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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.)

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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 $\nu$.

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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
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