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I have been thinking about which kind of wild non-measurable functions you can define. This led me to the question:

Is it possible to prove in ZFC, that if a (Edit: measurabel) set $A\subset \mathbb{R}$ has positive Lebesgue-measure, it has the same cardinality as $\mathbb{R}$? It is obvious if you assume CH, but can you prove it without CH?

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    $\begingroup$ Life without GCH is hard business. That's why, you should always remember to bring your GCH along with you wherever you go. $\endgroup$
    – Harry Gindi
    Commented Dec 16, 2009 at 9:13
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    $\begingroup$ Amusing, Harry, but not apt. The answer to Sune's question is yes! $\endgroup$
    – Jonas Meyer
    Commented Dec 16, 2009 at 9:22
  • $\begingroup$ A related post on Mathematics: Cardinality of a set of positive Lebesgue measure. $\endgroup$
    – Martin Sleziak
    Commented Feb 22, 2020 at 6:29

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I found the answer in the paper "Measure and cardinality" by Briggs and Schaffter. In short: not if I interpret positive measure to mean positive outer measure. A proof is given that every measurable subset with cardinality less than that of $\mathbb{R}$ has Lebesgue measure zero. However, they then survey results of Solovay that show that there are models of ZFC in which CH fails and every subset of cardinality less than that of $\mathbb{R}$ is measurable, and that there are models of ZFC in which CH fails and there are subsets of cardinality $\aleph_1$ that are nonmeasurable. So it is undecidable in ZFC.

If it was intended that our sets are assumed to be measurable, then the answer would be yes by the first part above.

Edit: In light of the comment by Konrad I added a couple of lines to clarify.

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    $\begingroup$ I'm not sure I understand your answer. Don't you say that if $A\subset \mathbb{R}$ has positive measure, it cannot have the same cardinality as $\mathbb{R}$? $\endgroup$
    – Sune Jakobsen
    Commented Dec 15, 2009 at 11:57
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    $\begingroup$ But doesn't asserting that a set has positive Lebesgue measure tacitly assume that it is measurable? $\endgroup$
    – Konrad Swanepoel
    Commented Dec 15, 2009 at 12:03
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    $\begingroup$ Konrad, I feared there might be confusion about that. Strictly speaking that is what it should mean, but I was taking it loosely and thinking of outer measure. If only measurable sets were intended, what I described above shows that the answer is "yes". Sune, no. But perhaps I caused confusion for the reason Konrad mentions. $\endgroup$
    – Jonas Meyer
    Commented Dec 15, 2009 at 12:10
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I am not sure if this answer can be helpful or not, but since it is a very elemantary approach to the problem, it might be useful. Assume we are given a subset of R such as A with cardinality smaller than R. Then you can show that the cardinality of A and A+A for any infinite set A is the same. Hence the cardinality of A+A is the same as the cardinality of A which is smaller than R. But you can prove that for any set A of positive measure, A+A has at lease one open interval as a subset. This will be a contradiction with the fact that cardinality of A+A is smaller than R.

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    $\begingroup$ That's neat! The proof that Briggs and Schaffter give is also elementary but very different, so it's useful to see this one too. $\endgroup$
    – Jonas Meyer
    Commented Dec 16, 2009 at 8:49
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    $\begingroup$ How do you prove that if A has positive measure, A+A contains an open interval? $\endgroup$
    – Sune Jakobsen
    Commented Dec 16, 2009 at 13:25
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    $\begingroup$ Let's prove that if A is of positive measure then A+A contains an interval. First show that if m(A)>0 then there is an open interval L such that m(A intersect L)>(3/4)m(L). Now use this to show that A-A contains the interval K=(-0.5m(A),0.5m(A)). For the last part let b be a number inside K. Consider all the pairs inside L that their subtraction is equal to b. Prove that A contains at least one of those pairs otherwise the inequality at the begining of the proof can not be true. $\endgroup$
    – Ehsan
    Commented Dec 16, 2009 at 16:16
  • $\begingroup$ Sorry that I proved above that A-A contains an interval. This is also true for A+A but the proof will be slightly different. Anyways we could start with A-A in the proof of the original problem. $\endgroup$
    – Ehsan
    Commented Dec 16, 2009 at 16:29
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    $\begingroup$ 1-This is one of the fundamental theorems about the measurable sets that you can approximate their measure from above by open sets. You can find a proof in real analysis books like Folland. 2-Yes the proposition will be still true if you consider any positively measurable sets A and B. Thanks for the comment, i just wanted to make it as easy as possible. $\endgroup$
    – Ehsan
    Commented Dec 19, 2009 at 1:13
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The answer to the question is that it is independent of ZFC, if one is speaking of outer measure.

The right context for the question and its answer is the very active research area known as Cardinal Characteristics of the Continuum. The point of this subject is to investigate exactly how the dichotomy between countable and continuum plays out in situations when CH fails. For example, in this area, researchers define a number of cardinal invariants:

The bounding number b is the size of the smallest unbounded family of functions from ω to ω. There is no function that bounds every member of the family.

The dominating number d is the size of the smallest dominating family of functions ω to ω. Every function is dominated by a member of the family.

The additivity number for measure is the smallest number of measure zero sets whose union is not measure zero.

The covering number for measure is the smallest number of measure zero sets whose union is all of R.

The uniformity number for measure is the size of the smallest non-measure zero set.

The cofinality number for measure is the smallest size of a family of measure zero sets, such that every measure zero set is contained in one of them.

Remarkably, none of these numbers is provably equal to any other. In addition, there are models of set theory separating each of them both from ω1 and from the continuum. In the case of the uniformity number, this is the answer that Jonas Meyer has pointed to above.

One can define similar numbers using the ideal of meager sets in place of the ideal of measure zero sets, and the relationships between all these cardinal characteristics are precisely expressed by Cichon's diagram. In particular, no two of them are provably equal, and there are models of set theory exhibiting wide varieties of possible relationships.

There are dozens of other cardinal characteristics, whose relationships are the focus of intense study by set theorists working in this area. The main tool for separating these cardinal characteristics is the method of forcing and especially iterated forcing.

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    $\begingroup$ And if Joel's answer isn't enough to put the bed all the comments that foundational questions don't have any relationship to "real" mathematics,nothing will. Thanks,Joel,for the very knowledgable and tantalizing response. $\endgroup$
    – The Mathemagician
    Commented May 29, 2010 at 20:28
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    $\begingroup$ Belated thanks for the vote of confidence, Andrew! But to be honest, in the math circles with which I am familiar, one doesn't really seem to hear such comments as those to which you refer. But I've heard that things used to be different... $\endgroup$
    – Joel David Hamkins
    Commented Jan 29, 2012 at 23:10
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I'm interpreting the question as: Measurable, with positive measure, not as "having positive outer measure" (for which the answer is independent of the basic axioms of set theory, as pointed out by Joel).

The answer is yes. By elementary properties of Lebesgue measure (regularity), for any $\epsilon>0$, any set $C$ of positive measure contains compact subsets $C_\epsilon$ of measure within $\epsilon$ of the measure of $C$ (interpret this as "arbitrarily large" if $C$ has infinite measure). Any set of positive measure is obviously uncountable. It is straightforward to see that a compact uncountable set of the reals contains a perfect set, and that perfect sets have the same size as the reals. Therefore, $C$ must also have the size of the reals. (I guess the last step uses the Schroeder-Bernstein theorem.)

(On a side note, Cantor proved the result that closed uncountable subsets of ${\mathbb R}$ have the size of the reals. This extends to larger collections of sets, e.g., to all uncountable Borel sets. The first approach to the continuum hypothesis was to try to keep on extending this result.)

To see that perfect sets have the size of the reals: Check that any perfect set has a "copy" of Cantor's set; this is standard; baby Rudin essentially shows how in an exercise in Chapter 1 or 2. Cantor's set, by construction, obviously has size $2^{|{\mathbb N}|}$. Check that the reals also have this size, e.g., by noticing that ${\mathbb R}$ and $(0,1)$ have the same size, and identifying reals in $(0,1)$ with their infinite binary expansion. I suppose this may also use Schroeder-Bernstein depending on how one fleshes this outline out.

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A set of positive measure contains a closed subset of positive measure. It is known that closed subset of reals may have cardinality either continuum or at most countable (http://en.wikipedia.org/wiki/Perfect_set_property)

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