Probability of an event that is not measureable We know from measure theory that there are events 
that cannot be measured, ie they are not Lebesgue 
measureable. What do we call an event with a probability
that the probability measure is not defined on? What
types of statements would we make about such an event?
 A: As I stated in the comments how to deal with these types of events (non-measurable sets) is described in book: Weak convergence and empirical processes by A. van der Vaart and A. Wellner. You can browse the first few pages. 
The solution how to deal with these sets is quite simple. Approximate them with measurable sets. So suppose we have a probability space $(\Omega,\mathcal{A},P)$. For any set $B$ define outer probability (it is in page 6 in the book):
$$P^*(B)=\inf\{(P(A), B\subset A, A\in \mathcal{A}\}$$
It turns out that you can build a very fruitful theory with this sort of definition. 
A: Edit: In light of cardinal's comment: All I say below is implicitly about the Lebesgue measure (a complete measure). Rereading your question, it seems that that is also what you are asking about. In the general Borel measure case, it might be possible to extend the measure to include your set (something which is not possible with the Lebesgue measure because it is already as big as can be).
The probability of such an event would not be defined. Period. Much like a real valued function is not defined for a (non-real) complex number, a probability measure is defined on measurable sets but not on the non-measurable sets.
So what statements could we make about such an event? Well, for starters, such an event would have to be defined using the axiom of choice. This means that all sets which we can describe by some rule are excluded. I.e., all the sets we are generally interested in are excluded.
But couldn't we say something about the probability of a non-measurable event? Put a bound on it or something? Banach-Tarski's paradox shows that this will not work. If the measure of the finite number of pieces that Banach-Tarski decomposes the sphere into had an upper bound (say, the measure of the sphere), by constructing enough spheres we would run into a contradiction. By a similar argument backwards, we see that the pieces cannot have a non-trivial lower bound.
I haven't shown that all non-measurable sets are this problematic, although I believe that a cleverer person than I should be able to come up with an argument showing that we cannot in any consistent way put any non-trivial boundon the "measure" of any non-measurable set (challenge to the community). 
In summary, we can not make any statement about the probability measure of such a set, this is not the end of the world because all relevant sets are measurable.
A: There are already good answers, but let me contribute with another point. The Lebesgue measure is often considered on the Lebesgue $\sigma$-algebra, which is complete, and, as already pointed out, we need the axiom of choice to establish Lebesgue non-measurable sets. In general probability theory, and, in particular, in relation to stochastic processes, it is far from obvious that you can make a relevant completion of the $\sigma$-algebra, and non-measurable events are less exotic. In some sense, the gap between the Borel $\sigma$-algebra and the Lebesgue $\sigma$-algebra on $\mathbb{R}$ is more interesting than the exotic sets not in the Lebesgue $\sigma$-algebra. 
The problem that I mostly see, that is related to the question, is that a set (or a function) may not be obviously measurable. In some cases you can prove that it actually is, but it may be difficult, and in other cases you can only prove that it is measurable when you extend the $\sigma$-algebra by the null sets of some measure. To investigate the extensions of Borel $\sigma$-algebras on topological spaces you will often encounter so-called Souslin sets or analytic sets, which need not be Borel measurable. 
