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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 ?

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up vote 4 down vote accepted

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 $p_0=p_2=1/2$, $p_1=0$.)

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What may have confused the OP is that this doesn't depend on $1/3$ being $1/3$; you could do it for any $\epsilon > 0$ and get $\delta$ so that whenever an interval $I$ satisfies $|I| < \delta$, $\mu(I) < \epsilon$. If you knew the same for all measurable $I$, this would be the statement that $\mu$ is absolutely continuous with respect to Lebesgue measure. I think there's nothing wrong: I don't see how to get it for all measurable sets from knowing it for intervals. (Who knew: it's not enough to check the condition for absolute continuity on a generating set for the algebra.) Am I right? –  anon Apr 28 '11 at 0:37
    
Actually, there are examples of NOT-absolutely continuous non-atomic measures on $S^1$, so I am not sure whether your statement is correct. Passing from intervals to ANY measurable sets would not be pissible i would guess. –  Mathmath May 4 '11 at 0:49
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http://en.wikipedia.org/wiki/Cantor_distribution

It's on $[0,1]$ instead of $S^1$, so glue the endpoints together.

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Can somebody answer the first question ? –  Mathmath Apr 27 '11 at 4:44
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