Let $(X, \tau)$ be a topological space, and $\mathcal{P}(X, \tau)$ be the Borel probability measures living on $X$. Can we recover $(X, \tau)$ from $\mathcal{P}(X, \tau)$?
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3$\begingroup$ What do you "know" about $\mathcal{P}(X,\tau)$? To be more precise, I assume you want a theorem of the form "Suppose $(X_1, \tau_1), (X_2, \tau_2)$ are two topological spaces. Suppose there is a map $T : \mathcal{P}(X_1, \tau_1) \to \mathcal{P}(X_2, \tau_2)$ having properties A,B,C. Then $(X_1, \tau_1)$ and $(X_2, \tau_2)$ are homeomorphic." But what should go in for A,B,C? Just a bijection? (Obviously won't work). Preserves convex combinations? Isometry of the total variation metrics? Homeomorphism of the weak topologies? $\endgroup$– Nate EldredgeOct 5, 2015 at 18:39
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2$\begingroup$ So do you "know" what $\mathcal{P}(X,\tau)$ is as a set? Do you know its convex structure? Do you know its total variation metric / topology or its weak topology? Or something else? $\endgroup$– Nate EldredgeOct 5, 2015 at 18:44
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3$\begingroup$ But a fact that will probably be useful is that, under mild assumptions, if you equip $\mathcal{P}(X,\tau)$ with the weak topology, then the set of point masses is homeomorphic to $(X,\tau)$. And the point masses are exactly the extreme points. So if you know the convex structure and the weak topology, then the answer is morally "yes". The total variation topology will not be enough as it can only "see" the Borel structure of $X$, not the topology. Is that close to what you are looking for? $\endgroup$– Nate EldredgeOct 5, 2015 at 19:25
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4$\begingroup$ The title asks about a topological space; the question in the text asks about a probability space. I assume the former is intended, because it makes a bit more sense. The "topological space" question has an easy negative answer. The real line, the circle, the plane, and lots of other non-homeomorphic spaces have isomorphic $\sigma$-algebras of Borel sets, and so their Borel probability measures match in any of the ways that Nate suggested. $\endgroup$– Andreas BlassOct 5, 2015 at 19:26
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2$\begingroup$ To add one more to the comments already present. If you take any topological space $X$ and a Borel set $B$ which is not open, form $X'$ by extending the topology of $X$ in such a way that $B$ is open in $X'$. Then $X$ and $X'$ have the same Borel algebra but different topologies. $\endgroup$– François G. DoraisOct 5, 2015 at 21:09
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1 Answer
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I write this as an answer due mainly to space constraints. I want to elaborate a bit on Nate Eldredge's useful comments.
Think of Gelfand's classical theorem stating that a compact space $X$ is determined by the space $C(X)$ of continuous complex valued functions on $X$. This statement is more precise than this.
The space $C(X)$ is given an algebraic-topologic structure (commutative $C^*$-algebra). This structure alone completely determines the space $X$ as the spectrum of this algebra equipped with an appropriate topology. Continuous maps between two compact spaces $X_1$ and $X_2$ induce continuous morphisms between the corresponding $C^*$-algebras, and the isomorphisms of $C^*$-algebras induces homeomorphisms between the spectra. $\newcommand{\bR}{\mathbb{R}}$
It may help to assume first that your topological space $(X,\tau)$ is compact, for then the space of finite Borel measures can be identified with the space of bounded continuous linear maps $C(X)\to\bR$. We denote $\newcommand{\eP}{\mathscr{P}}$ by $\eP(X)$ the space of Borel probability measures on $X$. Then $\eP(X)$ is a closed convex subset of the dual $C(X)^*$. The map
$$ X\ni x\mapsto \delta_x \in \eP(X) $$
maps $X$ bijectively onto the set of extremal points of $\eP(X)$. (Above $\delta_x$ denotes the Dirac $\delta$-measure supported at $x$.) This map is continuous (w.r.t. to the weak topology on $C(X)^*$) and thus establishes a homeomorphism between $X$ and the set of extremal points of $\eP(X)$. In this sense $\eP(X)$ determines $X$ but this answer is far less satisfactory than Gelfand's theorem mentioned above.
answered Oct 5, 2015 at 19:27
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$\begingroup$ It should be noted that "compact" here presumably means "compact Hausdorff". $\endgroup$ Oct 5, 2015 at 19:29
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$\begingroup$ @NateEldredge Yes, compact =compact Hausdorff. $\endgroup$ Oct 5, 2015 at 19:52
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$\begingroup$ For spaces $X$ that are not first countable, I think it is necessary to restrict to regular measures. I am worried about something like the Dieudonné measure which is an extremal point but not a Dirac measure. $\endgroup$ Oct 5, 2015 at 19:56
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$\begingroup$ @NateEldredge Thanks for pointing this out. I was not aware of the Dieudonne measure. $\endgroup$ Oct 5, 2015 at 20:03