Is it consistent with ZFC that some translation-invariant extension of Lebesgue measure assigns nonzero measure to some set of cardinality $<\frak c$?

Question

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.)

Answer 1

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 $\nu$.

Answer 2

For what it's worth, here is a slightly more elementary proof.

If there is a translation-invariant measure on $\mathbb R$ assigning a non-zero value to some set of cardinality less than $\frak c$ and finite on finite intervals, there is a finite translation-invariant measure on $S^1 = \mathbb R/\mathbb Z$ that assigns a non-zero value to some set of cardinality less than $\frak c$.

Lemma: For $A\subset S^1$, if $|A|<\frak c$, then there is an $x\in S^1$ such that $A \cap (x+A) = \varnothing$.

Proof: If there is no such $x$, then for any $x\in S^1$ there are $a$ and $b$ in $A$ such that $a-b=x$. In other words, subtraction restricted to $A^2$ is surjective onto $S^1$. Hence $|A|=|A^2|\ge |S^1|=\frak c$.

Given this, if $|A|<\frak c$, choose a sequence $(x_n)$ in $S^1$ such that $x_0=0$ and $x_n+A$ is disjoint from $\bigcup_{i=0}^{n-1} (x_i+A)$. Then the $x_i+A$ are an infinite disjoint sequence of translates of $A$, so the measure of $A$ must be zero under any translation-invariant finite measure.

[The above argument basically shows that given any infinite group $G$ acting transitively on a space $X$, any finite finitely-additive $G$-invariant measure on $X$ assigns zero measure to any set of cardinality less than $|G|$.]