A “measure” on the rational numbers?
When thinking about measures and rational numbers, we’ve (by “we” I mean “people who have thought about measures”) all had the general sort of thought described in this math stackexchange question, right? https://math.stackexchange.com/questions/1084030/measure-on-the-set-of-rationals
The “can’t we just get the total length of the intervals” thing.
And the answers to that ask explain why “no, you really can’t have a countably additive uniform measure on the rationals, because if you had one, then you could take these countably infinite unions and get nonsense results”.
Of course, it still works for finite additivity. And that’s nice.
But as I assume is common, I was playing and hoping it was possible to extend this a little further, and find a way to get additivity for some sorts of countably infinite unions, not just finite ones.
Apologies if these are trivial, but I couldn’t find anywhere where they were documented. I don’t know how to search for these things.
What if instead of requiring that the measure of the countably infinite disjoint union be the infinite sum of measures of the things being unioned together, we just require that the measure of the union of the sets be at least the sum of the measures of the sets (and that for finite disjoint unions, that it be exact)
I notice that the order of the sets for the counterexamples seems kind of wild. What if we place some restrictions on how the sets in the infinite union can be ordered?
What if we be careful to not try to assign a measure to all the sets, just some of them?
Now, with just the first and third idea, we might say, something like:
Any interval in the rational numbers with rational endpoints is “measurable” and has length equal to the difference between its endpoints.
The complement of a “measurable” set of rationals with respect to some interval in the rational numbers with rational endpoints, that it is a subset of, is also a “measurable” set of rationals.
Some particular interval with irrational endpoints is “measurable”.
Then you could derive that the “measure” of that interval is in fact what you would expect it to be, because it must be at least that amount by the first idea, and its complement in a larger interval with rational endpoints must be at least the amount you would expect it to be also, and the sum of the two has to be exactly the length of the larger interval with rational endpoints, and this forces the two sets to be exactly what they should be.
All nice and fun, not too difficult, and not really impressive either.
Let’s think about that second idea some.
It feels like it should be safe to take an infinite union of intervals if each interval is entirely after all the previous ones. Like, you should be able to just transform them into the corresponding intervals in the real numbers, and then that should work fine, right?
When trying to look into that, I came across a paper by Paul Erdős and András Hajnal called “On a classification of denumerable order types and an application to the partition calculus” which describes a, well, it describes a classification of denumerable (countable) order types, and then an application to something which I didn’t really read that part.
Anyway, it shows that the countable order types with no dense suborder can all be put into this hierarchy thing, somewhat similar to the birthday thing for the Surreal Numbers.
The way it works is that, at level 0, there are the order types 0 (the order type of the empty set) and 1 (the order type of a set with one thing in it).
And then for each countable ordinal after 0, that level introduces the order types which are the well ordered sums of any of the order types from any lower levels, as well as the order types which are the reverse well ordered sums of any of the order types from any lower levels.
And it turns out that any total order which does not have a dense suborder (by which I mean, an ordered set with the given order type does not have a subset which has an order type which is dense) (henceforth “discrete order type”, also called “scattered order”) is found at some level in that hierarchy.
And the other ones are (in some sense) “based on” the order type of the rationals.
So, suppose you have some set of “measurable” sets of rational numbers such that for each pair of these sets, for one of them, all of the elements of it are less than all the elements in the other set. This defines a total order on these sets.
What I want to say is that, if the order type of this set of “measurable” sets is a discrete order type, then the (infinite) union of these “measurable” sets is also “measurable”, and that the measure is the sum of the measures of the individual sets.
I think that this should follow from the following assumptions:
If the order type of these set of “measurable” sets is either ω or ω* (the order types of the positive integers, and of the negative integers, respectively), then the infinite union of them is also “measurable”.
Some other assumptions described earlier in this post.
I think it should follow in a fairly straightforward way by transfinite induction (induction over the countable ordinals) that, given that assumption, that if the order type of the set is at any level in the hierarchy described, then the union will keep the “measurability”, and the measure of the union will be the sum of the measures.
And therefore, because the hierarchy contains all the discrete order types, then it should work for all the discrete order types.
I haven’t totally written out the proof yet, but there’s a very rough proof sketch there, and I’m pretty sure it should work.
Then all that would be left to show would be that it is consistent to assume the assumptions (that they don’t cause a contradiction), and then that would, I think, establish that there is a “uniform” “measure” on the rationals, with the exception that measure of the union only has to equal the sum of the measures if the sets being unioned together have a total order which is discrete.
