Finite-window pair counts are measurable

For a measurable rectangle $B\times C$, the map $\eta \mapsto (\eta \otimes \eta )(B\times C)=\eta (B)\eta (C)$ is measurable by [F1]. Let $\mathcal{D}:=\{A\in {\mathcal{B}}_W^{\otimes 2}:\eta \mapsto (\eta \otimes \eta )(A)\text{ is measurable}\}\mathrm{.}$ The whole space belongs to $\mathcal{D}$ because its value is $\eta (W{)}^2$: for each integer $k\ge 0$, the level set $\{\eta (W{)}^2\ge k\}$ is a countable union of evaluation level sets $\{\eta (W)=n\}$. The total product mass is finite. For $A\in \mathcal{D}$, finite additivity gives $(\eta \otimes \eta )((W\times W)\setminus A)=\eta (W{)}^2-(\eta \otimes \eta )(A)$; this finite-valued measurable difference is measurable by [F4], so complements belong. Countable disjoint unions belong by countable additivity, since the partial sums are measurable by [F4] and their pointwise limit is measurable by [F5]. Thus $\mathcal{D}$ is a lambda-system containing the rectangles.

Rectangles form a pi-system and generate the product Borel sigma-algebra by [F6] and [F7]. Dynkin's theorem [F3] applied to step 1.1 therefore gives that $\eta \mapsto (\eta \otimes \eta )(A)$ is measurable for every Borel $A\subseteq W\times W$. The set $D_R$ is closed, so its product-measure evaluation is measurable.

If $\eta ={\sum }_{i=1}^n{\delta }_{x_i}$, then $(\eta \otimes \eta )(D_R)={\sum }_{i,j=1}^n{1}_{\{\mathrm{\parallel }{x}_i-x_j{\mathrm{\parallel }}_2\le R\}}$. The $n$ terms with $i=j$ all occur since $R\ge 0$; every remaining unordered pair of distinct occurrences contributes exactly two terms. Hence the displayed formula for $S_R$ is a finite nonnegative integer and is measurable by step 2.1 and [F1].