The projection of a Borel set of $\mathbb{R}^n$ needn't be Borel, although the projection of a closed set is a countable union of compact sets, hence Borel.

My question is if $A,B \subset \mathbb{R}^n$ are Borel such that $\pi_k(A), \pi_k(B)$ are known to be Borel, is it true $\pi_k(A \cap B)$ is Borel?

I haven't found an easy way to prove this, and can't find an answer, but it feels possible because intersecting seems limited in how much it can sabotage the projections. I proved a geometric theorem about such sets/projections, but lack that the projections are Borel.


1 Answer 1


Negative example.

Let $T \subseteq [0,1]\times[0,1]$ be a Borel set with non-Borel projection $\pi_1(T)$ on the first coordinate. Let $S = [0,1]\times[0,1]$ be the quare. Our sets are made up of vertical translates of these two: $$ A = T \cup (S + (0,2)), \\ B = T \cup (S + (0,5)), $$ so that $\pi_1(A) = [0,1]$ is Borel, $\pi_1(B) = [0,1]$ is Borel, but $\pi_1(A\cap B) = \pi_1(T)$ is not Borel.


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