Suppose, with $0<b<1$, we have an AR(1) process $$ X_{n+1} = b X_n + \epsilon_n $$
where the $\epsilon_n$ are i.i.d. random variables with distribution corresponding to the probability measure $\mu$ (that is to say, if $F$ is the distribution function of $\epsilon_1$, then $F(t) = \mu((-\infty, t))$).
For any set $B \in \mathcal{P}(\mathbb{R})$ and $x\in \mathbb{R}$, we define $x + B = \lbrace x + y \,|\, y \in B \rbrace$. I thought the following definition would define a Markov kernel $$ P: \mathbb{R} \times \beta(\mathbb{R}):P(x, B) = \mu(B - b x) $$
To see that this is well defined, recall that when showing that lebesgue measure is invariant under translation, it had to be proved that sets of the form $B + x$ are indeed elements of $\beta(\mathbb{R})$.
It is straightforward to prove that for any fixed $x \in \mathbb{R}$ , $P_x : B \mapsto P(x, B)$ is a measure. But I do not see why for a fixed $B \in \beta(\mathbb{R})$, $P_B: x \mapsto P(x,B)$ is measurable. I checked that it is measurable if $\mu$ is dirac measure $\mu = \delta_0$, but $\mu = \lambda_{[0,1]}$ already seemed tricky. Are additional conditions on $\mu$ needed for $P$ to be a Markov kernel (or is the definition of $P$ not useful)?
Edit
Considering the results below, the new question is, for the $F$ defined below, how do we show that $F \in \beta(\mathbb{R}^2)$?
One strategy to prove measurability directly (as opposed to using some general theorem about existence of Markov kernels with certain properties) would be as follows. Consider the following corollary of Theorem 13.5 (existence of product measures) in the book Measures, Integrals and Martingales by Schilling.
Corollary
Let $\nu$ be a measure on $(\mathbb{R},\beta(\mathbb{R}))$. Then for all $E \in \beta(\mathbb{R}^2)$, $\phi: x \mapsto \int \mathbb{1}_E(x,y) \nu(dy)$ is measurable.
Let
\begin{equation}
F = \lbrace (x,y) \in \mathbb{R}^2 \,|\, y \in B + x \rbrace
\end{equation}
We set $\phi_B$ to be $$ \phi_B(x) = \mu(B + x) $$ We have $\mathbb{1}_F(x,y) = \mathbb{1}_{B + x}(y)$ So we have $$ \phi_B(x) = \mu(B + x) = \int \mathbb{1}_{B + x}(y) \mu(dy) = \int \mathbb{1}_F(x,y) \mu(dy) $$ So if $F \in \beta(\mathbb{R}^2)$, then $\phi_B$ is measurable, so that $P_B: x \mapsto \phi_B(-b x) $ is measurable, as it is a composition of measurable functions.
Here is an image of how we can see that $F$ is a countable union in the case $B$ is an interval (blue region is $F$), but we want $F$ to be measurable for all $B \in \beta(\mathbb{R})$.
