Measures with specified support and arbitrary Assouad dimensions

TL;DR

This paper demonstrates that for any compact set in the real line with positive upper Assouad dimension, one can construct measures supported on it with prescribed upper or lower Assouad dimensions, extending the understanding of measure-dimension relationships.

Contribution

It establishes the existence of measures with arbitrary prescribed upper or lower Assouad dimensions supported on sets with positive upper Assouad dimension.

Findings

Existence of measures with any upper Assouad dimension greater than the set’s dimension.

Existence of measures with any lower Assouad dimension less than the set’s lower dimension.

Extension of measure construction techniques to arbitrary Assouad dimensions.

Abstract

We show that if the upper Assouad dimension of the compact set $E\subseteq \mathbb{R}$ is positive, then given any $D>\dim_{A}E$ there is a measure with support $E$ and upper Assouad (or regularity) dimension $D$. Similarly, given any $0\leq d<\dim_{L}E,$ there is a measure on $E$ with lower Assouad dimension $d$.

Equations

$${\dim }_{A}E=\inf \left\{\alpha :(\exists C_1,C_2>0)(\forall 0<r<R<C_1)\underset{x\in E}{\sup }{N}_r(B(x,R)\cap E)\le C_2{\left(\frac{R}{r}\right)}^{\alpha }\right\}$$

$${\dim }_{L}E=\sup \left\{\alpha :(\exists C_1,C_2>0)(\forall 0<r<R<C_1)\underset{x\in E}{\inf }{N}_r(B(x,R)\cap E)\ge C_2{\left(\frac{R}{r}\right)}^{\alpha }\right\}$$

$${\dim }_A\mu =\inf \left\{\alpha :(\exists C_1,C_2>0)(\forall 0<r\le R\le C_1)\underset{x\in supp\mu }{\sup }\frac{\mu (B(x,R))}{\mu (B(x,r))}\le C_2{\left(\frac{R}{r}\right)}^{\alpha }\right\}$$

$${\dim }_L\mu =\sup \left\{\alpha :(\exists C_1,C_2>0)(\forall 0<r<R<C_1)\underset{x\in supp\mu }{\inf }\frac{\mu (B(x,R))}{\mu (B(x,r))}\ge C_2{\left(\frac{R}{r}\right)}^{\alpha }\right\}$$

$${\dim }_{L}E\le {\dim }_{H}E\le {\underline{\dim }}_{B}E\le {\overline{\dim }}_{B}E\le {\dim }_{A}E$$

$${\dim }_L\mu \le {\dim }_H\mu \le {\dim }_A\mu \text{,}$$

$$\mu (B(x,R))\le C\mu (B(x,R/2))$$

$$\mu (B(x,\tau R))\ge C\mu (B(x,R))$$

$${\dim }_A\text{supp}\mu \le {\dim }_A\mu ,$$

$${\dim }_L\mu \le {\dim }_L\text{supp}\mu .$$

$$\frac{\mu (I_{0^{(n)}})}{\mu (I_{0^{(n+k)}})}=p^{-k}\text{ while }\frac{|I_{0^{(n)}}|}{|I_{0^{(n+k)}}|}=3^k,$$

$$\bigcup _{w:|w|=k}{I}_w\bigcap E=E$$

$$2c{s}^k\le \ell (I_w)\le 2C{s}^k,$$

$$\text{KRS}(s,c,C)$$

$$I_w\cap E=\bigcup _{j=1}^{N_w}(I_{wj}\cap E).$$

$$I_w,I_{w{j}_1},I_{w{j}_1{j}_2,}...,I_{wu},$$

$$s^{-({\dim }_{L}E-\epsilon )}\le N_w\le s^{-({\dim }_{A}E+\epsilon )}.$$

$$N_r(B(x,R))\le A{\left(\frac{R}{r}\right)}^{{\dim }_{A}E+\epsilon /2}\text{ .}$$

$$N_r(B(x_w,2C{s}^k))\ge N_w.$$

$$N_r(B(x_w,2C{s}^k))$$

$$N_w\le s^{-({\dim }_{A}E+\epsilon )}.$$

$$N_r(B(x,R))\ge A^{\prime }{\left(\frac{R}{r}\right)}^{{\dim }_{L}E-\epsilon /2}$$

$$N_w\ge N_{2C{s}^{k+1}}(B(x_w,c{s}^k))\ge A^{\prime }{\left(\frac{c{s}^k}{2C{s}^{k+1}}\right)}^{{\dim }_{L}E-\epsilon /2}\ge s^{-({\dim }_{L}E-\epsilon )}$$

$$\mu (I_{wj})=\Delta (I_{wj})\mu (I_w).$$

$$P_j:=I_v^{(j)},I_{v{j}_1}^{(j)},I_{v{j}_1{j}_2}^{(j)},...,I_{v{v}^{\prime }}^{(j)},$$

$$I_w^{(j)},I_{w{i}_1}^{(j)},I_{w{i}_1{i}_2}^{(j)},...,I_{w{w}^{\prime }}^{(j)}$$

$$\frac{\mu (I_{\sigma })}{\mu (I_{\eta })}=\frac{\mu (I_{\tau })\Delta (I_{\tau i_1})\cdot \cdot \cdot \Delta (I_{\sigma })}{\mu (I_{\tau })\Delta (I_{\tau j_1})\cdot \cdot \cdot \Delta (I_{\eta })}=\frac{\Delta (I_{\tau i_1})}{\Delta (I_{\tau j_1})}.$$

$$a\mu (I_{\eta })\le \mu (I_{\sigma })\le \frac{1}{a}\mu (I_{\eta })$$

$$\frac{c}{2}{s}^{J+1}\le R<\frac{c}{2}{s}^J$$

$$2C{s}^K\le r<2C{s}^{K-1}\text{.}$$

Full text

The research of K. Hare is partially supported by NSERC 2016:03719. The research of F. Mendivil is partially supported by NSERC 2012:238549. K. Hare and L. Zuberman thank Acadia University for their hospitality when some of this research was done.

This paper is in final form and no version of it will be submitted for publication elsewhere.

1. Introduction

The upper and lower Assouad dimensions of measures are analogues of the Assouad dimensions of metric spaces and provide quantitative information about the extreme local behaviour of the measures. These dimensions were extensively studied by Fraser and Howroyd in [3] and Käenmäki et al. in [7] and [8], where they were called the upper and lower regularity dimensions. It is known that the upper Assouad dimension of a measure $\mu,$ $\dim_{A}\mu,$ is finite if and only if the measure is doubling and that the upper Assouad dimension of $\mu$ always dominates the upper Assouad dimension of the support of $\mu$. Similarly, a measure $\mu$ has positive lower Assouad dimension, $\dim_{L}\mu$, if and only if $\mu$ is uniformly perfect and the lower Assouad dimension of a doubling measure is always dominated by the lower Assouad dimension of its support.

In [11], Volberg and Konyagin proved that any compact, doubling, metric space $E$ supports a measure whose upper Assouad dimension is arbitrarily close to that of the space $E$. This was extended to complete doubling spaces in [10]. The analogue for lower Assouad dimension was proved by Byland and Gudayol in [1] and Käenmäki and Lehrbäck in [7]. Volberg and Konyagin also showed that there is a complete, doubling, metric space $E$ with the property that every measure supported on $E$ has upper Assouad dimension strictly greater than that of $E$.

Fraser and Howroyd in [3] proved that if $E$ is the closure of the discrete set $\{q^{n}:n=1,2,...\}\subseteq\mathbb{R}$ where $q<1,$ then the measure $\mu=\sum p^{n}\delta_{q^{n}}$, where $p<1,$ has upper Assouad dimension $\log p/\log q$. Thus there are measures $\mu$ with support $E$ and $\dim_{A}\mu$ equal to any given value greater than $\dim_{A}E$ (which is zero, in this case). It is natural to ask if this property holds more generally. The purpose of this note is to show that it does hold for any compact subset of $\mathbb{R}$ with positive upper Assouad dimension. More specifically, if $\dim_{A}E>0,$ then given any $D>\dim_{A}E$ (including $D=\infty)$ there is a measure $\mu$ with support $E$ and $\dim_{A}\mu=D$. A key ingredient in our argument is the generalised nested cubes construction of Käenmäki et al in [9].

Similarly, given any $0\leq d<\dim_{L}E$ there is a measure $\nu$ with support $E$ and $\dim_{L}\nu=d$. If $d>0,$ we can achieve both specified dimensions with a single measure. We also show that these results need not be true if we only assume $E$ is a compact, infinite subset of $\mathbb{R}$.

2. Definitions

Given $E$, a compact subset of a metric space, we write $N_{r}(E)$ for the least number of sets of diameter at most $r$ that are required to cover $E$.

Definition 2.1**.**

The upper and lower Assouad dimensions of $E,$ denoted $\dim_{A}E$ and $\dim_{L}E$ respectively, are given by

[TABLE]

and

[TABLE]

One can similarly define the Assouad dimensions of any finite, positive Borel measure defined on the metric space.

Definition 2.2**.**

The upper and lower Assouad dimensions of measure $\mu,$ denoted $\dim_{A}\mu$ and $\dim_{L}\mu$ respectively, are given by

[TABLE]

and

[TABLE]

The upper and lower Assouad dimensions of a measure have also been referred to as the upper and lower regularity dimensions.

The Assouad dimensions quantify the extreme local behaviour of the set or measure. Indeed, the following relationships are known for all compact metric spaces $E$ and measures $\mu$:

[TABLE]

and

[TABLE]

where $\dim_{H}E,$ \dim$${}_{B}E and $\overline{\dim}_{B}E$ denote the Hausdorff, lower and upper box dimensions of $E$ respectively and $\dim_{H}\mu$ is the Hausdorff dimension of $\mu$. The upper Assouad dimension of a subset of $\mathbb{R}^{n}$ is at most $n$. More generally, the upper Assouad dimension of $E$ is finite if and only if $E$ is a doubling metric space. Similarly, the upper Assouad dimension of measure $\mu$ is finite if and only if the measure is doubling, meaning there is a constant $C$ such that

[TABLE]

whenever $x$ belongs to the support of $\mu$ and $R>0$. Unlike the case for sets, the upper Assouad dimension of a measure on $\mathbb{R}^{n}$ can be infinite. Dually, the lower Assouad dimension of a set (or measure) is positive if and only if the set (resp., measure) is uniformly perfect, which for measures means there are constants $C>1,\tau>1$ such that

[TABLE]

for all $x\in$supp$\mu$ and $0<R\leq diam($supp$\mu)/(2\tau)$. It is not difficult to see that for any measure $\mu$,

[TABLE]

and if $\mu$ is a doubling measure, then

[TABLE]

For more background on these dimensions and proofs of these basic facts we refer the reader to [2]-[8].

As the upper Assouad dimension of a union of two sets is the maximum of their upper Assouad dimensions, one might expect that the upper Assouad dimension of the sum of two measures is their maximum upper Assouad dimension. This is not true as the following example illustrates.

Example 2.3**.**

Measures $\mu,\nu$ with $\dim_{A}(\mu+\nu)<\max(\dim_{A}\mu,\dim_{A}\nu)$: The two measures will be supported on the classical middle third Cantor set. Pick $0<p<1/2$. We will define the measures by specifying them on the Cantor intervals of the standard Cantor set construction. We label these intervals as $I_{w}$ where $w$ is a finite word on the letters $0,1$, where $I_{w0}$ is the left child of interval $I_{w}$ and $I_{w1}$ is the right child.

Let $w$ be such a word. If $w=0^{(n)}$, then define $\mu(I_{w})=p^{n}$. If $w=0^{(n)}1\sigma$ where $n\geq 0$ and $\sigma$ is a word of length $\left|\sigma\right|=k$ on the letters $0,1,$ then put $\mu(I_{w})=p^{n}(1-p)2^{-k}$. The measure $\nu$ will be defined similarly, but with the roles of $0,1$ interchanged. It is easy to see that $\mu,\nu$ are probability measures.

As $p<1/2,$ if $w=0\sigma,$ then $\mu(I_{w})\leq(1-p)2^{-|\sigma|}$, while $\nu(I_{w})=(1-p)2^{-|\sigma|}$. Consequently, $(\mu+\nu)(I_{w})=O(2^{-|w|})$. Similarly, if $w=1\sigma,$ then $\nu(I_{w})\leq(1-p)2^{-|\sigma|}$, while $\mu(I_{w})=(1-p)2^{-|\sigma|}$, so again, $(\mu+\nu)(I_{w})=O(2^{-|w|})$. It follows easily from this that $\dim_{A}(\mu+\nu)=\log 2/\log 3$.

But,

[TABLE]

thus $\dim_{A}\mu\geq|\log p|/\log 3>\log 2/\log 3$. (In fact, we have equality.) A similar statement holds for $\nu$.

3. Upper Assouad Dimension

We begin by reviewing the construction introduced by Käenmäki, Rajala and Suomala in [9], which we will call the KRS construction. This construction will be used in the remainder of the paper to construct measures which have the required upper and/or lower Assouad dimensions. We remark that we have made slight changes to some of their notation and we will also introduce further terminology that will be helpful for our application.

KRS Construction: By the KRS construction of a set $E\subseteq[0,1]$, we will mean the nested, dyadic-like ‘cubes’ construction described in [9]. This construction was shown to hold for any metric space which is doubling. In the case of subsets of $\mathbb{R}$, the generalized cubes of their construction can be taken to be intervals.

To be more specific, Käenmäki et al have shown that for any $0<s<1/3$ and for each $k\in\mathbb{N}$ there are disjoint intervals, $I_{w},$ labelled by words $w$ of length $\left|w\right|=$ $k,$ with

[TABLE]

and which have the properties listed below:

Properties. (i) There are constants $c,C>0$ such that if $\left|w\right|=$ $k$ (we call $k$ the **level **of $I_{w}$), then

[TABLE]

where $\ell(I_{w})$ denotes the length of the interval. By choosing $s$ sufficiently small, we can assume $c=3/8$ and $C=9/8$, so $C=3c$. We will write

[TABLE]

if we want to emphasize the particular choice of parameters $s,c,C$.

(ii) Each $I_{w}$ contains a distinguished point, $x_{w},$ such that $d(x_{w},I_{w}^{c})\geq cs^{k}$ if $\left|w\right|=k.$

(iii) Each $I_{w}$ has at least one child. Moreover, $I_{w}$ decomposes into its children, $I_{wj},$ $j=1,...,N_{w},$ where $N_{w}$ $\leq N=N(s,E)$. By this, we mean that $I_{wj}\subseteq I_{w}$, $I_{wi}\bigcap I_{wj}=\phi$ if $i\neq j$ and

[TABLE]

We will call $I_{wj}$ a boundary child (or boundary interval) of the interval $I_{w}$ of level $k$ if $d(I_{wj},I_{w}^{c})<cs^{k+1}$. As $\ell(I_{wj})\geq 2cs^{k+1},$ there can be at most two boundary children for each $I_{w},$ one on each side of $I_{w}$. Any child of $I_{w}$ that is not a boundary child will be called an interior child (or interior interval). Two boundary intervals that do not have the same parent (that is, they are not siblings) will be called cousins. We will write $I_{w^{-}}$ for the parent of $I_{w}$.

We will say that two level $k$ intervals, $I_{w}$ and $I_{v},$ are adjacent if the distance between them is at most $cs^{k}$. This ensures there is no interval of level $k$ between them and consequently the open interval in-between contains no points of $E$. If adjacent intervals are not siblings, they must both be boundary intervals of (different) parents that are adjacent to each other. Any interval can be adjacent to at most two other intervals (one on each side). At most one can fail to be a sibling.

Notice that the distinguished point, $x_{w},$ does not belong to a boundary child of $I_{w}$. Consequently every $I_{w}$ has a child that is an interior interval. The interior child of $I_{w}$ containing $x_{w}$ will be called the distinguished **interior child **/ interval.

We will say $I_{w}$ splits if it has at least two children. Notice that if $I_{w}$ has a boundary child it must split.

By a path of length $L$, we will mean a sequence of nested intervals,

[TABLE]

where $\left|u\right|=L$. Here each interval in the sequence is a child of the previous interval. We will call it a** boundary path** if all the descendents of $I_{w}$ are boundary intervals. A boundary path is a splitting path (meaning each interval, other than possibly the last, splits) because $I_{w}$ and each $I_{wj_{1}...j_{k}}$ (except possibly $I_{wu})$ has a boundary child and hence at least two children.

Lemma 3.1**.**

Let $E\subseteq[0,1]$ and $\varepsilon>0$. There is a choice of $s(\varepsilon)<1/3$ and integer $k(\varepsilon)$ such that the following holds: Take any KRS$(s,c,3c)$ construction of $E$ with $s\leq s(\varepsilon)$. If $N_{w}$ is the number of children of interval $I_{w}$ and $\left|w\right|\geq k(\varepsilon),$ then

[TABLE]

Proof.

From the definition of the upper Assouad dimension we can find constants $A$ and $\rho>0,$ depending on $\varepsilon,$ such that for all $r\leq R\leq\rho$ and $x\in E,$ we have

[TABLE]

Choose $s(\varepsilon)$ so small that $A(3s^{-1})^{\dim_{A}E+\varepsilon/2}\leq s^{-(dim_{A}E+\varepsilon)}$ for all $s\leq s(\varepsilon)$. Choose $k(\varepsilon)$ such that $2Cs(\varepsilon)^{k}<\rho$ for all $k\geq k(\varepsilon)$ where $C=3c$.

Take any interval $I_{w}$ from a KRS$(s,c,3c)$ construction of $E$ where $\left|w\right|\geq k\geq k(\varepsilon)$. Let $x_{w}$ be the distinguished point of $I_{w}.$ Consider $B(x_{w},2Cs^{k})\supseteq I_{w}$. As the children of $I_{w}$ have length at least $2cs^{k+1},$ if $r=2cs^{k+1}$ we clearly have

[TABLE]

But from (3.1), we also have

[TABLE]

Consequently,

[TABLE]

Similarly, choose positive constants $A^{\prime},\rho^{\prime},$ depending on $\varepsilon,$ such that

[TABLE]

for all $r\leq R\leq\rho^{\prime}$. As $B(x_{w},cs^{k})\subseteq I_{w},$ we have

[TABLE]

for large enough $k,$ provided we choose $s\leq s(\varepsilon)$ and $s(\varepsilon)>0$ sufficiently small. ∎

The main goal of this section is to prove the following theorem.

Theorem 3.2**.**

Let $E\subseteq[0,1]$ be compact. If $\dim_{A}E>0,$ then given any $D>\dim_{A}E,$ there is a probability measure $\mu$ with support $E$ and $\dim_{A}\mu=D$.

Our basic strategy is to use the KRS construction to define a probability measure $\mu$, with support equal to $E,$ by specifying $\mu$ on each $I_{w}$ in a consistent fashion. Indeed, we will assign positive weights, $\Delta(I_{wj}),$ so that $\sum_{j=1}^{N_{w}}\Delta(I_{wj})=1$ and define

[TABLE]

The task in proving the theorem is to define these weights appropriately.

There are two cases to consider for the proof of the theorem. The easier case is when there are arbitrarily long boundary paths, in which case there are also arbitrarily long boundary paths beginning at arbitrarily deep levels.

Lemma 3.3**.**

Assume $\dim_{A}E<D<\infty$ and let $\varepsilon=(D-\dim_{A}E)/2$. If there is a KRS$(s,c,3c)$ construction of $E,$ with the choice of $s\leq s(\varepsilon)$ from Lemma 3.1, which admits arbitrarily long boundary paths, then there is a probability measure $\mu$ with support $E$ and $\dim_{A}\mu=D$.

Proof.

For each $j>k(\varepsilon)$, as defined in Lemma 3.1, choose a boundary path beginning at level $n_{j}$ and having length $l_{j}\geq j,$ where $n_{j+1}>n_{j}+l_{j}$. Assume this is the path

[TABLE]

where $\left|v\right|=n_{j}$ and $\left|v^{\prime}\right|=l_{j}$. Consider, also, the adjacent boundary path (should it exist)

[TABLE]

where each $I_{vj_{1}j_{2}...j_{k}}^{(j)}$ is adjacent to $I_{wi_{1}i_{2}...i_{k}}^{(j)}$ and $\left|w^{\prime}\right|\leq\left|v^{\prime}\right|$ (the inequality being strict if for some $m<\left|v^{\prime}\right|,$ the interval $I_{wi_{1}...i_{m}}^{(j)}$ has no boundary child adjacent to $I_{vj_{1}...j_{m+1}}^{(j)}$). Notice that while $I_{v}^{(j)}$ and $I_{w}^{(j)}$ may be siblings, all other pairings are cousins.

We define the measure $\mu$ as follows: The weights associated with each of the intervals $I_{vj_{1}}^{(j)},I_{vj_{1}j_{2},}^{(j)}$, $...,I_{vv^{\prime}}^{(j)}$ and $I_{wi_{1}}^{(j)},I_{wi_{1}i_{2},}^{(j)}$, $...,I_{ww^{\prime}}^{(j)}$ will be the value $a=s^{D}$. Each sibling of these intervals, other than the distinguished sibling, will be assigned weight $p=s^{D-\varepsilon}>a$. Each of the distinguished siblings (which are not any of the intervals in the paths) will be assigned the weight which ensures that summing over all children of a given parent is $1$. This weight will be at least $p$ since, as shown in Lemma 3.1, $p=s^{D-\varepsilon}\leq 1/(\max N_{w})$. The paths arising from different $j$ come from different levels, hence this specification is well defined.

This procedure assigns weights to (precisely) all the children of each of $I_{v}^{(j)},I_{vj_{1}}^{(j)}$, $I_{vj_{1}j_{2},}^{(j)}$, $...,I_{(vv^{\prime})^{-}}^{(j)}$ and $I_{w}^{(j)},I_{wi_{1}}^{(j)}$, $I_{wi_{1}i_{2},}^{(j)}$, $...,I_{(ww^{\prime})^{-}}^{(j)}$. For all remaining intervals, $I_{\sigma},$ of level at least $k(\varepsilon)$, we assign weight $p$ to all children of the parent $I_{\sigma}$, except the distinguished child which is assigned the weight $1-p(N_{\sigma}-1)$, so the sum of the weights of all the children of $I_{\sigma}$ is $1$. We also assign equal weights (adding to $1$) to the intervals of level $k(\varepsilon)$. This defines the probability measure $\mu$.

One important observation is that $a$ is the minimum weight assigned to any interval.

Another important observation is that any two adjacent intervals that are not siblings are either both assigned weight $a$ or both assigned weight $p$, and in either case, have equal weights. To see why this is relevant, suppose that $I_{\sigma}$ and $I_{\eta}$ are adjacent cousins. Let $I_{\tau}$ be their nearest common ancestor and assume that $I_{\tau},I_{\tau i_{1}},...,I_{\tau i_{1}...i_{l}}=I_{\sigma}$ and $I_{\tau},I_{\tau j_{1}},...,I_{\tau j_{1}...j_{l}}=I_{\eta}$ are paths. Then $I_{\tau i_{1}}\neq I_{\tau j_{1}}$. Furthermore, while $I_{\tau i_{1}}$ and $I_{\tau j_{1}}$ may be siblings, all other pairs $I_{\tau i_{1}...i_{m}},$ $I_{\tau j_{1}...j_{m}}$ are adjacent cousins as they do not have the same parent. Consequently, their weights are the same. Hence

[TABLE]

As $\Delta(I_{\tau i_{1}}),\Delta(I_{\tau j_{1}})\in[a,1],$

[TABLE]

for all adjacent cousins. A similar statement holds for the measures of any two siblings.

We are now ready to compute the upper Assouad dimension of $\mu$. First, consider any $r\leq R$ and $x\in E$. Choose integers $J,K$ such that

[TABLE]

and

[TABLE]

Since $2Cs^{K}\leq(c/2)s^{J}$ and $C>c,$ we have $K>J$. Choose $v,w=v\tau$ such that $\left|v\right|=J,$ $\left|w\right|=K$ and $x\in I_{w}\subseteq I_{v}$. As $\ell(I_{w})\leq 2Cs^{K},$ we have $I_{w}\subseteq B(x,r)$, and so $\mu(B(x,r))\geq\mu(I_{w})$. Since the minimum value of any weight is $a$, it follows that $\mu(I_{w})\geq a^{K-J}\mu(I_{v})$.

Consider $B(x,R)$. If this set was to intersect three intervals of level $J$, then it would contain one of them. But $2R<cs^{J}<$ length of any interval of level $J$, hence this is impossible. Thus it can intersect at most two such intervals, one of which must be $I_{v}$.

If $B(x,R)$ only intersects one interval of level $J$, then $B(x,R)\cap E\subseteq I_{v}$. Hence $\mu(B(x,R))\leq\mu(I_{v})$ and

[TABLE]

for a suitable constant $A_{1}$.

If $B(x,R)$ intersects two intervals, $I_{v}$ and $I_{\sigma}$, then the distance between $I_{v}$ and $I_{\sigma}$ can be at most the diameter of the ball, $2R<cs^{J},$ so these are adjacent intervals. As noted in (3.2), this means $\mu(I_{v})+\mu(I_{\sigma})\leq(1+1/a)\mu(I_{v})$. Thus we also have

[TABLE]

It follows from these computations that $\dim_{A}\mu$ $\leq D$.

To see the equality, consider one of the special boundary paths,

[TABLE]

where $I_{v}^{(j)}$ is of level $n_{j}$ and $I_{vv^{\prime}}^{(j)}=I_{w}^{(j)}$ is of level $n_{j}+l_{j}$. Then $\mu(I_{w}^{(j)})=\mu(I_{v}^{(j)})a^{l_{j}}$. Let $x_{j}$ be the distinguished point of $I_{w}^{(j)}$ and choose $r_{j}=cs^{n_{j}+l_{j}}$ so that $B(x_{j},r_{j})\subseteq I_{w}^{(j)}$. Take $R_{j}=2Cs^{n_{j}}$ so that $B(x_{j},R_{j})\supseteq I_{v}^{(j)}$. Then

[TABLE]

As $R_{j}\rightarrow 0$ and $R_{j}/r_{j}\rightarrow\infty$ as $j\rightarrow\infty,$ it follows that $\dim_{A}\mu\geq D,$ proving the equality. ∎

The more difficult case is when there are not arbitrarily long boundary paths. For this, it is convenient to introduce further terminology.

Notation 3.4**.**

Assume we have fixed a KRS construction. For each $n\in\mathbb{N}$, let

[TABLE]

Of course, $\zeta_{n}\in[0,1]$ and decreases as $n$ increases. We will set

[TABLE]

*and call $\zeta$ the **proportionality constant *of the KRS construction.

Lemma 3.5**.**

For any set $E$ of positive upper Assouad dimension and any KRS construction with parameter $s<1/3,$ the proportionality constant $\zeta$ is strictly positive.

Remark 3.6**.**

Under the much stronger hypothesis, $\dim_{L}E>0,$ we actually have $\zeta=1$ for small enough $s$. This can be seen in the proof of Theorem 4.1.

Proof.

Let $0<t<\dim_{A}E$. We can find $x_{n}\in E,$ $R_{n}$ and $r_{n}<R_{n}$ with $R_{n}\rightarrow 0$ and $R_{n}/r_{n}\rightarrow\infty,$ such that

[TABLE]

Choose integer $k_{n}$ such that

[TABLE]

Without loss of generality we can assume that $R_{n}<s^{n}$ and $k_{n}\geq n$.

Temporarily fix $n$ and choose the largest integer $N$ such that $cs^{N}/2>R_{n}$. At most two level $N$ intervals from the KRS$(s,c,C)$ construction can intersect $B(x_{n},R_{n})$ and hence one of them, call it $I_{w},$ must satisfy

[TABLE]

Consider the optimal covering of $I_{w}$ by intervals of radius $r_{n}$. By ordering these intervals left-to-right in $\mathbb{R}$ and keeping every third interval, we see there are at least

[TABLE]

disjoint intervals $\{J_{l}\}$ of radius $r_{n}$ and separated by at least $r_{n}$, each of which contains at least one point of $E\cap I_{w}$.

Let $M$ be the smallest integer such that $2Cs^{M}<r_{n}$ and for each $J_{l}$ choose a subinterval $I_{wu_{l}}$ of level $M,$ which intersects $J_{l}$ and no other $J_{m}$. The definitions of $N,M$ ensure that

[TABLE]

Combined with (3.3), this implies

[TABLE]

hence for large enough $n,$ $n\leq k_{n}\leq M-N+1\leq 2k_{n}$.

Now consider all the paths beginning at level $N$ with interval $I_{w}$ and ending at level $M$, hence of length $M-N$. As each KRS interval can have at most $(C/c)s^{-1}$ children, an easy induction argument shows that if every such path had at most $\varepsilon(M-N)$ splittings, then there would be a total of at most $\left((C/c)s^{-1}\right)^{\varepsilon(M-N)}$ of these paths. But as the intervals $I_{wu_{l}}$ are all distinct, there are at least $ns^{-(k_{n}-1)t}/6$ of these paths. Since $M-N\sim k_{n},$ we deduce that $\varepsilon\geq\gamma t$ for a suitable constant $\gamma>0.$ It follows that $\zeta>0$. ∎

We are now ready to complete the proof of Theorem 3.2.

Proof.

[of Theorem 3.2] First, suppose $D<\infty$, say $D=\dim_{A}E$ $+2\varepsilon$ for $\varepsilon>0$. Choose $s=s(\varepsilon)$ as given in Lemma 3.1 and consider a KRS$(s,c,3c)$ construction of $E$. If this construction admits arbitrarily long boundary paths, then Lemma 3.3 can be applied to produce a measure supported on $E$ with upper Assouad dimension $D$.

Thus we can assume there exists some $M>0$ such that there are no boundary paths of length exceeding $M$. We will take $a>0$ such that $a^{\zeta}=s^{D},$ where $\zeta$ is the proportionality constant of this KRS construction, guaranteed to be positive by Lemma 3.5. Notice that Lemma 3.1 gives

[TABLE]

We begin by defining the measure. Choose $n_{1}>M$ such that $\left|\zeta_{n_{1}}-\zeta\right|<\zeta/2$. Choose a path $P_{1}$ beginning at level at least $n_{1}$ and with length $L_{1}\geq$ $n_{1},$ which splits on at least $\,L_{1}\zeta/2$ levels. The choice of $\zeta_{n_{1}}$ also ensures no path beginning at level at least $n_{1}$ and length $L_{1}\geq$ $n_{1}$ splits on more than $L_{1}3\zeta/2$ levels.

Assuming $n_{1}<n_{2}<\cdot\cdot\cdot<n_{j-1}$ and paths $P_{1},..,P_{j-1}$ have been identified (with the properties as described in the induction step outlined below), choose $n_{j}$ much greater than the level of the last interval in path $P_{j-1},$ such that $\left|\zeta_{n_{j}}-\zeta\right|<\zeta 2^{-j}$. Choose a path $P_{j}$ beginning at level at least $n_{j}$, length $L_{j}\geq$ $n_{j},$ and which splits on at least $L_{j}\zeta(1-2^{-j})$ levels. Furthermore, notice no such path can split on more than the proportion $\zeta(1+2^{-j})$ levels. That is the induction step.

Next, consider these special paths $P_{j}:I_{v},I_{vi_{1}},...,I_{vi_{1},..,i_{L_{j}}}$. If the interval $I_{vi_{1}...i_{l}}$ is an only child, we assign weight $1$ to it. Otherwise, we assign it the weight $a$ and distribute the remaining $1-a$ equally among its siblings. Notice these remaining weights will have value at least $a$.

For all remaining intervals in the KRS construction (those that are not siblings of intervals in one of the special paths), we simply assign equal weight to all children of a given parent so that the sum of the weights of the sibling group is $1$. This completes the definition of the measure $\mu$. It is clear from the construction that the support of $\mu$ is the set $E$. It is important to observe that$\ a$ is the minimum weight.

To show that $\dim_{A}\mu=D,$ we will actually verify that for every $\varepsilon>0$ there are constants $C^{\prime},C^{\prime\prime}$ and $\rho$ (depending on $\varepsilon)$ such that for all $r\leq R\leq\rho$ and $x\in E$, we have

[TABLE]

and that for some sequence $r_{n}\leq R_{n}\leq\rho$ and $x_{n}\in E$ with $R_{n}/r_{n}\rightarrow\infty,$ we have

[TABLE]

Fix $\varepsilon>0$ and choose $J$ such that $2^{-J}<\varepsilon$. Put $\rho=(c/2)s^{n_{J}}$. Take any $x\in E$, $r\leq R\leq\rho$ and assume

[TABLE]

for suitable $j\geq n_{J}$ and $k>j.$

As in the proof of Lemma 3.3, there are intervals $I_{w}\subseteq I_{v}$ such that $x\in I_{w},$ $\left|w\right|=k$, $\left|v\right|=j$, $B(x,r)\supseteq I_{w}$ and $B(x,R)\subseteq I_{v}\cup I_{\sigma}$ where $I_{\sigma}$ is also of level $j$ and is adjacent to $I_{v}$. If $I_{v}$ and $I_{\sigma}$ are siblings, their measures are comparable. So assume their common ancestor, $I_{\tau},$ is $m$ generations earlier. Since the path from this ancestor to either of $I_{v}$ or $I_{\sigma}$ is a boundary path (except possibly at the first descendent), $m\leq M+1$. But then

[TABLE]

so again their measures are comparable. Hence

[TABLE]

Now consider the path from $I_{v}$ to $I_{w}$. This path begins at level at least $n_{J}$. If the path length, $k-j,$ is at least $n_{J},$ then by the definition of $\zeta_{n_{J}}$ there would be at most $\zeta(1+2^{-J})(j-k)$ levels where the path splits. On the levels where it does not split, the assigned weight is $1$ and, of course, all weights are at least $a$. Hence, in this case,

[TABLE]

Consequently, for suitable constants $C_{1},C_{2}$ we have

[TABLE]

Otherwise, $k-j<n_{J}$ and then we have $\mu(B(x,r))\geq a^{k-j}\mu(I_{v})\geq a^{n_{J}}\mu(I_{v})$. Thus

[TABLE]

for some constant $C_{3}$. This establishes (3.4).

To see (3.5), consider the special path $P_{j}$ for $j\geq J$. Suppose this path begins at $I_{v}^{(j)}$ of level $M_{j}$ $\geq$ $n_{j}$ and ends at $I_{w}^{(j)},$ having length $L_{j}\geq n_{j}.$ This path splits on at least $\zeta(1-2^{-j})L_{j}\geq\zeta(1-\varepsilon)L_{j}$ levels (and the weights are $a$ on the splitting children, $1$ on the non-splitting levels). Thus

[TABLE]

Take $x_{j}$ to be the distinguished point of $I_{w}^{(j)}$ and let $r_{j}=cs^{M_{j}+L_{j}},$ so $B(x_{j},r_{j})\subseteq I_{w}^{(j)}$. Take $R_{j}=\left|I_{v}^{(j)}\right|\sim s^{M_{j}}$. Then $B(x_{j},R_{j})\supseteq I_{v}^{(j)},$ thus

[TABLE]

Since $R_{j}/r_{j}\sim s^{-L_{j}}\rightarrow\infty$, this establishes (3.5) and that completes the proof for $D<\infty.$

Now suppose $D=\infty$. Choose a sequence $\{D_{n}\}\rightarrow\infty$ with $\dim_{A}E<D_{n}<\infty$. Assume $D_{1}=\dim_{A}E+2\varepsilon$ and take a KRS$(s(\varepsilon),c,3c)$ construction.

If this KRS construction admits arbitrarily long boundary paths, choose paths $P_{j}$ as in the proof of Lemma 3.3 and assign weights $a_{j}$ $=s^{D_{j}}$ to the intervals in the paths $P_{j}$. All other intervals will be assigned equal weights chosen to ensure that the sum over all children of a given parent is $1$. Let $\mu$ be the associated measure. This measure may no longer have the property that the measure of adjacent intervals are comparable, but this is not necessary for obtaining the lower bound on the upper Assouad dimension. As seen in the final paragraph of the proof of Lemma 3.3, it is enough to consider the starting and ending intervals on the path $P_{j},$ say $I_{v}^{(j)}$ and $I_{w}^{(j)},$ and the measure of the balls $B(x_{j},R_{j})\supseteq I_{v}^{(j)}$ and $B(x_{j},r_{j})\subseteq I_{w}^{(j)},$ to deduce that $\dim_{A}\mu\geq D_{j}$. Since $D_{j}\rightarrow\infty,$ we conclude that $\dim_{A}\mu=\infty$.

Otherwise, there are not arbitrarily long boundary paths. Then we modify the argument given in the first part of the proof of this theorem in a similar fashion. Instead of assigning weight $a=s^{D/\zeta}$, we assign weight $a_{j}=s^{D_{j}/\zeta}$ to the intervals in the paths $P_{j}$ that split and assign the remaining intervals the suitable weights so that the sum of all children of a given parent is $1$. Again we do not need the comparability of the measure of adjacent intervals as consideration of the intervals in the special paths is enough to see that $\dim_{A}\mu=\infty$. ∎

Remark 3.7**.**

We remark that the same proof works for any compact set $E\subseteq\mathbb{R}$ for which there is a KRS construction with positive proportionality constant and this can happen without $\dim_{A}E>0$. For example, let $E$ be the closure of the set $\{\alpha^{n}:n=0,1,2,...\}$ with $\alpha<1$. We can take a KRS construction with $s=\alpha$ that has $\zeta_{n}=1$ for all $n$. It is shown in [3] that if $\mu$ is the discrete measure $\sum_{n=1}^{\infty}\alpha^{\lambda n}\delta_{\alpha^{n}}$ having support $E,$ then $\dim_{A}\mu=\lambda$. Our methods also give this.

If, however, $\zeta=0$ for all KRS constructions, then it need not be the case that the set $E$ supports measures with all possible upper Assouad dimensions. One trivial example is to take $E$ to be a finite set. Then any measure supported on $E$ will have upper Assouad dimension zero.

More interestingly, one can construct infinite compact sets which support only measures with upper Assouad dimension either zero or infinity, with both values arising.

Proposition 3.8**.**

There is an infinite compact subset $E$ of $\mathbb{R}$ such that

[TABLE]

Proof.

Let $\alpha<1$ and $M>1$. Define $x_{0}=0$ and $x_{n}=\alpha^{M^{n}}$ for $n\in\mathbb{N}$. Set $E=\{x_{n}:n\geq 0\}$. We will see this set has the desired properties. Note that any measure supported on $E$ is of the form $\mu=\sum_{n=0}^{\infty}p(n)\delta_{x_{n}}$ for a nonegative, summable sequence $\{p(n)\}_{n=0}^{\infty}$.

Choose $n$ large enough that $3x_{n}<x_{n-1}$ and $x_{n}>2x_{n+1}$. Then $\mu(B(x_{n},2x_{n}))=\mu([0,3x_{n}))\geq p(0),$ while $\mu(B(x_{n},x_{n}/2))=\mu((x_{n}/2,3x_{n}/2))=p(n)$. Since $p(n)\rightarrow 0$ as $n\rightarrow\infty,$ it follows that if $p(0)\neq 0,$ then $\mu$ is not doubling and hence its Assouad dimension is infinite.

Thus we now assume $p(0)=0$. Put $t_{N}=\sum_{n=N}^{\infty}p(n)$. We will distinguish between three cases.

Case (i). There is a subsequence with $\lim_{j\rightarrow\infty}t_{n_{j}}/t_{n_{j}+1}=1$. Pick $n_{j}$ so that $3x_{n_{j}}<x_{n_{j}-1}$ and $x_{n_{j}}>2x_{n_{j}+1}$. Then $\mu(B(x_{n_{j}},2x_{n_{j}}))=t_{n_{j}}$ and

[TABLE]

Since the subsequence satisfies $t_{n_{j}}/(t_{n_{j}}-t_{n_{j}+1})\rightarrow\infty$, we again conclude that $\dim_{A}\mu=\infty.$

Case (ii). There is a subsequence with $\lim_{j\rightarrow\infty}t_{n_{j}}/t_{n_{j}+1}=\infty$. If $2x_{n_{j}}<x_{n_{j}-1}$ we have that $\mu(B(0,2x_{n_{j}}))=t_{n_{j}}$ and $\mu(B(0,x_{n_{j}}))=t_{n_{j}+1}$. Thus

[TABLE]

and again we deduce that $\dim_{A}\mu=\infty.$

Case (iii). Otherwise, there are constants $1<\lambda\leq\Lambda$ such that $\lambda\leq t_{n}/t_{n+1}\leq\Lambda$ for all $n$. Let $\rho=x_{1}-x_{2}$. Fix $x\in E$ and $0<r<R\leq\rho$. If $x=x_{1},$ then choose the minimal $L\geq 1$ such that $x_{L+1}\leq x-r$. Then

[TABLE]

Otherwise, $x\leq x_{2}$ and we can choose $N\geq 1$ such that $x_{N+1}<x+R\leq x_{N}$, so $\mu(B(x,R))\leq\mu([0,x_{N}])\leq t_{N}$. Choose $K$ such that $x\leq x_{K}<x+r\leq x_{K-1}$. If $x-r>0$, pick $L\geq K$ such that $x_{L+1}\leq x-r<x_{L}$ $\leq x$. In this case,

[TABLE]

so we have the bound

[TABLE]

If $x-r\leq 0$, then $\mu(B(x,r))\geq t_{K}$ and a similar estimate is still valid.

Put $\gamma:=\log\Lambda/\log M>0$. From the relations $\alpha^{M^{N+1}}=x_{N+1}\leq x+R$ and $x+r\leq x_{K-1}=\alpha^{M^{K-1}}$ we obtain

[TABLE]

Using (3.7) and these bounds on $\Lambda^{-N}$ and $\Lambda^{K}$ we deduce that

[TABLE]

for a suitable constant $C_{0}$.

Consider the function $F(x)=\frac{\log(x+r)}{\log(x+R)}$, whose derivative is

[TABLE]

As the function $\varphi(t)=t\log t$ is decreasing if $t<e^{-1},$ $F^{\prime}<0$ for small enough $R$. Thus $F(x)^{\gamma}\leq F(0)^{\gamma}$ for small $x\geq 0$. Moreover, for any $s>0,$ the function $\varphi_{s}(t)=t^{s}\log t$ is decreasing if $t<e^{-1/s}$ and hence $\log r/\log R\leq\left(R/r\right)^{s}$ for small enough $R$. Consequently

[TABLE]

for any $s>0.$ Together with (3.6) this proves $\dim_{A}\mu=0$. ∎

4. Lower Assouad Dimension

The corresponding result for lower Assouad dimension is much easier. We can even arrange to simultaneously achieve both the prespecified upper and lower Assouad dimensions when the lower Assouad dimension of $E$ is positive.

Theorem 4.1**.**

Given $0<d<\dim_{L}E\leq\dim_{A}E<D<\infty,$ there is a measure $\mu$ with support $E$, $\dim_{L}\mu=d$ and $\dim_{A}\mu=D.$

Proof.

Fix $0<\varepsilon<\min(\dim_{L}E-d,D-\dim_{A}E)/2$. Choose $s$ $>0$ so small that $s^{-\varepsilon}-2s^{d}\geq 1$, $s^{\varepsilon}+s^{d}<1$, $s^{-(d+\varepsilon)}\geq 3$ and $s\leq s(\varepsilon)$ from Lemma 3.1. Consider a KRS($s,c,3c)$ construction of $E$ with this choice of $s$. According to Lemma 3.1, for all long enough words $w,$

[TABLE]

Thus every (deep enough) interval from the KRS construction splits and if we put $p=s^{D-\varepsilon}$, then $s^{D}\leq p\leq 1/N_{w}$ for all such $w$.

Now choose a rapidly increasing sequence $\{n_{j}\},$ with $n_{1}$ sufficiently large that (4.1) holds for $\left|w\right|\geq n_{1}$. Select an infinite path consisting of interior intervals.

For even values of $j,$ we will assign weight $s^{d}$ to the interior intervals of the path at levels $n_{j},...,n_{j+1}-1$. The siblings of each of these intervals will be assigned equal weights so the sum of the weights over all the children of a given parent is $1$. We will denote these weights by $z_{w}$ if the interior interval is $I_{w}$.

For $j$ odd, we do the same thing, but with weight $s^{D}$ assigned to the interior intervals of the path and denote the weights assigned to siblings of $I_{w}$ as $y_{w}$.

All boundary intervals from the KRS construction that are not siblings of the intervals in the path will be assigned weight $p.$ We will assign equal weights to the siblings of these boundary intervals or to sibling groups which have no sibling boundary interval so that the sum of the weights of any sibling group is $1$. Call these weights $x_{w}$ (where $I_{w}$ is any member of the sibling group). Note that $x_{w}\geq p\geq s^{D}$ as $p\leq 1/N_{w}$ and $x_{w}\leq 1/(N_{w}-\eta_{w})$ where $\eta_{w}=$# boundary siblings of $I_{w}$. Since $N_{w}\geq s^{-(d+\varepsilon)}$ and $s^{-\varepsilon}-2s^{d}\geq 1,$ it is easy to see that $x_{w}\leq s^{d}$.

One can similarly use the bounds on $N_{w}$ and the relations $s^{-\varepsilon}-2s^{d}\geq 1$ and $s^{\varepsilon}+s^{d}<1$ to see that $s^{D}\leq y_{w},z_{w}\leq s^{d}$. For example,

[TABLE]

since $1+s^{D}\geq s^{\varepsilon}+s^{d}$. Thus $s^{D}$ is the minimim weight assigned and $s^{d}$ the maximum.

We also observe that a boundary interval will never have a child that is assigned weight $s^{d}$ or $s^{D},$ hence all boundary intervals that are children of boundary intervals are assigned the weight $p$.

Suppose $I_{w}$ and $I_{v}$ are adjacent intervals at a given level. If they are siblings, their $\mu$-measures are obviously comparable. If they are not siblings, the paths from their nearest common ancestor to $I_{w}$ and $I_{v}$ consists of only boundary intervals after the first descendent. Consequently, the intervals in these paths all have weight $p$ except possibly the first two generations, hence the $\mu$-measures of adjacent intervals are comparable.

Since there are arbitrarily long paths with weights $s^{d}$ and $s^{D}$ respectively, standard arguments now show that $\dim_{L}\mu=d$ and $\dim_{A}\mu=D$. ∎

Remark 4.2**.**

We note that it is easy to create a measure with given support $E$ and lower Assouad dimension [math]. Just take any measure $\mu$ with support $E$ and suppose $e\in E$. Then $dim_{L}(\mu+\delta_{e})=0$. Observe that if $e$ is not an isolated point in $E,$ then $\dim_{A}(\mu+\delta_{e})=\infty$.