Is a $\sigma$-algebra generated by complete independent $\sigma$-algebras also complete?

Question

$ \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\sP}{\mathscr{P}} $

Let $(\Omega, \cA, \mu)$ be a probability space and $\cA_1, \cA_2$ sub $\sigma$-algebras of $\cA$. Let $\cB := \sigma (\cA_1 \cup \cA_2)$, i.e., $\cB$ is the $\sigma$-algebra generated by $\cA_1 \cup \cA_2$. I would like to ask if the following statement is true or not:

Assume that $(\Omega, \cA_1, \mu), \Omega, \cA_2, \mu)$ are complete and that $\cA_1, \cA_2$ are independent. Then $(\Omega, \cB, \mu)$ is also complete.

Thank you so much for your elaboration! Below is my failed attempt:

Let $\cA_3 := \{A_1 \cap A_2 : A_1 \in \cA_1 \text{ and } A_2 \in \cA_2\}$. Then $\cB = \sigma (\cA_3)$. Let $\sP (\Omega)$ be the power set of $\Omega$. Let $\Sigma$ be the collection of sets $N$ in $\cB$ with the property that if $\sP (\Omega) \ni C \subset N$ and $\mu (N)=0$ then $C \in \cB$. It suffices to prove $\Sigma = \cB$. We proceed by Dynkin's $\pi$-$\lambda$ theorem theorem. Clearly, $\cA_3$ is a $\pi$-system.

Let's prove that $\cA_3 \subset \Sigma$. Let $A_1 \in \cA_1$ and $A_2 \in \cA_2$ such that $\sP (\Omega) \ni C \subset (A_1 \cap A_2)$ and $\mu (A_1 \cap A_2)=0$. By independence of $\cA_1$ and $\cA_2$, we get $\mu (A_1) \mu(A_2)=0$. WLOG, we assume $\mu (A_1)=0$. Because $(\Omega, \cA_1, \mu)$ is complete, $C \in \cA_1 \subset \cB$.

Let's prove that $\Sigma$ is a $\lambda$-system. Let $A_1, A_2 \in \Sigma$ such that $A_1 \subset A_2$. First, we need to verify that $A_2 \setminus A_1 \in \Sigma$. Let $\sP (\Omega) \ni C \subset (A_2 \setminus A_1)$ and $\mu (A_2 \setminus A_1)=0$. We need to prove $C \in \cB$. WLOG, we assume $\mu (A_2) >0$. Then $\mu (A_2) = \mu (A_1) >0$.

Answer 1

If the space is too large, this may fail. We have independence postulated, so let's make it explicit as a product measure space.

[See some relevant discussion about $\Delta$ HERE ]

Let $K$ be $\{0,1\}^T$ where $|T| > \mathfrak c$, let $\nu$ be the product measure on the completed Borel sets $\mathcal G$. Let $\mathcal T$ be the trivial sigma-algebra on $K$, namely $\{\varnothing, K\}$.

Let $\Omega = K \times K$, let $\mu = \nu \otimes \nu$ be the product measure, and let $\mathcal A$ be the completed Borel sets on $\Omega$. Now $\mathcal A_1 = \mathcal G \otimes \mathcal T$ and $\mathcal A_2 = \mathcal T \otimes \mathcal G$ are independent with respect to the measure $\mu$. And they are both complete.

The diagonal $\Delta = \{(x,x) : x \in K\}$ does not belong to the product sigma-algebra $\mathcal B = \sigma(\mathcal A_1 \sup \mathcal A_2) = \mathcal G \otimes \mathcal G$. Even though $\Delta$ has outer measure zero (by appropriate versions of Fubini's theorem), it does not belong to $\mathcal B$.