# Cardinality of non-measurable sets

Science Advisor
Gold Member
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}##?

## Answers and Replies

FactChecker
Science Advisor
Gold Member
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}##

Demystifier
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.

Last edited by a moderator:
Demystifier
FactChecker
Science Advisor
Gold Member
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}##.

Demystifier and member 587159
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]

Last edited by a moderator:
FactChecker
Science Advisor
Gold Member
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
Science Advisor
Gold Member
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.

Last edited: