1
$\begingroup$

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 $A \times \mathbb{R}$?

$\endgroup$

1 Answer 1

0
$\begingroup$

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

https://unapologetic.wordpress.com/2010/07/19/sections-of-sets-and-functions/

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.