IID sequences as Markov chains with state-independent kernel

For each $x$, $A\mapsto K(x,A)=\nu (A)$ is a probability measure; for each [given] $A$, $x\mapsto \nu (A)$ is constant and measurable. Thus $K$ is a probability kernel, including $\nu (A)=0$ and $1$ and a one-point state space.

Put ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$. By [F1]--[F2], [F1, F2, F3, step 1.1] $\sigma (X_{n+1})$ is independent of ${\mathcal{F}}_n$. Hence, for $B\in {\mathcal{F}}_n$ and $A\in \mathcal{E}$, $\mathbb{E}[1_B{1}_{\{X_{n+1}\in A\}}]=\mathbb{P}(B)\nu (A)=\mathbb{E}[1_{B}K(X_n,A)]\mathrm{.}$ Therefore the constant $K(X_n,A)=\nu (A)$ is a version of the indicated conditional probability. By [F3], equivalently $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=\int fd\nu$ for every bounded measurable $f$. This verifies the claim at $n=0$ and all later times. Choice is used only for those conditional-expectation classes.