde{E}_\alpha$, where
$$E_\alpha=\begin{cases}
\{0\}, & \alpha\in \tilde{E}\\
\{1\}, & \alpha\notin \tilde{E}\end{cases}$$.
Then the element of $\prod_{\alpha\in\mathbb R^1} \tilde{E}_\alpha$ has uncountably many $0$s and uncountably many $1$s. However, for every $\pi_{\alpha}^{-1}(E_\alpha)$, it has at most one $0$ or $1$, then for the set in $\bigotimes_{\alpha\in\Lambda} \mathcal M_\alpha$, it has at most countably many $0$s or $1$s, since the $\sigma$-algebra is closed only for countable union and intersection. Therefore, $\prod_{\alpha\in\mathbb R^1} \tilde{E}_\alpha$ is not contained in the product $\sigma$-algebra.
But I'm not sure if it is correct, or if the "countably many $0$s or $1$s" stuff is strictly correct, could you please help me?
Best Answer
In the case that you are looking at every singleton subset of $\prod_{\alpha\in\mathbb{R}}X_{\alpha}$ is an element of the $\sigma$-algebra on RHS so it is enough to prove that singletons cannot be elements of $\bigotimes_{\alpha\in\mathbb{R}}\mathcal{M}_{\alpha}$.
We can think of $\prod_{\alpha\in\mathbb{R}}X_{\alpha}$ as the set of function $f:\mathbb{R}\to\left\{ 0,1\right\} $ and let $\pi_{\alpha}:\prod_{\alpha\in\mathbb{R}}X_{\alpha}\to X_{\alpha}=\left\{ 0,1\right\} $ be prescribed by $f\mapsto f\left(\alpha\right)$.
Let $\mathcal{C}$ denote the collection of countable subsets of $\mathbb{R}$.
For $C\in\mathcal{C}$ let $\pi_{C}:\prod_{\alpha\in\mathbb{R}}X_{\alpha}\to\prod_{\alpha\in C}X_{\alpha}$ be prescribed as the function that sends every function $f:\mathbb{R}\to\left\{ 0,1\right\} $ to its restriction on $C$.
Then the collection $\mathcal{A}$ of subsets of $\prod_{\alpha\in\mathbb{R}}X_{\alpha}$ that can be written as $\pi_{C}^{-1}\left(B\right)$ for some $C\in\mathcal{C}$ and some $B\subseteq\prod_{\alpha\in C}X_{\alpha}$, i.e:$$\mathcal{A}:=\left\{ \pi_{C}^{-1}\left(B\right)\mid C\in\mathcal{C},\;B\subseteq\prod_{\alpha\in C}X_{\alpha}\right\} $$ can be shown to be a $\sigma$-algebra.
Evidently every $\pi_{\alpha}$ is measurable wrt to $\mathcal{A}$ so that: $$\bigotimes_{\alpha\in\mathbb{R}}\mathcal{M}_{\alpha}\subseteq\mathcal{A}$$
If $f\in\pi_{C}^{-1}\left(B\right)\in\mathcal{A}$ then - because $C\neq\mathbb{R}$ - we can find a function $g:\mathbb{R}\to\left\{ 0,1\right\} $ with $f\neq g$ and $\pi_{C}\left(f\right)=\pi_{C}\left(g\right)$.
This shows that no element of $\mathcal{A}$ is a singleton.
Consequently no element of $\bigotimes_{\alpha\in\mathbb{R}}\mathcal{M}_{\alpha}$ is a singleton.