Give a counterexample of the compactness property I want to refute that if $\Omega$ is complete metric space, $F$ is Borel $\sigma$-algebra $\Omega$, then for any positive $\varepsilon$ there is a compact set $K$ such that $\mathbb{P}(K)>1−\varepsilon$.
I know that for this limitation to be true, I need need separability of $\Omega$ or at least separable support of $\mathbb P$. Therefore, I need to give a counterexample, but unfortunately I cannot come up with one. I would be very grateful for help
 A: One particularly simple example we might hope for is if we let $\Omega$ be a discrete metric space. Then the compact sets are all finite, so as long as finite sets are measure-zero, we're done. Of course the simplification to the compact sets comes with a price elsewhere: now the Borel algebra is the full power set and thus we need a (countably-additive) probability measure that is defined on the whole power set and such that finite sets have measure zero. This is the measure problem. The nonexistence of such a set and measure is consistent with ZFC, and the existence of one is consistent with ZFC if and only if a measurable cardinal is consistent. So we have a "trivial" counterexample, but at a high consistency strength.
But it turns out any counterexample requires a measurable cardinal, so the discrete counterexample isn't suboptimal in terms of consistency strength.

Addition
Since the reference given to this second claim is an encyclopedic 900-page volume that doesn't take a particularly direct route toward the result, I'll include a more streamlined version of Fremlin's proof here. This is my first pass at understanding this theorem, so I may have introduced some errors and there may well be some additional simplifications I missed.
First, a definition.

A cardinal $\kappa$ is measure-free if it does not admit a non-zero finite measure on all of $P(\kappa)$ such that finite sets have measure zero.

If a there is a cardinal that is not measure-free, the least such cardinal is real-valued measurable, and the existence of a real-valued measurable cardinal is equiconsistent with the existence of a measurable cardinal. (See, e.g. Jech Set Theory 3rd edition, chapter 10 and the beginning of chapter 22.) The main theorem is

Let $\Omega$ be a complete metric space with weight $w(\Omega)=\kappa$ (i.e. $\kappa$ is the minimum cardinality of a base for $\Omega$'s topology). If $\kappa$ is measure-free, then for any finite Borel measure $(\Omega, \mathcal B(\Omega), \mu)$ and $\epsilon > 0$, there is a compact $K\subseteq \Omega$ with $\mu(\Omega\setminus K) < \epsilon.$

First, note that if we can find a closed, separable $\Omega_0\subseteq \Omega$ such that $\mu(\Omega\setminus\Omega_0)=0,$ we're done: $\Omega_0$ is a complete and separable metric space and the restriction of $\mu$ is a finite Borel measure on $\Omega_0.$ Thus, by the result OP cites that additionally assumes separability, there is a compact set $K\subseteq \Omega_0$ with $\mu(\Omega_0\setminus K)<\epsilon,$ and since $\mu(\Omega\setminus\Omega_0)=0,$ we have $\mu(\Omega\setminus K)<\epsilon.$
So, it suffices to prove

If $\Omega$ is a complete metric space, $w(\Omega)$ is measure-free, and $\mu$ is a finite Borel measure on $\Omega$, then there is a closed, separable $\Omega_0\subseteq \Omega$ such that $\mu(\Omega\setminus \Omega_0)=0.$

First a couple lemmas:

Lemma 1. Let $\Omega$ be a metric space with $w(\Omega)$ measure-free, $\mu$ a finite Borel measure on $\Omega,$ and let $\mathcal F$ be some collection of disjoint open subsets of $\Omega$. Write $$\mathcal F^0 := \{U\in\mathcal F: \mu(U)=0\}\\\mathcal F^{>0}:=\{U\in \mathcal F: \mu(U)> 0\}.$$ Then $\mathcal F^{>0}$ is countable and $\mu\left(\bigcup \mathcal F^{0}\right) =0.$
Proof. It's clear that $\mathcal F^{>0}$ is countable since the measure is finite and the sets in $\mathcal F$ are disjoint.
For the other claim, say that to the contrary, $\mu\left(\bigcup \mathcal F^{0}\right) > 0.$ We will show this means $w(\Omega)$ is not measure-free. Write $\mathcal F^0$ as an indexed set $\mathcal F^0 = \{U_i: i\in I\}.$ Then, we can define a non-zero finite measure space $(I, P(I), \nu)$ by letting $\nu(J) = \mu\left(\bigcup_{i\in J} U_i\right)$ for $J\subseteq I.$ Since the $U_i$ are disjoint, it's easy to verify countable additivity and show that $\nu$ is a measure on the full power set of $I$ with $\nu(\{i\}) = \mu(U_i) = 0.$ Thus $|I|$ is not measure-free.
Since the $U_i$ are disjoint, for any base for $\Omega$'s topology, there will be some distinct basis element inside each $U_i.$ Thus, we have $|I|\le w(\Omega).$ But it's well-known that if a cardinal $\kappa$ is not measure-free, then neither is any $\lambda>\kappa$: we can extend a witnessing measure on $\nu$ $P(\kappa)$ to one on $P(\lambda)$ by letting $\nu'(X) = \nu(X\cap\kappa).$ So $w(\Omega)$ is not measure-free.


Lemma 2. If $\Omega$ is a metric space, there is a base $\mathcal B$ for its topology that is a countable union of disjoint collections. In other words, $\mathcal B = \bigcup_{n=1}^\infty \mathcal B_n$ where each $\mathcal B_n$ is a disjoint collection of open sets.
Proof Sketch. Let $(q_n,q_n')$ be an enumeration of $\{(q,q')\in\mathbb Q^2: 0<q<q'\}.$ Enumerate $\Omega=\{x_\xi: \xi < \kappa\}$ where $\kappa=|\Omega|.$ Let $$G_{n\xi} = \{x\in \Omega: d(x,x_\xi) < q_n\land \inf_{\eta < \xi}d(x,x_\eta)>q_n'\}$$ and then take $\mathcal B_n=\{G_{n\xi}: \xi<\kappa\}.$  and verify the $G_{n\xi}$ are open sets, disjoint for fixed $n,$ and that $\bigcup_n \mathcal B_n$ is a base. For more info, see Fremlin's Measure Theory, Volume 4, section 4A2L(g)(ii)

Then we can put it together as follows. Let $\mathcal B=\bigcup_n\mathcal B_n$ be a base as in the second lemma. For each $n,$ similarly to the first lemma, let $\mathcal B_n^{>0}$ denote the collection of positive-measure sets in $\mathcal B_n$ and $\mathcal B_n^0$ the null sets. By the first lemma, each $\mathcal B_n^{>0}$ is countable and the union of the sets in $\mathcal B_n^0$ is null. Define $\Omega_0 = \Omega\setminus \bigcup_n \left(\bigcup \mathcal B_n^0\right).$ $\Omega_0$ is clearly closed with $\mu(\Omega\setminus \Omega_0)=0,$ so we just must show that it is separable.

Claim: If $U\in\mathcal B$ and $U\cap\Omega_0\ne \emptyset,$ then $U\in \bigcup_n\mathcal B_n^{>0}.$
Proof. Let $x\in U\cap\Omega_0.$ $U\in \mathcal B,$ so $U\in \mathcal B_n$ for some $n$. Since $x\in \Omega_0,$ it follows that $U\notin \mathcal B_n^0$ (or we'd have $x\in \bigcup \mathcal B_n^0$), and thus $U\in \mathcal B_n^{>0}.$

So, noting that $\bigcup_n \mathcal B^{>0}_n$ is countable, $\{U\cap \Omega_0:U\in\mathcal B\}$ is a countable base for $\Omega_0,$ so $\Omega_0$ is separable.
A: In addition to spaceisdarkgreen's answer:
In  Convergence of probability measures, 1999, by Billigsley, p.12, there's the next place:
(in other words): Let $S$ be uncountable discrete space. Then it is complete and not separable. If $S$ has the power of continuum, if we assume the axiom of choice and the continuum hypothesis, it may be shown that all probability measures are tight, i.e. for all $\epsilon$ there exists compact $K$ such that $P(K) > 1- \epsilon$.
The idea of the proof: if there is a not tight measure on such $S$ then it's easy to get a noatomic probability measure on $S$ and then work with uncountable measurable cardinal, see also  math.stackexchange.com/questions/2511273
It also shows that your question refers to deep problems of set theory, because even in "not very complicated" cases like $[0,1]$ with metric $d(x,y) = \mathbf{1}_{\{ x \ne y \}}$ the question about existence of not tight measures implies reference to continuum hypothesis.
