Consider the Lebesgue measure space $\mathscr M^1=(\mathbb R,\mathcal M,m)$. So, the Product space is given by $$\mathscr M^2=(\mathbb R^2, \mathcal M\times \mathcal M, m_2).$$
Even if $\mathscr M^1$ is complete, the space $\mathscr M^2$ is not complete. For instance, if $\mathcal N$ is the Vitali set, then $\{0\}\times \mathcal N$ is not in $\mathcal M\times \mathcal M$. But I'm not really sure how to show that. My question are the following :
1) I know that $$\mathcal M\times \mathcal M=\sigma \{A\times B\mid A,B\in \mathcal M\}.$$ But do we have that $$\{A\times B\mid A,B\in \mathcal M\}=\sigma \{A\times B\mid A,B\in \mathcal M\} \ \ ?$$ I guess it's not true (I can't find a counter example, do you have one ?)
2) So if $\mathcal M\times \mathcal M\neq \{A\times B\mid A,B\in \mathcal M\}$, an element $C\times D$ can be in $\mathcal M\times \mathcal M$ but neither $C$ nor $D$ is in $\mathcal M$. So how can I prove that $\{0\}\times \mathcal N\notin \mathcal M\times \mathcal M$ ?
If $A$ is any set in $\mathbb R$ which is not Lebesgue measurable than $A\times \mathbb \{0\}$ does not belong to $\mathcal M \times \mathcal M$. [This is because its 'section' $A$ is not Lebesgue measurable]. But it is it is contained in $ \mathbb R \times \{0\}$ which has measure $0$. This proves that the product measure is not complete. In 2) you seem to assume that any set in the product is a product set. That is not true. If you have a product set $C\times D$ which belongs to the product sigma algebra then $C$ and $D$ necessarily belong to $\mathcal M$. [An example of a subset of $\mathbb R^{2}$ which is not of the form $C \times D$ is the diagonal $\{(x,y)\in \mathbb R^{2}:x=y\}$].