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Prompted by this question I was looking for $A \subset (0,1)$ such that for any interval $(a,b)\subset (0,1), A \cap (a,b)$ and $A^c \cap (a,b)$ are both uncountable. One such $A$ is the set of all numbers that have a finite number of $1$'s in their base $3$ expansion. As no choice was used in the construction, it should be Lebesgue measurable, but I can't prove it. How is it proved?

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For each $n$, let $A_n$ be the subset of elements of $A$ that have at least a $1$ in the first $n$ digits of their ternary expansion, but no $1$s after the $n$th digit. Your set $A$ is equal to the union of the $A_n$s and the Cantor set. Each $A_n$ is a finite union of scaled translates of the Cantor set (take $3^{-n}C$, where $C$ is the Cantor set, to get all numbers that have ternary expansion with no $1$s and that have $0$ in the first $n$ positions; then you can "translate" by adding a suitable number that has a tail of $0$s and appropriate $1$s in the appropriate coordinates).

So each $A_n$ is a finite union of Lebesgue-measurable sets (scaled Cantor sets are Lebesgue-measurable, and translates of a Lebesgue-measurable set are Lebesgue-measurable), hence Lebesgue measurable. $A$ is a countable union of Lebesgue measurable sets, hence Lebesgue measurable.

Added. As Andres Caicedo points out in the comments below, the argument above shows that the set $A$ is in fact not merely Lebesgue measurable, but Borel.

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This would prove that $A$ is null, as we might expect. –  Ross Millikan Apr 4 '11 at 0:54
    
Indeed it would. –  Arturo Magidin Apr 4 '11 at 0:58
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@Ross, note that the description Arturo gives shows that, in fact, $A$ is Borel (this is much stronger than simply being Lebesgue measurable). Just about any set that you explicitly describe can be easily shown Borel by unraveling the given description: In this case, if only finitely many 1s are used, then your set is a countable union (depending on where the last 1 appears), etc. –  Andres Caicedo Apr 4 '11 at 1:39
    
@Andres: Thanks. Good point. –  Ross Millikan Apr 4 '11 at 2:02
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The number $x_n$ in the $n^\text{th}$ place after the radix point in the ternary expansion of $x$ is a measurable function of $x$. This is true because the floor function is Borel measurable, and $x_n=\left\lfloor 3\cdot\left(3^{n-1}x-\lfloor3^{n-1}x\rfloor\right)\right\rfloor$. The set $\{x:x_n\neq 1\}$ is measurable for each $n$, and so therefore is the set $$\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{x:x_k\neq 1\}.$$


Certain null sets of the second category give further examples of this. The complement of such a set can be contructed by taking a countable union of closed sets with empty interior whose complements have progressively smaller measure (e.g., using fat Cantor sets).

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Following on Andres' comment in Arturo's answer, you have also proved the set Borel, right? I only see countably many of the allowable operations here. –  Ross Millikan Apr 4 '11 at 2:05
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Yes, the functions involved are Borel measurable (floor is Borel, continuous functions are Borel, and compositions of Borel functions are Borel), and the inverse image of the open set $\mathbb{R}\setminus\{1\}$ under a Borel map is Borel. The Borel sets form a $\sigma$-algebra, so the countable union of countable intersections of Borel sets is Borel. (The set is even an $F_\sigma$, but that isn't as clear in my answer as in Arturo's.) The construction of null sets of the second category I mentioned also yields Borel sets ($G_\delta$s, even). –  Jonas Meyer Apr 4 '11 at 2:12
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