Show that cartesian product of a measurable set with $\mathbb{R}$ is measurable 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.
 A: 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 $M$?
A: Let's show first that if $A$ is measurable then $A\times \mathbb{R}$ is measurable.  It is enough to consider the case $A$ bounded.  Then it is enough to show that $A \times I$ is measurable for a bounded open interval. Now, for every $\epsilon > 0$ there exists  $K\subset A \subset U$, $K$ compact, $U$ open, such that $\mu(U \backslash K) < \epsilon$. Take $J$ compact, $J\subset I$. We get $J\times K \subset A\times I \subset U\times I$, $\ \ J\times K$ compact, and $U\times I$ open $\ldots$.
Assume now that $A$ is not measurable. Let us show that $A\times I$ is not measurable.  May assume that $A$ is bounded.  Take $M\subset A$, $M$ measurable, of largest possible measure, and $N\supset A$, $N$, measurable, of smallest possible measure. Since $A$ is not measurable, we have $\mu(M) < \mu (N)$.
Let us show that $M\times I \subset A\times I$ is a measurable subset of $A\times I$ of largest possible measure. For this, it is enough to show that every measurable subset $K$ of $(A\backslash M) \times I $ has measure zero. We may assume $K$ compact. The projection of $K$ onto the first component is a compact subset $L$ of $A\backslash M$, and so of measure $0$. Since $K \subset L\times I$, we get $\mu(K) \le \mu(L\times I) = 0\cdot |I| = 0$. Similarly we show that $N\times I$ is a measurable set of smallest measure containing $A\times I$.
From the above it follows that $A\times I$ is not measurable.
