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Let $X$ be metric compact space and $\mu$ be a positive, finite, atomless, regular measure on $\sigma$-algebra of Borel subsets of $X$. Does there exist a set $C\subset X$ such that $C$ is uncountable and of $\mu$-measure zero?


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This is a duplicate of – Carl Mummert Feb 14 '12 at 22:25
I imagine you want to start with an uncountable metric space? – David Mitra Feb 14 '12 at 22:25
Yes. For countable $X$ the measure $\mu$ would have an atom. – Richard Feb 14 '12 at 22:30
Ah, silly me... – David Mitra Feb 14 '12 at 22:33
@CarlMummert. This is a bit different since the other does not assume any regularity conditions on the measure. – azarel Feb 14 '12 at 22:38
up vote 4 down vote accepted

An uncountable compact metric space contains a homeomorphic copy $C$ of the Cantor space, and $C$ is Borel in the original space. The Cantor space can be decomposed into uncountably many disjoint, uncountable, Borel subsets (because $C \cong C\times C$). If all of these had positive measure, some infinite number would all have measure larger than $1/n$ for some fixed $n$, which is impossible because the original measure was finite.

(This is similar to the answer by azarel but easier.)

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I believe azarel answered first, so really the credit should be his. – Carl Mummert Feb 15 '12 at 20:58

There is a subset $A$ of the Cantor set $C$ of cardinality $\aleph_1$ so that $\mu(A)=0$ for all non-atmoic Borel measure on the Cantor set. On the other hand, since $X$ is uncountable, there is a homeomorphic copy of the Cantor set inside $X$ hence the copy of $A$ inside $X$ must have measure zero.

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Could you please provide a reference for the first sentence? – Jonas Meyer Feb 15 '12 at 2:16
@JonasMeyer. Lemma 2.1.21 in Walks on ordinals and their characteristics by Stevo Todorcevic – azarel Feb 15 '12 at 6:08
Thanks very much! – Jonas Meyer Feb 15 '12 at 15:04

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