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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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2 Answers2

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

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
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The following solution is based on the hint given by Robert Israel

By $\sigma$ finite of $\mathbb{R}^{n}$ , wlog, we can assume $m(D)<\infty$ , so $f(x)=m(D \Delta(D+x))$ is a continuous function. This can be seen by $m(A \Delta B)=\int_{\mathbb{R}^{n}} |1_{A}-1_{B}| \ dm$ and for $f \in L^{1}$ , we have $\lim\limits_{h \to 0 } \int_{\mathbb{R}^{n}} |f(x+h)-f(x)| \ dx =0$

Then, suppose $m(D)>0$ and $m(D^{c})>0$ . We prove this is impossible by contradiction. Then since both of them have measure greater than zero, them have Lebesgue density point $x,y$ respectively. Let $I(x,\epsilon)$ be the interval centered at $x$ with radius $\epsilon$ ,so $m(I(x,\epsilon))=2\epsilon$ .

Then by Lebesgue density point theorem, we have $$\begin{align*} m(D \cap I(x,\epsilon)) \geq \frac{2}{3} \cdot 2 \epsilon \iff m(D-x \cap I(0,\epsilon)) \geq \frac{4}{3} \epsilon \\ \\ m(D^{c} \cap I(y, \epsilon)) \geq \frac{2}{3} \cdot 2 \epsilon \iff m(D^{c}-y \cap I(0,\epsilon)) \geq \frac{4}{3} \epsilon \end{align*}$$ Then by $m(A)+m(B)- m(A\cup B)= m(A\cap B)$ , we have $$\begin{align*} m(D^{c} -y \cap D-x) &\geq m(D^{c}-y \cap I(0,\epsilon) \cap D-x \cap I(0,\epsilon)) \\ &= m(D^{c}-y \cap I(0,\epsilon) )+ m( D-x \cap I(0,\epsilon)) \\&- m((D-x \cap I(0,\epsilon))\cup(D^{c}-y \cap I(0,\epsilon)) ) \\ & \geq \frac{4}{3} \epsilon + \frac{4}{3} \epsilon - m(I(0,\epsilon)) \\ & \geq \frac{8}{3}\epsilon - 2 \epsilon \\ &= \frac{2}{3}\epsilon \\ &>0 \end{align*}$$Then notice $$\begin{align*} m(D^{c}-y \cap D-x) = m(D\cap(D^{c}+x-y))= m(D \backslash (D+x-y))\leq m(D \Delta (D^{}+x-y)) \end{align*}$$

Then by $\left\{ b_{n} \right\}\subset_{dense} \mathbb{R}^{n}$ , let $b_{n} \to x-y$ as $n\to \infty$, so by continuity, we have $$\begin{align*} m(D \Delta(D^{}+x-y))=\lim\limits_{n\to \infty } m(D \Delta (D^{}+x-y)) =0 \end{align*}$$Thus, we have $0<m(D^{c}-y \cap D-x) \leq 0$ which gives the contradiction.

M_k
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