Let $X$ be a locally compact Hausdorff space and $\mu$ a regular atom-free measure (i.e. $\mu(\{x\})=0$) for every $x\in X$. Show that if $B\subset X$ is a $\sigma$-finite Borel set and $a$ a real number between $0$ and $\mu(B)$, then there exists a Borel set $A\subset B$ with $\mu(A)=a$.
I thought of using Urysohn's Lemma: Let $X$ be a normal topological space, and let $E$ and $F$ be disjoint closed subsets of $X$. Then there is a continuous function $f$, such that $f(x)=0$ for $x\in E$ and $f(x)=1$ for $x\in F$.
I have different ideas but all of them revolve around that Lemma.
I plan to define
$$\mu(A)=\int_A f dx$$
Since $\mu$ is regular, I don't have to worry much about compact or open sets.
Then my $E$ can be a $x\in B$ (a point), moreover $\mu(x)=0$ and $\int_xfdx=0$, then $F$ is $\overline{B}-x$ (here $F$ is not closed, that might be a problem) and then by applying Urysohn's Lemma and using the continuity of $f$, I get the desired result (given a real number $a$, there exists a Borel set $A$, such that $A\subset B$ and $\mu(A)=a$).
There are some issues with my approach nonetheless, I think. For example, I don't have any guarantee that I can find an $f$ such that $\mu(A)=\int_Afdx$, or isssues with closed/open sets.
I also think there might be a more elegant or easier argument.
I'd appreciate any help!