Introducing the notion of a random measure, textbooks usually start with a locally compact second countable Hausdorff space. Where does this requirement come from?

I would like to have a motivation for this requirement. That is, I would like to define a random measure on a general measurable space $(X,\mathcal{B})$ simply as a kernel from a probability space to $(X,\mathcal{B})$. Additional structure of $(X,\mathcal{B})$ should then be motivated by e.g. counterintuitive examples. For example, in the book of Last and Penrose (2017) (link to pdf), Exercise 2.5 yields that point measures usually do not have the representation with Dirac measures.

Are there other examples, (intuitive) motivations and reasons to use a locally compact (!) second countable (!) Hausdorff space?

Once a locally compact second countable Hausdorff space $(X,\mathcal{T})$ is given, the theory of random measures just considers measures $\mu$ that are locally finite, that is, $\mu(K) < \infty$ for all relatively compact sets $K \subset X.$ Therefore, I wonder the following:

Let $\mathcal{B} = \sigma(\mathcal{T})$ be the Borel-$\sigma$-algebra on $X,$ let $M_X$ be the set of all measures on $(X,\mathcal{B})$ and let $\mathcal{M}_X$ be the $\sigma$-algebra generated by the evaluation maps $\mu \mapsto \mu(B)$ for all $B \in \mathcal{B}.$

Is the set of all locally finite measures $\mathcal{M}_X$-measurable?

  • 1
    $\begingroup$ If you want a $\sigma$-algebra for the set of probability measures on a measurable space, you are looking for this: ncatlab.org/nlab/show/Giry+monad $\endgroup$ Feb 21 '20 at 21:50
  • $\begingroup$ Thanks for the link! Actually I already have a $\sigma$-algebra, that is, I implicitly consider the $\sigma$-algebra on the set of all measures on $(X,\mathcal{B})$ that is generated by the evaluation maps $\mu \mapsto \mu(B)$ for all $B\in\mathcal{B}$ $\endgroup$
    – Henning
    Feb 21 '20 at 22:01
  • $\begingroup$ I think it may depend in part on what one wants to do with the random measure. Do you have some examples of textbooks where you've seen this, so we could see what the context is? $\endgroup$ Feb 21 '20 at 22:52
  • $\begingroup$ I think sometimes people use more general random measures. For example, Poisson point process of excursions (of a Markov process) is a random measure with values in the Skorohod space, a Polish space which is not locally 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. $\endgroup$ Feb 21 '20 at 23:08
  • 1
    $\begingroup$ @ Nate Eldredge: e.g. Kallenberg's books (Random Measures (1974), www4.stat.ncsu.edu/~boos/library/mimeo.archive/…, Foundations of Modern Probability Theory (2002), Random Measures and Applications (2017) ) and also tDaley and Vere-Jones "Introduction to the Theory of Point Processes" Vol. II other textbooks just use $\mathbb{R}^k$ or certain subsets $\endgroup$
    – Henning
    Feb 21 '20 at 23:16

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"


"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.)

  • $\begingroup$ Thank you for your extensive and convincing analysis of the situation. I was aware of the fact, that ''nice structure of the state space $(X,\mathcal{B})$'' implies ''nice structure of the measure space $(M_X,\mathcal{M}_X)$'', where $\mathcal{M}_X$ denotes the evaluation $\sigma$-algebra. My question, however, aims at the converse direction (not rigorously, but by examples), that is, "too less structure of the state space$(X,\mathcal{B})$" leads to "too less structure of the measure space $(M_X,\mathcal{M}_X)$." Are there any examples? Thank you! $\endgroup$
    – Henning
    Feb 22 '20 at 10:27
  • $\begingroup$ Well, certainly if $X$ is any non-metrisable then there does not exist a metrisable topology on $M_X$ for which the map $x\mapsto\delta_x$ is a topological embedding. So that's one obvious lack of nice structure. Now as for separability: if, say, $X$ is an uncountable set equipped with the discrete topology, then $\mathcal{B}(X)$ and hence $\mathcal{M}_X$ are not countably generated. (More generally, I think there's a result assuming the axiom of choice that the Borel $\sigma$-algebra of a non-separable metric space is never countably generated. $\endgroup$ Feb 22 '20 at 11:35
  • $\begingroup$ If $X$ is a non-Lusin separable metric space, then the topology of weak convergence on $M_X$ is not Lusin. I'm afraid I don't know of counterexamples to the disintegration theorem for non-Lusin separable metric spaces. By the way, non-separability of metric spaces causes all kinds of fundamental problems to working with probabilities in general: e.g. if I recall correctly, the diagonal in $X\times X$ cannot be $\mathcal{B}(X)\otimes\mathcal{B}(X)$-measurable, which is really bad! Likewise, I think the set of Dirac masses will not be $\mathcal{M}_X$-measurable. $\endgroup$ Feb 22 '20 at 11:58
  • $\begingroup$ Thank you for your hints and comments! I think your hint concerning the measurability of the set of Dirac measures is what I was looking for. (cf. my example below). $\endgroup$
    – Henning
    Feb 22 '20 at 15:35

Let $D$ be the set of Dirac measures. In fact, $D$ would constitut a reasonable event, regarding probabilistic considerations (since we want to talk about random measures). However, in general it is not an event.

Consider, e.g., $X = [0,1]$ with the cocountable $\sigma$-algebra $\mathcal{B} = \sigma (\{\{x\} : x \in \mathcal{X})$ and note that the $\sigma$-algebra $\mathcal{M}_X$ is countably determined. Then, there is no countable set system $\mathcal{E} \subset \mathcal{B}$ such that $\text{pr}_{\mathcal{E}}^{-1}\big(\text{pr}_{\mathcal{E}}(D)\big) = D$, where $\text{pr}_{\mathcal{E}} : M_X \to [0,\infty]^{\mathcal{E}}$ is the projection of the set functions on $\mathcal{E}.$ This is due to the fact that we can define a measure $\mu$ by $\mu(B) = 1,$ if $B^c$ is countable and else $\mu(B) = 0,$ where $B \in \mathcal{B}.$ Note, $\mu$ is in $\text{pr}_{\mathcal{E}}^{-1}\big(\text{pr}_{\mathcal{E}}(D)\big)$ since $\mu(B) \in \{0,1\}$ for all $B,$ but $\mu$ is not in $D$ since there is no $x\in X$ such that $\mu = \delta_x.$ Hence, $D$ is not $\mathcal{M}_X$-measurable.

Let $(X,\mathcal{T})$ be a locally compact second countable Hausdorff space, then the set of all locally finite measures on $(X,\mathcal{B})$ is $\mathcal{M}_X$-measurable.

To prove this, we first note that there is a countable basis $\mathcal{U}$ of $\mathcal{T}$ and a sequence $(G_k)_{k\in\mathbb{N}}\in\mathcal{T}^\mathbb{N}$ of relatively compact sets such that $G_k \uparrow X$ for $k \to \infty.$ Then, $\mathcal{E} :=\big\{ \bigcap \mathcal{O} : \mathcal{O} \subset \mathcal{U} \cup G(\mathbb{N}),\ |\mathcal{O}| < \infty \big\}$ is a countable and intersection stable generator of $\mathcal{B}.$ According to the measure uniqueness theorem, a locally finite measure $\mu$ is uniquely determined by $\mu|_{\mathcal{E}}$. Consequently, the set of all locally finite measures is countably determined, that is, $\mathcal{M}_X$-measurable.

  • $\begingroup$ Nice proof :) although perhaps a bit more should be said about why $\mu$ is in $\mathrm{pr}_\mathcal{E}^{-1}(\mathrm{pr}_\mathcal{E}(D))$: If we let $E_1$ be the intersection of all the sets in $\mathcal{E}$ with countable complement, and let $E_2$ be the union of all the countable sets in $\mathcal{E}$, then clearly $E_1\setminus E_2$ is non-empty, and for any $x\in E_1\setminus E_2$, $\mu$ agrees with $\delta_x$ on $\mathcal{E}$. $\endgroup$ Feb 22 '20 at 21:10
  • $\begingroup$ You are right, one has to be careful at this point. Thanks for completing the proof :) $\endgroup$
    – Henning
    Feb 22 '20 at 21:37

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