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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?

Thanks.

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    Every finite measure on the Borel sets of a metric space is automatically regular.2012-02-14

2 Answers 2

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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.2012-02-15
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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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    @JonasMeyer. Lemma 2.1.21 in Walks on ordinals and their characteristics by Stevo Todorcevic2012-02-15