$X$ is locally compact Hausdorff space, $\mu$ is Borel regular measure. How to prove $\mu$ is cover $[0,\mu(A)]$ I mean $X$ is locally compact Hausdorff space, $\mu$ is Borel regular measure, and $\mu(\{x\})=0$. For any subset $A$ with finite measure. How to prove for any $0<b<\mu(A)$, we always can find a Borel subset $B$ of $A$, such that $\mu(B)=b$?
 A: Maybe there are easier arguments, but it is a special case of the Lyapunov theorem:
If $\mu$ is non-atomic finite measure and $f \in L^1(X,\mathbb{R}^n)$ then the set $\left\{\int_{A} f \,d\mu : A \text{ measurable}\right\}$ is compact and convex in $\mathbb{R}^n$.
You can find a proof in Theorem 8.23, p.133 of these notes.
A: Edit:  I can only prove under the assumption that $X$ is first-countable. Thanks to @Displayname for pointing the flaw. :-)

We may assume that $X = A$, and $\mu$ is finite.
And since $\mu(X) = \sup_{K \subset X \text{: compact}} \mu(K)$,
it is enough to show that for any compact $K \subset X$,
$$
  \{\mu(D) \,|\, D \subset K \text{: measurable}\}
  =
  [0,\mu(K)].
$$
That is, we may assume that $X$ is compact (Hausdorff).
Take a descending sequence of neighborhoods of $x$,
$V_{x,1} \supset V_{x,2} \supset \dotsb$, such that
$$
  \{x\}
  =
  \bigcap V_{x,j}.
$$
Since $\mu$ is finite and $\mu(\{x\}) = 0$, we can conclude
that we have an open neighborhood $V$ of $x$ such that
$\mu(V)$ is as small as we wish.
Claim:
Let $B_n \subset X$ be such that
$\mu(B_n) \leq b$.
Then, there is a $B_{n+1} \supset B_n$ such that
$$
  b - \frac{1}{n}
  <
  \mu(B_{n+1})
  \leq
  b.
$$
In fact, cover $X$ with a finite number
(possibly 0) of open sets $V$ with $\mu(V) < \frac{1}{n}$.
From these, take $V_1, \dotsc, V_k$ such that
$0 \leq b - \frac{1}{n} < \mu(B_n \cup V_1 \cup \dotsb \cup V_k) \leq b$.
(Why can we do that?)
Make $B_{n+1} = B_n \cup V_1 \cup \dotsb \cup V_k$.
Now, just take $B = \bigcup B_n$.
In this case,
$$
  \mu(B)
  =
  \lim \mu(B_n)
  =
  b.
$$
A: Since $\mu$ is regular and $\mu(A)$ is finite, we can find a compact subet $B_{1}$ of $A$, s.t, $b< \mu (B_{1})\leq b+k$ for some positive $k$. Then, $\forall x \in B_{1}$, we chose an open neighborhood $O_{x}$ of $x$ with $\mu(O_{x})\leq \epsilon$. Thus $\left\{O_{x} \right\}$ forms an open cover of $B_{1}$. Thus, we have $B_{1}=\bigcup_{j=1}^{N} (B_{1}\bigcap O_{j} )$ and $b< \mu (B_{1})=\mu (\bigcup_{j=1}^{N} (B_{1}\bigcap O_{j} ))\leq b+k$. Let $z_{n}=\mu (\bigcup_{j=1}^{n} (B_{1}\bigcap O_{j} )), 1 \leq  n\leq N$. Note that $z_{n}$ is increasing and $z_{n}-z_{n-1} \leq \epsilon$ with $b< z_{N} \leq b+k$. Since $b,k$ are fixed, we can chose $\epsilon$ small enough so that there is a $1 \leq  m\leq N$, s.t, $b<z_{m} \leq b+\frac{k}{2}$. So $b< \mu (\bigcup_{j=1}^{m} (B_{1}\bigcap O_{j} ))\leq b+\frac{k}{2}$. Now we chose a compact set $B_{2}$ of $\bigcup_{j=1}^{m} (B_{1}\bigcap O_{j} )$, for which we have $b< \mu (B_{2})\leq b+\frac{k}{2}$. Repeat this we construct a decreasing seq $\left\{B_{i} \right\}$, s.t, $b< \mu (B_{i})\leq b+\frac{k}{i}$. Now, $B=\bigcap B_{i}$ is the set we need.
