Since R^n is separable ( has a countable dense subset ), the arbitrary union may be replaceable by a countable union. If so, then then set is Borel. I assume that each closed ball has nontrivial radius and can be approximated by a union from a countable collection of closed balls ( or their complements by a countable union of open balls ).

Note that separability is key here, as is the nontriviality of every closed ball in the collection.

Gerhard "Ask Me About System Design" Paseman. 2010.10.26

EDIT: While I say separability may be important in the answer below, I think having a countable basis (second-countable?) is even more important for the answer. END EDIT.

Since R^n is separable ( has a countable dense subset ), the arbitrary union may be replaceable by a countable union. If so, then then set is Borel. I assume that each closed ball has nontrivial radius and can be approximated by a union from a countable collection of closed balls ( or their complements by a countable union of open balls ).

Note that separability is key here, as is the nontriviality of every closed ball in the collection.

Gerhard "Ask Me About System Design" Paseman. 2010.10.26