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? :)




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**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.



Is the following product-like space a Polish space?
*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.)




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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
$$\mathbb{X} = \sum\_{a\in \mathbb A} \mathbb B(a) = \{(a,b) \mid a\in \mathbb A, b\in \mathbb B(a)\}.$$




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*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.*




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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.


3. The second projection is a homeomorphism for a fixed $a \in \mathbb A$,
$$\pi\_2(a,b)=b \quad :\quad \mathbb X\_a := \{b \mid (a,b)\in \mathbb X\} \to \mathbb B(a),$$
4. Our dense sequence $b\_n(a)$ embeds continuously
$$a,n \mapsto (a,b\_n(a)) \quad :\quad \mathbb A \times \mathbb N \to \mathbb X.$$
5. 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.




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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.)


