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 $k \geq 0$,
$$\{\mu \in M_{\mathrm{atom}}(E): \mu_2(A\times B)=k\}=\{\mu \in M_{\mathrm{atom}}(E): \mu(A \cap B)<\infty, \mu(A)\mu(B)-\mu(A \times B)=k\}.$$