Uniform laws on finite subset families

Let $B=\{0,1\}$. The shift $j\mapsto j-1$ is a bijection from the conventional index set $[n]=\{1,\dots ,n\}$ to the von Neumann natural $n=\{0,\dots ,n-1\}$; for $n=0$ both sets are empty. Thus $[n]$ is finite by [F1]. The set $B$ is bijective with $2$, so it is finite as well. For each $I\in {\mathcal{S}}_{n,r}$ define its indicator function ${\chi }_I:[n]\to B$ by ${\chi }_I(j)=1$ if $j\in I$ and $0$ otherwise. The map $I\mapsto {\chi }_I$ is injective because equal indicator functions have equal inverse images of $\{1\}$. Its image is a subset of the finite set $B^{[n]}$, which is finite by [F2] and [F3]. Therefore some natural $k$ is bijective with this image; composing that bijection with $I\mapsto {\chi }_I$ makes ${\mathcal{S}}_{n,r}$ bijective with $k$, so [F1] makes it finite.

If $r=0$, then $\mathrm{\varnothing }\in {\mathcal{S}}_{n,r}$. If $r>0$, then $\{1,\dots ,r\}\subseteq [n]$ because $r\le n$, and the shift $j\mapsto j-1$ is a bijection from $\{1,\dots ,r\}$ to the von Neumann natural $r$; hence this subset has cardinality $r$ by [F1, F5], so $\{1,\dots ,r\}\in {\mathcal{S}}_{n,r}$. Thus ${\mathcal{S}}_{n,r}$ is nonempty. By [F5], its cardinality $M:=\mathrm{\mid }{\mathcal{S}}_{n,r}\mathrm{\mid }$ is defined; by [F6] and nonemptiness, $M\ge 1$. Give each $I\in {\mathcal{S}}_{n,r}$ weight $q(I)=1\mathrm{/}M$. These weights are nonnegative and their finite sum is $M(1\mathrm{/}M)=1$.

The outcome set is finite by step 1.1 and its weights satisfy the conditions in step 1.2, so [F4] makes $({\mathcal{S}}_{n,r},q)$ a finite probability space. The argument includes $n=0$, $r=0$, and $r=n$.