When a set of functions with a property is a prevalent subset of some other set of functions, almost all functions with that property belong to that set of functions.
Is it true the set of all measurable functions (using the uniform probability measure for sets measurable in caratheodory sense) with infinite or undefined expected values form a prevalent subset of the set of all measurable functions?
Depends on what you want to use the function for. In QM you often want normalizable wave functions; the integral over probability density over all space as finite, or they fall off fast enough at infinity.
What do you call a measurable function. ? Yes, sometimes you cannot
define well enough. Depends on psi>.
I canot answer , unless I kow "measurable" . If it means having a finite norm?
There is no measure that I know of for integrable, let alone measurable-but-not-integrable functions. But there are lots of measurable-but-not-integrable functions. The functions f_{a,p}(x) = 1/|x-a|^p with p >= 1 are all measurable but non-integrable. As are non-degenerate linear combinations of these functions. Or multiply these functions by other functions bounded away from zero near x=a. The impression I get is that measurable-but-not-integrable functions vastly outnumber integrable functions. Comparing cardinalities is a little tricky: do you identify any two integrable or measurable functions that are equal almost everywhere? If you do, then both measurable and integrable functions have the cardinality of the continuum. If you don't then both have the cardinality of 2^continuum. Why? Because graphs of measurable (and also integrable) functions are measurable sets, and cardinality of measurable sets in R^n is 2^continuum. But there are uncountable sets of (Lebesgue) measure zero (e.g., Cantor set) and any function f:C->R can be extended to an integrable function f:R->R by zero. How many functions are there C->R? That contains function C->{0,1} and so there are 2^|C| = 2^continuum integrable functions. This is also the upper bound on both integrable and measurable functions' cardinality, so they must be the same.
According to a tutor I spoke to, if I use prevelance for estimating the “size” of the subset of a function space (in that space), then the hypothesis that the set of measurable functions with infinite or undefined expected values are a prevelant subset of the set of all measurable functions is not correct.
I’m not sure how to disprove him but if we cannot use prevelance, I have the following “intuitions”:
1. There is “high probability” that for every integer n, if we define the function f:A->B then the set of functions which totally disconnected for a subset of (A ∩ [-n,n]) x (B ∩ [-n,n]), as n-> infinity, has a positive 2-d Lebesgue measure.
2. Functions that satisfy 1. cannot approximate the subpace of (A ∩ [-n,n]) x (B ∩ [-n,n]) where f is totally disconnected, as n-> infinity, with simple functions. Therefore the function is not integrable w.r.t the measure derived from the radon-nikodym derivative of the uniform probability measure.
3. Since the functions satisfying 1. and 2. are not integrable and there is a “high probability” these functions exist and there is “high” probability that ”most” measurable functions have an undefined expected value. (I have to show that this “high probability” is the value 1).
Thus my hypothesis “almost all” measure functions (in the carathedory sense) have infinite or undefined expected values is true.
Am I correct?
Using prevalence (see Wikipedia article https://en.wikipedia.org/wiki/Prevalent_and_shy_sets), suppose integrable functions R->R is prevalent in the space of measurable functions R->R. Then there is a P, a finite-dimensional subspace of measurable functions, such that for any measurable f, f+p is integrable for almost all p in P. Now consider all functions f_a(x)=1/|x-a| for x not= a and 0 otherwise. Note that each f_a is measurable but not integrable (not even locally integrable). If f_a + p is integrable, then p(x)-1/|x-a| is locally integrable around a. Thus P must contain all functions of the form 1/|x-a|+integrable fn. This would make P infinite dimensional. Which is a contradiction. Thus: integrable functions are not a prevalent subset of measurable functions R->R.
With regard to your questions, I have some issues: Just considering functions [0,1]->R,
1. You talk of a set of functions having positive 2-D Lebesgue measure. The graph of a function R->R has a 2-D Lebesgue measure; how is this extended to a set of functions R->R? This leads to questions about constructing a measure over functions or subsets of RxR.
2. In what sense is the set of totally disconnected functions (or graphs of functions) [-n,n]->[-n,n] a subspace? Do you mean a vector space (I doubt this interpretation)? A measure space? Can you define the sigma-algebra?
3. To apply this, you need a measure on sets of measurable functions, which again is a problem.
This is why prevalence was formulated as a concept. Function spaces usually do not have measures defined on them (which is why the Wiener measure is so important). There is no need to create measures over large function spaces. Without a measure, mathematicians either turn to Baire category theory for complete metric spaces or something like prevalence which uses finite-dimensional Lebesgue measure.
So it's not that your intuition is so bad, it's that it is so hard to formulate what you are thinking into something that is provable.
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