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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.

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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.

  1. Divide $I_0$ into two equal intervals $I_1^{(1)}, I_1^{(2)}$. The doubling property implies $\mu(I_1^{(k)})\le A/(1+C^{-1})$.
  2. Continue the dyadic partition. At generation $m$ we have $2^m$ intervals of length $|I_0|/2^m$ and with measure at most $A/(1+C^{-1})^m$.
  3. 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.

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  • $\begingroup$ It seems that you are implicitly using the fact that $\mu(\left\{x\right\})=0$ for any $x\in\mathbb{R}$ in the dyadic partition. This isn't obvious too me, except as a consequence of what we're trying to prove. $\endgroup$ – Matt Rosenzweig May 31 '15 at 4:51
  • $\begingroup$ I didn't say those were closed intervals... make them half-open. $\endgroup$ – user147263 May 31 '15 at 4:57
  • $\begingroup$ Yes, but then you are no longer using the doubling property as stated. Perhaps, this is my fault. The definition of doubling measure used in the statement of (3) is that there exists a constant $C'>0$ such that $B(x,2r)\leq CB(x,r)$, where $B(x,r)$ denotes the open ball of radius $r$ centered at $x$. $\endgroup$ – Matt Rosenzweig May 31 '15 at 5:08
  • $\begingroup$ Okay. If $\mu$ is doubling and $\mu(\{x\})=a>0$ then for every positive integer $k$, the measure of $[x+2^{-2k}, x+2^{1-2k}]$ is at least $a/C$. 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. $\endgroup$ – user147263 May 31 '15 at 5:12

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