denote the set of feature labels for features to which $n$ belongs. Call this a *uniform random labeling*. $\indfar_{n}$ can be thought of as a (globally finite) simple point process on $[0,1]$. Again, we refer the reader to @kallenberg:2002:foundations [p. 178] for a discussion of measurability.

\[ex:feat\_6\_unif\_rand\] Again consider the feature allocation $$\detfa_{6} = \{\{2,5,4\},\{3,4\},\{6,4\},\{3\},\{3\}\}.$$ Now consider the random variables $$U_{1}, U_{2}, U_{3}, U_{4}, U_{5}, \randlabfa_{1}, \randlabfa_{2}, \randlabfa_{3}, \randlabfa_{4}, \randlabfa_{5}
 \iid \unif[0,1].$$ Recall from that $U_{1}, \ldots, U_{5}$ gave us the order-of-appearance labeling of the features. This labeling allowed us to index the features as in , copied here: $$A_{1} = \{2,5,4\}, A_{2} = \{3\}, A_{3} = \{3,4\}, A_{4} = \{3\}, A_{5} = \{6,4\}.$$

With this order-of-appearance labeling in hand, we can assign a uniform random label to each feature. In particular, we assign the uniform random label $\randlabfa_{k}$ to the feature with order-of-appearance label $k$: $A_{1} = \{2,5,4\}$ gets label $\randlabfa_{1}$, $A_{2} = \{3\}$ gets label $\randlabfa_{2}$, $A_{3} = \{3,4\}$ gets label $\randlabfa_{3}$, $A_{4} = \{3\}$ gets label $\randlabfa_{4}$, and $A_{5} = \{6,4\}$ gets label $\randlabfa_{5}$. Let $\indfar_{n}$ indicate the set of uniform random feature labels for the features to which index $n$ belongs. The uniform random labeling gives 