Completeness of product measure

Question

Given two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$, we can define the product measure $\mu\times\nu$ on the algebra $\mathcal{A}\times\mathcal{B}$ defined as the $\sigma$-algebra generated by all measurable rectangles in $X\times Y$.

Assuming $\mu$ and $\nu$ are complete, $\mu\times\nu$ need not be. For example, consider $A\in\mathcal{A}$ with $A\ne \emptyset$ but $\mu(A)=0$, and $B\in\mathcal{P}(Y)-\mathcal{B}$ (if $\mathcal{P}(Y)-\mathcal{B}\ne\emptyset$). Then $A\times B\notin \mathcal{A}\times\mathcal{B}$, but $A\times Y\supset A\times B$ is a null set.

How can we prove that $A\times B\notin \mathcal{A}\times\mathcal{B}$? It is not obvious to me that we cannot generate $A\times B$ from the measurable rectangles. I cannot think of a way to have $A\times B\in \mathcal{A}\times\mathcal{B}$, however I do not know how I would prove that it is impossible?

Answer 1

If a set $C\subset X\times Y$ is an element of $\mathcal A\times\mathcal B$, then the section $C_x:=\{y\in Y: (x,y)\in C\}$ is an element of $\mathcal B$, for each $x\in X$. Now take $C=A\times B$ and $x\in A$ (using your choice of $A$ and $B$).

Answer 2

Example.Let $(X,\mathcal A,\mu)=(Y,\mathcal B, \nu)=(R, \mathcal M,m)$ where $R$ is the reals, $m$ is Lebesgue measure, and $\mathcal M=$dom $(m).$

Let $T=S\times \{0\}$ where $S\in \mathcal P(R)\backslash \mathcal M.$

Let $D_0$ be the Boolean algebra on $R^2$ generated by the Cartesian product $\mathcal M\times \mathcal M$.

For ordinal $a\in \omega_1,$ let $D_{a+1}$ be the Boolean algebra on $R^2$ generated by unions of countable subsets of $D_a.$

For $0\ne a=\cup a\in \omega_1,$ let $D_a=\cup_{b\in a}D_b.$ $$ \text {Now }\quad \mathcal M^*= \cup_{a\in \omega_1}D_a$$ is the $\sigma$-algebra generated by the Cartesian product $\mathcal M\times \mathcal M$, because $\omega_1$ is a regular cardinal.

By transfinite induction on $a\in \omega_1,$ if $T\subset U\in M^*$ then $\{x:(x,0)\in U\}\in \mathcal M,$ so $T\not \in \mathcal M^*.$

But for every $r>0$ there exist subsets $\{u_n\}_{n\in N}, \{v_n\}_{n\in N}$ of $\mathcal M$ such that $\cup_{n\in N}(u_n\times v_n)\supset T$ and $\sum_{n\in N}m(u_n)m(v_n)<r.$