For any measurable set $A\subset\mathbb{R}$ and $r\in(0,\mu(A))$ we have $(\mu|_{2^A})^{-1}(r)\neq\emptyset$ Recently when I tried to prove a statement I needed to rely on the following fact that intuitively feels correct, but I wasn't able to prove it accurately. Here it is:

Consider a set $A\subset\mathbb{R}$ of positive measure $\mu(A)=a\in\mathbb{R}_+$ (standart Lebesgue) and a value $b\in(0,a)$. Can we always find such $B\subset A$ that $\mu(B)=b$?

I tried to use the concept of inner measure. Suppose for a moment that $A$ is bounded and so it lies in a segment $[x,y]\subset\mathbb{R}$. Then we can have $\overline{A}=[x,y]\setminus A$ and $\mu(\overline{A})=y-x-a$. For any $\varepsilon>0$ we can cover $\overline{A}$ with a union of pairwise disjoint intervals $J_\varepsilon$ such that $\mu(J_\varepsilon\setminus\overline{A})<\varepsilon$. This gives us $\overline{J_\varepsilon}\subset A$ and $a>\mu(\overline{J_\varepsilon})>a-\varepsilon$. So if we choose $\varepsilon=a-b$ we'll get $a>\mu(\overline{J_\varepsilon})>b$.
So the problem here is to choose $J_\varepsilon$ not for $\mu(J_\varepsilon\setminus\overline{A})<\varepsilon$ but for $\mu(J_\varepsilon\setminus\overline{A})=\varepsilon$. Can we do that?
 A: $\def\rr{\mathbb{R}}$Let $f(r) = μ( A \cap [-r,r] )$ for every $r \in \rr_+$. Then prove that $f$ is continuous on $\rr_+$. Note that $f(0) = 0$. Then prove that $f(r) \to μ(A)$ as $r \to \infty$ (say by MCT for sets). Then you're done because for any $b \in [0,μ(A))$ there is some $r \in \rr_+$ such that $f(r) = b$.
A: yeah this is what we call atomless measure for the case of Lebesgue measure the proof is simple you can take A as an interval $A=[c,d]$(because every borel set is a reunion of interval )
$$
B=[c,c+\frac{b}{a}A] 
$$
Let Now $A'$ a Lebesgue set then there exist a Borel set $A$ such that $\mu(A'-A)=0$ 
but  in every Borel $A$  set with positive measure we have a set of open disjoint interval $(I_n=]c_n,d_n[)$ such that $\cup_n( I_n) \subset A$ so we put $a_i=\mu(]c_i,d_d[)$, $a=\mu (\cup_n( I_n))=\sum_i a_i \leq \mu(A)$
since $\sum_i a_i$ congere and $b\leq a$ there exist $0\leq b_i\leq a_i$ such that $\sum_ib_i=b$we put then $V_i=]c_i,c_i+\frac{b_i}{a_i}[\subset I_i$, and $B=\cup_i V_i$
However for dirac measure this assumption isn't true, we speak about atomic measure.
we can prove similarly that every absolutely continuous measure is atomless. 
