# Smallest size of a non-measurable set of reals

The question is pretty much the title. I'm wondering if anything is known about the smallest size $$\kappa$$ of a non-measurable subset of the real numbers (regarding the Lebesgue measure). Since we have $$\kappa\geq\aleph_0$$ and $$\kappa\leq\mathfrak{c}$$ with $$\kappa=\mathfrak{c}$$ at least being consistent (under CH or MA), it might be an interesting cardinal invariant to look at.

• If I look at the Wikipedia articles Cardinal characteristic of the continuum (current revision) and Cichon's diagram (current revision), it seems that this is the cardinal denoted $\operatorname{non}(\mathcal N)$ and $\operatorname{non}(\mathcal L)$. – Martin Sleziak Jan 10 '20 at 21:36
• Knowing this notation might help when searching for results about this cardinal. (And perhaps also some of the references given in those Wikipedia articles might contains some pointers.) – Martin Sleziak Jan 10 '20 at 21:37
• I wasnt aware of the connection to $\text{non}(\mathcal{N})$. So does this just follow, because we can carry out the Vitali construction on any set of positive and finite measure and therefore obtain, for every non-nullset $M$, a non-measurable set $N\subseteq M$? – Hannes Jakob Jan 10 '20 at 21:55
• Well, a measurable set of positive measure necessarily has cardinality $\mathfrak{c}$. So if it has cardinality less than $\mathfrak{c}$ and isn't null, it must already be non-measurable. – Nate Eldredge Jan 11 '20 at 2:20
• @YCor Yes to both. – Andrés E. Caicedo Jan 27 '20 at 1:26

Claim: The smallest size of a non-measurable set is $$\text{non}(\mathcal{L})$$:
$$\geq$$: If $$A$$ is non-measurable, then $$A$$ is not null.
$$\leq$$: If $$A$$ is not null and not of size continuum, then $$A$$ has to be non-measurable, because the difference set of any set of positive measure has to contain an interval around $$0$$ (https://en.wikipedia.org/wiki/Steinhaus_theorem) and therefore be of cardinality $$\mathfrak{c}$$. Because furthermore, the cardinality of the difference set is less than or equal to the cardinality of $$A\times A$$ which is equinumerous with $$A$$, $$A$$ is of cardinality $$\mathfrak{c}$$.
Therefore either $$\text{non}(\mathcal{L})=\mathfrak{c}$$ and $$\leq$$ is trivial or there is a non-null set of cardinality $$<\mathfrak{c}$$ which, by the argument above, also is a non-measurable set.