Quote Originally Posted by Saposhiente View Post
http://en.wikipedia.org/wiki/Almost_surely , pi will be irrational in a given location.
Kolmogorov's axioms of probability don't work for all subsets of R, only Borel sets, which are generated by countably infinite unions/intersections of lengths. The Borel sets - even the extended Borel sets - are smaller in carnality than the power set of R, which means you can't make probability statements about a set without first ensuring it's in B. Strange things start to happen when you play with unmeasurable sets, like being able to dissemble solid sphere of radius one into disjoint pieces, move them through space without intersecting any of them, and assemble them into two or more disjoint solid spheres of radius one.

In this case, you can't easily. Writing I for the set of irrationals, it's not hard to see that I isn't a Borel set, since it contains an uncountable number of points - it follows that Q, the rationals, isn't either since the Borels are closed under complements and Q = I'.

With some limiting arguments that I honestly cannot remember, you can show that the Lebesque integral with respect to the Lebesque measure of f(x) = 0 (x rational), f(x) = 1 (x irrational) over a finite interval is zero. Because of the way the Lebesque integral is calculated, I'm fairly sure this can't be extended to (0, infinity) though.


Note however that showing an event has probability zero is somewhat different from saying it can't happen. Applied strictly, this would mean that any observation of a continuous variable has probability zero, despite having just been done. Usually people get around this by arguing that continuous random variables are really just approximations to a fabulously complicated discrete reality, but I don't think that really works here. If nothing else there's somewhere where gravitational curvature is continuous over an interval, which I think implies that circumference/diameter is also a continuous function. If it varies at all, it attains a rational value. You have probability zero of choosing a particular gravity such that this happens at random from all possible gravities , but that does not mean it doesn't happen.

For an easier example, consider the distribution f(x) = x over [0,1]. Y attains values 0, 1/2, and 1, but you have zero probability of choosing them at random from all values y attains.

(Unless one decides the entire universe is actually discrete, and continuity is always a vastly more tractable approximation, in which case all bets are off.)