Question on the proof of the simple Markov property of a Brownian motion

Question

Today we proofed the (simple) Markov property for the Brownian motion. But I really don't get a crucial step in the proof. The theorem states in particular that for $s\geq0$ fixed, the process $(C_t:=B_{t+s}-B_{s}, t\geq0)$ is independent from $\mathcal{F}_s=\sigma(B_u, 0\leq u\leq s)$.

The proof starts with the remark, that it suffices to show that $\forall n, 0\leq t_1<t_2\dots<t_n$ and $\forall m, 0\leq u_1<u_2\dots<u_m$ the two vectors $(C_{t_1},\dots,C_{t_n})$ and $(B_{u_1},\dots,B_{u_m})$ are independent. But I just cannot figure out why this is true?

Anyone got some advise? Thanks a lot!

Answer 1

If $D$ is a countable subset of $[0,s]$ and $S\in\mathcal B(\mathbb R^\infty)$, then $\{\omega,(B_u(\omega))_{u\in D}\in S)$ belongs to $\sigma(B_u,0\leqslant u\leqslant s)$. The collection of sets of this form is a $\sigma$-algebra.

So, after passing from countable intersections to finite ones, it's enough to prove that the process $(C_t,t\geqslant 0)$ is independent of $\{\omega,(B_u(\omega))_{u\in D}\in S)$ with $D$ finite ($|D|=d$) and $S\in\mathcal{B}(\mathbb R^d)$.

By a similar argument, we are reduced to check independence of sets of the form $\{\omega,(C_t(\omega)_{t\in D'}\in S'\}$, $S'\in\mathcal B(\mathbb R^{|D'|})$.

Answer 2

Because these sigma-algebras $\sigma(B_u;u\leqslant s)$ and $\sigma(B_{s+u}-B_s;u\geqslant0)$ are generated by the pi-systems that one suggested that you use hence if the pi-systems are independent, so are the sigma-algebras (a result often called Dynkin lambda-pi theorem).