A question about non-atomic measures on $S^1$ I have the following question : a detailed answer should be great !
Let $\mu$ be non-atomic probability measure on the unit circle $S^1$ . Prove that : there is a number $\delta > 0 $ such that when the Lebesgue measure of an interval I is < $\delta$ then $\mu(I) < 1/3 $.
The above is the actual question. However, it seems like that on the unit circle $S^1$, the non-atomic probability measures are similar to absolutely continuous measure with respect to the Lebesgue measure, is it correct ? If not, why not correct ? Is there an example of a non-atomic probability measure  on $S^1$ such that it is NOT absolutely continuous w.r.t. the Lebesgue measure ?
 A: Suppose not true, i.e. that there are arbitrarily small intervals with $\mu(I)\geq1/3$. Let $I_i$ be a sequence of such intervals with lengths converging to $0$. By compactness of $S^1$ we can choose a subsequence such that the centers of the intervals converge to some point $x_0\in S^1$. For any $n$, all the intervals in the subsequence, except for finitely many, are contained in $(x_0-1/n,x_0+1/n)$, hence $\mu((x_0-1/n,x_0+1/n))\geq 1/3$, hence $\mu(x_0)=\mu(\cap_n (x_0-1/n,x_0+1/n))\geq 1/3$ so $\mu$ is not non-atomic.
edit: if you want some more non-atomic singular measures: choose probabilities $p_0,\dots,p_9$ (so that $\sum p_i=1$; suppose non of $p_i$'s is $1$ if you want a non-atomic measure). The measure $\mu$ is as follows: say $\mu([0.326,0.327))=p_3p_2p_6$ (that is, informally, figure $0\leq i\leq9$ has probability $p_i$, and they are independent). Unless $p_i=0.1$ for all $i's$, the measure is singular (as follows from the strong law of large numbers). If all $p_i>0$ then $\mu(I)>0$ for all intervals. The graphs of resulting distribution functions are quite beautiful. (Cantor measure corresponds to base $3$, with $p_0=p_2=1/2$, $p_1=0$.) 
A: http://en.wikipedia.org/wiki/Cantor_distribution
It's on $[0,1]$ instead of $S^1$, so glue the endpoints together.
