Prove Property of Doubling Measure on $\mathbb{R}$ 
Problem. Let $\mu$ be a fixed finite measure on $\mathbb{R}$. $\mu$ is said to
be doubling if there exists a constant $C>0$ such that for any two
adjacent intervals $I=[x-h,x]$ and $J=[x,x+h]$,
$$\dfrac{\mu(I)}{C}\leq\mu(J)\leq C\mu(I)\tag{1}$$
Assuming that $\mu$ is doubling, show that there exist constants
$C>0,\delta>0$ such that for every interval $I$,
$$\mu(I)\leq B\left|I\right|^{\delta}\tag{2},$$
where $\left|\cdot\right|$ denotes the Lebesgue measure on
$\mathbb{R}$.

This problem comes from an old preliminary exam. A related question concerning this problem was asked before. In that question, a commenter said that it was a special case of the following lemma:

For a doubling measure $\mu$, there exist constants $C'>0,\delta>0$ such that for any balls $B'\subset B$,
$$\dfrac{\mu(B')}{\mu(B)}\leq C'\left(\dfrac{\left|B'\right|}{\left|B\right|}\right)^{\delta}\tag{3}$$

Although I don't quite have the details of the proof of preceding result worked out, I'm having difficulty seeing how to obtain inequality (2) as a special case of (3). If I knew that all balls of some fixed radius $r_{0}$ (say $r_{0}=1$) were uniformly bounded, then I see how to proceed. But I do not know how to establish such a bound. Any suggestions to get me on the right track would be appreciated.
 A: The problem is poorly written. There are no nontrivial finite doubling measures on $\mathbb{R}$. Indeed, if $\mu(\mathbb{R})<\infty$, then there is an interval $I$ with $\mu(I)>(1-\epsilon)\mu(\mathbb{R})$. An adjacent interval $J$ of equal length will have $\mu(J)<\epsilon\mu(\mathbb{R})$, and $(1-\epsilon)/\epsilon$ can be arbitrarily large. 
If we drop finite, then the statement is false because, e.g., $\mu(E)=\int_E x^2\,dx$ is a doubling measure, and $\mu([a,b])=(b^3-a^3)/3$ cannot be estimated in terms of $(b-a)$. 
What is true is that the decay estimate $\mu(I)\leq B\left|I\right|^{\delta}$ holds  for intervals $I$ contained in some fixed large  interval $I_0$, with $B$ dependent on $I_0$ but not on $I$. The proof goes like this: 
(0) Preparatory step: singletons have zero measure.  Indeed, if  $\mu(\{x\})=a>0$, then for every positive integer $k$,    $\mu([x+2^{-2k}, x+2^{1-2k}])\ge a/C$. These intervals are disjoint, and summing over $k$ leads to $\mu([x,x+1])=\infty$. This is something the definition doesn't allow: a doubling measure must be finite on compact sets.


*

*Divide $I_0$ into two equal intervals $I_1^{(1)}, I_1^{(2)}$. The doubling property implies $\mu(I_1^{(k)})\le A/(1+C^{-1})$.

*Continue the dyadic partition. At generation $m$ we have $2^m$ intervals of length $|I_0|/2^m$ and with measure at most $A/(1+C^{-1})^m$.

*Given $I\subset I_0$, pick $m$ such that $2^{-m-1}\le |I|/|I_0| \le 2^{-m}$. The interval $I$ is covered by two dyadic intervals of the $m$th generation. Hence, 
$$\mu(I)\le 2A/(1+C^{-1})^m \le B|I|^\delta$$
where $\delta $ has something to do with $\log (1+C^{-1})$ and $\log 2$. 



You may want to observe that the statement proved above is actually $(3)$, which is the correct way to express the decay property of doubling measures. 
