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It is well known that if $(M,\tau)$ is a compact Hausdorff topological space then (by Riesz–Markov–Kakutani representation theorem + Banach–Alaoglu theorem) we have that the space $$\mathcal M_1(M) :=\left\{\mu;\ \mu\ \text{is a }\tau\text{-Borel probability measure on }M\right\} $$ is compact in the weak$^*$ topology, $\textit{i.e}.$ the topology generated by the basis of neighborhoods $$V(\mu;f_1,\ldots,f_n;\varepsilon):=\left\{\lambda\in\mathcal M_1(M);\ \left|\int f_i\ \mathrm{d}\mu - \int f_i\ \mathrm{d}\lambda\right|<\varepsilon, \ \forall \ i\in\{1,\ldots,n\}\right\}, $$ where $f_1,\ldots,f_n \in C^0_b(M)=\{g: M\to\mathbb R; \ g \text{ is a continuous bounded function}\}$.

My question: Is there an example of a compact topological space $M$, such that $\mathcal M_1(M)$ is not compact in the weak${^*}$ topology?

I have searched online but I was not able to find any reference for this problem. Moreover, I have tried to construct a counter-example but I have failed miserably.

Can anyone help me?

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  • $\begingroup$ Do you mean if there is a non-Hausdorff compact topological space $M$ for which the weak-* compactness of the space of probability (Radon) measures on $M$ fails? (by the way, continuous real-valued functions on a compact topological space are always bounded, even without the Hausdorff property) $\endgroup$ May 16, 2020 at 5:21
  • $\begingroup$ I have only written continuous bounded function because I was writting the general definition (if it was need for some reason). About the first question. When I say "Borel probability" I mean a probability measure with respect to the $τ$ borel $σ$-algebra (the smallest $σ$-algebra that contains $\tau$). $\endgroup$ May 16, 2020 at 7:27
  • $\begingroup$ So, your underlying compact topological space $M$ is assumed to be Hausdorff, after all...? $\endgroup$ May 16, 2020 at 16:23
  • $\begingroup$ No, $M$ is not Hausdorff. M is only compact. Like $Z_2$ with the topology $\{Z_2,∅\}$. There are some books that say $X$ is compact if $X$ is Hausdorff and if each of its open covers has a finite subcover. In my case, $X$ compact does not imply that $X$ is Hausdorff, it only means that each open cover has a finite subcover. $\endgroup$ May 16, 2020 at 17:35
  • $\begingroup$ This should be stated more explicitly in the question, then, since this is absolutely critical. I've posted an answer below where I go through the steps of the proof in a bit more detail in the Hausdorff case, so it may become clearer in which of those the Hausdorff property of $M$ may be necessary. The answer is supposed to be updated in the future if a counterexample shows up, even if the bounty is expired or granted to someone else. $\endgroup$ May 16, 2020 at 17:47

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The answer to your question is no: $\mathcal{M}_1(M)$ is always weak-* compact, because one can always in a certain sense "recover" the Hausdorff property of $M$ as far as the properties of the space of continous functions on $M$ are concerned.

First of all, just a small remark: continuous real-valued functions whose domain $M$ is compact are always bounded, regardless of $M$ being Hausdorff or not. The proof is classical: let $M$ be a compact (being Hausdorff is not needed) topological space, $f:M\rightarrow\mathbb{R}$ continuous, and $\{U_j\ |\ j\in J\}$ be an open cover of $f(M)$. Since $f$ is continuous, $\{f^{-1}(U_j)\ |\ j\in J\}$ is an open cover of $M$. Since $M$ is compact, there is $I\subset J$ finite such that $\{f^{-1}(U_j)\ |\ j\in I\}$ is a subcover of $M$ and therefore $\{U_j\ |\ j\in I\}$ is a finite subcover of $\{U_j\ |\ j\in J\}$, hence $f(M)\subset\mathbb{R}$ is also compact. By the Heine-Borel theorem, $f(M)$ must be closed and bounded.

The (real) vector space $X=C(M,\mathbb{R})=C^0_b(M)$ of continuous ( $\Rightarrow$ bounded) (real valued) functions on a compact Hausdorff topological space $M$ is a Banach space when endowed with the supremum norm $\|f\|=\sup \{|f(x)|\ |\ x\in M \}$. The Hausdorff property of $M$ is also not needed to show completeness of $X$ in this norm - only the completeness of the codomain $\mathbb{R}$ is used. What is lost when one forfeits the Hausdorff property of $M$ is that one is no longer able to separate points of $M$ using continuous real-valued functions on it - more precisely, any (locally) compact Hausforff space is completely regular. However, the quotient space $\tilde{M}=M/\sim$ of $M$ modulo the equivalence relation $$x\sim y\Leftrightarrow f(x)=f(y)\text{ for all }f:M\rightarrow\mathbb{R}\text{ continuous}$$ is Hausdorff (and compact) with the quotient topology, is canonically homeomorphic to $M$ if the latter happens to be Hausdorff already (since then one can separate the points of $M$ with $X$ and therefore the equivalence class $[x]\in\tilde{M}$ of $x\in M$ modulo $\sim$ $$[x]=\{x'\in M\ |\ x'\sim x\}$$ equals $\{x\}$ for all $x\in M$), and we may identify $X=C(M,\mathbb{R})$ with $C(\tilde{M},\mathbb{R})$ (Edit: this can be done even if $M$ is not compact and always yields a completely regular - but then not necessarily compact - $\tilde{M}$, see e.g. Theorem 3.9, pp. 40-41 of the book by L. Gilman and M. Jerison, Rings of Continuous Functions, van Nostrand, 1960, as pointed in the answer to this MO question made by the OP's author, also linked in his comments below). Moreover, due to the compactness of $M$ and Hausdorffness of $\tilde{M}$ the quotient map $$M\ni x\mapsto[x]\in\tilde{M}$$ is even closed.

(Edit) The Riesz-Markov-Kakutani representation theorem, on its turn, actually holds even if $M$ is not Hausdorff (without uniqueness in this case, as pointed in the comments below). One is able to identify the topological dual $X'$ of $X$ with the space of signed, finite Baire measures on $M$ (recall that the Baire $\sigma$-algebra on $M$ is the $\sigma$-algebra $\mathfrak{Ba}(M)$ generated by the zero level sets of elements of $X=C(M,\mathbb{R})$, which is contained in the Borel $\sigma$-algebra $\mathfrak{Bo}(M)$ of $M$) see e.g. Lemma 8.25, pp. 293-295 of the book by V. Komornik, Lectures on Functional Analysis and the Lebesgue Integral, Springer-Verlag, 2016. This lemma relies only on a version of Dini's lemma (see e.g. Lemma 8.24, pp. 293 of the same book) and does not require the Hausdorff property. One can go even further and identify $X'$ with the space of signed, finite (but no longer necessarily regular) Borel measures on $M$ thanks to the Corollary to Theorem II.2.6.1, pp. 227-228 of the book by K. Fuchssteiner and W. Lusky, Convex Cones (North-Holland, 1981), as also pointed in the the answer to the aforementioned MO question. More precisely, the results quoted above tell us how to recover a measure from a positive linear functional on $X$ (which, by the way, happens to be continous). This measure, as pointed in the comments, is not necessarily unique - consider e.g. Dirac measures concentrated on distinct points of $[x]$ for some $x\in M$ with $[x]\neq\{x\}$. This happens, for instance, for $x\in M$ with $\{x\}\neq\overline{\{x\}}$ if $M$ is not $T_1$.

Conversely, any signed, finite Borel (resp. Baire) measure $\mu$ on $M$ yields an element of $X'$ once we identify each such $\mu$ with the associated integral, seen as a (bounded, hence continous) linear functional on $X$: $$\mu(f)=\int_M fd\mu\ ,$$ so that for each Borel (resp. Baire) subset $A\subset M$ we have $$\mu(A)=\int_M\mathbb{1}_A d\mu\ ,\quad\mathbb{1}_A(x)=\begin{cases} 1 & (x\in A) \\ 0 & (x\not\in A) \end{cases}\ .$$ As discussed above and in the comments below, this map is no longer injective if $M$ is not Hausdorff. However, one can still identify each element of $X'$ with a unique signed Radon measure on $\tilde{M}$ by the standard Riesz-Markov-Kakutani representation theorem, of course.

The weak-* topology on $X'$ is just the locally convex vector topology of pointwise convergence, determined by the seminorms $$\|\mu\|_f=|\mu(f)|=\left|\int_M fd\mu\right|\ ,\quad f\in X\ ,$$ hence a fundamental system of neighborhoods of zero in $X'$ in this topology is given by the neighborhoods $V(\mu;f_1,\ldots,f_n;\epsilon)$ you wrote above.

Finally, the Banach-Alaoglu theorem is an abstract result for Banach spaces which states that the unit ball $$B=\{\mu\in X'\ |\ \|\mu\|=\sup\{|\mu(f)|\ |\ \|f\|\leq 1\}\leq 1\}$$ in the norm topology of the topological dual $X'$ of a Banach space $X$ is compact in the weak-* topology. In our particular example, probability measures on $M$ are just the elements of $X'$ satisfying $$0\leq f\in X\Rightarrow\mu(f)\geq 0\ ,\,\mu(1)=1\ ,$$ the set $\mathcal{M}_1(M)$ of which is a closed subset of $X'$ in the weak-* topology (hence also in the norm topology of $X'$). Since $$|\mu(f)|\leq\mu(|f|)\leq\mu(1)=1$$ for all $f\in X$ with $\|f\|\leq 1$, $\mathcal{M}_1(M)$ is a (weak-* closed) subset of $B$, hence it must be weak-* compact as well.

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    $\begingroup$ I think he wants to know if the same holds without the assumption of $(M,\tau)$ being Hausdorff. It is clear from your proof that you have used the property of being Hausdorff a lot. Do you know if the same is true without using it? $\endgroup$ May 16, 2020 at 17:47
  • $\begingroup$ Yeah, that's the problem. I do not know in which of the steps above (if any) the Hausdorff property of $M$ is necessary. This is not needed to show that continuous real valued functions on $M$ are bounded, though. In any case, the answer will need to be updated in the future if this turns out to be the case. $\endgroup$ May 16, 2020 at 17:50
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    $\begingroup$ To make that a concrete example, if $M$ is the line with two origins $0$ and $0'$ then $\tilde{M} = \mathbb{R}$ (up to Homeomorphism). If the measure you get on $\tilde{M} = \mathbb{R}$ is $\delta_0$ then using the definition here $\psi^* \delta_0(\{0,0'\}) = \psi^* \delta_0(\{0\}) = \psi^* \delta_0(\{0'\}) = 1$ and so additivity clearly fails without further modification. This mathoverflow answer claims that the right additional information to use is a family of probability measures on the fibers of $\psi$ but I haven't verified the details of it. $\endgroup$ May 16, 2020 at 20:36
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    $\begingroup$ I'll add the above amendments to my answer when I can. $\endgroup$ May 16, 2020 at 21:37
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    $\begingroup$ Interesting... however, I'd like to go through the details of that to see whether I could make my answer self-contained. I'll try to do that tomorrow. $\endgroup$ May 17, 2020 at 19:24

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