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.


1 Answer 1


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 $\mu (E)=\frac 1 3$.

The conclusion is true for non-atomic measures and this is a standard theorem for such measures.

Ref.: https://en.wikipedia.org/wiki/Atom_(measure_theory)


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