# Thread: measure of set satisfying ~CH

1. ## measure of set satisfying ~CH

1) Supposing that the Continuum Hypothesis is false, and that S is a set whose cardinality is between those of the sets of naturals and reals. Then is the Lebesgue measure of S = 0?

2. ## Re: measure of set satisfying ~CH

Originally Posted by nomadreid
1) Supposing that the Continuum Hypothesis is false, and that S is a set whose cardinality is between those of the sets of naturals and reals. Then is the Lebesgue measure of S = 0?
Lebesgue measure is defined for subsets of ${\mathbb{R}}^{n}$. First step is how will you define an $S$ such that $S \in {\mathbb{R}}^{n}$ and ${\aleph}_{0}<|S|<|\mathbb{R}|$ ?

3. ## Re: measure of set satisfying ~CH

Originally Posted by godelproof
Lebesgue measure is defined for subsets of ${\mathbb{R}}^{n}$. First step is how will you define an $S$ such that $S \in {\mathbb{R}}^{n}$ and ${\aleph}_{0}<|S|<|\mathbb{R}|$ ?
Thanks for responding. There are two possible answers to this.

(1) The existence of S is consistent with ZF, and the Lebesgue measurability of all subsets of R (of which S would be one) is consistent with ZF plus another axiom (such as AD or the existence of an inaccessible cardinal), so the question would be in a model of sets fulfilling the appropriate axioms to have such an S. In other words, I am excluding the possibility that S is something like a Vitali set. I do not know whether the existence of S is sufficient to answer the question; if not, go to (2).
(2) Any appropriate forcing technique, such as Cohen's forcing, will do.

4. ## Re: measure of set satisfying ~CH

as i understand it, you'll want to assume the axiom of choice is false as well (so that you can be sure S is measureable). i don't think you lose any generality by considering subsets of [0,1]. but i'm not convinced that this alone is enough to guarantee that the lebesgue measure of S will be 0. imagine you have uncountably many translates of a cantor set. can we stack enough of these together to get "some" measure, but not "mostly everything"? (if so, then perhaps models exist in which the lebesgue measure of S is any number between 0 and 1).

if a suitable model exists (and i suspect its construction would be rather complicated) what would it "mean"? would it be useful? from a philosophical point of view, there are compelling reasons for me to believe the axiom of choice should be true. every image should have a pre-image.

i think there are two distinct, but related problems here: one is related to cardinality, and one is related to measureability. i don't think assuming the negation of the continuum hypothesis is sufficient for assuming the measureability of a given uncountable subset of the reals. and it's not at all clear that ZF + ~C + ~CH only admits one model.

5. ## Re: measure of set satisfying ~CH

you'll want to assume the axiom of choice is false as well (so that you can be sure S is measureable).
Although assuming ~AC and the existence of an inaccessible cardinal is a sure-fire way to make sure that there is a model in which S is measurable, it would be a further question as to the necessity of ~AC. That is, ZFC allows non-measurable sets, but I have not seen where it forbids the existence of a measurable uncountable set with cardinality less than that of the continuum.

). i don't think you lose any generality by considering subsets of [0,1]. but i'm not convinced that this alone is enough to guarantee that the lebesgue measure of S will be 0. imagine you have uncountably many translates of a cantor set. can we stack enough of these together to get "some" measure, but not "mostly everything"?
Depends on the uncountability. Obviously you can stack a continuum-number of copies of any non-empty set to get something that gives a non-zero measure. But whether you can stack uncountable but less-than-continuum number of copies of a set of measure zero to get a non-zero measure is precisely my question.

(if so, then perhaps models exist in which the lebesgue measure of S is any number between 0 and 1).
If S did not have a unique measure, then it would not be considered measurable. Or do you mean that for every r in [0,1] there can be an S_r so that the measure of S_r = r?

if a suitable model exists (and i suspect its construction would be rather complicated) what would it "mean"?
It might tell us something about the structure of the continuum.

would it be useful?
Depends what you mean by "useful"? Sometimes some odd spin-offs come from something that appears, at first glance, to be useless wondering. The continuum hypothesis is one such example.

from a philosophical point of view, there are compelling reasons for me to believe the axiom of choice should be true. every image should have a pre-image.
You are perhaps aware of the joke, "The Axiom of Choice is obviously true, the Well Ordering Theorem is obviously false, and Zorn's Lemma is obviously undecidable. " But even if a proposition is accepted as true, seeing what its negation or its absence can bring can nonetheless be worthwhile.

i think there are two distinct, but related problems here: one is related to cardinality, and one is related to measureability.
I am asking how they are related. Probably the answer to my original question is positive, but there are often strange exceptions with things like Vitali sets, singular cardinals, and other weirdos, so I am cautious.

i don't think assuming the negation of the continuum hypothesis is sufficient for assuming the measureability of a given uncountable subset of the reals.
Definitely not. In fact, I believe that ZF + ~CH & "there exists a Vitali set" is equiconsistent with ZF.

, and it's not at all clear that ZF + ~AC + ~CH only admits one model.
Right, it admits an infinite number of models. Take any of them for which Lebesgue measurability is meaningful.

6. ## Re: measure of set satisfying ~CH

Originally Posted by nomadreid
Although assuming ~AC and the existence of an inaccessible cardinal is a sure-fire way to make sure that there is a model in which S is measurable, it would be a further question as to the necessity of ~AC. That is, ZFC allows non-measurable sets, but I have not seen where it forbids the existence of a measurable uncountable set with cardinality less than that of the continuum.
right. if you do not assume ~AC, then you have to prove your set S is measureable first.

Depends on the uncountability. Obviously you can stack a continuum-number of copies of any non-empty set to get something that gives a non-zero measure. But whether you can stack uncountable but less-than-continuum number of copies of a set of measure zero to get a non-zero measure is precisely my question.
and i think it's a HARD question. if we had a more explicit description of S, perhaps "S copies" would do the trick (i am thinking of some kind of quotient here).

If S did not have a unique measure, then it would not be considered measurable. Or do you mean that for every r in [0,1] there can be an S_r so that the measure of S_r = r?
i meant the latter.

It might tell us something about the structure of the continuum.
this is possible. it is more likely to tell us something about measure.

Depends what you mean by "useful"? Sometimes some odd spin-offs come from something that appears, at first glance, to be useless wondering. The continuum hypothesis is one such example.
perhaps i may be missing the point, but it seems to me that while all can agree that "countably infinite" and "uncountably infinite" are different, we haven't made much headway on where we should draw the boundary. i did not mean to imply such musings are not interesting in and of themselves, perhaps you might find some new insight on the nature of continuity in doing so.

You are perhaps aware of the joke, "The Axiom of Choice is obviously true, the Well Ordering Theorem is obviously false, and Zorn's Lemma is obviously undecidable. " But even if a proposition is accepted as true, seeing what its negation or its absence can bring can nonetheless be worthwhile.
Jerry Bona, i believe

I am asking how they are related. Probably the answer to my original question is positive, but there are often strange exceptions with things like Vitali sets, singular cardinals, and other weirdos, so I am cautious.
i am inclined to believe you are correct almost everywhere.

Definitely not. In fact, I believe that ZF + ~CH & "there exists a Vitali set" is equiconsistent with ZF.
my point exactly.

Right, it admits an infinite number of models. Take any of them for which Lebesgue measurability is meaningful.
again, i don't think it is obvious that we will get the same measure in each model.

7. ## Re: measure of set satisfying ~CH

Thanks. Given your reference (Jerry Bona), I found the exact quote: "The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?"

As far as "for every r in [0,1] there can be an S_r so that the measure of S_r = r?", I don't think that the cardinality has anything to do with the non-zero size of a set. After all, even keeping the cardinality constant, that of the continuum, we have simply the usual measure on the collection of sub-intervals of [0,1] which fulfills this condition. So my question is simply whether all sets of less-than-continuum cardinality must have zero Lebesgue measure. True, different models will give different values for the measure, so I would like to know whether any of them will give a non-zero measure.

As far as S copies of S, how you would put the copies together would be crucial, since a straightforward bunch of copies would just give S X S, with the same cardinality of S, not getting us anywhere.

As far as a more explicit description of S: One could start by using the S from Cohen's proof, but if this gives a set of zero measure (and I believe that it does), then to find whether any S that satisfies ~CH AND that has a non-zero Lebesgue measure exists is part of the question. Yes, it is a hard question; that's why I don't know the answer.

You are right, the boundary between countable and uncountable is fuzzy, which is one of the reasons why CH has generated so much work.

I like your "correct almost everywhere". Now, I just have to find the right ultrafilter.... :-)