Fact: For any real-valued function f, measurable or not, on a probability space, there exists a largest measurable function fL such that fL≤f and a smallest measurable function fU such that fU≥f, and fL and fU are unique up to almost sure equality.
Definition: A set U in a probability space is maximally nonmeasurable providing all its measurable subsets have measure zero and all its measurable supersets have measure one.
Definition: A sequence X1,X2,... of independent identically distributed not necessarily measurable random variables will be a sequence of functions on an infinite product of copies of a probability space, such that Xn(w1,w2,...)=F(wn) for each n and a single fixed function F.
Henceforth suppose X1,X2,... are like that. Let Sn=X1+...+Xn.
Easy consequence of the Law of Large Numbers: If X1L and X1U have finite expectations, then almost surely E[X1L]≤ liminf Sn/n≤ limsup Sn/n≤E[X1U].
Can one strengthen this? E.g., can one hope that one of the inequalities is an equality? Yesterday I finished proving a negative answer.
Theorem: Suppose X1L and X1U are integrable. Let A be any proper non-empty subset of the interval [E[X1L],E[X1U]] (which implies that E[X1L]<E[X1U]). Consider the respective subsets of our probability space where:
- lim Sn/n exists
- lim Sn/n exists and is in A
- limsup Sn/n is in A
- liminf Sn/n is in A
- all the limit points of Sn/n are in A
This has a very interesting consequence for the philosophy of science, namely that unless we assume at the outset that what we are observing in the real world are measurable random variables, we can never come to that conclusion on the basis of observation of frequencies. For non-trivial cases (i.e., ones where E[X1L]<E[X1U]) of nonmeasurable random variables can equally well give neat limiting frequencies and not give them—any such limiting outcome is itself probabilistically maximally nonmeasurable.