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Assume that topological space $X$ contains subset $C$ homeomorfic to Cantor set (i.e there exists a set $C\subset X$ which is metrizable compact without isolated points and has closed-open basis). Assume that $\mu$ is a positive finite measure defined on Borel subsets of $X$ which is regular and atomless. Is it then $\mu(C)=0$ ?


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To make the question better, could you explain how you thought it up and what background material you are familiar with? – Carl Mummert Feb 14 '12 at 21:36
up vote 6 down vote accepted

Not necessarily: the closed unit interval contains Cantor sets of positive Lebesgue measure. One such example is known as the Smith-Volterra-Cantor set or fat Cantor set. It’s constructed in the same general way as the more familiar middle-thirds Cantor set, by removing open intervals from $[0,1]$, but the removed intervals are smaller. At stage $0$ you remove the middle $1/4$, leaving $[0,3/8]\cup[5/8,1]$. In general, at stage $n$ you have a union of $2^n$ closed intervals, and you remove the middle $2^{-2(n+1)}$ from each. An easy calculation shows that the total measure of the removed intervals is $1/2$, so the Cantor set that remains also has measure $1/2$.

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The Cantor set itself carries an atomless probability measure defined on all its Borel sets, so the answer is no.

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To expand on the answer of Carl Mummert, there is an increadibly useful theorem due to Kuratowski that says the following: There exists a measurable bijection with measurable inverse between any two Polish spaces of the same cardinality and this cardinality is either (at most) countable or the cardinality of the continuum. It follows that, as a measurable space, the Cantor set is really the same space a $\mathbb{R}$, $[0,1]$, or $[0,1]^\infty$. So the uniform distribution on $[0,1]$ corresponds to some atomless probability measure on the Cantor set. Now there is a result that says that every finite measure on the Borel sets of a metrizable metric space is automatically regular, so requiring regularity doesn't really change the problem.

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Thanks for this. I had in mind the "fair coin" measure on the Cantor space, but this achieves the same result. – Carl Mummert Feb 15 '12 at 13:27

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