Folland's Real Analysis 7.11

Suppose $\mu$ is a Radon measure on $X$ such that $\mu(\left\lbrace x \right\rbrace)=0$ for all $x \in X$, and $A \in \mathbb{B}_X$ satisfies $0 < \mu(A) < \infty$. Then for any $\alpha$ such that $0 < \alpha < \mu(A)$ there is a Borel set $B \subset A$ such that $\mu(B) = \alpha$

I'm not even sure how to start this problem. It seems like a nice result, but any hints on how to start?

Solution 1:

Here's the broad outline: fix $\alpha$ with $0<\alpha<\mu(A)$, and define $$ S_{\alpha}=\{B\subset A:B=U\cap A,\;U\; \mathrm{open},\;\mu(B)\leq \alpha\} $$

$S_{\alpha}$ is non-empty, and if $S_{\alpha}$ is partially ordered by inclusion, then one can show that it satisfies the hypotheses of Zorn's lemma. Therefore $S_{\alpha}$ has a maximal element $B$, which can be shown to satisfy $\mu(B)=\alpha$.