Metric spaces in which Prohorov's theorem is not valid

Let X be a Hausdorff topological space. By a measure on X we understand a tight Borel probability measure on X. The set of all measures on X is denoted by P(X); this set is a topological space with the usual topology. This topology can be described as follows: If #~ is a net in P(X) and # ~ P(X) then lim #~ = # if and only if liminf #~ (G) = # (G) for every open set G c X. For convenience a space X is called a Prohorov space if for every compact set M c P(X) and every e > 0 there exists a compact set A c X such that # (A) > 1 for each/~M. It is well known that every topologically complete space X (i. e. space which is a G~ subspace of some compact space) is a Prohorov space (see Corollary 1 of Theorem 1). Varadarajan [3] claimed to prove that a metric space X is a Prohorov space provided that every Borel measure on X is tight (consequently a separable metric space which is a Borel subset of its completion is a Prohorov space), but his proof is incorrect. An example of a K~ metric non-Prohorov space (and therefore the proof of non-validity of Varadarajan's theorem) was given by Davies [1]. In this note it is proved that a co-analytic separable metric space is a Prohorov space if and only if it is topologically complete (consequently a separable metric space which is a Borel subset of its completion is a Prohorov space if and only if it is a G~ subset of its completion). This theorem gives also a solution of the problem whether the space of rational numbers is a Prohorov space (see e.g. [1]). The reader, who is interesting only in this problem, can find its solution in part III which does not depend on topological results of part II. We begin with the following trivial lemma which will be used without special mention.