TITLE: Translates of measure zero set
QUESTION [11 upvotes]: 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.

REPLY [10 votes]: 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.