# Prokhorov theorem

Hi there. It is known that on a polish space, if a family of bounded positive measures (no need to be probabilities) is tight, then it is relatively compact in the space of positive measures with usual weak topology. Furthermore, on any metric space, if a family of probability measures is tight, then it is relatively compact in the space of probability measures with the usual weak topology. So the question is, whether it is the case that on a Borel space, if a family of bounded positive measures is tight, then it is relatively compact in the space of positive measures with usual weak topology? (I am aware of some more general results on this about family of signed measures on topological spaces, but let us be specific).

A related question, is the space of positive measures (on Borel spaces endowed with Borel sigma algebra) closed in the space of signed measures equipped with the usual weak topology, i.e. that generated by bounded continuous functions? Many thanks indeed.

• I am not sure what are you asking for. If $X$ is locally compact but not compact, then you may embed $X$ into the space of positive measures on $X$ just by $x\mapsto \delta_x$. Thus, you have a closed subset which is not compact. – Jan Veselý Apr 8 '12 at 23:58
• Your condition for relative compactness in Polish spaces is not sufficient, the measures have to be uniformly bounded. – Michael Greinecker Sep 20 '16 at 22:22

## 1 Answer

If $C_b$ is the space of bounded continuous real-valued functions on a topological space $T$, $M_t$ the space of bounded signed tight meaures on $T$, then each uniformly bounded (uniformly) tight subset of $M_t$ is relatively $\sigma (M_t,C_b)$-compact (see, for example: D.H. Fremlin, D.J.H. Garling, R.G. Haydon, Bounded measures on topological spaces, Proc. London Math. Soc. 25(1972), 115--136).

The answer to the second question is yes, bay the definition of a positive measure.