Let $\mathcal{M}_1(\mathbb R)$ denote the space of Borel probability measures on $\mathbb R$. The space is a Polish space (a space which admits a complete, separable, metric) using, say the Levy-Prokhorov metric. For $\mu \in \mathcal{M}_1(\mathbb R)$, let $L^1(\mu)$ denote the Banach space of $\mu$-integrable functions (mod $\mu$-null). Again, for each $\mu$, $L^1(\mu)$ is a Polish space.

Is the space $$\mathbb X = \left\{(\mu,f) \ \middle|\ \mu \in \mathcal{M}_1(\mathbb R),\ f\in L^1(\mu)\right\}$$ a Polish space?

Of course we need to specify a topology. I am assume the most natural topology. (I believe this is the topology generated by sets of the form $$\left\{(\nu,g) \in \mathbb X \ \middle|\ d(\mu,\nu)<\varepsilon \text{ and } \|g-f\|_{L^1(\nu)} < \varepsilon \right\}$$
where $f \in C_0(\mathbb R)$, $\mu \in \mathcal{M}_1(\mathbb R)$, $\varepsilon > 0$, and $d$ is a metric on $\mathcal{M}_1(\mathbb R)$ which makes it a Polish space.)

If $\mathbb X$ is a Polish space, is there a nice metric?

If $\mathbb X$ is not a Polish space, how nice is this space?

The motivation for my question is whether $\mathbb X$ is a computable metric space, so feel free to answer that question as well? :)

Terminology: Feel free to correct me if I am mistaken on the terminology. I originally called $\mathbb X$ a "coproduct" and wrote it as $$\coprod_{\mu \in \mathcal{M}_1(\mathbb R)} L^1(\mu),$$ but I was corrected in the comments. (However, the underlying set is a direct sum of sets, and I think similar notation $\sum_{x:A} B(x)$ is used in homotopy type theory to express the sum of dependent types.)

After doing some research I think I should say the map $\mathbb X \to \mathcal{M}_1(\mathbb R)$ given by $(\mu,f)\mapsto \mu$ is a fiber bundle where $\mathbb X$ is the total space and $\mathcal{M}_1(\mathbb R)$ is the base space. (Although, I could be mistaken.)

Topology: In case I gave the wrong basis for the topology above, I want the following maps to be continuous: $$(\mu,f)\mapsto \mu,\quad (\mu,f)\mapsto \|f\|_{L^1(\mu)},\quad (\mu,f)\mapsto \int fg\,d\mu$$ where $g$ is a bounded continuous function.

  • 1
    $\begingroup$ @SebastianGoette I changed my terminology. Hopefully it is (more) correct now. $\endgroup$
    – Jason Rute
    Feb 29, 2016 at 20:53

1 Answer 1


Yes, $\mathbb X$ is a Polish space, even a computable one, but it doesn't look like it has a nice metric.

(I figured out the answer on my own question, but any other insightful answers or references would still be welcome.)

Suppose $\mathbb A$ is a Polish space and suppose that $\mathbb B(a)$ is a parametrized family of Polish metric spaces, each with a metric $d_a$. Further assume that the $\mathbb B(a)$ are "uniformly presented" in that for each $a \in \mathbb A$ there is dense sequence of points $b_n(a) \in \mathbb B(a)$ such that the following map is continuous: $$a, m, n \mapsto d_a(b_m(a), b_n(a))\quad :\quad \mathbb A \times \mathbb N\times \mathbb N \to \mathbb R.\qquad(*)$$

(As a technical point, assume $d_a(\cdot,\cdot) \leq 1$ for all $a\in \mathbb A$.)

Our goal is to find the appropriate Polish topology on the dependent sum set $$\mathbb{X} = \sum_{a\in \mathbb A} \mathbb B(a) = \{(a,b) \mid a\in \mathbb A, b\in \mathbb B(a)\}.$$

The TL;DR version is that we embed $\mathbb B(a)$ (uniformly) into $[0,1]^\mathbb N$, associate $\mathbb{X}$ with the resulting $G_\delta$ (${\bf \Pi}^0_2$) subspace of $A \times [0,1]^\mathbb N$, and then apply Alexandrov's Theorem which says that all $G_\delta$ subsets of Polish spaces are Polish.

The appropriate topology on $\mathbb{X}$ is the weakest topology where

  1. The first projection is continuous, $$\pi_1(a,b)=a \quad :\quad \mathbb X \to \mathbb A,$$

  2. The distances to the dense set $b_n(a)$ are continuous: $$(a,b) \mapsto (d_a(b,b_n(a)))_{n\in\mathbb{N}} \quad :\quad \mathbb X \to \mathbb[0,1]^\mathbb{N}.$$

The main idea is we will replace $b$ with the more manageable object $(d_a(b,b_n(a)))_{n\in\mathbb{N}} \in \mathbb[0,1]^\mathbb{N}$. Let $f\colon \mathbb X \to \mathbb A \times [0,1]^\mathbb N$ be the injective map $$f(a,b) = (a,(d_a(b,b_n(a)))_{n\in\mathbb{N}}),$$ and give $\mathbb X$ the topology induced by this map. That is, $f$ is a homeomorphism from $\mathbb X$ to the range of $f$, which we will call $\mathbb Y$. This is weakest topology satisfying (1) and (2) above. Moreover, this topology has the following nice properties.

  1. The second projection is a homeomorphism for a fixed $a \in \mathbb A$, $$\pi_2(a,b)=b \quad :\quad \mathbb X_a := \{b \mid (a,b)\in \mathbb X\} \to \mathbb B(a),$$

  2. Our dense sequence $b_n(a)$ embeds continuously $$a,n \mapsto (a,b_n(a)) \quad :\quad \mathbb A \times \mathbb N \to \mathbb X.$$

  3. A subbasis is given by sets of the following form, $$\{(a,b)\in \mathbb X \mid d_\mathbb {A}(a,a_0)<\varepsilon \text{ and }d_a(b,b_n(a))<\varepsilon\}$$where $a_0 \in \mathbb A$ and $\varepsilon >0$.

Now, we show $\mathbb{Y}$, and therefore $\mathbb X$, is Polish. The set $\mathbb{Y}$ is a $G_\delta$ (${\bf \Pi}^0_2$) subspace of $A \times [0,1]^\mathbb N$. This can be seen by looking at the definition of $\mathbb{Y}$, $$\mathbb Y = \{(a,(x_n))\mid \forall \varepsilon\ \exists n\ (x_n < \varepsilon)\} \cup \{(a,(x_n))\mid \forall n,m\ (x_n + x_m \geq d_a(b_n(a),b_m(a)) \leq x_n - x_m)\}$$

Alexandrov's theorem says that every $G_\delta$ subset of a Polish space is Polish. Therefore $\mathbb Y$ (and $\mathbb X$) are Polish.

This proof is effective (including Alexandrov's theorem). So $\mathbb X$ is a computable metric space (effective Polish space) if $\mathbb A$ is a computable metric space and $a,m,n \mapsto d_a(b_n(a),b_m(a))$ is computable.

Unfortunately, however, there doesn't appear to be a nice metric since we applied Alexandrov's theorem. (The exception is when the $\mathbb B(a)$ are compact (in some uniform sense?). Then $\mathbb Y$ is closed in $A \times [0,1]^\mathbb N$ and can use the same metric.)


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