Measurability for countable sums of Dirac measures

Question

Let $(E,\mathcal{E})$ be any measurable space and denote by $M_{\mathrm{atom}}(E)$ the set of all measures $\mu$ of the form $$\mu=\sum_{i \in I} \delta_{x_i}$$ with $I$ at most countable and $x_i \in E$ for every $i \in I$ (the $x_i$ are not necessarily distinct). Equip $M_{\mathrm{atom}}(E)$ with the smallest $\sigma$-algebra for which all the functions $$\begin{array}{ccccc} & & M_{\mathrm{atom}}(E) & \to & \{0,1,2,\ldots,\infty\} \\ & & \mu & \mapsto & \mu(A) \\ \end{array}$$ for $A \in \mathcal{E}$ are measurable.

For $\mu=\sum_{i \in I} \delta_{x_i} \in M_{\mathrm{atom}}(E)$, set $$\mu_2=\sum_{i \neq j \in I} \delta_{(x_i,x_j)} \in M_{\mathrm{atom}}(E^2)$$ with $E^2$ equipped with the product $\sigma$-algebra.

Question 1. Is it true that $$\begin{array}{ccccc} & & M_{\mathrm{atom}}(E) & \to & M_{\mathrm{atom}}(E^2) \\ & & \mu & \mapsto & \mu_2 \\ \end{array}$$ is measurable?

Edit. Here is some motivation. Assume that the diagonal $D= \{(x,x): x \in E\}$ is measurable in $E \otimes E$ and consider a random measure $M$ with values in $M_{\mathrm{atom}}(E)$. Then "$M$ is simple" is an event (that is $\{ \forall x \in E: M(\{x\}) \leq 1 \}$ is an event) when the mapping $\mu \mapsto \mu_2$ is measurable, since $$\{ \mu \in M_{\mathrm{atom}}(E) : \forall x \in E: \mu(\{x\}) \leq 1 \}= \{ \mu \in M_{\mathrm{atom}}(E) : \mu_2(D)=0\}.$$

Edit 2. Reformulation of the motivation in measure-theoretical terms. Let $M_{\mathrm{atom}}^{\mathrm{simple}}(E)$ be the subset of $M_{\mathrm{atom}}(E)$ made of measures $\mu$ of the form $$\mu=\sum_{i \in I} \delta_{x_i}$$ with $I$ at most countable and $x_i \in E$ with $x_i \neq x_i$ for $i \neq j$.

Question 2. Assume that the diagonal $D= \{(x,x): x \in E\}$ is measurable in $E \otimes E$. Is it true that $M_{\mathrm{atom}}^{\mathrm{simple}}(E)$ is a measurable subset of $M_{\mathrm{atom}}(E)$?

If $\mu \mapsto \mu_2$ is measurable, then $M_{\mathrm{atom}}^{\mathrm{simple}}(E)= \{ \mu \in M_{\mathrm{atom}}(E): \mu_2(D)=0\}$, so the answer would be yes. But perhaps it is simpler to answer directly Question 2 than Question 1.

Partial answer for a separable metric space. If $(E,d)$ is a separable metric space equipped with its Borel $\sigma$-algebra, then the answer to both questions is affirmative: writing $D^{(1/n)}= \{z \in E^2: d(z,D)<1/n\}$, since $D= \cap_{n \geq 1} D^{(1/n)}$, it suffices to check that $\mu \mapsto \mu_2(O)$ is measurable for every open set $O \subset E^2$ [Edit: this is flawed: the measure $\mu_2$ is not necessarily finite...]. By separability, $O$ can be written as a countable disjoint union of measurable products $A \times B$, so it suffices in turn to check that $\mu \mapsto \mu_2(A \times B)$ is measurable for $A,B$ measurable. This readily follows from the fact that for every integer $k \geq 0$, $$\{\mu \in M_{\mathrm{atom}}(E): \mu_2(A\times B)=k\}=\{\mu \in M_{\mathrm{atom}}(E): \mu(A \cap B)<\infty, \mu(A)\mu(B)-\mu(A \times B)=k\}.$$

Answer 1

It's been a long time for me without using $\sigma$-algebra : I am feeling a bit rusty so please forgive (by order of gravity) any lengthy < naive < wrong remarks of mine. This is only a partial answer, unfortunately.

I assume you're using the discrete $\sigma$-algebra on $\overline{\mathbf{N}}:=\mathbf{N}\cup\{\infty\}$.

For any measurable space $(E,\mathcal{E})$ let's call $\mathfrak{B}(E,\mathcal{E})$ the $\sigma$-algebra that you use on $M_{atom}(E)$, that is \begin{align*} \mathfrak{B}(E,\mathcal{E}) = \sigma\big\{\chi_A^{-1}(\{n\}) \,:\, A\in \mathcal{E}, n\in\overline{\mathbf{N}}\big\}, \end{align*} where for $A\in\mathcal{E}$, $\chi_A$ denotes the evaluation map defined on $M_{atom}(E)$ by $\mu\mapsto \mu(A)$.

With the same notation as above, you choose to equip $M_{atom}(E^2)$ with $\mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E})$. I claim that \begin{align*} \mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E}) = \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2. n\in\overline{\mathbf{N}}\big\},\qquad (\star) \end{align*} One inclusion is a consequence (by minimality) of the definition of $\mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E})$ given above. The second inclusion which is needed to be proven is (the first equality is simply the definition) \begin{align*} \mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E}) = \sigma\big\{\chi_B^{-1}(\{n\}) \,:\, B\in \mathcal{E}\otimes\mathcal{E}, n\in\overline{\mathbf{N}}\big\}\subseteq \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2, n\in\overline{\mathbf{N}}\big\}. \end{align*} To establish this inclusion, introduce the subset $\mathfrak{S}\subseteq\mathscr{P}(E\times E)$ of all $B\in\mathcal{E}\otimes\mathcal{E}$ such that for all $k\in\overline{\mathbf{N}}$ \begin{align} \chi_B^{-1}(\{k\})\in \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2, n\in\overline{\mathbf{N}}\big\}. \end{align} One checks that $\mathfrak{S}$ forms a $\sigma$-algebra, containing obviously all $A_1\times A_2$ for $(A_1,A_2) \in \mathcal{E}^2$. Using that $\mathcal{E}\otimes \mathcal{E}$ is precisely the $\sigma$-algebra generated by these " tensor " elements we recover $\mathcal{E}\otimes\mathcal{E}\subseteq \mathfrak{S}$, which implies the desired inclusion, again by a minimality argument. $(\star)$ is thus proved.

Now, my hope was to obtain something stronger than $(\star)$, namely :

\begin{align*} \mathfrak{B}(E^2,\mathcal{E}\otimes \mathcal{E}) = \sigma\big\{\chi_{A_1\times A_2}^{-1}(\{n\}) \,:\, (A_1,A_2)\in \mathcal{E}^2,\, A_1\times A_2 \cap \Delta = \emptyset,\,n\in\overline{\mathbf{N}}\big\},\qquad (\star\star) \end{align*} where $\Delta:=\{(x,x)\,:\, x\in E\}$ is the diagonal of $E\times E$. To establish $(\star\star)$ the most natural thing would be to prove \begin{align*} \mathcal{E}\otimes\mathcal{E} = \sigma\big\{A_1\times A_2\,:\,(A_1,A_2)\in \mathcal{E}^2,\, A_1\times A_2 \cap \Delta = \emptyset\big\},\qquad(\star\star\star) \end{align*} and simply reproduce the above argument of minimality.

PROBLEM : It's not hard to check that a necessary condition for $(\star\star\star)$ to hold is that $\Delta$ itself belongs to $\mathcal{E}\otimes\mathcal{E}$. This is not an empty assumption (see Nedoma's pathology), but if you're working in a not-so-fat-and-ugly space, maybe you could cope with it. I have a strong feeling that if $\Delta$ does belong to the $\sigma$-algebra, then $(\star\star\star)$ holds -- and so does $(\star\star)$ in that case -- but I did not manage to write it rigorously. I expect a somehow general statement about generated $\sigma$-algebras, when you just erase a part of the generating system and yet recover all the $\sigma$-algebra ...

---

What follows is done under assumption $(\star\star)$.

With this characterization of the measurable sets at the arrival of your map $T:\mu\mapsto \mu_2$, the measurability of the latter boils down to establish, for any $(A_1,A_2)\in\mathcal{E}^2$ for which $A_1\times A_2\cap \Delta = \emptyset$ and $n\in\overline{\mathbf{N}}$, \begin{align*} T^{-1}\big(\chi_{A_1\times A_2}^{-1}(\{n\})\big) \in \mathfrak{B}(E,\mathcal{E}). \end{align*} $T^{-1}\big(\chi_{A_1\times A_2}^{-1}(\left\{n\right\})\big)$ is composed of measures $\mu \in M_{atom}(E)$ such that $\mu_2(A_1\times A_2)=n$. Now $\delta_{(x_i,x_j)}(A_1\times A_2) = \delta_{x_i}(A_1)\delta_{x_j}(A_2)$ and we are therefore asking (since $A_1\times A_2$ does not meet $\Delta$) \begin{align*} \sum_{i\neq j \in I} \delta_{x_i}(A_1)\delta_{x_j}(A_2)&=\sum_{i,j \in I} \delta_{x_i}(A_1)\delta_{x_j}(A_2)\\ &= \chi_{A_1}(\mu)\chi_{A_2} (\mu) \\ &= n. \end{align*} The set of such measures $\mu$ is indeed in $\mathfrak{B}(E,\mathcal{E})$ as it can be written for finite $n$ as \begin{align*} \bigcup_{d|n} \chi_{A_1}^{-1}(\{d\})\cap \chi_{A_2}^{-1}\left(\left\{\frac{n}{d}\right\}\right) \end{align*} and for infinite $n$ as \begin{align*} \chi_{A_1}^{-1}(\{\infty\})\cup \chi_{A_2}^{-1}(\{\infty\}). \end{align*}