TITLE: Models with fixed cardinality of non-Lebesgue measurable sets
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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 $\mathbb{R}$.