# Is it consistent with ZFC that there is a translation-invariant extension of Lebesgue measure that assigns nonzero measure to some set of measure less than c?

It is consistent with ZFC (but not ZFC+CH, of course) that there is a subset $A$ of nonzero outer Lebesgue measure that has cardinality less than $c$. There will then be an extension of Lebesgue measure that assigns non-zero measure to $A$ and there will be a translation-invariant extension of Lebesgue measure that assigns zero measure to $A$ (since there is an extension that assigns zero to all sets of cardinality less than $c$, using the uncountable cofinality of $c$ and the fact that any such set has null inner measure).

Question: Is it consistent with ZFC that there be a translation-invariant extension of Lebesgue measure that assigns nonzero measure to some set of cardinality less than $c$?

If yes, then it will be consistent with ZFC that there be a translation-invariant extension of Lebesgue measure which has a set of null measure whose complement has cardinality less than $c$, which will be rather amusing, I think.

(There are two kinds of proofs I've seen of the fact that every set of nonzero Lebesgue measure has cardinality $c$. One kind depends on there being a closed subset of nonzero measure and then a bunch of bisections. That won't work for extensions of Lebesgue measure. The other kind depends on the continuity of convolutions of characteristic functions, which then depends on the $L^1$-continuity of translation, which then depends on approximation by characteristic functions of intervals, and that won't work either.)

There can be no translation invariant extension of the Lebesgue measure which gives a set of cardinality less than continuum positive measure. Suppose that $\nu$ is a translation invariant extension of the Lebesgue measure with $\nu(A)>0$ for some set $A$ of cardinality less than continuum. Take note that $\mathbb{R}$ is a vector space over $\mathbb{Q}$. Let $B$ be the vector subspace of $\mathbb{R}$ generated by $A$. Then $B$ has cardinality less than continuum as well, so the quotient group $\mathbb{R}/B$ has cardinality continuum. Therefore there is an uncountable set $C\subseteq\mathbb{R}$ such that $c+B\neq d+B$ for distinct $c,d\in C$. Clearly $c+A\neq d+A$ for distinct $c,d\in C$. In particular, by translation invariance, $\mu(c+A)>0$ for all $c\in C$. However, this means that $\mathbb{R}$ has uncountably many pairwise disjoint sets of positive $\nu$-measure. This contradicts the $\sigma$-finiteness of $\nu$.
• Should the "vector subspace of $\mathbb R$ generated by $A$" just be the additive subgroup generated by $B$? Then (after fixing some typos) this looks like it should work. I think it's also simpler than the proofs that I've seen of the fact that every Lebesgue measurable subset of positive measure has cardinality $c$. Is there a reference for this proof? – Alexander Pruss Mar 25 '13 at 22:01
• You can also just take the additive subgroup of $B$. I just used vector spaces out of personal preference since $\mathbb{R}$ is a vector space over $\mathbb{Q}$. – Joseph Van Name Mar 25 '13 at 22:17
• The proof more generally shows that there is no quasi-translation-invariant (preserves null-measure under translations) $\sigma$-finite measure on $\mathbb R$ that assigns a nonzero measure to a set of cardinality less than $c$. Nice to know this fact. – Alexander Pruss Mar 26 '13 at 19:10