Is the smallest closed set with probability 1 the same as the intersection of all closed sets with prob. 1?

Question

I know that the intersection of closed sets is closed, but does the intersection of sets with probability one have probability one? (I guess not, because the sigma-algebra is not closed under uncountable intersection, so the probability is not even necessarily defined.)

(Of course, by sets, I mean subsets of R, and by the probability of a set S, I mean P(X is in S).)

If not, then what is a representation of the smallest closed set with probability one (the support of a random variable)?

Taranto: Thanks, though I don't think you're right. For example, if the measure is the Lebesgue-measure (uniform distribution) on [0,1] and 0 everywhere else then there are lots of closed sets with measure 1 (e.g. [0,2]).

Andrew: Thanks, but how do you show that the union of all open sets with zero measure is the union of all base sets with zero measure?

Andrew: by the countable open base in this case, do you mean the intervals of the type (a,b) where a and b are rational? How do I show that every open set can be generated from these by countable union?

Answer 1

Yes, this is true for every Borel measure (measure on the Borel sigma-algebra) on R or R^n, and more generally on any topological space with a *countable* open base. The intersection of all closed sets of full measure is the complement of the union of all open sets of zero measure, and that's the same as the union of all the base sets of zero measure (and there are only countably many of those by assumption).

Without a countable open base, I think this would be false.

Note: The two unions in this argument are equal simply because every open set is a union of base sets (by the def. of an open base for a topology).
Note: For the real line R, the intervals (a,b) with a and b rational form a countable open base, and similarly for R^n the products of such intervals form a countable base. (An open set in R^n can be defined as one whose every point x is the center of an n-dimensional "rectangle" contained in the set, and such a rectangle can be approximated by one with rational vertices.)

Answer 2

It's been years since I did this stuff -- but I think you can turn this problem around and look at unions of open sets.

Let's consider measurable subsets of [0,1]. There are lots of measurable sets with measure one. But I'm pretty sure that there is only one closed set with measure one. Why? Because if it is closed, then its complement is open. The only open set with measure zero is the null set. This is because a non-null open set has to contain an interval.

I don't know if this helps you -- but I believe that if you have a closed set with probability one, it has to be the whole thing.

Answer 3

What kind of sets and how do you define the openness and closedness? (Sets of real numbers or subsets of a metric space? In topology, it is open/closed by definition so the question is trivial.)