The steps to showing that a process $(W_t)_{t \geq 0}$ is a Brownian motion (BM) are as follows:
(1)$W_0 = 0$
(2) $ \forall t ~~~W_t$ is continuous
(3)$W_t \sim N(0,t)$
(4)$W_{t+s}-W_{s} \sim N(0,t)$
(5)$W_{t+s}-W_{s} \bot \mathscr F_s=(W_u)_{u\in(0,s)} $
For example, if we take $X_t = W_{2t}-W{t}$ we can show that if $X$ satisfies these 5 steps then it is also a BM.
My question is that if a Gaussian process is fully characterized by its covariance and mean functions, is it enough to show that $X$ has the same covariance and mean function as $W$ instead of having to go through the 5 steps, or is there a contradictory case in which $X$ has the same mean and covariance function but does not satisfy some other condition such as continuity or independent increments? Answers should hopefully provide that contradiction.
If $(X_t)_{t \geq 0}$ is a Gaussian process with the same mean and covariance function as a Brownian motion, then $(X_t)_{t \geq 0}$ satisfies (1),(3)-(5). Moreover, it follows from Kolmogorov's continuity theorem that $(X_t)_{t \geq 0}$ has a modification with (exclusively) continuous sample paths. However, the process $(X_t)_{t \geq 0}$ does not necessarily have continuous sample paths.