Measurability condition of point processes

Question

Let $(\Omega, F, \mathbb{P})$ be a probability space and let $(S, \Sigma)$ be a measurable space. Let $S^{\infty}$ be the collection of countable subsets of $S$. Assume that $\left\{x\right\} \in \Sigma$ for every $x \in S$. By a point process on $(S, \Sigma)$ we mean any random countable subset of $S$.

In my understanding, a random countable subset $X$ on $(S, \Sigma)$ is a random variable $$X: \Omega \rightarrow S^{\infty}.$$ But as a random variable, it should be a measurable function. However, it is unclear to me, with respect to what sigma algebra? I.e., what is the sigma algebra $S^{\infty}$ is equipped with?

Answer 1

Probably the most useful (and certainly the most common) way to proceed is as follows. Define a $\sigma$-field $\mathscr S$ of subsets of $S^\infty$ to be that generated by subsets of $S^\infty$ of the form $\{H\in S^\infty: \#(H\cap B) =n\}$, for $B\in\Sigma$ and $n\in\{0,1,2,\ldots,\infty\}$. (Here $\#(C)$ denotes the cardinality of $C$.) With this definition of $\mathscr S$, a map $X:\Omega\to S^\infty$ is a random countable set (i.e., a random element of $S^\infty$) if and only if $$ \{\omega\in\Omega: \#(X(\omega)\cap B) =n\}\in\mathcal F, $$ for all $B\in\Sigma$ and all $n\in\{0,1,2,\ldots,\infty\}$.

Under minimal conditions on $(S,\Sigma)$, the points of $X$ can be enumerated in a measurable way: There is a sequence of $S$-valued random variables, $\xi_1, \xi_2,\ldots$ and an $\{0,1,2,\ldots,\infty\}$-valued random variable $N$ (all defined on $(\Omega,\mathcal F)$) such that $$ \#(X(\omega)\cap B) =\sum_{n=1}^{N(\omega)}1_B(\xi_n(\omega)), $$ for all $B\in\Sigma$ and all $\omega\in\Omega$. ($N$ is there to account for the possibility that $X(\omega)$ is a finite set for some $\omega$s.)

Answer 2

More detail on my comments: It depends on what aspect of "the set of all countable subsets of $S$" you want to capture. Note that this set is distinct from the simpler set $$ S^{\mathbb{N}} = \{(s_1, s_2, s_3, ...): s_i \in S \quad \forall i \in \mathbb{N}\}$$ where $\mathbb{N}=\{1, 2, 3, ...\}$ is the set of natural numbers. The set $S^{\mathbb{N}}$ is simpler because it has the structure of a countably infinite ordered list, whereas "the set of all countable subsets of $S$" could contain finite sets and does not necessarily order the elements of the sets.

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If you really want to define a random element that specifies all points, you could do this:

Define a new element called "null" that is distinct from all elements of $S$. Then define: $$S' = S \cup \{null\}$$ and define $\Sigma'$ as the set of all subsets of $S'$ of the form $A$ or $A \cup \{null\}$ for some $A \in \Sigma$. You can show that $\Sigma'$ is indeed a sigma algebra on $S'$.

Now for $i \in \{1, 2, 3, ...\}$ define random elements $$X_i:\Omega\rightarrow S'$$ that are measurable with respect to the input space $(\Omega, F)$ and the output space $(S', \Sigma')$. Then $(X_1, X_2, X_3, ...)$ specifies the countable collection of points in $S$, and if $X_i=null$ then there is no point included for index $i$. This allows having a finite set of points, meaning that $X_i=null$ for all but finitely many $i$.

Then you can define $$X=(X_1, X_2, X_3, ...)$$ which can either be viewed as a sequence of random elements, or can be viewed as a random element itself: $$X:\Omega\rightarrow (S')^{\mathbb{N}}$$ equipped with the product sigma algebra $$ \otimes_{i=1}^{\infty} \Sigma' = \Sigma' \otimes \Sigma' \otimes \Sigma'...$$ which is the sigma algebra generated by all sets of the form $$ A_1 \times A_2 \times ... \times A_n \times S' \times S' \times S' ...$$ for some positive integer $n$ and some sets $A_i\in \Sigma'$ for $i \in\{1, ..., n\}$.