Wiener process related counterexample The Wiener process is defined by the three properties:
1. $W(0) = 0$, 
2. $W(t)$ is almost surely continuous, and 
3. $W(t)$ has independent increments with $W(t) - W(s) \sim N(0, t-s)$  (for $0 ≤ s < t$).
What would be an example of a process which satisfies 1) and 3), but not 2) ?
I am going to teach an introductory class on Brownian motion at advanced undergrad level.
Just wanted to make sure that all the conditions are mutually independent.
 A: To continue on Byron's answer, properties 1) and 3) specify the law of the process, but not the topological features of a given trajectory $\omega$. In a sense, 1) and 3) are enough to define the Wiener measure, since generating n points from a brownian path only needs those. Byron exhibited another 'version' of the process that is not continuous by changing the process on a set of measure 0 (his stopping time will never hit a specific point $t$ taken in advance because the law of T has a density).
A typical verification that needs to be done is if a process defined by its law has continuous versions, which is what entails Kolmogorov continuity theorem (link wikipedia page). Basically, the idea is that if $X_t$ and $X_s$ are close on average when $t$ and $s$ are close, you can change the process on a set of zero measure to get something continuous.
I can also recommend reading the beginning of the classical Revuz and Yor about definitions of 'undistinguishable processes' and 'versions of the same process'.
Cheers
A: This is not hard to find such an example. Let $P$ be Wiener measure on the space  $\Omega = C([0,\infty))$ of continuous functions $t\mapsto \omega(t)$. Then the process $\omega(t)$ satisfies all three conditions of a Brownian motion.
Now let's define a new process $W(t)$ that is "almost" equal to $\omega(t)$, but where we deliberately wreck the sample path continuity. 
Take any random time $T:\Omega\to [0,\infty)$ that has a continuous distribution on $(\Omega, P)$,
and let $W(t,\omega)=\omega(t)$ when $t\not=T(\omega)$, but $W(t,\omega)=\omega(t)+1$ otherwise. The process $W(t)$ still satisfies 1 and 3 but the sample path continuity fails at exactly at the time point $T(\omega)$ for each $\omega$. 
There are many such random times $T$, for example you could use $T(\omega):=\inf [t>0: \omega(t)=1 ]$, i.e. the hitting time of 1.
A: The author of this question might be more pleased with the following answer. Let $U$ be a  uniform(0,1) random variable, independent of a Brownian motion $W$. Then, the process $W'$ defined by $W'(t) = W(t) + {\mathbf 1}(t=U)$, where ${\mathbf 1}$ denotes indicator function, is discontinuous at time $U$. However, for any choice of (fixed) times $t_i$, $i=1,...,n$, we have, almost surely,  $W'(t_i) = W(t_i)$ for all $i$, and hence, trivially, $W'$ has the same distributional properties stated for $W$. Furthermore, if we define $W'$ by $W'(t) = W(t) + {\mathbf 1}(t \in UA)$, where $A$ is a dense set in $(0,\infty)$ of measure zero (and where $UA:= \{Ua: a \in A\}$), then $W'$ is nowhere continuous (since $UA$ is dense in $(0,\infty)$); nevertheless, as before, almost surely $W'(t_i)  = W(t_i)$ for all $i=1,...,n$ (since ${\rm P}(t \in UA) = {\rm P}(t/U \in A) = 0$).
Side notes: 1) Actually, as follows from the theory of Lévy processes, the almost sure continuity in the definition of Brownian motion is equivalent to almost sure cadlaguity (right-continuity with left limits); 2)  The answer can be adapted to Lévy processes in general ($W$ is a special case), showing that the almost sure cadlaguity in the definition of Lévy process is not implied by the other conditions. 
Finally, the author of this question ``wanted to make sure that all the conditions are mutually independent.'' This is, however, not the case, if we split condition 3) into subconditions. See this thread: link text
