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](https://en.wikipedia.org/wiki/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 $\mathbb{R}$?
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> 
> 


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



Models with fixed cardinality of non-Lebesgue measurable sets
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 $\mathbb{R}$.


