# Are Borel sets whose projections are Borel closed under intersection?

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.

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.