Almost sure weak convergence of random probability measures

Given a sequence (μ n ) of random probability measures on a metric space S, consider the conditions: (i) μ n →μ (weakly) a.s. for some random probability measure μ on S; (ii) μ n (f) converges a.s. for all f∈C b (S). Then, (i) implies (ii), while the converse is not true, even if S is separable. For (i) and (ii) to be equivalent, it is enough that S is Radon (i.e. each probability on the Borel sets of S is tight) or that the sequence (P μ n ) is tight, where Pμ n (·)=E(μ n (·)). In particular, (i)⇔(ii) in case S is Polish. The latter result is still available if a.s. convergence is weakened into convergence in probability. In case S=T ∞ with T Radon, a.s. convergence of μ n (f), for those f∈C b (S) which are finite products of elements of C b (T), is sufficient for (i). In case and the limit μ is given in advance, a.s. convergence of characteristic functions is enough for μ n →μ (weakly) a.s. Almost sure weak convergence of random probability measures.