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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.

Vulcan
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1 Answers1

2

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 $m(D \Delta (D + x - y))$?

Robert Israel
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  • I am engaging measure theory for the first time, and I am unfamiliar with the term "Lebesgue point of a set," and thus I don't see the connection you intend in 2). Could you clarify what this term means? Then perhaps I can make sense of 2). – Vulcan Jan 26 '12 at 21:00
  • Actually I should have said "point of density" - or "Lebesgue point of the indicator function". The Lebesgue density theorem says that for almost every $x \in D$, $\lim_{\varepsilon \to 0} \frac{m(D \cap (x-\varepsilon,x+\varepsilon))}{2\varepsilon} = 1$. – Robert Israel Jan 27 '12 at 02:32
  • For a further hint, if $m(D \cap (x-\varepsilon, x+\varepsilon))$ and $m(D^c \cap (y-\varepsilon, y+\varepsilon))$ are both very close to $2 \varepsilon$, what can you say about $m(D \Delta (D + x - y))$? – Robert Israel Jan 27 '12 at 02:39
  • I am not yet familiar enough with Lebesgue differentiation to make the argument go through. However, a fellow graduate student in my program suggested that the above result follows via contradiction by making smart uses of an interesting result:

    Let $A$ be Lebesgue measurable, with $m(A)>0$ (here $m$ denotes the Lebesgue measure). Then for any $0<\rho<1$, there exists an open interval $I$ such that $m(A \cap I)> \rho \cdot m(I)$. I was wondering if you would be up for offering some hints on how this argument might proceed.

     – Vulcan Jan 29 '12 at 17:18
  • 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 $m((D-x) \cap (D^c - y))$? – Robert Israel Jan 29 '12 at 21:46