# 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 $\operatorname{non}(\mathcal L)$. Jan 10, 2020 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.) Jan 10, 2020 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 $N\subseteq M$? Jan 10, 2020 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. Jan 11, 2020 at 2:20
• @YCor Yes to both. Jan 27, 2020 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.