$A$ not Lebesgue-measurable in $\Bbb{R} \implies A \times \Bbb{R}$ is not Lebesgue-measurable in $\Bbb{R}^2$?

Question

How do I show that if $A$ is not Lebesgue-measurable in $\Bbb{R}$ then $A \times \Bbb{R}$ is not Lebesgue-measurable in $\Bbb{R}^2$?

Answer 1

For any set $E \subset \mathbb{R}^2$, you can define a section of $E$ as $$ E_y := \{x \in \mathbb{R} : (x,y) \in E\} $$ Similarly define $E^x$, and let $\mathcal{M}$ be the collection of all subsets $E$ of $\mathbb{R}^2$ such that $E_y$ and $E^x$ are both measurable subsets of $\mathbb{R}$.

Now prove that this collection $\mathcal{M}$ is a $\sigma$-algebra that contains all rectangles. Hence, $\mathcal{M}$ must contain all measurable subsets in $\mathbb{R}^2$. In particular, if $A \subset \mathbb{R}$ is such that $E := A\times \mathbb{R}$ is measurable, then $$ A = E_0 $$ must be measurable in $\mathbb{R}$.