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In a previous post I asked the following question:

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 $\mathbb{R}$).

Robert Israel gave hints to one neat strategy for proving the above result. But now I want to see whether anyone visiting can prove the above result without using the Lebesgue density theorem. In particular, a fellow graduate student in my program suggested that the result follows via contradiction by making smart uses of an interesting result I asked about earlier today:

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 $m(A \cap I)> \rho \cdot m(I)$.

While I am curious to see alternative strategies for proof, I would be particularly interested to see a proof using this last result, if anyone's game to try!

Community
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Vulcan
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1 Answers1

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The assumption gives that $$\chi_D(x)=\chi_D(x+b_n)\mbox{ for almost every }x\tag{1}$$ (because the integral of the non-negative measurable function $|\chi_D-\chi_{D+b_n}|$ is $0$).

Assume that $\mu(D)\gt 0$. Our goal is to prove that $\mu(D^c)=0$. Fix $\rho\in (0,1)$. There is an interval $I$ such that $\lambda(D\cap I)\gt \rho\lambda(I)$. From (1), we get $\lambda((D+b_n)\cap I)\gt \rho\lambda(I)$, and by density of $\{b_n,n\geqslant 1\}$, that for each $t\in\mathbb R$, $\lambda((D+t)\cap I)\geqslant \rho\lambda(I)$. Taking $N:=\lfloor\frac 1{\lambda(I)}\rfloor$ and using the fact that $\lambda(D\cap (k\lambda(I),(k+1)\lambda(I))\geqslant\rho\lambda(I)$ we get that $\lambda(D\cap (0,1))\geqslant \rho$. Indeed, $$\lambda(D\cap (0,1))\geqslant \lambda\left(\bigsqcup_{k=0}^ND\cap (k\lambda(I),(k+1)\lambda(I)\right)\geqslant (N+1)\rho\lambda(I)\geqslant \rho.$$

As $\rho$ was arbitrary, we obtain that $\lambda(D\cap (0,1))=1$. A similar reasoning gives $\lambda(D\cap (n,n+1))=1$ for all $n\in\mathbb Z$, hence $\lambda(D^c\cap (n,n+1))=0$ for $n$ integer.

Davide Giraudo
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  • Almost everything is clear. What I don't understand is "Taking $N:=\lfloor\frac 1{\lambda(I)}\rfloor$ and using the fact that $\lambda(D\cap (k\lambda(I),(k+1)\lambda(I))\geqslant\rho\lambda(I)$".

    a) you don't use $N$ later. b) from what follows the "using the fact" inequality? Many thanks.

     – vesszabo Oct 01 '13 at 09:54
  • $N$ didn't appeared because I didn't detailed enough. I've edited. – Davide Giraudo Oct 01 '13 at 09:58
  • (This is a comment by a new user @Eduardo, which was put as an answer) Beware with the answer by Davide Giraudo, it is flawed. He is assuming that $(0,1)\supset\bigsqcup_{k=0}^N(k\lambda(I),(k+1)\lambda(I))$, which makes no sense. –  Jun 17 '15 at 10:23
  • @Eduardo Can you tell me why it does not make sense? Maybe the union should be stopped at $N-1$ instead of $N$, but I think it is correct. – Davide Giraudo Jun 17 '15 at 11:58
  • @DavideGiraudo Where did you use your first argument: "for each $t\in \mathbb{R}$, $\lambda((D+t)\cap I)\ge \rho \lambda(I)$? Also, how do you get this result by $\lambda((D+b_n)\cap I)\ge \rho \lambda(I)$? Can we take the limit as $b_n\to t$? – Hermi Feb 06 '24 at 01:42
  • @DavideGiraudo Can you also please explain how do you use (1) to get $\lambda((D+b_n)\cap I)\ge \rho \lambda(I)$? – Hermi Feb 06 '24 at 01:43