Suppose that $B_t : t \geq 0$ is a stochastic process with:
1) $B(0) = 0$
2) For $0 \leq t_0 < t_1 \leq t_2 < t_3$, $B(t_1) - B(t_0)$ and $B(t_3) - B(t_2)$ are independent.
3) $B(t + h) - B(t) \sim N(0,h)$, for all $t \geq 0$ and $h > 0$
4) a.s. $t \to B(t)$ is continuous.
Is $B(t)$ necessarily a Brownian motion? If not, what is a counter example? Note that the difference is that for $2)$, the usual definition requires independence among the adjacent differences for any finite set of times.
I am aware that pairwise independent normal do not have to be normal, so I guess that could figure in here. However, I feel that there might be enough leftover structure to force the stronger independence statement.
Also, I saw the same definition here: http://wwwf.imperial.ac.uk/~mdavis/course_material/sp1/BROWNIAN_MOTION.PDF