# Representating Borel probability measures as closed subset of $\mathbb{R}^{\infty}$

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!

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

# 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.

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)

• 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 $\mathbb{R}^\infty$. Mar 19 '20 at 20:32