Brownian motion has a bunch of different definitions. My question is about showing the property in the title using a certain definition of BM and nothing else.
The (partial) definition I am given is that for any $t > s \geq 0$, $B_t - B_s$ is independent of $\mathcal{F}_s$, the filtration to which $B$ is adapted, and is normally distributed with mean $0$ and variance $t-s$.
So now using this property I need to show that for any $0 \leq s_0 < s_1 <\ldots <s_n$ the sigma algebras $\mathcal{F}_{s_0}, \sigma(B_{s_1}-B_{s_0}), \ldots, \sigma(B_{s_n}-B_{s_{n-1}})$ are independent.
It is trivial to show that these sigma algebras are pairwise independent. But how do I show mutual independence, i.e. $$P(\bigcap_{i=0}^nA_i) = \prod_{i=0}^nP(A_i)$$ for any $A_0 \in \mathcal{F}_{s_0}, A_1 \in \sigma(B_{s_1}-B_{s_0}),\ldots$.
The next question is showing that $(B_{t+s_1}-B_{t+s_0},B_{t+s_2}-B_{t+s_1}\ldots,B_{t+s_n}-B_{t+s_{n-1}})$ has the same distribution for all $t \geq 0$. If I could first show that this vector is Gaussian for $t = 0$, then I would argue as follows. Using the result above I could factor the characteristic function of this vector. Then for the characteristic function of each increment $B_{s_j}-B_{s_{j-1}}$, I would substitute the characteristic function of $B_{t+s_j}-B_{t+s_{j-1}}$ for some arbitrary $t \geq 0$ since the two characteristic functions are identical. Using the first result again I could say the product of the characteristic functions of $B_{t+s_j}-B_{t+s_{j-1}}$ is equal to the characteristic function of $(B_{t+s_1}-B_{t+s_0},B_{t+s_2}-B_{t+s_1}\ldots,B_{t+s_n}-B_{t+s_{n-1}})$. Since the characteristic functions of $(B_{t+s_1}-B_{t+s_0},B_{t+s_2}-B_{t+s_1}\ldots,B_{t+s_n}-B_{t+s_{n-1}})$ and $(B_{s_1}-B_{s_0},B_{s_2}-B_{s_1}\ldots,B_{s_n}-B_{s_{n-1}})$ are equal, the two vectors must have the same distribution. Actually, while writing this I realized this has nothing to do with $(B_{s_1}-B_{s_0},B_{s_2}-B_{s_1}\ldots,B_{s_n}-B_{s_{n-1}})$ being a Gaussian vector. Is my argumentation correct then?
