Mt+(S) is a cone ofM(S) if S is complete, because thenMt+(S) = Ms+(S). In general,Mt+(S) is not a cone ofM(S) because it is not closed. It follows from Remark 4.20 thatMt+(S) is dense inMs+(S). Assume thatMt+(S) is closed and S is separable. Then,Mt+(S) =M+(S). This implies that S is universally measurable; however, not all separable metric spaces are universally measurable ([1, Sect. 11.5]).
ASJC Scopus subject areas
- Theoretical Computer Science