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I am considering the set related to the Cantor ternary set.Let A be the set of numbers in [0,1] whose ternary expansions have only finitely many 1's.Prove that $\lambda(A)=0$ where $\lambda$ is the Lebesgue measure.

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  • $\begingroup$ What's $\lambda$? The (Lebesgue) measure? $\endgroup$ – Gerry Myerson Oct 17 '12 at 1:30
  • $\begingroup$ @GerryMyerson Yes $\endgroup$ – 89085731 Oct 17 '12 at 1:33
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Let $C$ be the Cantor set. Let $B$ be the set of numbers in $[0, 1]$ whose ternary expansions consist of finitely many $1$s (the rest are $0$s). Since this set consists of rational numbers only, it's countable.

Let $x \in B$. We know that $\lambda(C) = 0$. We also know that the Lebesgue measure is translation invariant. Therefore, $\lambda(C + x) = 0$. Hence: $$ \lambda \left(\bigcup_{x \in B}(C + x)\right) \le \sum_{x \in B} \lambda(C + x) = 0 $$

Each number in your set $A$ can be written as the sum of a number in the Cantor set and another in $B$. Hence $A \subset \bigcup_{x \in B}(C + x)$ and $\lambda(A) = 0$.

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Hint: For positive integer $m$ and subset $S$ of $\{1,2,\ldots,m\}$, let $A_{m,S}$ be the set of numbers such that of the first $m$ ternary digits $d_1 d_2, \ldots, d_m$, only $d_j$ for $j \in S$ are 1. Find $\lambda(A_{m,S})$. Then take $m$ to infinity for fixed $S$. Then take the union over all $S$.

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  • $\begingroup$ I don't know what you mean by taking all S $\endgroup$ – 89085731 Oct 17 '12 at 1:54
  • $\begingroup$ Sorry,could you be more detailed,thanks. $\endgroup$ – 89085731 Oct 17 '12 at 2:25
  • $\begingroup$ Let $S$ be a subset of $\{1,2,\ldots,m\}$ with $k$ elements. Show that exactly $2^{m-k}$ of the $3^m$ possible digit-strings $d_1d_2\ldots d_m$ with $m$ ternary digits have $d_j = 1$ for $j \in S$ and $d_j \ne 1$ for $j \notin S$. So what is $\lambda(A_{m,S})$? Let $A_S = \bigcap_{m \ge \max(S)} A_{m,S} $. What is $\lambda(A_S)$? Now note that $A$ is the union of the sets $A_S$ for all finite subsets $S$ of the positive integers. $\endgroup$ – Robert Israel Oct 17 '12 at 6:06

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