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Quantization dimension of probability measures A. V. Ivanov Institute of Applied Mathematical Research of the Karelian Research Centre of the Russian Academy of Sciences, Petrozavodsk, Russia
Bibliographic databases:
   § 1. IntroductionQuantizing a probability measure means approximating this measure by measures with finite support. Studies of quantization problems form a meaningful branch of probability theory (see [1]). Within quantization theory, the quantization dimensions (the upper one $\overline{D}(\mu)$ and the lower one $\underline{D}(\mu)$) of a probability measure $\mu$ are specified. These dimensions characterize the asymptotic rate at which this measure is approximated by measures with finite support. The quantization dimensions of a probability measure $\mu$ defined on a metric compact space $(X,\rho)$ are known not to exceed the corresponding box dimensions $\dim_B$ of its support, namely
$$
\begin{equation}
\overline{D}(\mu)\leqslant\overline{\dim}_B(\mathrm{supp}(\mu))\quad\text{and} \quad \underline{D}(\mu)\leqslant\underline{\dim}_{\,B}(\mathrm{supp}(\mu))
\end{equation}
\tag{1}
$$
(see [2]; for probability measures in $\mathbb{R}^n$ with the metric induced by the norm, see [1]). Inequalities (1) specify an upper bound for the quantization dimensions of measures with a given support and give rise to the following natural problems concerning intermediate values of quantization dimensions, which were stated in [3]. Is it true that for any number $b\in[0,\overline{\dim}_BX]$ ($b\in[0,\underline{\dim}_{\,B}X]$) there is a probability measure $\mu_b$ on the compact space $X$ such that $\overline{D}(\mu_b)=b$ ($\underline{D}(\mu_b)=b$)? The intermediate value problem for upper dimensions was solved in the positive in [3]. Its solution was based on the existence of a closed subset $F$ in $X$ of prescribed upper box dimension $\overline{\dim}_BF=b$ for any $b\in[0,\overline{\dim}_BX]$. Such an approach is impossible for lower quantization dimensions since an example of a compact metric space $X$ of dimension $\dim_BX=1$ such that all of its nonempty proper closed subsets have dimension $\underline{\dim}_{\,B}$ equal to zero was constructed in [4]. However, our paper shows that the above problem of lower dimensions is also solved in the positive (see Theorem 4). Thus, values of the lower quantization dimension of probability measures on $X$ cover the entire range $[0,\underline{\dim}_{\,B}X]$ regardless of the presence of ‘gaps’ in the set of values of the lower box dimension of closed subsets of the compact space $X$. In addition, we prove (Theorem 3) that if $X$ is a metric compact space and $\dim_BX=a\leqslant\infty$, then for any numbers $b\in[0,a]$ and $c\in[b,a]$ there exists a measure $\mu\in P(X)$ such that $\underline{D}(\mu)=b$ and $\overline{D}(\mu)=c$. § 2. Definitions and lemmasIn what follows $(X,\rho)$ is a metric compact space and $P(X)$ is the space of probability measures on $X$ with the Kantorovich–Rubinstein metric $\rho_P$ defined by
$$
\begin{equation*}
\rho_P(\mu,\nu)=\sup\bigl\{|\mu(f)-\nu(f)|\colon f\in\mathrm{Lip}_1(X)\bigr\},
\end{equation*}
\notag
$$
where $\mathrm{Lip}_1(X)$ is the set of real functions on $X$ satisfying the Lipschitz condition with constant $1$ and $\mu(f)=\displaystyle\int f\, d\mu$. For each measure $\mu\in P(X)$ its support $\mathrm{supp}(\mu)$ is defined to be a minimal closed subset of full measure of $X$. It is known (see [5], Ch. 7) that the set1
$$
\begin{equation*}
P_n(X)=\bigl\{\mu\in P(X)\colon|\mathrm{supp}(\mu)|\leqslant n\bigr\}
\end{equation*}
\notag
$$
is closed in $P(X)$ for any $n\in \mathbb{N}$ and $\bigcup_{n\in\mathbb{N}}P_n(X)$ is everywhere dense in $P(X)$. Hence for any measure $\mu\in P(X)$ and each $\varepsilon>0$ the quantity
$$
\begin{equation*}
N(\mu,\varepsilon)=\min\bigl\{n\colon \rho_P(\mu,P_n(X))\leqslant\varepsilon\bigr\}
\end{equation*}
\notag
$$
is well defined. If $\mu\not\in \bigcup_{n\in\mathbb{N}}P_n(X)$, then $N(\mu,\varepsilon)$ increases unboundedly as $\varepsilon\to 0$. The rate of this increase is characterized by the quantization dimension $D(\mu)$ of $\mu$ (see [1] and [2]) as follows:
$$
\begin{equation*}
D(\mu)=\lim_{\varepsilon\to 0}\frac{\log N(\mu,\varepsilon)}{-\log\varepsilon}.
\end{equation*}
\notag
$$
If this limit does not exist, then the upper and lower limits are considered and we arrive at the upper $\overline{D}(\mu)$ and lower $\underline{D}(\mu)$ quantization dimensions of $\mu$, respectively. Clearly, it is always true that $\underline{D}(\mu)\leqslant\overline{D}(\mu)$. The notation $D(\mu)$ means that ${\overline{D}(\mu)=\underline{D}(\mu)}$. For $x\in X$, we let $B(x,\varepsilon)$ denote a closed $\varepsilon$-ball: $B(x,\varepsilon)=\{y \colon \rho(x,y)\leqslant\varepsilon\}$. The closed $\varepsilon$-neighbourhood of a subset $A\subset X$ is denoted similarly: $B(A,\varepsilon)=\{y\colon \rho(y,A)\leqslant\varepsilon\}$, where $\rho(y,A)=\inf\{\rho(y,x)\colon x\in A\}$. A subset $A$ is called an $\varepsilon$-net in $X$ if $X=B(A,\varepsilon)$. For $\varepsilon>0$ we let $N(X,\varepsilon)$ denote the least number of points in an $\varepsilon$-net in $X$. The definition of the box dimensions of a compact space $X$ (the upper dimension $\overline{\dim}_BX$ and the lower one $\underline{\dim}_{\,B}X$) is similar to the definition of the quantization dimensions; more precisely,
$$
\begin{equation*}
\overline{\dim}_BX=\varlimsup_{\varepsilon\to 0}\frac{\log N(X,\varepsilon)}{-\log\varepsilon}\quad\text{and} \quad \underline{\dim}_{\,B}X=\varliminf_{\varepsilon\to 0}\frac{\log N(X,\varepsilon)}{-\log\varepsilon}.
\end{equation*}
\notag
$$
When $\overline{\dim}_BX{\kern1pt}{=}\,\underline{\dim}_{\,B}X$, the notation $\dim_BX$ is used. (The functorial generality of box and quantization dimensions was discussed in [2]; the theory of box dimensions is presented in [6].) It is known that if $A$ is a finite subset in $X$, then for any measure $\mu\in P(X)$, where $P(A)$ is the set of measures whose support lies in $A$ (see [2], Proposition 2).A subset $A\subset X$ is said to be $\varepsilon$-separated if $\rho(x,y)>\varepsilon$ for any two distinct points $x,y\in A$. Any $\varepsilon$-separated subset of a compact space is finite. It is obvious that a maximal (with respect to inclusion) $\varepsilon$-separated set is an $\varepsilon$-net in $X$. Thus, for any $\varepsilon >0$ there is an $\varepsilon$-separated $\varepsilon$-net in any metric compact space. In what follows we need the following assertions borrowed from [4] and [3], respectively. Theorem 1. Let $X$ be a metric compact space, and let $\overline{\dim}_BX=a\leqslant\infty$. Then for any number $b\in[0,a]$ there is a closed subset $F$ of $X$ such that Theorem 2. Let $X$ be a metric compact space and $\overline{\dim}_BX=a\leqslant\infty$. Then for any number $b\in[0,a]$ there is a probability measure $\mu$ on $X$ such that $\overline{D}(\mu)=b.$ The following assertion holds. Proposition 1. If a sequence $\varepsilon_n>0$, $n\in \mathbb{N}$, converges to zero monotonically ($\varepsilon_{n+1}\leqslant\varepsilon_n$) and $\lim_{n\to\infty}{\log\varepsilon_n}/{\log\varepsilon_{n+1}}=1$, then Proof. We obviously have
$$
\begin{equation*}
\overline{D}(\mu)\geqslant\varlimsup_{n\to\infty}\frac{\log N(\mu,\varepsilon_n)}{-\log\varepsilon_n}.
\end{equation*}
\notag
$$
We prove the reverse inequality. Assume that a sequence $\delta_k>0$ is such that
$$
\begin{equation*}
\overline{D}(\mu)=\lim_{k\to\infty}\frac{\log N(\mu,\delta_k)}{-\log\delta_k}.
\end{equation*}
\notag
$$
If the intersection $\{\varepsilon_n\colon n\in\mathbb{N}\}\cap\{\delta_k\colon k\in\mathbb{N}\}$ is infinite, then
$$
\begin{equation*}
\varlimsup_{n\to\infty}\frac{\log N(\mu,\varepsilon_n)}{-\log\varepsilon_n}\geqslant\lim_{k\to\infty}\frac{\log N(\mu,\delta_k)}{-\log\delta_k}.
\end{equation*}
\notag
$$
It remains to consider the case when $\varepsilon_n\not=\delta_k$ for all $n,k\in\mathbb{N}$. Then for any $k$ above some $k_0$ there is a natural number $n(k)$ such that $\delta_k\in(\varepsilon_{n(k)+1},\varepsilon_{n(k)})$. Therefore,
$$
\begin{equation*}
\begin{aligned} \, \lim_{k\to\infty}\frac{\log N(\mu,\delta_k)}{-\log\delta_k} &\leqslant \varlimsup_{k\to\infty}\biggl( \frac{\log N(\mu,\varepsilon_{n(k)+1})}{-\log\varepsilon_{n(k)+1}} \frac{\log\varepsilon_{n(k)+1}}{\log\varepsilon_{n(k)}}\biggr) \\ &\leqslant\varlimsup_{n\to\infty}\frac{\log N(\mu,\varepsilon_n)}{-\log\varepsilon_n}. \end{aligned}
\end{equation*}
\notag
$$
So the first equality is proved. The remaining equalities can be proved similarly.
The proposition is proved. Lemma 1. Assume that a sequence $\{\varepsilon_n\colon n\in\mathbb{N}\}$ satisfies the assumptions of Proposition 1 and $H_n$ is a sequence of $\varepsilon_n$-separated $\varepsilon_n$-nets in a metric compact space $X$. Then Proof. Since $H_n$ is an $\varepsilon_n$-net in $X$, it is true that We also have
$$
\begin{equation}
|H_n|\leqslant N\biggl(X,\frac{\varepsilon_n}2\biggr).
\end{equation}
\tag{4}
$$
In fact, since $H_n$ is an $\varepsilon_n$-separated set, we have $|B(x,\varepsilon_n/2)\cap H_n|\leqslant 1$ for any point $x\in X$. Therefore, if $A$ is an $(\varepsilon_n/2)$-net in $X$, then $|H_n|\leqslant |A|$. Inequalities (3) and (4) yield the assertion of the lemma. Let $\mu,\nu\in P(X)$, and let $p,q>1$ be numbers such that $p+q=1$. Clearly, $p\mu+q\nu\in P(X)$. The following lemma was proved in [3]. Lemma 3. It is true that If the dimension $D(\mu)$ exists for $\mu$ and $D(\mu)\geqslant\underline{D}(\nu)$, then $\underline{D}(p\mu+q\nu)=D(\mu)$. Proof. Let $\xi\in P(X)$ be an $\varepsilon$-approximation of the measure $p\mu+q\nu$, and let ${A=\mathrm{supp}(\xi)}$. Then $\varepsilon\geqslant (p\mu+q\nu)(\rho(x,A))\geqslant p\mu(\rho(x,A))$; hence $\mu(\rho(x,A))\leqslant (\varepsilon/p)$. So we have It immediately follows from this inequality that $\underline{D}(\mu)\leqslant\underline{D}(p\mu+q\nu)$. In a similar way we obtain $\underline{D}(\nu)\leqslant\underline{D}(p\mu+q\nu)$.
Now we assume that the dimension $D(\mu)$ exists and $D(\mu)\geqslant\underline{D}(\nu)$. Let $\xi$ and $\eta$ be $\varepsilon$-approximations of $\mu$ and $\nu$, respectively, let $\mathrm{supp}(\xi)=A$, $\mathrm{supp}(\eta)=B$, $|A|=N(\mu,\varepsilon)$, and $|B|=N(\nu,\varepsilon)$. Then
$$
\begin{equation*}
(p\mu+q\nu)(\rho(x,A\cup B))\leqslant p\mu(\rho(x,A))+q\nu(\rho(x,B))\leqslant\varepsilon.
\end{equation*}
\notag
$$
Consequently,
$$
\begin{equation}
N(p\mu+q\nu,\varepsilon)\leqslant N(\mu,\varepsilon)+N(\nu,\varepsilon).
\end{equation}
\tag{5}
$$
We choose a sequence $\varepsilon_n\to 0$ so that
$$
\begin{equation*}
\underline{D}(\nu)=\lim_{n\to\infty}\frac{\log N(\nu,\varepsilon_n)}{-\log\varepsilon_n}.
\end{equation*}
\notag
$$
From inequalities (5) and $D(\mu)\geqslant\underline{D}(\nu)$ we deduce that The lemma is proved. § 3. Basic resultsTheorem 3. Let $X$ be a metric compact space, and let $\dim_BX=a\leqslant\infty$. Then for any numbers $b\in[0,a]$ and $c\in[b,a]$ there is a measure $\mu\in P(X)$ such that $\underline{D}(\mu)=b$ and $\overline{D}(\mu)=c$. Proof. We show that for any $b\in[0,a]$ there is a measure $\mu'\in P(X)$ such that $D(\mu')=b$.
For $b=0$, the required measure $\mu'$ is any measure with finite support. Now assume that $b>0$. Case 1. $b<a<\infty$. We denote the ratio $b/(a-b)$ by $p$. Then We set $\varepsilon_n=2^{-pn}$. Using induction we construct a sequence $H_n$, $n\in\mathbb{N}$, of $\varepsilon_n$-separated $\varepsilon_n$-nets in $X$ as follows. As $H_1$ we take a maximal (with respect to inclusion) $\varepsilon_1$-separated subset of $X$. Assume that for $i<k$ the sets $H_i$ have already been constructed. We choose a maximal $\varepsilon_k$-separated subset $C_k$ of $X\setminus B(H_{k-1},\varepsilon_k)$ and set $H_k=H_{k-1}\cup C_k$. Proceeding in this way, we arrive at the required sequence $H_n$, for which $C_n=H_n\setminus H_{n-1}$ for $n>1$. We set $C_1=H_1$ and define $\mu'$ by the formula
$$
\begin{equation}
\mu'=\sum_{n\in\mathbb{N}}\frac{1}{2^n}\sum_{x\in C_n}\frac{1}{|C_n|}\delta_x,
\end{equation}
\tag{7}
$$
where $\delta_x$ is the Dirac measure.
We prove that
$$
\begin{equation}
N\biggl(\mu',\frac{\varepsilon_n}{2^{n+2}}\biggr)\geqslant\frac{1}{2}\,|H_n|
\end{equation}
\tag{8}
$$
for any $n$. Assume that a measure $\nu\in P(X)$ is such that $\rho_P(\mu',\nu)\leqslant{\varepsilon_n}/{2^{n+2}}$ and $A=\mathrm{supp}(\nu)$. We show that $|A|\geqslant |H_n|/2$. We assume the opposite. Since $H_n$ is an $\varepsilon_n$-separated set, for $x\in H_n$ the balls $B(x,\varepsilon_n/2)$ are pairwise disjoint. We set
$$
\begin{equation*}
H'_n=\biggl\{x\in H_n\colon B\biggl(x,\frac{\varepsilon_n}2\biggr)\cap A=\varnothing\biggr\}.
\end{equation*}
\notag
$$
By assumption $|A|<|H_n|/2$; therefore, $|H'_n|>|H_n|/2$. We have $\rho(x,A)\geqslant\varepsilon_n/2$ for any point $x\in H'_n$. Consequently,
$$
\begin{equation*}
\begin{aligned} \, \rho_P(\mu',\nu) &\geqslant\rho(\mu',P(A))=\mu'(\rho(x,A))\geqslant\frac{1}{2^n}\sum_{k\leqslant n}\sum_{x\in C_k}\frac{1}{|H_n|}\rho(x,A) \\ &\geqslant\frac{1}{2^n}\sum_{x\in H'_n}\frac{1}{|H_n|}\rho(x,A)>\frac{\varepsilon_n}{2^{n+2}}. \end{aligned}
\end{equation*}
\notag
$$
Thus, inequality (8) is proved. By virtue of (6), (8), Proposition 1, Lemma 1 and the condition $\dim_BX=a$ we derive the following lower estimate for the lower quantization dimension of $\mu'$:
$$
\begin{equation}
\begin{aligned} \, \notag \underline{D}(\mu') &=\varliminf_{n\to\infty}\frac{\log N(\mu',\varepsilon_n/2^{n+2})}{-\log (\varepsilon_n/2^{n+2})} \\ &\geqslant\varliminf_{n\to\infty}\frac{\log (|H_n|/2)}{-\log \varepsilon_n} \lim_{n\to\infty}\frac{\log \varepsilon_n}{\log (\varepsilon_n/2^{n+2})}=\frac{p}{p+1}a=b. \end{aligned}
\end{equation}
\tag{9}
$$
Now we prove that
$$
\begin{equation}
N\biggl(\mu',\frac{\varepsilon_n}{2^n}\biggr)\leqslant |H_n|
\end{equation}
\tag{10}
$$
for each $n$. By construction, $H_n$ is an $\varepsilon_n$-net in $X$. Therefore, $\rho(x,H_n)\leqslant\varepsilon_n$ for each $x\in X$. Thus, we have
$$
\begin{equation*}
\rho_P(\mu',P(H_n))=\mu'(\rho(x,H_n))=\sum_{k>n}\frac{1}{2^k}\sum_{x\in C_k}\frac{1}{|C_k|}\rho(x,H_n)\leqslant\frac{\varepsilon_n}{2^n},
\end{equation*}
\notag
$$
which implies (10). By analogy with the reasoning in (9) we conclude that ${\overline{D}(\mu')\,{\leqslant}\, b}$. So $D(\mu')=b$.
Case 2. $b=a<\infty$. We set $\varepsilon_n=2^{-n^2}$. As above, we construct a sequence of $\varepsilon_n$-separated $\varepsilon_n$-nets $H_k$ and define $\mu'$ using (7). Since the sequence $\varepsilon_n/2^{n+2}$ satisfies the assumptions of Proposition 1, by analogy with (9) we obtain
$$
\begin{equation*}
\underline{D}(\mu') \geqslant\varliminf_{n\to\infty}\biggl( \frac{\log (|H_n|/2)}{-\log \varepsilon_n} \frac{\log \varepsilon_n}{\log (\varepsilon_n/2^{n+2})}\biggr) =a.
\end{equation*}
\notag
$$
Since $\overline{D}(\mu')\leqslant\dim_BX$, it follows that $D(\mu')=a$.
Case 3. $b=a=\infty$. We set $p=1$ and repeat the reasoning conducted in Case 1. As a result, we obtain a measure $\mu'$ such that
$$
\begin{equation*}
\underline{D}(\mu')=\varliminf_{n\to\infty}\frac{\log N(\mu',\varepsilon_n/2^{n+2})}{-\log (\varepsilon_n/2^{n+2})} \geqslant\varliminf_{n\to\infty}\biggl( \frac{\log (|H_n|/2)}{-\log \varepsilon_n} \frac{\log \varepsilon_n}{\log(\varepsilon_n/2^{n+2})}\biggr)=\infty
\end{equation*}
\notag
$$
by (9).
Case 4. $a=\infty$ and $b<a$. We set
$$
\begin{equation*}
f(\varepsilon)=\inf\biggl\lbrace \frac{\log N(X,\delta)}{-\log \delta}\colon 0<\delta\leqslant\varepsilon\biggr\rbrace .
\end{equation*}
\notag
$$
Clearly, and $f(\varepsilon_1)\geqslant f(\varepsilon_2)$ for $\varepsilon_1<\varepsilon_2$. In addition,
$$
\begin{equation}
\frac{\log N(X,\varepsilon)}{-\log\varepsilon}\geqslant f(\varepsilon)
\end{equation}
\tag{11}
$$
for each $\varepsilon>0$.
Using induction we construct the following sequences $\varepsilon_k$ and $p_k$. Assume that $\varepsilon_0>0$ is such that $f(\varepsilon_0)>b$. We set $p_1=b/f(\varepsilon_0)$ and $\varepsilon_1=2^{-p_1}$. Step $k$. We set
$$
\begin{equation}
p_k=\max\biggl\lbrace \frac{b}{f(\varepsilon_{k-1})},\frac{k-1}{k}p_{k-1}\biggr\rbrace \quad\text{and} \quad \varepsilon_k=2^{-p_kk}.
\end{equation}
\tag{12}
$$
It follows from (12) that $kp_k\geqslant(k-1)p_{k-1}$ for any $k\in\mathbb{N}$. Hence the sequence $\varepsilon_k$ is monotonically decreasing ($\varepsilon_k\leqslant\varepsilon_{k-1}$). We show that $\lim_{k\to\infty}kp_k=\infty$ and $\lim_{k\to\infty}\varepsilon_k=0$. Assume the opposite: $\lim_{k\to\infty}\varepsilon_k\,{=}\,h\,{>}\,0.$ Then ${b/f(\varepsilon_{k-1})\,{\geqslant}\, b/f(h)}$ and, owing to (12), $p_k\geqslant b/f(h)$ for all $k$. Consequently, $\lim_{k\to\infty}kp_k=\infty$, which is a contradiction.
Now we show that
$$
\begin{equation}
\lim_{k\to\infty}p_k=0\quad\text{and} \quad \lim_{k\to\infty}\frac{p_{k+1}}{p_k}=1.
\end{equation}
\tag{13}
$$
Since $f(\varepsilon_{k-1})\leqslant f(\varepsilon_k)$, the sequence $p_k$ is monotonically decreasing ($p_k\leqslant p_{k-1}$). If ${p_k=p_{k-1}(k-1)/k}$ for all $k$ above some $k_0$, then $p_k=p_{k_0}k_0/k$ and $kp_k=k_0p_{k_0}$, which is impossible in view of the fact that $\lim_{k\to\infty}kp_k=\infty$. Therefore, there is a sequence $k_n$ such that $p_{k_n}=b/f(\varepsilon_{k_n-1})$; thus, $\lim_{k\to\infty}p_k=0$. Furthermore, we have
$$
\begin{equation*}
1\geqslant\frac{p_{k+1}}{p_k}\geqslant \frac{k}{k+1},
\end{equation*}
\notag
$$
which yields the second equality in (13).
We derive from (13) that
$$
\begin{equation*}
\lim_{n\to\infty} \frac{\log\varepsilon_n}{\log\varepsilon_{n+1}}=1.
\end{equation*}
\notag
$$
Hence the sequence $\varepsilon_k$ satisfies the assumptions of Proposition 1. By virtue of (11) and (12),
$$
\begin{equation*}
\frac{\log N(X,\varepsilon_k)}{-\log\varepsilon_k}\geqslant\frac{b}{p_{k+1}}.
\end{equation*}
\notag
$$
Therefore, for any $k$ there is a natural number $q_k\leqslant N(X,\varepsilon_k)$ such that
$$
\begin{equation}
\frac{\log q_k}{-\log\varepsilon_k}\leqslant\frac{b}{p_{k+1}}\leqslant\frac{\log (q_k+1)}{-\log\varepsilon_k}.
\end{equation}
\tag{14}
$$
Now using induction we construct an increasing sequence (with respect to inclusion) of $\varepsilon_k$-separated subsets $H_k\subset X$ of cardinality $|H_k|=q_k$. Step 1. We take a maximal $\varepsilon_1$-separated subset $D_1$ of $X$. Since $|D_1|\geqslant N(X,\varepsilon_1)$, we can find a subset $H_1$ of $D_1$ with cardinality $|H_1|=q_1$. Step $k$. We take a maximal $\varepsilon_k$-separated subset $D_k$ of $X\setminus B(H_{k-1},\varepsilon_k)$. The set $H_{k-1}\cup D_k$ is an $\varepsilon_k$-separated $\varepsilon_k$-net in $X$; therefore, $|H_{k-1}\cup D_k|\geqslant N(X,\varepsilon_k)\geqslant q_k$. Thus, we can find a subset $C_k$ in $D_k$ such that the set $H_k=H_{k-1}\cup C_k$ is of cardinality $q_k$. Now we define the measure $\mu'$ using (7). From (8) we obtain the inequality
$$
\begin{equation*}
N\biggl(\mu',\frac{\varepsilon_k}{2^{k+2}}\biggr)\geqslant\frac{1}{2}q_k.
\end{equation*}
\notag
$$
The sequence ${\varepsilon_k}/{2^{k+2}}$ satisfies the assumptions of Proposition 1; consequently,
$$
\begin{equation}
\begin{aligned} \, \notag \underline{D}(\mu') &=\varliminf_{k\to\infty}\frac{\log N(\mu',\varepsilon_k/2^{k+2})}{-\log (\varepsilon_k/2^{k+2})} \\ &\geqslant\varliminf_{k\to\infty}\frac{\log (q_k/2)\cdot p_{k+1}}{-\log \varepsilon_k} \frac{\log \varepsilon_k}{\log (\varepsilon_k/2^{k+2})\cdot p_{k+1}}. \end{aligned}
\end{equation}
\tag{15}
$$
Owing to (13) and (14), we have
$$
\begin{equation*}
\lim_{k\to\infty}\frac{\log \varepsilon_k}{\log (\varepsilon_k/2^{k+2})\cdot p_{k+1}}=1\quad\text{and} \quad \lim_{k\to\infty}\frac{\log (q_k/2)\cdot p_{k+1}}{-\log \varepsilon_k}=b.
\end{equation*}
\notag
$$
So
We prove that
$$
\begin{equation}
N\biggl(\mu',\frac{\mathrm{diam}(X)}{2^k}\biggr)\leqslant q_k.
\end{equation}
\tag{16}
$$
In fact,
$$
\begin{equation*}
\begin{aligned} \, \rho_P(\mu',P(H_k)) &=\mu'(\rho(x,H_k)) \\ &=\sum_{n=k+1}^\infty\frac{1}{2^n}\sum_{x\in C_n} \frac{\rho(x,H_k)}{|C_n|} \leqslant\frac{\mathrm{diam}(X)}{2^k}, \end{aligned}
\end{equation*}
\notag
$$
which yields (16).
By analogy with (15), we derive from (16) that
$$
\begin{equation*}
\begin{aligned} \, \overline{D}(\mu') &=\varlimsup_{k\to\infty}\frac{\log N(\mu',\mathrm{diam}(X)/2^k)}{-\log (\mathrm{diam}(X)/2^k)} \\ &\leqslant\varlimsup_{k\to\infty}\frac{\log q_k\cdot p_{k+1}}{-\log \varepsilon_k} \frac{\log \varepsilon_k}{\log (\mathrm{diam}(X)/2^k)\cdot p_{k+1}}=b. \end{aligned}
\end{equation*}
\notag
$$
Thus, $D(\mu')=b$.
Hence the existence of a measure $\mu'$ of dimension $D(\mu')=b$ has been proved for all $a$ and $b$. Now assume that $c\in[b,a]$. According to Theorem 1, there is a closed subset $F\subset X$ such that $\underline{\dim}_{\,B}F=0$ and $\overline{\dim}_BF=c$. By Theorem 2 there is a probability measure $\nu\in P(F)$ such that $\overline{D}(\nu)=c$. We also have $\underline{D}(\nu)=0$ since $\underline{D}(\nu)\leqslant\underline{\dim}_{\,B}F=0$. In addition, we know that the quantization dimensions of $\nu$ with respect to $F$ coincide with the quantization dimensions of this measure with respect to $X$ (see [2], Proposition 5). We set By Lemmas 2 and 3 the measure $\mu$ is the required one.Theorem 3 is proved. Corollary 1. If $X$ is a metric compact space and $\dim_BX=a\leqslant\infty$, then for any $b\in[0,a]$ there is a measure $\mu\in P(X)$ such that $D(\mu)=b$. The following theorem, which gives an answer to the question of intermediate values of the lower quantization dimension, is also true. Theorem 4. Let $\underline{\dim}_{\,B}X=a<\infty$. Then for any number $b\in[0,a]$ there is a probability measure $\mu\in P(X)$ such that $\underline{D}(\mu)=b$. Proof. If $a=\infty$, then the assertion of the theorem is a direct consequence of Theorem 3. For $b<a<\infty$ it is straightforward to verify that the measure $\mu'$ constructed in the proof of Theorem 3 (Case 1) has lower quantization dimension In a similar way, for $b=a<\infty$, the measure $\mu'$ (Case 2) has dimension $\underline{D}(\mu')=a$. The theorem is proved. Remark 5. In Theorem 3, Corollary 1 and Theorem 4 we can additionally assume that $\mathrm{supp}(\mu)=X$. In fact, we take a countable everywhere dense set $\{x_i\colon i\in\mathbb{N}\}$ in $X$ and define $\xi\in P(X)$ by It is known that $D(\xi)=0$ and $\mathrm{supp}(\xi)=X$ (see [2], Proposition 8). We set By Lemmas 2 and 3 the support of $\mu'$ is $X$, and the quantization dimensions of $\mu'$ coincide with the quantization dimensions of $\mu$.
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