A question over product measure and completion

Question

Supposing two $\sigma$-finite measure spaces not necessarily complete as $(X, B_X,m_X)$ and $(Y, B_Y,m_Y)$, take their completions as $(X,\overline{B_X},\overline{m_X}), $and $(Y,\overline{B_Y},\overline {m_Y})$. Then take the product measure space of the former two spaces as $(X\times Y,B_X\times B_Y,m_X\times m_Y) $ and the latter two as $ (X\times Y, \overline{B_X}\times \overline{B_Y},\overline{m_X}\times \overline{m_Y})$. Since they are $\sigma$-finite, then they are uniquely defined and these two product measure spaces are not generally identical(this is obvious by taking $B_X,B_Y$ as Borel $\sigma$-algebra of real line). However if we further take completion of these two product measure spaces are they equal? I.e., $$(X\times Y,\overline {B_X\times B_Y}, \overline {m_X\times m_Y})=(X\times Y, \overline {\overline {B_X} \times \overline {B_Y}},\overline {\overline{m_X}\times \overline {m_Y}})?$$ This is true if $B_X$ and $B_Y$ are Borel $\sigma$ algebra with $X$ and $Y$ as the real line, with $m_X,m_Y$ as Lebesgue measure. My question is that is it true in general.

Answer 1

Since product of $\sigma$-algebras need no to be $\sigma$-algebra, it is better to show that $\overline{\sigma(M\times N)}=\overline{\sigma(\overline{M}\times\overline{N})}$.

That $M\subseteq\overline{M}$ and $N\subseteq\overline{N}$ implies that $M\times N\subseteq\sigma(\overline{M}\times\overline{N})\subseteq\overline{\sigma(\overline{M}\times\overline{N})}$ and then $\sigma(M\times N)\subseteq\overline{\sigma(\overline{M}\times\overline{N})}$ and then $\overline{\sigma(M\times N)}\subseteq\overline{\sigma(\overline{M}\times\overline{N})}$.

Now we prove that $\sigma(\overline{M}\times\overline{N})\subseteq\overline{\sigma(M\times N)}$. It suffices to prove that $\overline{M}\times\overline{N}\subseteq\overline{\sigma(M\times N)}$. For $E\times F\in\overline{M}\times\overline{N}$, then there are $A\in M$, $B\in N$ and $S$, $T$, and $U\in M$, $V\in N$, $U,V$ are null, $S\subseteq U$, $T\subseteq V$, $E=A\cup S$, $F=B\cup T$.

Now define $P_{1,1}=S$, $P_{2,1}=E$, $P_{1,2}=F$, $P_{2,2}=T$, $Q_{1,1}=U$, $Q_{2,1}=A\cup U$, $Q_{1,2}=B\cup V$, $Q_{2,2}=V$, then \begin{align*} E\times F&=(A\cup S)\times(B\cup T)\\ &=(A\times B)\cup(P_{1,1}\times P_{1,2})\cup(P_{2,1}\times P_{2,2}), \end{align*} and \begin{align*} & P_{1,1}\times P_{1,2}\subseteq Q_{1,1}\times Q_{1,2}\\ & P_{2,1}\times P_{2,2}\subseteq Q_{2,1}\times Q_{2,2}, \end{align*} and $Q_{1,1}\times Q_{1,2}$ and $Q_{2,1}\times Q_{2,2}$ are null in $M\times N$, so $E\times F\in\overline{\sigma(M\times N)}$, we are done.