This topic came out today during a discussion with a colleague. I realized that a counter-example to his claim could be constructed if there exits a subset $A \subset [0,1]$ such that $0 < \mu(A) < 1$ (where $\mu$ is Lebesgue measure), $A$ is dense in $[0,1]$ (with respect to the standard topology on $\mathbb{R}$), $\textit{and } A \textit{ does not contain an interval}$. However, I am not sure whether such a set exists. So the question is: do such sets exist? If so, is there an explicit construction, if not, how does one prove non-existence?
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asked Mar 20, 2015 at 0:20
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2 Answers
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By the Central Limit Theorem, something like this should have the property that $0<\mu([a,b]\cap A)<b-a$ for all $0\leq a<b\leq 1$. Let $0.a_1a_2\cdots$ be the binary expansion of $x\in[0,1]$. Let $A$ consist of all $x$ such that for all $n$ sufficiently large, $$ -1 < \frac{\frac n2-(a_1+a_2+\cdots+a_n)}{\sqrt{n}}<1. $$ The bounds $-1$ and $1$ could be replaced by any $\alpha<\beta$ in $\mathbb{R}$.
For another solution see https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval.
answered Mar 20, 2015 at 1:12
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If you allow $A$ to contain an interval, let $$A = (\mathbb Q\cap [0,1])\cup [0,1/2].$$ If not, consider $q_i$ the $i$th rational number in some enumeration of $\mathbb Q\cap [0,1]$, and let $$A=\bigcup^\infty_{i=1} (q_i-4^{-i}, q_i+4^{-i})\cap (\mathbb R\setminus\mathbb Q)$$
answered Mar 20, 2015 at 0:32
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$\begingroup$ My apologies, I intended to not consider trivial sets which contain intervals $\endgroup$ Commented Mar 20, 2015 at 0:40
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2$\begingroup$ The second example of Kjos-Hanssen does contain an interval. For an example not containing an interval, simply take $A=([0,1/2]-\mathbb{Q})\cup([1/2,0]\cap \mathbb{Q})$. However, I am guessing that the condition you actually want is that for all $0\leq a<b\leq 1$, we have $0<\mu([a,b]\cap A)<b-a$. $\endgroup$ Commented Mar 20, 2015 at 0:47
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3$\begingroup$ One may exhibit cantor subsets of $[0,1]$ with any measure between $0$ and $1$ inclusive, by an explicit construction where at stage $n$ one removes the central part of proportion $a_n$, for a suitable sequence. Take the union of such a set with $\mathbb{Q}\cap[0,1]$. $\endgroup$ Commented Mar 20, 2015 at 0:53
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1$\begingroup$ Constructing a measurable subset $A$ of $[0,1]$ such that $0 < \mu(A\cap [a,b]) < b-a$ for any $0\le a < b\le 1$ is an old and well-known elementary exercise in Measure Theory; see Rudin's Real and Complex Analysis. $\endgroup$ Commented Mar 20, 2015 at 2:32