It is known that for sufficiently "nice" space $S$, and a transition kernel $p$ on $S$: the consistent collection of finite dimensional distribution defined by
$$\mathbb P_n(X_0\in B_0,\cdots,X_n\in B_n)=\int_{B_0}\mu(\mathrm dx_0)\int_{B_1}p(x_0,\mathrm dx_1)\cdots\int_{B_n}p(x_{n-},\mathrm dx_n)$$
may be extended to a probability measure $\mathbb P_\mu $ on $S^{\mathbb{N}}$.
I would like to show that $$\mathbb P_\mu(A) = \int \mu(\mathrm dx) P_{\delta_x}(A),$$ where $\mathbb P_x $ are the measures introduced similarly from
$$\mathbb P_n(X_0 \in B_0, \cdots, X_n \in B_n) = \int_{B_0} \delta_x(\mathrm dx_0) \int_{B_1} p(x_0, \mathrm dx_1) \cdots \int_{B_n} p(x_{n-1}, \mathrm dx_n).$$ That is for each $x$ we have changed the initial distribution from $\mu$ to the point mass $\delta_x$, and then we integrate those new probabilities of $A$ over $x$ with respect to $\mu$.
Would it be sufficient to show that for any $n $ and any $B_0, \cdots, B_n$, $$\int_{B_0} \mu(\mathrm dx_0) \int_{B_1} p(x_0, \mathrm dx_1) \cdots \int_{B_n} p(x_{n-1}, \mathrm dx_n)\\ = \int \mu(\mathrm dx) \int_{B_0} \delta_x(\mathrm dx_0) \int_{B_1} p(x_0, \mathrm dx_1) \cdots \int_{B_n} p(x_{n-1}, \mathrm dx_n)?$$
Much grateful for any help provided!
EDIT: I see that I haven't written anything about the sequence $(X_n) $ above. I believe if we want the first equation to hold we may take $(X_n)$ to be some kind of canonical process?
$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Suppose the σ-algebra associated with $S$ is $\mathscr{S}$, then for any $B_0, \cdots, B_n \in \mathscr{S}$,\begin{align*} &\peq \int_S μ(\d x) \int_{B_0} δ_x(x_0) \,\d x_0 \int_{B_1} p(x_0, \d x_1) \cdots \int_{B_n} p(x_{n - 1}, \d x_n)\\ &= \int_S μ(\d x) \int_S \left( \int_{B_1} p(x_0, \d x_1) \cdots \int_{B_n} p(x_{n - 1}, \d x_n) \right) I_{B_0}(x_0) δ_x(x_0) \,\d x_0\\ &= \int_S μ(\d x) \left( \int_{B_1} p(x, \d x_1) \cdots \int_{B_n} p(x_{n - 1}, \d x_n) \right) I_{B_0}(x)\\ &= \int_{B_0} μ(\d x) \int_{B_1} p(x, \d x_1) \cdots \int_{B_n} p(x_{n - 1}, \d x_n), \end{align*} which implies$$ P_μ(A) = \int_S μ(\d x) P_x(A). \quad \forall A \in \mathscr{S}^{\mathbb{N}} $$