# Prokhorov's theorem in non separable metric spaces

Recently, working in some calculations I needed to use the Prokhorov's theorem about compactness for probability measures. However, a friend warned me that I had not the hypotesis of separability required by the theorem.

After some searching, over the books which I have reach, this is the version of the theorem that I found:

Let ${\displaystyle (S,\rho )}$ be a separable metric space. Let ${\displaystyle {\mathcal {P}}(S)}$ denote the collection of all probability measures defined on ${\displaystyle S}$ (with its Borel σ-algebra).

Theorem (Prokhorov). A collection ${\displaystyle K\subset {\mathcal {P}}(S)}$ of probability measures is tight if and only if the closure of ${\displaystyle K}$ is sequentially compact in the space ${\displaystyle {\mathcal {P}}(S)}$ equipped with the topology of weak convergence.

However, in a internet research I found the following document,

where no separability is required to get one of the directions of the theorem, here a description of the result contained in this notes:

Theorem. Let $S$ be a metric space. If collection ${\displaystyle K\subset {\mathcal {P}}(S)}$ of probability measures is tight then the closure of ${\displaystyle K}$ is sequentially compact in the space ${\displaystyle {\mathcal {P}}(S)}$ equipped with the topology of weak convergence. Conversely if $S$ is separable and complete, then each relatively compact set is tight.

Question: Is this version correct? If so, can someone provide me some reference?

• The first statement is not correct without either having $S$ be complete as well, or $\mathcal{P}(S)$ being Radon probability measures instead of probability measures. A counterexample is a Lebesgue unmeasurable subset $S$ of $[0,1]$ with outer measure 1 equipped with the subspace topology - the restriction of Lebesgue measure to $S$ is a singleton in $\mathcal{P}(S)$, and therefore compact, but it is not tight. – Robert Furber Mar 13 '17 at 20:29