If U is open in R, then is open, hence Borel, in S. Now suppose that are sets in R such that is Borel for each n. Then , which is a countable union of Borel sets and in therefore Borel in S. By a similar argument, is Borel.

That shows that the class of subsets V ⊆ R for which is Borel in S is a σ-algebra containing all the open sets in R. Therefore this class contains all the Borel sets in R. Thus f is Borel measurable.