TL;DR : the question is, assuming continuum hypothesis is false is there a non Lebesgue-measurable set with cardinality strictly smaller than $\mathbf R$ ?
Let $A$ be a subset of $\mathbf R$, i will note $\lambda$ the Lebesgue measure and $\mathscr L(\mathbf R)$ the Lebesgue sigma algebra. I was wondering about the link between the cardinality of $A$ and its properties with respect to measure theory. Some facts are well known :
if $\mathrm{Card}(A) \leq \mathrm{Card}(\mathbf N) $ then $A$ is measurable and has measure $0$.
if $\mathrm{Card}(A)=\mathrm{Card}(\mathbf R)$ then everything is possible : we can find $A\in\mathscr L(\mathbf R)$ with $\lambda(A)=0$ (e.g. Cantor set) or $\lambda(A)>0$ (e.g. [0;1]) and we can find $A\notin \mathscr L(\mathbf R)$ (e.g. Vitali set).
Now assume that $\mathrm{Card}(\mathbf N)<\mathrm{Card}(A)<\mathrm{Card}(\mathbf R)$, what could happen now ? That is, what happen if we assume continuum hypothesis is false ?
From the Steinhaus theorem we know that, for a measurable subset $B$ of $\mathbf R$, if $\lambda(B)>0$ then $B+B$ contains an open interval. So $\mathrm{Card}(B)=\mathrm{Card}(B+B)=\mathrm{Card}(\mathbf R)$ and thus we know that if $A$ is measurable then $\lambda(A)=0 $.
Now let $C$ be the usual Cantor set, there exists an injection of $A$ in $C$ so there exists a subset of $C$ (hence measurable) with the same cardinal as $A$, so it is possible to have that $A$ is measurable.
So it remains to know if there exists a non measurable set with cardinal strictly smaller than $\mathfrak c$.
Bonus : What could be the Hausdorff dimension of a subset of $\mathbf R$ with cardinality strictly smaller than $\mathfrak c$ ? what about it's inner and outer measure ?