# Intersection of Borel subset with line

Let $y \in \mathbb{R}$. If $A$ is a Borel subset of $\mathbb{R}^2$, then $$A(y) = \{x \in \mathbb{R}| (x,y) \in A\}$$ is a Borel subset of $\mathbb{R}$.

I think that I have to show that $A(y)$ is in Borel-sigma algebra in $\mathbb{R}$. But the sigma algebra itself has no clear form, I mean I just know that it is the sigma algebra generated by all open sets, there is no clear form of its elements. Another thing is I am not sure if Borel "subset" means the same as Borel 'set' (which simply mean it is in the Borel sigma algebra). Any idea or hint about how to begin ? It is very good if anyone can list some important steps in proving this statement.

You can prove the statement by considering the set $M = \{ A\subseteq \mathbb{R}^2\,|\, \forall y\in\mathbb{R} \ A(y) \ \text{is Borel}\}$ and showing that $M$ contains all open sets and $M$ is a $\sigma$-algebra. Then $M$ must contain all Borel sets, because the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing all open sets.${}$
• @Carl Mummert: Please have a look at the question again. It isn't asking about the projection of the Borel set, but about the sets $A(y)$ for various $y\in\mathbb{R}$, there is a difference. – user35359 Sep 13 '15 at 0:25
• I did misread that. There is an alternate, easier proof in that case: just intersect $A$ with a vertical line, and the result will be a Borel set. I had to make a meaningless edit to the answer to make it possible to vote on it again. Thanks for the polite correction. – Carl Mummert Sep 13 '15 at 0:28
• Shouldn't the defintion be $M=\{A \subset \mathbb {R}^2 : A(y) \text { is a Borel subset of } \mathbb {R}\}?$ – zhw. Sep 13 '15 at 1:08