Deleting one occurrence from a locally finite counting measure is measurable

On the parameter space $S\times {\mathcal{N}}_{\mathrm{l}\mathrm{f}}(S)$ define $Q((x,\mu ),B):=\mu (B)$. For each parameter this is a measure, its evaluations are measurable by [F2], and $(K_n)$ is a common finite-mass exhaustion because $\mu (K_n)<\mathrm{\infty }$ for every locally finite $\mu$. Thus $Q$ is a uniformly sigma-finite kernel.

The diagonal $\mathrm{\Delta }=\{(x,y):x=y\}$ is closed in $S\times S$ because $S$ is Hausdorff. The product topology has the countable basis consisting of rectangles from countable bases of the two copies of $S$; every open set is a countable union of these rectangles. Hence the product Borel sigma-algebra equals $\mathcal{B}(S)\otimes \mathcal{B}(S)$, so $1_{\mathrm{\Delta }}$ is product measurable by [F4].

For $(x,\mu )$, $\mu (\{x\})=\int 1_{\mathrm{\Delta }}(x,y)Q((x,\mu ),dy)$. The measurability theorem in [F3] makes $(x,\mu )\mapsto \mu (\{x\})$ measurable; therefore its inverse image $D$ of $[1,\mathrm{\infty }]$ is measurable.

On $D$, $\mu -{\delta }_x$ is again a locally finite counting measure: in a representation of $\mu$ as a countable sum of point masses, remove one occurrence at $x$; no compact-set count increases. For every Borel $B$, $R(x,\mu )(B)=\mu (B)-1_B(x),$ where $\mathrm{\infty }-1$ is interpreted as $\mathrm{\infty }$. This evaluation is measurable on $D$, since both terms are measurable and on $D$ the subtraction is nonnegative. The evaluations generate the target sigma-algebra [F2], so $R$ is measurable.

Steps 2.1-3.1 prove measurability of the domain and deletion map, including the multiplicity convention.