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, 


[Prokhorov Theorem](http://www.maths.lth.se/matstat/staff/bengtr/mathprob/prohorov.pdf)
.


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?



Prokhorov's theorem in non separable metric spaces
It is correct, see Theorem 8.6.7 in volume 2 of Bogachev's "Measure Theory" monograph. See also his Theorem 8.6.8 for a version of the second statement which covers the case of a non-separable space.


