Proving continuity of Brownian paths Maybe I'll make myself do a routine exercise by posting it here and then later posting my own solution after someone else posts one.  Or maybe not.  Maybe it's not as routine as I think it might be.
H. P. McKean's book Stochastic Integrals has this on page 5:


1.2 CONSTRUCTION OF THE BROWNIAN MOTION
Consider the space of continuous paths $t\to b(t)\in R^1$ with $b(0)=0$ and impose the probabilities:
    \begin{align} & P\left[\bigcap_{k\le n} (a_k\le b(t_k) \le b_k) \right] \\[15pt] & {}\quad = [\cdots\cdots\text{[etc.]}\cdots\cdots] \end{align}


He's assuming at the outset that the paths are continuous.
Suppose we didn't make that assumption.
Suppose $b(t)$ depends on $t$ and for $0<t<s$, $b(t)-b(s)\sim N(0,s-t)$ and we have independent increments, i.e. for any finite number of non-intersecting intervals $0<t_k<s_k$, the increments $b(t_k)-b(s_k)$ are mutually independent.  And assume $b(0)=0$ if you like.
Can we prove P(b is continuous)=1?
Can we prove $P(b\text{ is almost everywhere continuous})=1$?
Can we prove the yet stronger statement $P(\text{For some continuous function $c$, $b=c$ almost everywhere})=1$?
More generally, how much can be said about continuity?
 A: This question is very similar to the one that Stefan linked in the comments.  The issue here basically boils down to the fact that continuity of stochastic processes is a pretty thorny question unless you work on one of a few very nice measure spaces (like the space of real valued continuous functions in the compact open topology), where continuity or something like it is forced on all of the paths.  
If you just want to consider a stochastic process to be a collection of random variables on some arbitrary measure space indexed by $\mathbb{R}^+$, then the most natural $\sigma$-algebra to ask questions about continuity in is the product algebra.  If we take a moment to think, it is clear that there is no hope of saying anything about path continuity in such a space.  Continuity at a point $t_0$ depends on the value that the path takes along all sequences of points $t_n \to t_0$.  The collection of such sequences is uncountable and there is really no way to reduce the dependence to a countable one.  Next we want to argue that simultaneously, for all $t$ (another uncountable set) we can say something about continuity at $t$ and that is clearly a substantially more difficult problem.  Up to here, the non-measurability of these events has all been shown carefully.  I think Durrett gives a reference in his graduate probability textbook.
George Lowther gives a really nice example in the comments to Byron's answer in the link I posted.  We can construct an almost nowhere continuous Gaussian Process equivalent to Brownian Motion by taking a continuous Brownian Motion $\omega(t)$ and setting $f(t) = \omega(t) 1_{\{t: \hspace{.1pc}\omega(t) \text{ is irrational}\}}(t)$.  It is not hard to see that the continuity set of this function is exactly the set where $\omega(t) = 0$, which is well known to have Lebesgue measure zero.  We had better hope that the event that the process is continuous (or a.e. continuous or even not a.e. discontinuous) is not measurable, since we can find two processes satisfying the conditions you have listed where the support of the continuity set contains a measure one set or is contained in a measure zero set.
It appears to me that the non-measurability problem actually gets worse as you add in more conditions like "there exists a continuous function such that...", since you are now adding in a couple of extra layers of unions you have to take, including one over Lebesgue measure zero sets (another uncountable set).  I don't know how to prove that the event is not measurable rigorously though.
Let me say something about what is well known though.  If we are given a stochastic process $W(t)$ with the correct finite dimensional distributions, then there is a single stochastic process $\tilde{W}(t)$ which is pathwise continuous (we construct it by hand from $W(t)$ to be pathwise continuous, so there is no need to worry about measurability of the continuity event--it is the whole space) so that for all $t \geq 0$, $P \left(W(t) = \tilde{W}(t) \right) = 1$.  Among other things, this gives that for any fixed countable collection of time points, $P\left(W(t_n) = \tilde{W}(t_n)\right) = 1$ which appears to me to be the best we could realistically hope for.
To prove this, one uses Kolmogorov's continuity theorem.
