I've been reading [1] and attempting to prove statements given without proof. In the paper the authors construct a measurable space of measures over a base space, and as an aside show an elegant way to lift measurable functions from the base space into measurable functions in this new space. Perhaps a bit worrying is that I have little idea how to see these functions are in fact measurable, and the authors give this fact without proof!


Still, I'd like to press on and see this, since similar steps are used throughout proofs in the paper. I'd like to present the function, along with my attempt so far to prove it is measurable, and ask for help and perhaps some references to study in order to make the paper more accessible in general. Please forgive me if this question is too basic or pedantic.


### Point Measures


Let $X = (E, S)$ be a measurable space. For each $x \in E$ define a primitive
point measure $\mu\_x(F) = 1\_F(x)$. A point measure is an at most countable sum $\sum \mu\_{x\_i}$ of primitive point measures (with multiplicity allowed.) A point measure is essentially a set of points with non-negative integer masses.


Let $M$ be a set of point measures over $X$. Define a measurable space $Y = (M, T)$. Here $T$ is the sigma algebra generated by the family $\{(F)\_k : F \in S, 0 \leq k \leq \infty\}$, where $(F)\_k = \{\mu \in M : \mu(F) \leq k\}$.


Now let $f$ be a measurable function on $X$ and define a function $$[f] = \mu \mapsto \int\_E f\ d\mu = \sum f(x\_i)$$ Thus we lift each measurable function on $X$ into a function on $Y$ (with the same target space in each case). In fact $[f]$ is measurable.


I'd like to verify that $[f]$ is measurable. The paper leaves this fact as obvious to the reader, and uses similar conclusions in some of it's proofs. I believe this point is both important, and relatively easy to see with the proper tools. What follows is my somewhat ham fisted attempt so far.


### What I have so far


I've been working in the case where the target space of $f$ is $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$. Here I know it suffices to show that the intervals $(-\infty, k]$ have measurable preimages under $[f]$. I've worked out a few interesting cases for simple functions.


1. *Indicator functions*. If $F \in S$ we can see that $[1\_F](\mu) = \int\_E 1\_F\ d\mu = \int\_F\ d\mu = \mu(F)$, and hence $[1\_F]^{-1} (-\infty, k] = (F)\_k$. (Similarly we see that $[a\*1\_F]^{-1} (-\infty, k] = (F)\_{k/a}$.)
2. *Two valued simple functions*. Let $F, G \in S$ be disjoint, and say $f = a\*1\_F + b\*1\_G$. We see $[f](\mu) = a\mu(F) + b\mu(G)$, so that for every $\alpha, \beta$ such that $a\alpha + b\beta \leq k$, we get $[f]\(F)\_\alpha \cap (G)\_\beta \leq k$. It's not a stretch to show that in fact $$[f]^{-1} (-\infty, k] = \bigcup \{(F)\_\alpha \cap (G)\_\beta : a\alpha + b\beta \leq k\}$$ Since $M$ is a set of point measures we have $(F)\_\alpha = (F)\_{floor(\alpha)}$. Thus the union ranges over at most countable unique entries, and thus is measurable.
3. *Simple functions*. Let $F\_i \subset S$ be a finite family of disjoint sets and let $f = \sum a\_i \* 1\_{F\_i}$. It's perhaps a little more difficult to see that $$[f]^{-1} (-\infty, k] = \bigcup \{\cap (F\_i)\_{\alpha\_i} : \Sigma a\_i \alpha\_i \leq k \}$$ which is again a measurable union.


At this point I'd like to pick an arbitrary measurable function $f$ and exhibit a sequence $f\_i$ of simple functions such that $f\_i \rightarrow f$ pointwise (or perhaps $\bigwedge f\_i = f$.) From here we'd have a sequence $[f\_i]$ of measurable functions, which might be shown to converge in some suitable way to $[f]$. I believe it is possible to give $M$ (or the measurable functions over $Y$) some metric or topology. However, I am unsure how to proceed.


In another direction, I'd also like to find a more elegant tool that can let me work the proof with other measurable target spaces. I wonder if there some more abstract structure that we can give $Y$ that would grant this.


### Questions


1. Is $T$ in some sense natural as a sigma-algebra for $M$? I wonder if it arises as the Borel algebra of a natural topology for a space of measures, or something like this.
2. Do we need to work directly with generators of $T$ and the target space of $f$ to prove measurability of $[f]$? I'd like to see whether $[f]$ is measurable for target spaces other than the reals, and a more abstract tool for doing this would be helpful.
3. Similarly, is there a concise way to show measurability when the target space is the real?
4. On the other hand, is it hard to fill in the gap at the end of my work approach so far?
5. Just out of curiosity, can we see this proof by using abstract nonsense and diagram chasing? (I imagine the function lifting scheme might gives us something like a functor.)
6. Finally, would the result still hold if $Y$ were a space or more general measures? (I am pretty sure it fails even if we consider point masses with non-negative real mass.)


Any insight, hints, discussion or references will be very helpful. Thanks all!


**Reference**: [1] Finkelstein, Tucker, Veeh. "[Point Processes Without Topology](https://projecteuclid.org/ebooks/institute-of-mathematical-statistics-lecture-notes-monograph-series/Statistics-probability-and-game-theory/chapter/Point-processes-without-topology/10.1214/lnms/1215453565)", *Statistics, probability and game theory: Papers in honor of David Blackwell*, 1996.



Measurable functions lifted onto a space of point measures are measurable
As you have shown the case of indicator functions you get the case of positive measurable functions $f$ by linearity of the integrals and Levi's theorem on monotone convergence: There is a sequence $0\le f\_n= \sum\limits\_{k=1}^{m(n)} a\_{n,k} 1\_{F\_{n,k}}$
such that $f\_n \le f\_{n+1}\to f$. Hence you get that
$$
[f](\mu)= \int f d\mu = \lim\_{n\to\infty} \int f\_n d\mu = \lim\_{n\to\infty} \sum\_{k=1}^{m(n)} a\_{n,k} [1\_{F\_{n,k}}](\mu)
$$
is a monotone limit of sums of measurable functions and hence measurable. For not necessarily positive $f$ you already have a problem to *define* $[f]$ because the integrals might not exist.


