I'm trying to generalize this theorem. I'm mot sure if the generalization is indeed true, and if my proof is correct. Could you have a check on my attempt?
Let $(\Omega_i, \mathcal{F}_i, \mu_i)_{i \in I}$ be a collection of probability spaces. Here $I$ is uncountable. Let $\Omega :=\prod_i \Omega_i$ and $\bigotimes_i \mathcal{F}_i$ be the smallest $\sigma$-algebra on $\Omega$ such that all projection maps $\pi_i: \Omega \to \Omega_i$ are measurable. Let
$$
\mathcal C := \left\{ \prod_i C_i \,\middle\vert\, C_i \in \mathcal F_i, \text{the set } \{i \in I \mid C_i \neq \Omega_i\} \text{ is finite} \right\}.
$$
Then $\mathcal C$ is an algebra on $\Omega$ and
$$
\bigotimes_i \mathcal{F}_i = \sigma (\mathcal C).
$$
We define $\nu: \mathcal C \to [0, 1]$ as follows. For $C = \prod_i C_i \in \mathcal C$, there is a finite subset $J$ of $I$ such that $C_i = \Omega_i$ for all $i \notin J$. Let $\nu (C) := \prod_{i \in J} \nu_i (C_i)$. Clearly, $\nu$ is finitely additive. Let's prove that $\mu$ is $\sigma$-additive.
Let $(C_n) \subset \mathcal C$ be a decreasing sequence such that $C_n \searrow \emptyset$. It suffices to prove that $\inf_n \nu (C_n) =0$. We assume $C_n = \prod_i C_{n,i}$. Let $I_n := \{ i \in I \mid C_{n, i} \neq \Omega_i \}$. Then $I_n$ is finite for all $n$. Let $J := \bigcup_n I_n$. Then $J$ is a countable subset of $I$. Let $C_n' := \prod_{i \in J} C_{n,i}$ for all $n$. Here we have the same situation as in the case where $I$ is countable. This completes the proof.
Remark: The proof of $\sigma$-additivity of $\mu$ is inspired by this note in which the author uses the following lemma.
Lemma: Let $X$ be a set, $\mathcal{A}$ a non-empty collection of subsets of $X$, and $\sigma(\mathcal{A})$ the sigma-algebra generated by $\mathcal{A}$. Then for any $A \in \sigma(\mathcal{A})$ there is a countable subset $\mathcal{A}^{\prime}$ of $\mathcal{A}$ such that $A \in \sigma\left(\mathcal{A}^{\prime}\right)$.
Best Answer
My definition of $\mathcal C$ does not imply that $\mathcal C$ is an algebra on $\Omega$. Instead, it should be defined as $$ \mathcal C := \left \{ A \times \prod_{i \notin J} \Omega_i \,\middle\vert\, J \text{ a finite subset of } I, A \in \bigotimes_{i \in J} \mathcal F_i \right \}. $$
Then definition of $\nu: \mathcal C \to [0, \infty)$ is adjusted accordingly. If $C = A \times \prod_{i \notin J} \Omega_i$ with $J$ a finite subset of $I$ and $A \in \bigotimes_{i \in J} \mathcal F_i$, then $$ \nu (C) := (\otimes_{i \in J} \mu_i) (C). $$