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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 $\mu(B)=b$?

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If I understand correctly, $X$ is locally compact and Hausdorff, $\mu$ is a regular Borel measure that vanishes on singletons, and you want to prove that if $0<b<\mu(A)<\infty$, there is a Borel set $B\subseteq A$ such that $\mu(B)=b$. Is that correct? –  Brian M. Scott Oct 31 '12 at 17:29
    
yes, that is what I want to prove. –  user46007 Oct 31 '12 at 17:32
    
When $X=\mathbb{R}^n$ you can solve this by looking at the function $x\mapsto \mu(A\cap [-x,x))$. Maybe this can be used in a similar fashion in this setup. –  Stefan Hansen Oct 31 '12 at 18:42

2 Answers 2

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.

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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). In particular, $X$ is first countable. (why?)

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, $$ \mu(B) = \lim \mu(B_n) = b. $$


PS: If you don't know where the assumption that $\mu$ is finite was used... read it again. (hint: one place is to find $K$, but there is another one) :-)

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