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Let E be a measurable set. $x$,$y$ $\in E$ are $\delta$ equivalent if $x=2^{n}y$ for some integer $n$. The $\delta$-index of a point $x$ in $E$ is the number of elements in its $\delta$ equivalent class and is denoted by $\delta{_{E}} (x)$. Let $E(\delta; k) = \{x\in E:\delta_{E}(x)=k\}$.Then E is the disjoint union of the sets $E(\delta; k)$.

My question is:

If E is a Lebesgue measurable set, then each $E(\delta,k)$, $(k\geq 1)$ is also Lebesgue measurable. How?

Sunder Deep
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  • Isn't $E(\delta; k)$ empty for all $k$ other than $1$ and $\aleph_0$? – Jack M Jul 25 '19 at 11:24
  • No, we have some examples where $E(\delta , k)$ is nonempty for some $k>1$. Let $E=[-3 \pi , -\pi) \cup [\pi , 2 \pi)$. For this $E$, $E(\delta , 2)$ is non empty. – Sunder Deep Jul 25 '19 at 11:45

1 Answers1

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$E(\delta;k)=\cup_F \cap_{n \in F} (E\cap {2^{-n}} E)$ where the union is taken over all finite subsets of $\mathbb Z$ with cardinality $k$.

Kavi Rama Murthy
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