Radon measure on a metric space X 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 $0<a< \mu(X)$?
Or more generally does there always exist a set  $X_1$ such that $\mu(X_1) = a$
 A: First note that your measure has the following property, a consequence of the fact that Radon measures are locally finite:

($*$) If $x\in X$, then $\lim_{r\to 0^+}\mu(B(x,r)) = 0$.

With this, you can prove what you want with a Zorn's lemma argument (sorry, there is almost certainly a better way), which goes as follows. First, let $T$ be the collection of all open subsets $U$ of $X$ with measure $\mu(U)\leq a$. Partial order $T$ by inclusion. First we must show $T$ is nonempty. If $x$ lies in the support of $\mu$, then every ball around $x$ has positive measure (by definition of the support of $\mu$), and property ($*$) tells us that there are balls around $x$ of as small (positive) measure as we like. Thus $T$ is nonempty. If $\{U_\alpha\}$ is a totally ordered subset of $T$, then $\bigcup U_\alpha$ is an upper bound of $\{U_\alpha\}$ in $T$. Thus we can apply Zorn to get a maximal element $U$ of $T$. I claim that $\mu(U) = a$. If $\mu(U)<a$, then, since $\mu(U)<\mu(X)$, there is a point $x\notin U$ that lies in the support of $\mu$. By choosing $r>0$ small enough, $\mu(U) + \mu(B(x,r))<a$ by property ($*$). But then $U\cup B(x,r)\in T$ strictly contains $U$, contradicting the maximality of $U$.
This argument proves that you can find an open set $U$ with $\mu(U) = a$, and that in fact you can choose $U$ to be a maximal open set with this property.
