7
$\begingroup$

In the usual $\mathsf{ZFC}$, we know that there are $2^\mathfrak{c}$ many subsets of $\mathbb{R}$ that are not (Lebesgue) measurable. On the other hand, the Solovay model also provides us a model of $\mathsf{ZF}$ which has $0$ non-measurable subsets of $\mathbb{R}$.

I would like to know if the following in-between assertion has been established or is currently being researched on:

For any cardinal $\kappa < 2^\mathfrak{c}$, is it consistent with $\mathsf{ZF}$ that there are exactly $\kappa$ non-measurable subset of $\mathbb{R}$?

If this is false, can we classify the cardinals in which the assertion above holds?

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ If $X\subseteq (0,1)$ is a non-measurable subset and $Y\subseteq (1,2)$ is a null set, then $X\cup Y$ is a non-measurable subset of $\mathbb{R}$, and there are $2^\mathfrak{c}$ null subsets of $Y$. $\endgroup$
    – Hanul Jeon
    Commented May 15, 2021 at 6:32
  • $\begingroup$ @HanulJeon I was initially skeptical on if having one non-measurable set implies the existence of a bounded non-measurable set, but it is not difficult to obtain one. See my answer below. Thanks/ $\endgroup$
    – Clement Yung
    Commented May 15, 2021 at 6:47

1 Answer 1

Reset to default
6
$\begingroup$

Actually, $\mathsf{ZF}$ easily proves that if there exists at least one non-measurable subset of $\mathbb{R}$, then there must be $2^\mathfrak{c}$ non-measurable subsets of $\mathbb{R}$.

Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = (X \cap [0,\infty)) \cup (X \cap (-\infty,0)) =: A \cup B$. If both $A$ and $B$ are measurable, then $X$ must also be measurable, at it is a finite union of measurable sets. Thus, we must have that either $A$ or $B$ is non-measurable.

WLOG say $A$ is not measurable. Now there are $2^\mathfrak{c}$ measurable sets in $(-\infty,0)$, and for any of such set $C$, we have that $A \sqcup C$ is not measurable. This gives $2^\mathfrak{c}$ distinct non-measurable subsets of $\mathbb{R}$.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ I'm not sure that ZF proves that countable union of (Lebesgue-)measurable sets is measurable. $\endgroup$
    – Wojowu
    Commented May 15, 2021 at 6:57
  • 2
    $\begingroup$ Wojowu’s objection is warranted, however, countable union is not needed for the argument. ZF certainly proves that the union of two measurable sets is measurable, thus if $X$ is non-measurable, then $X\cap[0,+\infty)$ or $X\cap(-\infty,0]$ is non-measurable, and this gives plenty of room to make the argument work. $\endgroup$
    – Emil Jeřábek
    Commented May 15, 2021 at 8:11
  • 2
    $\begingroup$ Thank you all for the comments. With that in mind, is it consistent with $\mathsf{ZF}$ that there exists a non-measurable set but all bounded sets are measurable? $\endgroup$
    – Clement Yung
    Commented May 15, 2021 at 8:18
  • $\begingroup$ Thank you. Note that you don’t actually need $C$ measurable: $A\cup C$ is non-measurable for every subset $C\subseteq(-\infty,0)$. $\endgroup$
    – Emil Jeřábek
    Commented May 15, 2021 at 9:18
  • $\begingroup$ You can make this much simpler with symmetric differences: if $X$ is a subset of the Cantor set (hence null) then $X\triangle A$ is measurable iff $A$ is. $\endgroup$
    – Noah Schweber
    Commented Jan 3, 2022 at 3:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .