# Finding measurable subsets of any given value?

I'm not sure if the following is true, but I would hope it is with the regularity properties of $$\mu$$.

Let $$X$$ be a locally compact Hausdorff space with $$\mu$$ a nonzero Radon measure on $$X$$. Then given any $$\alpha \in (0,\mu(X))$$, there exists measurable $$E \subset X$$ with $$\mu(E)=\alpha$$.

My idea was to define something like $$\Lambda = \{A \subset X \: | \: \text{A measurable and \mu(A) \leq \alpha}\}$$ and try to use sequences in $$\Lambda$$ or even Zorn's lemma to get such a thing (Zorn's lemma seems harder to use as measurability only plays nice with countable unions).

However I'm kind of stumped on how to actually go about showing this. I know that for $$X = \mathbb{R}$$ and $$\mu=m$$, you can do this trivially.

Let $$X=\{1,2\}$$ with the usual metric. This is a compact Hausdorff space. Let $$\mu$$ be the uniform measure on $$X$$. Then $$0<\frac 1 3 <1=\mu (X)$$ but there is no set $$E$$ with $$\mu (E)=\frac 1 3$$.