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Please forgive me if this is a very easy question.

Let $A \subset [0,1]$ be a Borel-measurable set with strictly positive Lebesgue measure. Does there necessarily exist a countable subgroup $G$ of $\mathbb{R}$ such that the set $$ \{ t \in [0,1] : (t+G) \cap A \neq \emptyset \} $$ has Lebesgue measure 1?

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    $\begingroup$ Take $G=\mathbb{Q}$. Then $A-G$ has full measure (say, by the Lebesgue density theorem), so $(t+G)\cap A\neq\varnothing$ for almost all $t$. $\endgroup$ – Pablo Shmerkin Jun 28 '15 at 2:45
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    $\begingroup$ By monotone convergence, we can compute the measure of the displayed set as a limit where we replace $A$ by $U_n$, if $\chi_{U_n}$ decreases to $\chi_A$ a.e. Now use outer regularity to approximate $A$ by open sets, and take any dense $G$. $\endgroup$ – Christian Remling Jun 28 '15 at 5:15
  • $\begingroup$ @PabloShmerkin: Thank you very much for this, this is a nice simple solution. $\endgroup$ – Julian Newman Jun 28 '15 at 10:42
  • $\begingroup$ @ChristianRemling: Thank you for your answer. This is also a very nice answer. $\endgroup$ – Julian Newman Jun 28 '15 at 11:09
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    $\begingroup$ @ChristianRemling: In other words, having $U_n \downarrow A$ does not imply $U_n - G \downarrow A - G$. $\endgroup$ – Nate Eldredge Sep 15 '15 at 19:02
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As an exercise to myself, I thought I would write out the Lebesgue density argument suggested by Pablo Shmerkin.

Let $C$ be any countable dense subset of $\mathbb{R}$, such as $\mathbb{Q}$ (it doesn't have to be a subgroup). Note that $\{t \in [0,1] : (t+C) \cap A \ne \emptyset\}$ can be written more concisely as $(A-C) \cap [0,1]$.

For a measurable set $B$, Let $L_B(x) = \liminf_{r \to 0} \frac{1}{2r}m(B \cap (x-r, x+r))$ be the lower density of $B$ at $x$. Lebesgue's theorem asserts $L_B = 1_B$ almost everywhere. Since $A$ has positive measure, let us choose $x_0 \in A$ with $L_A(x_0) =1$. In particular, there exists $r_0 > 0$ so small that $m(A \cap (x_0-r, x_0+r)) \ge \frac{2r}{2} = r$ for all $0 < r < r_0$.

Now let $y \in \mathbb{R}$ be arbitrary. I will show $L_{A-C}(y) > 0$. Choose any positive $r < r_0$. Since $C$ is dense, we may choose $q \in C$ such that $|(y+q)-x_0| < r/2$. This means that $(x_0 - r/2, x_0 + r/2) \subset (y+q-r, y+q+r)$. So we have $$\begin{align*} m((A-C) \cap (y-r, y+r)) &\ge m(A-q \cap (y-r, y+r)) \\ &= m(A \cap (y+q-r, y+q+r)) \\ &\ge m(A \cap (x_0 - r/2, x_0 + r/2)) \\ &\ge \frac{r}{2}. \end{align*}$$ Since $r < r_0$ was arbitrary, this shows $L_{A-C}(y) \ge \frac{1}{4}$. Since $y$ was arbitrary, $L_{A-C} > 0$ everywhere. Since $1_{A-C} = L_{A-C}$ almost everywhere, $A-C$ has full measure. In particular $m((A-C) \cap [0,1]) = 1$.

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