Mecke identity for a finite-intensity Poisson point process

The addition map is measurable because for every Borel $B$ its output evaluation is $(\mu +{\delta }_x)(B)=\mu (B)+1_B(x)$, a measurable function of $(x,\mu )$; therefore $(x,\mu )\mapsto h(x,\mu +{\delta }_x)$ is measurable. The left integral is measurable by the evaluation-kernel integration assertion [F4].

If $\lambda =0$, then $\nu =0$ and [F1] makes $\eta$ the zero measure almost surely, so both sides vanish. Suppose $\lambda >0$ and write $Q=\nu \mathrm{/}\lambda$, $p_n=e^{-\lambda }{\lambda }^n\mathrm{/}n!$. By [F1], the left side equals ${\sum }_{n\ge 0}{p}_n{\int }_{S^n}{\sum }_{i=1}^{n}h(x_i,{\sum }_{j=1}^n{\delta }_{x_j})Q^{\otimes n}(d(x_1,\dots ,x_n))$.

For each $n\ge 1$, every summand indexed by $i$ has the same integral: coordinate permutation sends the $i$th coordinate to the first, preserves $Q^{\otimes n}$ by [F2], and leaves the sum of Dirac masses unchanged. Tonelli [F3] and $n{p}_n=\lambda p_{n-1}$ therefore turn step 1.2 into $\lambda {\sum }_{m\ge 0}{p}_m{\int }_{S^{m+1}}h(x_1,{\delta }_{x_1}+{\sum }_{j=2}^{m+1}{\delta }_{x_j})Q^{\otimes (m+1)}(d(x_1,\dots ,x_{m+1}))$.

Integrating first in $x_1$ in step 2.1 and using $\nu =\lambda Q$, the result is ${\int }_S{\sum }_{m\ge 0}{p}_m{\int }_{S^m}h(x,{\delta }_x+{\sum }_{j=1}^m{\delta }_{x_j})Q^{\otimes m}(d(x_1,\dots ,x_m))\nu (dx)$. By the mixed-binomial representation [F1], the sum and inner integral are $\mathbb{E}[h(x,\eta +{\delta }_x)]$ for each fixed $x$, so this is the right side of the claimed identity.