$P_{U}\times\lambda_{\mathbb{R}_{\ge0}}$, $X\sim P_{X}$ independent of the process, and $U:=\tilde{U}_{P_{U|X}(\cdot|X)}$. Then $(U,X)\sim P_{U,X}$. Hence we can express $U$ as a function of $X$ and $\{\bar{U}_{i},T_{i}\}$ (which is independent of $X$). This fact will be used repeatedly throughout the proofs in this paper.

For two different distributions $P$ and $Q$, $\tilde{U}_{P}$ and $\tilde{U}_{Q}$ are coupled in such a way that $\tilde{U}_{P}=\tilde{U}_{Q}$ occurs with a probability that can be bounded in terms of $dP/dQ$. We now present the core lemma of this paper. The proof is given in Appendix \[subsec:pf\_phidiv\].

\[Poisson matching lemma\]\[lem:phidiv\]Let $\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}$ be the points of a Poisson process with intensity measure $\mu\times\lambda_{\mathbb{R}_{\ge0}}$, and $P,Q$ be probability measures on $\mathcal{U}$ with $P,Q\ll\mu$. Then we have the following almost surely: $$\mathbf{P}\left\{ \left.\tilde{U}_{Q}\neq\tilde{U}_{P}\,\right|\,\tilde{U}_{P}\right\} \le1-\left(1+\frac{dP}{dQ}(\tilde{U}_{P})\right)^{-1},\label{eq:phidiv}$$ where we write $(dP/dQ)(u)=(dP/d\mu)(u)/((dQ/d\mu)(u))$ as in (we do not require $P\ll Q$). The right hand side of is considered to be 1 if $(dP/d\mu)(\tilde{U}_{P})>0$ and $(dQ/d\mu)(\tilde{U}_{P})=0$.

The exact expression for the left hand side of is in .

We usually do not apply the Poisson matching lemma on fixed $P,Q$, but rather on conditional distributions. The following conditional version of the Poisson matching lemma follows directly from applying the lemma on $(P,Q)\leftarrow(P_{U|X}(\cdot|X),Q_{U|Y}(\cdot|Y))$. The proof is given in Appendix \[subsec:phidiv\_cond\] 