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.
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.
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.)
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.