Cardinality of non-measurable sets

[Demystifier]

TL;DR Summary

Is cardinality of non-measurable sets $\aleph_0$ or $2^{\aleph_0}$?

The interval $[0,1]$ of real numbers has a non-zero measure. The set of all rational numbers in the interval $[0,1]$ has zero measure. But there are also sets that are somewhere in between, in the sense that their measure is neither zero nor non-zero. They are sets for which measure is not defined. Such sets, for instance, appear in the Banach-Tarski paradox. With a desire to get some better intuition of such sets, I ask about their cardinality. Are such sets continuous with cardinality $2^{\aleph_0}$, or discrete with cardinality $\aleph_0$? My intuition tells me that they should be continuous, but I want a confirmation.

Or is it perhaps undecidable in the ZFC axioms, in the same sense in which it is undecidable whether exist sets with cardinality bigger than $\aleph_0$ and smaller than $2^{\aleph_0}$?

 

[FactChecker]

Any countable set can be covered by a set of open intervals whose total length is less than any given $\epsilon > 0$. So the measure of any countable set is zero. The Continuum Hypothesis is that there is no set with cardinality between $\aleph_0$ and $2^{\aleph_0}$. If we accept the Continuum Hypothesis, then any unmeasurable set on the real line must have cardinality $2^{\aleph_0}$

 

[member 587159]

Have you tried googling? This is the first result:

https://math.stackexchange.com/a/1367140/661543
It should be noted that the existence of a non-measurable set follows from the axiom of choice. There is no need to invoke the continuum hypothesis here.

 

[member 587159]

Any countable set can be covered by a set of open intervals whose total length is less than any given $\epsilon > 0$. So the measure of any countable set is zero. The Continuum Hypothesis is that there is no set with cardinality between $\aleph_0$ and $2^{\aleph_0}$. If we accept the Continuum Hypothesis, then any unmeasurable set on the real line must have cardinality $2^{\aleph_0}$

I think the question asks about the size of the set of all non-measurable sets, not the size of a non-measurable set. But anyway, given that a countable set is the countable union of singeltons, it is trivial that any countable set is measurable. Thus any non-measurable set is uncountable.

 

[FactChecker]

I think the question asks about the size of the set of all non-measurable sets, not the size of a non-measurable set.

I just re-read it and am still not sure. I do not interpret it the way you do.

But anyway, given that a countable set is the countable union of singeltons, it is trivial that any countable set is measurable.

I agree.

Thus any non-measurable set is uncountable with cardinality $|\mathbb{R}|$. There really is no need to use the continuum hypothesis here.

It just guarantees a cardinality greater than $\aleph_0$. The continuum hypothesis says that it must be $2^{\aleph_0}$.

 

[member 587159]

I just re-read it and am still not sure. I do not interpret it the way you do.I agree.It just guarantees a cardinality greater than $\aleph_0$. The continuum hypothesis says that it must be $2^{\aleph_0}$.

Ok, the question is multi-interpretable. Forgive my stuborness. This happens when you post after a 12h flight on a plane. You are also right about the continuum hypothesis part.

[Edited my previous post to avoid confusion]

 

[Demystifier]

I think the question asks about the size of the set of all non-measurable sets, not the size of a non-measurable set.

Just to avoid confusion, I asked about the size of a non-measurable set.

 

[WWGD]

Maybe to add or put the nail in the coffin, using completeness of measure, every subset of a set of measure 0 is measurable with measure zero. Edit: and every measure can be completed.