Suppose that $\mu$ is a Radon measure on $X$ such that $\mu(\{x\})=0$ for all $x\in X$, and $A$ is a Borel set satisfying $0<\mu(A)<\infty$. I am trying to show that for any $\alpha$ such that $0<\alpha<\mu(A)$, there exists a Borel set $B\subset A$ such that $\mu(B)=\alpha$.
I tried using inner and outer regularity of $\mu$ but I probably did not do the right estimations so I'm stuck. I said that given $\varepsilon>0$, there exists a compact set $K$ such that $K\subset A$ and $\mu(A\setminus K)<\varepsilon$. For each $x\in K$, there exists an open set $U_{x}\subset X$ such that $x\in U_{x}$ and $\mu(U_{x})<\varepsilon$ (since $\mu(\{x\})=0$). Then $K$, being compact, is covered by finitely many of these open sets, say $U_{x_{1}},\ldots,U_{x_{m}}$. Since $K\subset A$, we also have $K\subset (\bigcup_{i=1}^{m}U_{x_{i}}\cap A)\subset A$. The middle set is a Borel set. But I don't see what I can say about its measure in terms of $\varepsilon$. Moreover, I haven't brought $\alpha$ into the picture.
