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.

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    $\begingroup$ 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 $\operatorname{non}(\mathcal L)$. $\endgroup$ Jan 10, 2020 at 21:36
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    $\begingroup$ 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.) $\endgroup$ Jan 10, 2020 at 21:37
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    $\begingroup$ 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 $N\subseteq M$? $\endgroup$ Jan 10, 2020 at 21:55
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    $\begingroup$ 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. $\endgroup$ Jan 11, 2020 at 2:20
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    $\begingroup$ @YCor Yes to both. $\endgroup$ Jan 27, 2020 at 1:26

1 Answer 1


I am just going to compile the comments into an answer so i can close this question.

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.


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