Continuous a.s. process In Ross's Stochastic processes:

  
*
  
*A stochastic process $\{X(t), t \geq     0\}$ is said to be a Brownian motion
  process if $X(0) = 0$, $\{X(t), t    \geq 0\}$ has stationary independent
  increments, and for every t > 0,
  $X(t)$ is normally distributed with
  mean 0 and variance $c^2t$.
  
*Brownian motion could also be
  defined as a Gaussian process having
  $E(X(t)) = 0$ and, for $s < t, \text{Cov}(X(s), X(t))= s$.
  
*The Brownian Bridge can be denned as
  a Gaussian process with mean value 0
  and covariance function $s(l - t), s     \leq t$.
  

I was wondering 


*

*if Brownian motion and Brownian
Bridge are both continuous a.s.?

*If the above three definitions for
Brownian motion and Brownian Bridge
already implicitly imply that the
processes such defined are continuous a.s.? Or do these definitions miss the continuity a.s. requirement?

*if a Gaussian process is always
continuous a.s.?


Thanks and regards!
 A: *

*They are commonly defined to be continuous a.s.; 2. No; 3. No (since any non-random function corresponds to a Gaussian process). 


Elaborating. 
Let $X = \{X_t:t \geq 0\}$ be a continuous Brownian motion (the same idea will be true for Brownian bridge) and $U$ an independent uniform$(0,1)$ variable. Define a process $\tilde X$ by $\tilde X_t = X_t$ if $t \neq U$, and $\tilde X_t = X_t + 1$ if $t=U$. Then $\tilde X$ is discontinuous at $t=U$; nevertheless it satisfies the conditions in Definitions 1 and 2 above. So, $\tilde X$ is a Brownian motion in law (it has the finite dimensional distributions of a Brownian motion), but is not a Brownian motion, which is commonly defined to be continuous a.s.
As for the last question, let $f:I \to \mathbb{R}$ be an arbitrary non-random function, and define a stochastic process $Y$ by $Y_t=f(t)$, $t \in I$. Then, by definition, $Y$ is a Gaussian process, since the joint distribution of  $(Y_{t_1},\ldots,Y_{t_n})$ is Gaussian for any $t_1,\ldots,t_n \in I$. So if $f$ is discontinuous, so is $Y$.
EDIT (more details). The law of an arbitrary stochastic process $Z=\{Z_t:t \in I\}$ is determined by its finite-dimensional distributions, that is, the distributions of $(Z_{t_1},\ldots,Z_{t_n})$ for all $n \geq 1$ and $t_1,\ldots,t_n \in I$. If $Z$ is a Gaussian process (say on an interval $I$), then its law is completely determined by its mean function $m(t)={\rm E}(Z_t)$ and covariance function $c(s,t)={\rm Cov}(Z_s,Z_t)$ (for all $s,t \in I$). Hence definitions 2 and 3 above characterize Brownian motion in law and Brownian bridge in law, respectively. In the example of the process $\tilde X$, for any $n \geq 1$ and fixed times $t_1,\ldots,t_n \geq 0$, the random vectors $(X_{t_1},\ldots,X_{t_n})$ and $(\tilde X_{t_1},\ldots,\tilde X_{t_n})$ are identically distributed (indeed, they are equal with probability $1$), and hence, in particular, the conditions in definitions 1 and 2 above are satisfied for the process $\tilde X$, which is a discontinuous Brownian motion in law. (Remark: a Brownian motion in law can moreover have sample paths which are nowhere continuous.)
EDIT. In view of definitions 1 and 2, it is interesting to note that a Brownian motion can be defined without requiring that the (one-dimensional) marginal distributions be normal (with variance proportional to $t$). Indeed, the following statement holds: A stochastic process $X = \{X_t:t \geq 0\}$ is a Brownian motion if $X_0 = 0$, $X$ has stationary independent increments, $X$ is a.s. continuous, and for every $t > 0$, $X_t$ has mean $0$.   
