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.
 A: 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.
A: 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.
