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Let $\{b_n\}_{n=1}^\infty$ be a dense subset of $\mathbb{R}$ and let $D \subseteq \mathbb{R}$ be a measurable set such that $m(D \triangle (D + b_n))=0$ for all $n \in \mathbb{N}$ (here, the $\triangle$ denotes the symmetric difference of the two sets, $D+ b_n = \{d + b_n : d \in D\}$, and $m$ stands for the Lebesgue measure). Prove that $m(D)=0$ or $m(D^c)=0$ (here $D^c$ is the complement of $D$ in $\mathbb{R}$).

I am having trouble getting a proof off the ground! In particular, it is not clear to me how and where the dense hypothesis would come in. Any help would be greatly appreciated.

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Hints:

1) $m(D \Delta (D + x))$ is a continuous function of $x$.

2) If $x$ and $y$ are Lebesgue points of $D$ and $D^c$ respectively, what can you say about $m(D \Delta (D + x - y))$?

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    An even further hint: using the Lebesgue density theorem, there are $x \in D$ and $y \in D^c$ and \epsilon > 0 such that m(D \cap (x-\epsilon, x+\epsilon)) > \epsilon and m(D^c \cap (y-\epsilon,y+\epsilon)) > \epsilon. What does this say about $m((D - x) \cap(-\epsilon,\epsilon))$ and $m((D^c-y) \cap (-\epsilon,\epsilon))$? What can you conclude about $m((D-x) \cap (D^c - y))$?2012-01-29