Is the separability of the space needed in the proof of the Prohorov's theorem? The Section 5 of the book:
Billingsley, P., Convergence of Probability Measures, 1999, 
studies Prohorov's theorem. A short reminder is given below.
Let $\Pi$ be a family of probability measures on $(S,\mathcal{F})$. We call $\Pi$ relatively compact if every sequence of elements of $\Pi$ contains a weakly convergent subsequence. The family $\Pi$ is tight if for every $\epsilon$ there is a compact set $K$ such that $P(K)>1-\epsilon$ for every $P$ in $\Pi$.
The direct half of the Prohorov's theorem is given in the Theorem 5.1: If $\Pi$ is tight, then it is relatively compact.
The converse half of Prohorov's theorem is given in the Theorem 5.2: Supose that $S$ is separable and complete. If $\Pi$ is relatively compact, then it is tight.
My question: In the proof of the Theorem 5.2 (i.e. relatively compact $\Rightarrow$ tight), we use separability and completness of the space $S$. On the other hand, in the proof of the Theorem 5.1 (i.e. tight $\Rightarrow$ relatively compact), I know that we do not need completness of $S$, but I do not know if we do need separability. I didn't find the place where separability is used in the proof of the Theorem 5.1. So my question is do I need or not the separability of the space $S$ in the direct part of the Prohorov's theorem?
Remarks: 


*

*I know the  proofs of the same theorem that use separability (e.g. Note). 

*Prohorov's theorem in most books is given as one theorem on Polish spaces, so they assume separability in both halfs. It goes like this usually: Let $S$ be a Polish space and $\Pi$ a collection of probability measures on $S$. Than $\Pi$ is tight if and only if it is relatively compact. 


The reason I am asking is that I would like to use the direct half of Prohorov's theorem on the problem I am currently working. Space $S$ in my case is complete but not separable.
Help with this would be great and needed. Thanks in advance.
 A: Separability is not necessary. In fact, tightness of a family of Borel probability measures implies relative compactness in the vague/weak-* topology on any completely regular space. For instance, this can be found in volume 4 of Fremlin's Measure Theory. Specifically Proposition 437U (b) shows that tight families are compact in the narrow topology, and 437K (c) shows that for completely regular spaces, the narrow topology agrees with the weak-* topology.

My original answer below answers the wrong question - the question is about whether tight implies relatively compact, rather than the other way.

Let $\kappa$ be a real-valued measurable cardinal, and $\mu : \mathcal{P}(\kappa) \rightarrow [0,1]$ a probability measure vanishing on singletons. Consider $\kappa$ to be a discrete metric space. Then the 1-element family $\{\mu\}$ is compact, because it is a singleton, but it is not tight because all compact subsets of $\kappa$ are finite sets, so have measure zero. 
You say that completeness is not necessary, but (unless you are making an extra assumption) it is, for essentially the same reason - there are separable metric spaces with Borel measures on them that are not tight.
A: You are correct that separability is not needed.  However, there is also not really any loss of generality in assuming it.  For suppose that $\Pi$ is tight.  Then for every $n$ there exists a compact set $K_n$ such that $\mu(K_n) > 1-\frac{1}{n}$ for all $\mu \in \Pi$.  So if we set $S_0 = \bigcup_{n=1}^\infty K_n$, then $S_0$ is separable and $\mu(S_0) = 1$ for all $\mu \in \Pi$.  We can now view $\Pi$ as a set of probability measures on $S_0$, and it is still tight (since the $K_n$ are also compact in $S_0$).  The separable case of the theorem then implies that $\Pi$ is weakly relatively compact in $\mathcal{P}(S_0)$, i.e. every sequence in $\Pi$ has a subsequence converging weakly in $\mathcal{P}(S_0)$, and you can easily check that such a subsequence also converges weakly in $\mathcal{P}(S)$.  So $\Pi$ is weakly relatively compact in $\mathcal{P}(S)$, as desired.
In other words, once you have a tight family, then all those measures live on a separable subset of $S$ anyway, so the rest of the space is irrelevant and might as well not be there.
