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Is there a measure $\mu$ with support $S \subseteq [0,1]$ such that it satisfies:

(i) $S$ has Lebesgue measure zero but is dense on $[0,1]$ with respect to the standard metric;

(ii) $\mu(S)<\infty$; and

(iii) $\forall \epsilon>0$, $\forall a,b \in [0,1]$ such that $b-a\geq \epsilon >0$, $\mu((a,b])\geq k(\epsilon)>0$.

The Cantor distribution fails (i). It is unclear to me that taking the route of the answers here ensures (iii).

The main reason I am asking this is that this possibility is mentioned in Diaconis and Freedman, 1990, p. 1317, but I'm struggling to construct such a measure.

Edit: (iii) should hold $\forall \epsilon>0$. Apologies for the imprecision.

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  • $\begingroup$ In fact the support is closed by definition, so if it's dense then it's $[0,1]$. You meant to say "... a measure $\mu$ concentrated on $S$". $\endgroup$ – David C. Ullrich Jul 3 '20 at 1:00
  • $\begingroup$ $S=\mathbb{Q}\cap[0,1]$ is dense in $[0,1]$ but $S \ne [0,1]$. $\endgroup$ – user_newbie10 Jul 3 '20 at 1:07
  • $\begingroup$ And that set $S$ is also not closed. $\endgroup$ – David C. Ullrich Jul 3 '20 at 1:13
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let $S$ be the set of dyadic rational numbers in $[0,1]$ (i.e. all rational numbers of the form $a/2^k$, with $0\leq a\leq 2^k$ and $a$ odd), and let $\mu(\{a/2^k\})=1/2^{2k}$. Property (i) is immediate. Further,$\mu$ is finite because there are at most $2^k$ dyadic rationals in $[0,1]$ with denominator $2^k$, so the measure of all such rationals is at most $2^k/2^{2k}=1/2^k$, and summing over all possible $k$ gives (ii).

to see $(iii)$, suppose $b-a>\epsilon>0$. Choose $k(\epsilon)$ so that $2^{-k(\epsilon)}<\epsilon$. The crucial observation is that there must exist a dyadic rational number in $(a,b)$ with denominator $2^{k(\epsilon)}$ or less. Then $(iii)$ follows immediately from this observation, since $\mu([a,b])$ would then be at least $2^{-2*k(\epsilon)}$.

To see why the observation is true, observe that the set of all such dyadic rationals is equal to $0, 1/2^{k( \epsilon)}, 2/2^{k(\epsilon)}, \dots, 1$, i.e. they are evenly spaced with the gap between successive terms being $1/2^{k(\epsilon)}$. Since the distance between $a$ and $b$ is larger than this gap, the observation follows.

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  • $\begingroup$ Curiously, (almost) exactly the same as my example (I was typing while yours appeared...) $\endgroup$ – David C. Ullrich Jul 3 '20 at 1:11
  • $\begingroup$ Edit: my mistake. I misread. $\endgroup$ – user_newbie10 Jul 3 '20 at 1:16
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Say $S_n=\{1/n,2/n,\dots,(n-1)/n,1\}$, and let $\mu_n$ be the probability measure uniformly distributed on $S_n$. It's easy to verify that $$\mu=\sum2^{-n}\mu_n$$works.

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  • $\begingroup$ Edit: My mistake. Terribly sorry. $\endgroup$ – user_newbie10 Jul 3 '20 at 1:14
  • $\begingroup$ @user_newbie10 Did you read the other answer? He gives a detailed explanation of why (iii) holds for every $\epsilon>0$. $\endgroup$ – David C. Ullrich Jul 3 '20 at 1:16

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