Prove that for each $y$ the y-section set $A_y=\{x\in \mathbb{R}: (x,y)\in A\}$ is a Borel measurable subset of $\mathbb{R}$.

Question

Suppose $A$ is a Borel measurable subset of $\mathbb{R}^2$. Prove that for each $y$ the y-section set $A_y=\{x\in \mathbb{R}: (x,y)\in A\}$ is a Borel measurable subset of $\mathbb{R}$.

I feel like that is not easy to show $A_y$ is closed in $\mathbb{R}$. Any other methods?

Answer 1

The claim is wrong.

Pick any set $B\subset \mathbb{R}$ which is not measurable.

Let $A = \{0\} \times B \subset \{0\}\times \mathbb{R}$, where the latter is Lebesgue measurable in $\mathbb{R^{2}}$, with zero measure, therefore the first is also Lebesgue measurable in $\mathbb{R^{2}}$ (Lebesgue measure is complete).

By the choice of $B$, we get $A_0$ is not measurable in $\mathbb{R}$.

Answer 2

Hint: If a property holds on the generating set, then the property holds on all Borel sets. The open boxes $(a,c)\times(b,d)$ generate $\mathcal{B}(\mathbb{R}^2)$.