# Why is it a borel set on the boundary of the unit ball of $E^n$?

Given $C$ convex body (compact convex set with non-empty interior points) in $E^n$ symmetric about the origin and containing the unit ball.

Let $A(r)$ denote ,for every real $r >1$, the subset of the boundary of the unit ball of $E^n$ obtaind by projecting on the boundary of the unit ball of $E^n$ from the origin ,the boundary points of $C$ which belong to $rBn$ (ball in $E^n$ of radius $r$ ).

I want to prove that $A(r)$ is a borel set on the boundary of the unit ball of $E^n$

Because the set is equal to $\frac{1}{r} \cdot(C\cap (r\cdot B_n ))$ and hence it is compact.
• i think that this set can't be written in this form since i want projection of the boundary points of C which belong to r$B_n$ i.e. that their norms are <=r so why you divide (C intersect r*$B_n$) by r – Roba Jul 14 '14 at 12:08