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Let $\mathcal{P}_d$ be the space of Borel probability measures on $\mathbb{R}^d$ endowed with the topology of weak convergence of probability measures. Does there exist a map $G: \mathcal{P}_d \to \mathbb{R}^{\infty}$ (the latter with the usual topology of pointwise convergence) $\textbf{which is one-to-one and continuous}$ such that $G(\mathcal{P}_d) \subseteq \mathbb{R}^{\infty}$ is closed?

Some thoughts on this: It is known that for $\{\mu_n\}_n \in \mathcal{P}_d$, the existence of lim$_n$$\int f d\mu_n$ in $\mathbb{R}$ for each $f \in C_b(\mathbb{R}^d)$ yields the existence of a unique $\mu \in \mathcal{P}_d$ such that $\mu_n \to \mu$ weakly. Hence, if one could choose $G$ as $$G(\mu) = \bigg(\int f d\mu\bigg)_{f \in C_b},$$ the answer to my question would be affirmative. Since $C_b$ is neither countable nor separable, such a choice of $G$ is not possible. Replacing $C_b$ by a dense, countable subset of $C_c(\mathbb{R}^d)$ does not work either, since then the limit object $\mu$ will in general only be a sub-probability measure. This allows only to conclude that this map $G$ would turn $G(\mathcal{M}^+_{\leq 1}) \subseteq \mathbb{R}^{\infty}$ into a closed set, where $\mathcal{M}^+_{\leq 1}$ denotes the set of all Borel sub-probability measure endowed with the vague topology.

Is there a way to solve this issue? I was thinking of replacing $C_b$ by the union of a dense, countable subset of $C_c$ and a sequence $(f_l)_{l \geq 1}$ of $C_b$-functions such that $f_l \uparrow 1$ uniformly. Then, the existence of a sub-probability measure $\mu$ as the vague limit of $(\mu_n)_n$ follows and my hope is to obtain $$\mu(\mathbb{R}^d) = lim^l\int f_l d\mu = lim^l \, lim^n \int f_l d\mu_n = lim^n \, lim^l \int f_l d\mu_n = lim^n \mu_n(\mathbb{R}^d) = 1,$$where the interchange of limits is justified by the uniform convergence of $f_l \to1$. However, the second equality uses that the integrals $\int f_l d\mu$ are represented by $lim^n \int f_l d\mu_n$. But the existence of $\mu$ is deduced from the Riesz-Markov-Kakutani representation and from there I do not know how to conclude that the integral representation of the limit object $\mu$ also holds for functions $f$ other than $f \in C_c$.

Is this approach feasible at all? If not, is there any other way to answer my question affirmatively? Thanks a lot in advance!

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  • $\begingroup$ I am unable to understand the question. Why can't you take $G(P)=(0,0,..0)$ for all $P$? $\endgroup$ Mar 19 '20 at 10:17
  • $\begingroup$ Sorry for being sloppy. Of course, I forgot to mention one crucial requirement: $G$ should be one-to-one. I will edit the post. Thanks! $\endgroup$
    – Marco
    Mar 19 '20 at 10:41
  • $\begingroup$ You probably want some more properties of $G$. Do you want it to be continuous and linear? $\endgroup$ Mar 19 '20 at 11:52
  • $\begingroup$ Sorry for again being imprecise. I've edited the post again. Linearity is, for my concerns, not necessarily needed I suppose. $\endgroup$
    – Marco
    Mar 19 '20 at 11:57
  • $\begingroup$ I was also sloppy :-). Linearity doesn't even make sense. I should have mentioned convexity instead of linearity. $\endgroup$ Mar 19 '20 at 11:59
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Yes.

In general, every Polish space is homeomorphic to a closed subset of $\mathbb{R}^\infty$; see Kechris, Classical Descriptive Set Theory, Theorem 4.17 (where the space is denoted $\mathbb{R}^{\mathbb{N}}$). And $\mathcal{P}_d$ with its weak topology is Polish; Kechris Theorem 17.23.

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I'm a little stumbled, but I cannot find a mistake in the following proof, which would somehow prove the statement of the above answer by Nate Eldredge without the assumption of the space under consideration being Polish (which I did not expect to be true!):

Let $X$ be a separable, metrizable topological space. Fix any metric $d$, which metrics the given topology on $X$. Now define

$$F: X \to \mathbb{R}^{\infty}, F(x) := (d(x,x_1),d(x,x_2),\dots), $$where $(x_n)_n$ is a fixed dense subset of $X$ (such set does not depend on the actual choice of $d$). $\mathbb{R}^{\infty}$ is endowed with the metric $\rho(\alpha,\beta) := \sum_{k\geq1} 2^{-k}|\alpha_k-\beta_k|$, which induces the topology of pointwise convergence. Let's show that $F$ is one-to-one and a homeomorphism between $X$ and $F(X) \subseteq \mathbb{R}^{\infty}$:

Firstly, for $d(x,y) = c > 0$ we find $x_k$ such that $d(x,x_k) < \frac{c}{3}$, thus $d(y,x_k) > \frac{c}{3}$ and hence $F(x) \neq F(y)$, whereby $F$ is one-to-one.

Secondly, for $\epsilon > 0$ and $x \in X$, if $d(x,y)< \epsilon$, then $|d(x,x_i)-d(y,x_i)| < \epsilon$ for each $i$, so that $\rho(F(x),F(y)) < \epsilon$, whereby $F$ is continuous.

Finally, let $\epsilon > 0$ and fix $F(x) \in F(X)$. There exists $x_n$ such that $d(x,x_n) < \frac{\epsilon}{3}$. If $y \in X$ with $d(y,x_n) > \frac{2}{3}\epsilon$, then $|d(x,x_n)-d(y,x_n)| > \frac{\epsilon}{3}$, hence $\rho(F(x),F(y)) > \frac{\epsilon}{3\cdot2^{n}}$. Thereby, $\rho(F(x),F(y)) \leq \frac{\epsilon}{3\cdot2^{n}} \implies d(x,y) < \epsilon, $whereby $F^{-1}$ is continuous from $F(X)$ to $X$.

Since $X$ is closed (as a subset of itself) and $F^{-1}$ is continuous, also $F(X) = (F^{-1})^{-1}(X)$ is closed.

Is there a mistake in my reasoning? (Sorry for posting this question as an answer, it was certainly too long for a comment and I did not want to open a new post for this)

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  • $\begingroup$ Your second-to-last paragraph only proves that $F(X)$ is closed in $F(X)$ (i.e. with respect to the subspace topology induced from $\mathbb{R}^\infty$), which is trivial. It does not prove that it is closed in $\mathbb{R}^\infty$. $\endgroup$ Mar 19 '20 at 20:32

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