# Singular Measure with Dense Support

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.

• 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 $S$". – David C. Ullrich Jul 3 '20 at 1:00
• $S=\mathbb{Q}\cap[0,1]$ is dense in $[0,1]$ but $S \ne [0,1]$. – user_newbie10 Jul 3 '20 at 1:07
• And that set $S$ is also not closed. – David C. Ullrich Jul 3 '20 at 1:13

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.

• Curiously, (almost) exactly the same as my example (I was typing while yours appeared...) – David C. Ullrich Jul 3 '20 at 1:11
• Edit: my mistake. I misread. – user_newbie10 Jul 3 '20 at 1:16

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.

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