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?

• This doesn't compute. Maybe I need coffee or I'm misreading this. There is a difference between a measure function not being defined and a set being non-measurable. If the question is related to the function, then it's simply a point at which the function is undefined. That doesn't preclude the possibility of a function that is defined and is a valid probability measure. – Iterator Aug 3 '11 at 18:41
• If you cannot establish a non-Lebesgue-measurable set without the axiom of choice, how do you propose to know whether a particular event with a non-measurable probability has happened or not? – Henry Aug 3 '11 at 19:12
• @Henry: The OP may be referring to just the terminology. As for how I might refer to such an event, I would have to invoke Douglas Adams' Infinite Improbability Drive. Or call it a White Queen phenomenon, as she could believe 6 impossible things before breakfast. :) – Iterator Aug 3 '11 at 22:16
• As cardinal pointed out, nonmeasurable sets are used very widely in probability theory. The book Weak convergence and empirical processes by van der Vaart, gives a very good introduction. The reading of this book requires quite a good background in mathematics, but the theory presented is in my opininion beautiful. – mpiktas Aug 4 '11 at 7:58
• Are you interested only in results involving Lebesgue measure or more generally within the framework of probability theory? There seems to be some doubts about this among the participants here. – cardinal Aug 11 '11 at 23:20

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.

• although I am not an expert on empirical process theory, it is my impression that the use of outer probabilities is not really based on a desire to assign probabilities to non-measurable sets, but because you don't want to go through the hassle of actually proving measurability all the time. And if you can live without things like Fubini's theorem then you basically don't loose anything by just computing outer probabilities. – NRH Aug 4 '11 at 11:31

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.

• This is an interesting answer and informative answer. But, you might be overly focused on Lebesgue measurability. Nonmeasurable sets are much more prevalent in probability theory. – cardinal Aug 4 '11 at 1:53

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.