If $A \times B$ is Lebesgue measurable in $\mathbb{R}^2$ and $B$ is Lebesgue measurable in $\mathbb{R}$ then $A$ is so?

Question

Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial: Write $A= A \times \{0\}= \cup_i(A_{i} \times B_{i})$,

$m_1(A)=m_2(A \times \{0\})=\cup_i m_1(A_i)m_1(B_i)$

I don't know what to do next! Is my approach correct?

where $m_1,m_2$ are 1 and 2 dimensional measures in $\mathbb{R},\mathbb{R^2}$ respectively.

Answer 1

The statement is not true.
Hint: Consider a set $B$ with measure 0.

The statement is true if $B$ has positive measure. To prove this, show that $\mu(A \times B) = \mu^*(A)\mu^*(B) = \mu_*(A) \mu_*(B)$ where $\mu^*$ is outer Lebesgue measure and $\mu_*$ is inner Lebesgue measure. This then forces $A$ to be measurable.

Answer 2

This is false. For a counterexample, take $B=\mathbb Q\ $ (or even $B=\emptyset)$ and $A=$ a Vitali set. Then, $\lambda^*_2(A\times B)=0$ so $A\times B\in \mathscr M(\mathbb R^2)$ but $A\notin \mathscr M(\mathbb R).$

Answer 3

As a complement to the answers showing that the statement can be false if $B$ has measure 0: let us see how we can prove the statement is true if $B$ has positive measure.

Consider the characteristic function $f(x,y) = \chi_{A \times B}(x, y) = \chi_A(x) \chi_B(y)$. By the Fubini-Tonelli theorem, the fact that $f$ is measurable and nonnegative implies that for almost every $y$, the slice $f_y : \mathbb{R} \to \mathbb{R}, x \mapsto f(x,y)$, is measurable. Since $B$ has positive measure, there must be some $y \in B$ such that this condition holds. But then, $f_y(x) = \chi_A(x)$ so the measurability of $f_y$ implies the measurability of $A$.