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If $X$ is a metric space and $0<\mu(X)$, where $\mu$ is a radon measure and $\mu(\{x\})=0$. Can we always split $X = X_1 \sqcup X_2$ into two disjoint sets where $\mu(X_1) = a$, for any $0?

Or more generally does there always exist a set $X_1$ such that $\mu(X_1) = a$

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    No, for this we need a criterion for the measure, called sometimes "atomless".2012-11-28
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    What about if we let $\mu({x}) = 0$ for every $x\in X$2012-11-28
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    In a metric space (where every singleton is a measurable set), $\mu(\{x\}) = 0$ for every $x$ is known as "atomless". Now in a really big metric space, you could have only measures $0$ and $\infty$, but I believe that is not allowed when you say "Radon measure".2012-11-28

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