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I'm having difficulty solving an exercise (#8) in chapter 3 of Stein's Real Analysis.

Exercise 8 Suppose $A$ is a measurable set in $\mathbb R$ with $m(A) > 0$. Does there exist a sequence {$s_n$} such that complement of $\cup_{n=1}^{\infty}(A+s_n)$ in $\mathbb R$ has measure zero?

Hint: For every $\epsilon > 0$, find an interval $I_{\epsilon}$ of length $l_{\epsilon}$ such that $m(A \cap I_{\epsilon}) \ge (1-\epsilon)m(I_{\epsilon})$. Then, consider $\cup_{k=-\infty}^{\infty}(A+t_k)$ with $t_k = kl_{\epsilon}$, and vary $\epsilon$.

I've guessed the first hint $m(A \cap I_{\epsilon}) \ge (1-\epsilon)m(I_{\epsilon})$ has to do with a point of (Lebesgue) density. But, I can't associate $A \cap I_{\epsilon}$ to $A+t_k$, as well as $\cup_{n=1}^{\infty}(A+s_n)$ in the question..

More detailed explanation about the given hint, or any other comments regarding the exercise would be appreciated. Thank you.

jason 1
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  • I think density theorem is overkill here. By just definition of measure, we can find an open set that approximates $A$ with factor $(1 - \epsilon)$, this open set consists of open intervals, at least one of them will be as good approximation too. – mihaild May 08 '23 at 08:35
  • Your comment is related to finding the proper $I_{\epsilon}$ I guess. (BTW, I may also refer to the exercise 28 of chapter 1 in Stein's Real Anaysis, after letting $\alpha$ be $1-\epsilon$.) Then, how could I proceed after defining the $I_{\epsilon}$? $A \cap kI_{\epsilon}$ or $A \cup kI_{\epsilon}$ seem not related to the conclusion of the exercise. – jason 1 May 08 '23 at 11:48

1 Answers1

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Fix a Lebesgue measurable $A\subseteq\Bbb R$ with positive measure. Let $\mu$ denote the Lebesgue measure.

Hint $1$: can you try, following the author’s hint, to find some union of unions of translates of $A$ (ignore the bit about the sequence $s$ for now) where there are successively smaller ‘gaps’ for each new layer (“varying $\epsilon$”)? That way the union covers $\Bbb R$ very well and the complement has small measure. If the measure is arbitrarily small, then it is zero!

More hints and a hopefully correct, full solution are left in stages below. I found both this problem and the thought process behind my solution confusing and confused: if you catch any errors please let me know!

For every $n\in\Bbb N$ there exists an open interval $J_n$ of measure less than $1$ satisfying: $$\mu(A\cap J_n)\ge(1-2^{-n})\mu(J_n)$$

Define for every $n\in\Bbb N$ the measurable sets:

$$T_n:=\bigcup_{k\in\Bbb Z}(A+k\cdot\mu(J_n))$$

And define: $$T:=\bigcup_{n\in\Bbb N}T_n$$

Using a bijection $\Bbb N\times\Bbb Z\cong\Bbb N$, we can find a sequence $(s_n){n\in\Bbb N}$ such that we may express $T$ as $\bigcup{n\in\Bbb N}(A+s_n)$. We then can be happy if we can show that $\mu(\Bbb R\setminus T)=0$.

Hint $2$: $\mu(\Bbb R\setminus T)=0$ iff. the intersection of $\Bbb R\setminus T$ with every element of some countable partition of $\Bbb R$ has measure zero. Moreover (if $B$ denotes some measurable set of our partition), $\mu((\Bbb R\setminus T)\cap B)\le\mu((\Bbb R\setminus T_n)\cap B)$ for any $n\in\Bbb N$. What if we could show $\mu((\Bbb R\setminus T_n)\cap B)\to0$ as $n\to\infty$?

Fix $k\in\Bbb Z,n\in\Bbb N$. In what is to follow, the following fact is used implicitly: for some interval $(a,b)$ of length $\ell:=b-a=\mu((a,b))$, we have that $\{(a,b)+m\cdot\mu((a,b)):m\in\Bbb Z\}$ form a disjoint almost-partition of the real line. $\mu((a,b))$ is the unique number which allows us to do this: the authors of your exercise didn't mention specifically translating by multiples of $\ell$ for no reason.

There are (by disjointness) at most $1+\mu(J_n)^{-1}$ translates $J_n+m\cdot\mu(J_n),m\in\Bbb Z$ having nonempty intersection with $(k,k+1)$. So, where $F$ is some set of integers of cardinality at most $1+\mu(J_n)^{-1}$: $$\begin{align}\mu((\Bbb R\setminus T_n)\cap(k,k+1))&=\mu\left((k,k+1)\cap\left(\Bbb R\setminus\bigcup_{m\in\Bbb Z}(A +m\cdot\mu(J_n))\right)\right)\&\le\mu\left(\bigsqcup_{j\in F}(J_n+j\cdot\mu(J_n))\setminus\bigcup_{m\in\Bbb Z}(A+m\cdot\mu(J_n))\right)\&\le\sum_{j\in F}\mu((J_n+j\cdot\mu(J_n))\setminus(A+j\cdot\mu(J_n)))\&\le2^{-n}\mu(J_n)\cdot(1+\mu(J_n)^{-1})\&\le2^{-(n-1)}\end{align}$$

Hence: $\mu((\Bbb R\setminus T)\cap(k,k+1))\le\mu((\Bbb R\setminus T_n)\cap(k,k+1))\le2^{-(n-1)}$ for all $n,k$. It follows that $\mu((\Bbb R\setminus T)\cap(k,k+1))=0$ for all $k$ and then that $\mu(\Bbb R\setminus T)=0$, as desired.

FShrike
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