Independence of Brownian motion

While by definition the increments of a Brownian motion are independent, it is unclear to me whether (that implies that) the random variables $W_t$ and $W_s$ are independent for $t \neq s$. While these random variables have different density functions, they are defined on the same state space, and I am not sure whether they are independent.

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"..whether they are independent for a given $\omega$" does not make sense. It is $W_t$ and $W_s$ that are random variables (as they are functions of $\omega$), and not $W_t(\omega)$ and $W_s(\omega)$ which are just real numbers. –  Stefan Hansen Nov 17 '12 at 11:56
Right, poorly formulated. –  lodhb Nov 17 '12 at 12:00

Suppose $W_t$ and $W_s$ are independent for $t\neq s$. Then for $0\leq s<t$ we would have that $W_t$ is independent of $\mathcal{F}^W_s=\sigma(W_u\mid u\leq s)$. Now the martingale property of the Brownian motion yields $$W_s=E[W_t\mid\mathcal{F}^W_s]=E[W_t]=0\quad\text{a.s.},$$ which certainly isn't true.
Take $t<s$, then you can write $W(s)=W(t)+(W(s)-W(t))$, from which you immediately see that they are not independent.
It is the increments which are independent. Take e.g. $t<s$, then $W(t)-W(0)$ and $W(s)-W(t)$ are independent.