# $\sigma$-algebra generated by measurable rectangles is not complete

Part of an exercise I'm trying involves the problem of showing that the $$\sigma$$-algebra generated by measurable rectangles in $$\mathbb{R}^2$$ is not complete and I'm having some trouble thus far. The problem setup is as follows:

For $$E \subseteq \mathbb{R}^2$$ set the vertical section $$E_x = \{y \in \mathbb{R} \mid (x,y) \in E\}$$ and let $$\mathcal{E}$$ be all $$E$$ such that $$E_x$$ is Lebesgue measurable for every $$x$$. I then show $$\mathcal{E}$$ is a $$\sigma$$-algebra containing all measurable rectangles. Now let $$\mathscr{I}$$ be the $$\sigma$$-algebra generated by the collection of measurable rectangles in $$\mathbb{R}^2$$ and $$\mathscr{C}$$ the product measure obtained from the Caratheodory extension theorem. The next part says that if $$A \in \mathscr{M}$$ is a subset with positive Lebesgue measure and $$P \subseteq A$$ is a non-measurable subset, show that $$P \times \{0\} \subseteq A \times \{0\}$$ has measure $$0$$ but is not in $$\mathcal{E}$$ and hence deduce .

Showing it has measure $$0$$ is fine: it just follows by definition more or less, but I really don't see how to show it is not in $$\mathcal{E}$$. By definition (if I'm not mistaken): $$(P \times \{0\})_x = \{y \in \mathbb{R} \mid (x,y) \in P \times \{0\}\} = \begin{cases} \{0\} & \text{ if } x \in P \\ \emptyset & \text{ if } x \notin P \end{cases}$$ But both of these are Lebesgue measurable, which contradicts the question.

Naturally I must be missing something, most likely involving the CET - any help would be appreciated!

Theorem: let $$\mathcal{S}$$ and $$\mathcal{T}$$ some $$\sigma$$-algebras of some spaces $$X$$ and $$Y$$. Then if $$A\in \mathcal{S}\otimes \mathcal{T}$$ (where $$\otimes$$ denote the product $$\sigma$$-algebra) then $$[A]_a:=\left\{x\in Y :(a,x)\in A\right\}\in\mathcal{T}\quad\text{ and }\quad [A]^b:=\left\{x\in X :(x,b)\in A\right\}\in \mathcal{S}$$ for any chosen $$a \in X$$ or $$b\in Y$$.
Then choosing some $$G\in \mathcal{L}$$ with positive measure, where $$\mathcal{L}$$ is the Lebesgue $$\sigma$$-algebra in $$\Bbb R$$, you know that there is some non-measurable $$P\subset G$$. Now choose some non-empty null set $$N\in\mathcal{L}$$.
Then by definition of the product measure you knows that $$\lambda _2(G\times N)=0$$, so $$P\times N$$ is measurable in the completion of $$\mathcal{L}\otimes \mathcal{L}$$. But note that $$[P\times N]^x=P$$ for any chosen $$x\in N$$, then by the theorem stated above you find that $$P\times N\notin \mathcal{L}\otimes \mathcal{L}$$.