Let $S\subset \mathbb{R}$ be measurable. Show that $S$ is measurable if and onlt if $S\times \mathbb{R}$ is measurable.
My attempt:
If $S$ is measurable, there exist open $U\subset \mathbb{R}$, closed $V\subset \mathbb{R}$ s.t.
$$U\supset S \supset V$$
s.t. $\lambda(U\setminus V)<\epsilon$. Also $U\times \mathbb{R} \supset S\times \mathbb{R} \supset V\times \mathbb{R}$ but I'm not sure how to show the difference in Lebesgue measure can be controlled part. Do I need to take any subset of $S\times \mathbb{R}$ and look at top and bottom approximations and then estimate the difference in Lebesgue measure?
I think this will help me figure out how to prove the other direction as well.
Best Answer
Consider this set $M = \{A\in\cal{B}(\mathbb{R})| A\times \mathbb{R} \in \cal{B}(\mathbb{R}^2)\}.$ If $A$ is open, $A\in M$. What else can you say about $M$?