Klenke's construction of Brownian motion Why does Klenke's concise construction of Brownian motion via probability transition kernels satisfy the motion's characterizing properties, equations (14.17) and (14.18)?
(results referenced in the construction)
 A: Let $0=t_0<t_1<\cdots <t_n \in \mathbb R$ ($0<n$) and define $J_0:=\left\{t_0, t_1, \dots, t_n\right\}$, $J:=\left\{t_1, \dots, t_n\right\}$.
Define $\underline{X}_0 := \left(X_{t_0}, X_{t_1}, X_{t_2}, \dots , X_{t_n}\right)$, $\underline X := \left(X_{t_1}, X_{t_2}, \dots , X_{t_n}\right)$, $\Delta\underline{X}_0 := \left(X_{t_1}-X_{t_0}, X_{t_2}-X_{t_1}, \dots, X_{t_n}-X_{t_{n-1}}\right)$. Note that, since $\mathbb P_{X_0}=\mathcal N_{0,0}$ (i.e. $X_0=0\space\space a.s.$), $\mathbb P_{\Delta\underline{X}_0}=\mathbb P_{\left(X_{t_1}, X_{t_2}-X_{t_1}, \dots, X_{t_n}-X_{t_{n-1}}\right)}$.
Let $\underline Y := \left(Y_1, Y_2, \dots, Y_n\right)$ be independent and such that $\mathbb{P}_{Y_i} = \mathcal{N}_{0, t_{i-1}-t_i}, \space i=1,\dots,n$. I'll show that $\mathbb{P}_{\Delta\underline{X}_0}=\mathbb{P}_\underline{Y}$.
From Corollary 14.44, $\mathbb{P}_{\underline{X}_0}=\delta_0\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}$. To each rectangle $Q_0=B_{t_0}\times \underbrace{B_{t_1}\times\cdots\times B_{t_n}}_{=:Q}\in \times_{i=0}^n \mathcal B$, we get  $\mathbb{P}_{\underline{X}_0}\left(Q_0\right)=\mathbb 1_{B_{t_0}}(0)\cdot\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\left(0, Q\right)$. So, based on Theorem 14.28, $\mathbb{P}_{\underline{X}_0}\left(Q_0\right)=\mathbb 1_{B_{t_0}}\left(0\right)\cdot\mathbb{P}_\underline{S}\left(Q\right)$, where $\underline{S}=\left(S_1,S_2,\dots,S_n\right), \space S_m:=\sum_{i=1}^m Y_i \space\space \left(m=1,\dots,n\right)$. Hence $\mathbb P_{\underline{X}}\left(Q\right)=P_{\underline{X}_0}\left(\mathbb R \times Q\right)=P_{\underline{S}}\left(Q\right)$. As the rectangles $Q$ comprise a $\pi$-system that generates $\mathcal B_J$, $\mathbb P_{\underline{X}}=P_{\underline{S}}$ (cf. Lemma 1.42, "Uniqueness by an $\cap$-closed generator", p. 20).
So $\mathbb{P}_{\Delta \underline{X}}=\mathbb{P}_{\left(S_1, S_2-S_1, \dots, S_n-S_{n-1}\right)}=\mathbb{P}_\underline{Y}$.
