Gaussian AR(1) chain

Let $\gamma =N(0,1)$. For Borel $A$, [F1, F3] $K(x,A)=\int 1_A(ax+\sigma z)\gamma (dz)\mathrm{.}$ For fixed $x$ this is the affine pushforward in [F1], hence a probability measure. The integrand is Borel on ${\mathbb{R}}^2$, so [F3], applied to the constant kernel $x\mapsto \gamma$, makes $x\mapsto K(x,A)$ measurable. Thus $K$ is a probability kernel. For $\sigma =0$ it is the deterministic kernel ${\delta }_{ax}$; empty/full $A$ give zero/one.

The recursion makes [F2, F3, F4, F5, step 1.1] ${\mathcal{F}}_n^X\subseteq \sigma (X_0,Z_1,\dots ,Z_n)$. By [F2], $Z_{n+1}$ is independent of this larger past and hence of ${\mathcal{F}}_n^X$. For a bounded Borel $f$, set ${\varphi }_f(x)=\int f(ax+\sigma z)\gamma (dz),$ which is measurable by [F3]. For $B\in {\mathcal{F}}_n^X$ and Borel $A,C$, independence applied to $B\cap \{X_n\in A\}$ gives $\mathbb{P}(B,X_n\in A,Z_{n+1}\in C)={\int }_B{1}_A(X_n)\gamma (C)d\mathbb{P}\mathrm{.}$ A pi--lambda argument [F4] extends this from rectangles $A\times C$ to every Borel subset of ${\mathbb{R}}^2$, and bounded simple approximation extends it to the function $(x,z)\mapsto f(ax+\sigma z)$. Therefore $\mathbb{E}[f(a{X}_n+\sigma Z_{n+1})\mid {\mathcal{F}}_n^X]={\varphi }_f(X_n)\mathrm{.}$ Since the left side is $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n^X]$ and ${\varphi }_f=Kf$, [F5] proves the Markov claim. The calculation includes $a=0$, $\sigma =0$, $f=0,1$, and $n=0$. Choice is used in [F1] and in the conditional expectations.