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Suppose $X \subseteq \mathbb{R}$ has measure zero. Can we find an uncountable $A \subseteq \mathbb{R}$ such that $X + A = \bigcup \{a + x: a \in A, x \in X\}$ has measure zero?

Clearly, the answer is yes under Martin's axiom and $\mathfrak{c} > \aleph_1$. But can we do this without additional assumptions.

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    $\begingroup$ You'll be able to make comments and edits if you register your account. $\endgroup$
    – Nate Eldredge
    Commented Oct 25, 2017 at 16:24

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Claim: For every null G-delta set $X$, there exists a perfect set $P$ such that $X + P$ is null.

Proof sketch: First assume MA plus $2^{\aleph_0} > \aleph_2$. Choose $A \subseteq \mathbb{R}$ of size $\aleph_2$ and a null G-delta set $G$ such that $X + A \subseteq G$. It follows that $\{y: X + y \subseteq G\}$ is a coanalytic set of size $> \aleph_1$ and hence contains a perfect set $P$. The statement "there exist a perfect set P and a null G-delta set G such that $X + P \subseteq G$" is $\Sigma^1_2(X)$. Apply Shoenfield's absoluteness.

Note that the proof is identical to Martin's proof of Claim 1 here.

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    $\begingroup$ The assumption that $X$ is $G_\delta$ is not needed: an arbitrary null $X$ is included in a null $G_\delta$ set anyway. $\endgroup$
    – Emil Jeřábek
    Commented Oct 27, 2017 at 11:52

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