# Projection of a measurable cylinder of the product $\sigma$-algebra

Take some set $A \subset \mathbb{R}$ such that $A \times \mathbb{R}$ is measurable in $\mathrm{B}(\mathbb{R}) \times \mathrm{B}(\mathbb{R})$, the product $\sigma$-algebra of the borel $\sigma$-algebra of $\mathbb{R}$ by itself (which is also $\mathrm{B}(\mathbb{R^2})$, the borel $\sigma$-algebra of $\mathbb{R}^2$ for its product topology). Then must $A$ be measurable in $\mathrm{B}(\mathbb{R})$?

I know the converse is true, and that the image of a measurable set for the product $\sigma$-algebra by a projection isn't necessary measurable, but what happens for a just cylinder $A \times \mathbb{R}$?

Alright, it turns out that it was true, $A$ needs to be measurable. Someone posted on another page here a link to the solution: