Why a change of a brownian motion does not depend on the past values of it? $(B_t)_{t \in \mathbb R_0^+}$ are random variables on $(\Omega,\mathcal A,P)$. 
$\forall r \le s, t > s, B_t-B_s,B_r$ are independent (i.e. $\sigma(B_t-B_s)$ and $\sigma(B_r)$ are independent, where with $\sigma(X)$ I denote the $\sigma$-algebra produced by the variable $X$).  
Can one show that $B_t-B_s$ and $\sigma(\{B_i| r \le s\})=\sigma(\bigcup_{r \le s}\sigma(B_r))$ are independent?
Does one need to use the fact that any endless sequence of random variables from $(B_t)_{t \in \mathbb R_0^+}$ is multivariate normally distributed?
 A: To show independence you just need to show that
$\Bbb{P} (B_t- B_s \in A_0, B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k) = \Bbb{P} (B_t- B_s \in A_0)\Bbb{P} ( B_{r_1} \in A_1,B_{r_2} \in A_2\ldots B_{r_k} \in A_k)$ for every $r_1, \ldots r_k \leq s$ and $A_1, \ldots, A_k \in \mathcal{B}$.
This is because $\sigma(\{B_r, r \leq s\})$ is generated by $E=\{(B_{r_1}, \ldots, B_{r_k}) \in A_1\times\ldots \times A_k\}$ and you will have shown that
$\Bbb{E} (1_{A_0}(B_t- B_s)  1_E) = \Bbb{E} (1_{A_0}(B_t- B_s))\Bbb{E} ( 1_E)$ 
Use a monotone class theorem and you get the result for arbitrary $E\in \sigma(\{B_r, r \leq s\})$
The monotone class theorem goes like this:
Consider $\mathcal{C} = \bigg\{E \in \sigma(\{B_r, r \leq s\}); \Bbb{E} (1_{A_0}(B_t- B_s)  1_E) = \Bbb{E} (1_{A_0}(B_t- B_s))\Bbb{E} ( 1_E)\bigg\}$
Note that 
1) $\Omega \in \mathcal{C}$
2) $A \in \mathcal{C}\Rightarrow A^c \in \mathcal{C}$
3) $A_1 \subset A_2 \subset \ldots $, $A_i \in \mathcal{C} \Rightarrow \cup_i A_i \in \mathcal{C} $ (this is were one usualy uses the monotone convergence theorem and that motivates the name monotone class theorem)
Every set $\mathcal{C}$ with properties 1),2),3) that contains the elementary sets $E = \{(B_{r_1}, \ldots, B_{r_k}) \in A_1\times\ldots \times A_k\}$ contain the sigma algebra generated by all the $E$'s of the form $E=\{(B_{r_1}, \ldots, B_{r_k}) \in A_1\times\ldots \times A_k\}$
