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!


1 Answer 1


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 $b\in Y$.

(If you dont know it you can easily prove the theorem stated above using the monotone class theorem.)

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 $P\times N\notin \mathcal{L}\otimes \mathcal{L}$.

  • $\begingroup$ Thanks! I guess that it was more or less just needing to consider the analogous horizontal slicing to get it then, for which the arguments apply in the same way. $\endgroup$
    – user385631
    Oct 29, 2019 at 11:11

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