The finite Cartesian product of algebras is an algebra In constructing the countable product of probability measures, I have come across this result.

Theorem: Let $\mathcal F_i$ be an algebra on $\Omega_i$ for $i=1, \ldots, n$. Then
$$
\mathcal F(n) := \left \{ \prod_{i=1}^n A_i \,\middle\vert\, A_i \in \mathcal F_i \right \}
$$ is an algebra on $\Omega (n) := \prod_{i=1}^n \Omega_i$.

Could you have a check on my attempt?

Proof: We proceed by induction on $n$. The statement clearly holds for $n=1$.  Let's prove it for $n=2$. Clearly, $\emptyset \in \mathcal F(2)$ and $\mathcal F(2)$ is closed under finite intersection. Let's prove that $\mathcal F (2)$ is closed under complement. Indeed,
$$
(A_1 \times A_2)^c = (A_1^c \times \Omega_2) \cup (\Omega_1 \times A^c_2) \cup (A^c_1 \times A^c_2).
$$
Assume the statement holds for all $p \le n$. Let's prove it for $n+1$. We have
$$
\mathcal F(n+1) =  \left \{ \prod_{i=1}^{n+1} A_i \,\middle\vert\, A_i \in \mathcal F_i \right \} \cong  \left \{ A \times B \,\middle\vert\, A \in \mathcal F(n), B \in \mathcal F_{n+1} \right \}.
$$

*

*By inductive hypothesis with $p=n$, $\mathcal F (n)$ is an algebra on $\Omega (n)$.

*By inductive hypothesis with $p=2$, $\mathcal F(n+1)$ is an algebra on $\Omega (n) \times \Omega_{n+1} \cong \Omega (n+1)$.

This completes the proof.
 A: Given a universe $\Omega$, if we identify $A\subset\Omega$ with its characteristic function taking values in the boolean ring $\mathbb Z/2\mathbb Z$, then $\mathcal F$ is an algebra in the sense of measure theory iff it's literally a ring with identity in the sense of abstract algebra. Indeed we have the correspondence $$\begin{cases} A\cap B\leftrightarrow \mathbb 1_A\mathbb 1_B \\A\triangle B\leftrightarrow 1_A+1_B \\ A^c\leftrightarrow 1_{\Omega}-1_A=1_{\Omega}+1_A \\ A\cup B\leftrightarrow 1_A+1_B+1_A1_B\end{cases}$$
Under this point view, since $\mathcal F_i$'s are indeed (Boolean) rings with identity, $\prod_{i=1}^n\mathcal F_i$ must be also a ring with identity, but it doesn't correspond to the the algebra generated by $\mathcal F_i$ inside the product $\prod_{i=1}^n \Omega_i$. Then to which (if any) algebra in meaure theory does it correspond? The answer is the disjoint union $\sqcup \Omega_i$.
To generate the boolean ring corresponding to the algebra generated by $\mathcal A_i, A_i\in\mathcal F_i$, we should take the tensor product $\otimes_{i=1}^n \mathcal F_i$, just like the notation used in usual measure theory. Now it should be clear why arguing $\otimes_{i=1}^n \mathcal F_i$ is full of monomial tensors won't work, since not every tensor can be expressed as a product, but only sums of product.
A: As pointed out by @KurtG. The "theorem" is plainly wrong. We don't know if
$$
(A_1^c \times \Omega_2) \cup (\Omega_1 \times A^c_2) \cup (A^c_1 \times A^c_2).
$$
belongs to $\mathcal F(2)$ or not.
