**By way of introduction:**

As expressed in some of the comments, I find the "locally compact" assumption possibly a bit too strong.

A weaker assumption than having a locally compact second-countable Hausdorff space would be that the space is a *Lusin space*, i.e. a separable metrisable topological space that satisfies the following beautifully equivalent statements:

- $X$ can be topologically identified with a Borel subset of a completely metrisable topological space;
- for any metrisation $d$ of the topology of $X$, $X$ is a Borel subset of the $d$-completion of $X$;
- if $X$ is uncountable then $(X,\mathcal{B}(X))$ is measurably isomorphic to $([0,1],\mathcal{B}([0,1]))$.

[I emphasise that this equivalence assumes that $X$ is a separable metrisable space.]

If $X$ is a Lusin space then we have some very nice properties in regards to random measures. But I will also explain some of the important properties more generally of second countable spaces, and of separable metrisable spaces.

**1. Countably generated Borel space**

The first thing to say is that everything is pretty much hopeless if you're working on a space $X$ for which the Borel $\sigma$-algebra $\mathcal{B}(X)$ is not countably generated. Much of probability theory is about almost sure statements, and it's often really important to be able to move logically from

*"for each $A \in \mathcal{B}(X)$, some assertion is almost surely true"*

to

*"it's almost surely true that for every $A \in \mathcal{B}(X)$ the assertion holds".*

The ability to do this often relies on $\mathcal{B}(X)$ being countably generated. This is fundamental to why second countability (or equivalently, for metrisable spaces, separability) is assumed - it guarantees that the Borel $\sigma$-algebra is countably generated.

**2. Measurable structure on the space of probability measures**

It would be really good if, on the space $M_X$ of Borel probability measures on $X$, the evaluation $\sigma$-algebra $\sigma(\mu \mapsto \mu(A) : A \in \mathcal{B}(X))$ is a "nice" $\sigma$-algebra to work with. We have the following:

**Theorem 1.** *Let $X$ be a Lusin space (resp. any separable metrisable space). Then $M_X$ equipped with the topology of weak convergence is also a Lusin space (resp. a separable metrisable space), and the Borel $\sigma$-algebra of the topology of weak convergence is precisely the evaluation $\sigma$-algebra.*

An immediate corollary is that if $X$ is a separable metrisable space then **the evaluation $\sigma$-algebra is countably generated.**

[Weak convergence of probability measures on a separable metric space has several equivalent definitions, one being that for every bounded continuous $g \colon X \to \mathbb{R}$, $\int g \, d\mu_n \to \int g \, d\mu$. The topology of weak convergence is a very nice and physically natural topology. One of its useful properties is that weak convergence can be determined using only countably many bounded continuous functions $g \colon X \to \mathbb{R}$.]

Sorry I don't have a good reference off hand for the above theorem, but I imagine it should be easy to find the result (maybe not stated as one single theorem) in textbooks or lecture notes on descriptive set theory and/or measure theory. The fact that the topology of weak convergence for a Lusin space is Lusin might not be stated explicitly, but what will probably be stated is that the topology of weak convergence for a compact metrisable space is compact; and since $[0,1]$ is obviously compact, combining this statement with all the other parts of the statement of Theorem 1 will then yield that the topology of weak convergence for a Lusin space is Lusin.

**3. Disintegration of measures, and conditional expectation of random measures**

For me, one of the marvels of Lusin spaces is the following disintegration theorem. Let $M_X$ be the set of Borel probability measures on $X$.

**Theorem 2.** *Let $(\Omega,\mathcal{F},\mathbb{P})$ be an arbitrary probability space, and let $X$ be a Lusin space. Let $M_{\Omega,X}$ be the set of functions $\dot{\mu}\colon \Omega \to M_X$ such that $\omega \mapsto \dot{\mu}(\omega)(A)$ is an $\mathcal{F}$-measurable map for all $A \in \mathcal{B}(X)$, and let $M_{\Omega,X;\mathbb{P}}$ be the set of equivalence classes of $M_{\Omega,X}$ under the equivalence relation*
$$ \dot{\mu}_1 \sim \dot{\mu}_2 \quad \Leftrightarrow \quad \textrm{for $\mathbb{P}$-a.a. $\omega \in \Omega$, $\ \dot{\mu}_1(\omega)=\dot{\mu}_2(\omega).$} $$
*Now let $M_{\Omega \times X;\mathcal{F},\mathbb{P}}$ be the set of probability measures on the product space $(\Omega \times X, \mathcal{F} \otimes \mathcal{B}(X))$ with the property that $\mu(E \times X)=\mathbb{P}(E)$ for all $E \in \mathcal{F}$. Then $M_{\Omega,X;\mathbb{P}}$ and $M_{\Omega \times X;\mathcal{F},\mathbb{P}}$ are in exact one-to-one correspondence with each other, via the identification*
$$ \mu(A) \ = \ \int_{\Omega \times X} \mathbf{1}_A(\omega,x) \, \dot{\mu}(\omega)(dx) \, \mathbb{P}(d\omega). $$

I don't know off hand any good textbook for the proof (the famous textbook on random dynamical systems by Ludwig Arnold gives the statement but I think it cites some other book - possibly not in English - for the proof). However, if you can't find the proof easily online, it is proved in my PhD thesis at https://spiral.imperial.ac.uk/handle/10044/1/39569 (Lemma 3.27 / Remark 3.28).

**Corollary.** *For any $\dot{\mu} \in M_{\Omega,X}$ and any sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{F}$, there exists an element $\mathbb{E}[\dot{\mu}|\mathcal{G}]$ of $M_{\Omega,X}$ with the property that for all $A \in \mathcal{B}(X)$, the function $\,\omega \mapsto \mathbb{E}[\dot{\mu}|\mathcal{G}](\omega)(A)$ is a version of the conditional expectation $\mathbb{E}[\omega \mapsto \dot{\mu}(\omega)(A)|\mathcal{G}]$.*

This is proved by taking the $\mu \in M_{\Omega \times X;\mathcal{F},\mathbb{P}}$ associated to $\dot{\mu}$, and then regarding $\mu$ as an element of $M_{\Omega \times X;\mathcal{G},\mathbb{P}|_\mathcal{G}}$ to go back in the other direction to get $\mathbb{E}[\dot{\mu}|\mathcal{G}]$. (Again, this can be found in Arnold's book.)

notlocally compact. That said, weak convergence of measures is much simpler on locally compact Polish spaces. And I think it gets very difficult to work with on non-separable metric spaces. $\endgroup$1more comment