Empirical-law bootstrap draws are conditionally IID

The shift $j\mapsto j-1$ is a bijection from $[n]$ to the von Neumann natural $n$. By [F10]–[F12], $[n]$ is finite with cardinality $n$. Let ${\mathrm{\Omega }}_n=\{f:[n]\to [n]\}$, the set of index vectors. By [F4], $\mathrm{\mid }{\mathrm{\Omega }}_n\mathrm{\mid }=n^n$. Assign weight $n^{-n}$ to each $\omega \in {\mathrm{\Omega }}_n$. The weights are nonnegative and sum to $n^n{n}^{-n}=1$, so [F2] gives a finite probability space $({\mathrm{\Omega }}_n,w)$ whose event sigma-algebra contains every subset. Each coordinate map $I_j(\omega )=\omega (j)$ and record map $Z_j^{\ast }(\omega )=z_{\omega (j)}$ is measurable, because every preimage is a subset of ${\mathrm{\Omega }}_n$; hence each $Z_j^{\ast }$ is a random element of $(\mathcal{Z},\mathcal{F})$ by [F6].

For $A_j\in \mathcal{F}$, put $B_j:=\{i\in [n]:z_i\in A_j\}$. Each $B_j$ is a subset of the finite set $[n]$, so [F13] makes it finite. The event $\{Z_j^{\ast }\in A_j\text{ for every }j\}$ consists of the functions $g:[n]\to [n]$ with $g(j)\in B_j$ for every $j$. The shift $\tau :[n]\to n$, $\tau (j)=j-1$, from step 1.1 has inverse $\rho :n\to [n]$, $\rho (k)=k+1$; thus $g\mapsto g\circ \rho$ is a bijection from this event's index functions onto ${\prod }_{k<n}{B}_{k+1}$. By [F5] this set has cardinality ${\prod }_{k<n}\mathrm{\mid }{B}_{k+1}\mathrm{\mid }$, which is the same product as ${\prod }_{j=1}^n\mathrm{\mid }{B}_j\mathrm{\mid }$ under that reindexing. Its probability under the uniform weights is therefore ${\prod }_{k<n}(\mathrm{\mid }{B}_{k+1}\mathrm{\mid }\mathrm{/}n)={\prod }_{j=1}^n(\mathrm{\mid }{B}_j\mathrm{\mid }\mathrm{/}n)$. Taking all events except the $j$th to be $\mathcal{Z}$ gives ${\mathbb{P}}^{\ast }(Z_j^{\ast }\in A_j)=\mathrm{\mid }{B}_j\mathrm{\mid }\mathrm{/}n=P_n(A_j)$ by [F3] and [F7]. Hence the joint event probability is the product of its marginal probabilities.

The one-coordinate case of step 2.1 gives each $Z_j^{\ast }$ law $P_n$. To check independence for any finite subfamily indexed by $J\subseteq [n]$, take arbitrary $A_j\in \mathcal{F}$ for $j\in J$ and set $A_j=\mathcal{Z}$ for every omitted coordinate. Since $P_n(\mathcal{Z})=1$ and ${\mathbb{P}}^{\ast }(Z_j^{\ast }\in \mathcal{Z})=1$, step 2.1 then gives ${\mathbb{P}}^{\ast }({\bigcap }_{j\in J}\{Z_j^{\ast }\in A_j\})={\prod }_{j\in J}{P}_n(A_j)$. This is the finite-intersection factorization in [F8], so the coordinate random elements are independent. They have a common measurable target and the same law, hence they are IID by [F9].