I am currently working through "Brownian Motion" by Morters and Peres and I am confused by something in their proof of the Strong Markov Property (Theorem 2.16) (the text can be found here: https://people.bath.ac.uk/maspm/book.pdf).
They define $B_k=\{B_k(t):t\geq 0\}$ where $B_k(t)= B(t+k/2^n)-B(k/2^n)$. In the middle of the proof they say $P\{B_k\in A\}=P\{B\in A\}$ and that this follows from the simple Markov property (theorem 2.3) but I cannot see why $B_k$ and $B$ have the same distributions. I know that $B_k$ is a Brownian motion starting at $0$ and that $B(t+k/2^n)-B(k/2^n) \sim N(0,t)$ so I believe that $P(B_k\in A) = P(B\in A)$ if the Brownian motion was assumed to be a standard Brownian motion, but that is not the case here. It seems to me that if $B(0)=x \in \mathbb{R}^d$ then $P(B_k\in A) = P(B-x\in A)$ (where $B-x = \{B(t)-x :t\geq 0\})$. I feel like I may not be understanding the simple Markov property properly.
Any clarification is greatly appreciated.