$\mu$ is $\sigma$-finite). For $P\ll\mu$ a probability measure over $\mathcal{U}$, define $$\tilde{U}_{P}\left(\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}\right):=\bar{U}_{K_{P}(\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}})},$$ where $$K_{P}\left(\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}\right):=\underset{i:\,\frac{dP}{d\mu}(\bar{U}_{i})>0}{\arg\min}T_{i}\left(\frac{dP}{d\mu}(\bar{U}_{i})\right)^{-1},$$ with arbitrary tie-breaking (a tie occurs with probability 0). We omit $\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}$ and only write $\tilde{U}_{P}$ if the Poisson process is clear from the context. If the Poisson process is $\{\bar{X}_{i},T_{i}\}_{i\in\mathbb{N}}$ instead of $\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}$, then the Poisson functional representation is likewise denoted as $\tilde{X}_{P}$. If $\bar{U}_{i}=(\bar{X}_{i},\bar{Y}_{i})$ is multivariate, and $P$ is a distribution over $\mathcal{X}\times\mathcal{Y}$, the Poisson functional representation is denoted as $(\tilde{X},\tilde{Y})_{P}$. We write its components as $(\tilde{X},\tilde{Y})_{P}=(\tilde{X}_{P},\tilde{Y}_{P})$.

Note that while $dP/d\mu$ is only uniquely defined up to a $\mu$-null set, changing the value of $dP/d\mu$ on a $\mu$-null set will only affect the values of $\tilde{U}_{P}$ on a null set with respect to the distribution of $\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}$, since the probability that there exists $\bar{U}_{i}$ on that $\mu$-null set is zero. Therefore $\tilde{U}_{P}$ is uniquely defined up to a null set.

By the mapping theorem [@kingman1992poisson; @last2017lectures] (also see Appendix A of [@sfrl_trans]), we have $\tilde{U}_{P}\sim P$. This is termed Poisson functional representation in [@sfrl_trans] since it can be regarded as a construction for the functional representation lemma [@elgamal2011network]. Consider the distribution $P_{U,X}$. Let $\{\bar{U}_{i},T_{i}\}_{i\in\mathbb{N}}$ be the points of a Poisson process with intensity measure 