Exercise 1.1.14 (Metric entropy formulation of Jordan measurability)

Question

Let $E\subset \mathbb{R}^d$ be a bounded set. For each integer $n$, let $\mathcal{E}_{*}(E,2^{-n})$ denote the number of dyadic cubes of sidelength $2^{-n}$ that are contained in $E$, and let $\mathcal{E}^{*}\left(E, 2^{-n}ight)$ be the numebr of dyadic cubes of sidelength $2^{-n}$ that intersect $E$ (i.e., metric entropy of $E$ at scale $2^{-n}$). Show that $E$ is Jordan measurable if and only if$$\lim_{n	o\infty}2^{-dn} \cdot \left(\mathcal{E}^{*}\left(E, 2^{-n}ight) - \mathcal{E}_{*}\left(E, 2^{-n}ight) ight) = 0$$in which case one has$$m(E) = \lim_{n	o\infty} \mathcal{E}_{*}\left(E, 2^{-n}ight) = \lim_{n	o\infty} \mathcal{E}^{*}\left(E, 2^{-n}ight).$$

Elu Thingol · Accepted Answer

Note that for any collection $\mathbf{C}$ of dyadic cubes, by properties of elementary measure we have $m(\bigcup \mathbf{C}) = \mathcal{E} \cdot 2^{-dn}$ where $\mathcal{E} = \# \mathbf{C}$. We now show both directions of equivalence. \begin{itemize}\item[$\implies$] Suppose that $\lim_{n\to\infty}2^{-dn} \cdot \left(\mathcal{E}^{*}\left(E, 2^{-n}\right) - \mathcal{E}_{*}\left(E, 2^{-n}\right) \right) = 0$. We use the criteria from Exercise 1.1.5 to prove the Jordan measurability. Pick an arbitrary $\epsilon>0$. By assumption, we can find $n \in \mathbb{N}$ large enough so that $$2^{-dn} \cdot \left(\mathcal{E}^{*}\left(E, 2^{-n}\right) - \mathcal{E}_{*}\left(E, 2^{-n}\right) \right) \leq \epsilon$$ We look closely at the definition of metric entropy. Notice that \begin{itemize}\item em the collection of all dyadic cubes that intersect $E$ form an elementary outer cover of $E$. (Pick an arbitrary $x\in E$. Then for $x=(x_1,\ldots,x_d)$ we can find integers $i_1,\ldots,i_d$ such that $i_1 \leq x_1\cdot 2^n < i_1 + 1, \ldots, i_d \leq x_d\cdot 2^n < i_d + 1$, i.e., there exists a dyadic cube $\left[\frac{i_1}{2^n},\frac{i_1+1}{2^n} \right) \times \ldots \times \left[\frac{i_d}{2^n},\frac{i_d+1}{2^n} \right)$ which contains $x\in E$, i.e., intersects $E$.)\item Similarly, the dyadic cubes that are contained in $E$ by definition form an inner cover of $E$.\end{itemize} Since each of the dyadic boxes has a Jordan measure of $2^{-nd}$, the volume of the outer cover formed by this boxes is $2^{-dn} \cdot \mathcal{E}^{*}\left(E, 2^{-n}\right)$. Similarly, the volume of an inner cover formed by the contained dyadic boxes is $2^{-dn} \cdot \mathcal{E}_{*}\left(E, 2^{-n}\right)$. But the Jordan measure of their set difference is the difference in their Jordan measures; thus, by our choice of $n_\epsilon \in \mathbb{N}$ it cannot exceed $\epsilon$, as desired.\item[$\impliedby$] Now suppose that $E$ is Jordan measurable. We use the following strategy to prove this assertion: \begin{enumerate}\item item first, we demonstrate that for large enough scale $n\in\mathbb{N}$ the measure of dyadic cubes intersecting a box $B$ and a measure of dyadic cubes contained in $B$ approximate $B$ very good;\item this we extend without difficulty to elementary sets;\item if elementary sets can be approximated by dyadic cubes, and Jordan measurable sets can be approximated by elementary sets, then Jordan measurable sets can be approximated by dyadic cubes\end{enumerate} Thus, let $B=\left[ a^{(1)}, b^{(1)} \right] \times \ldots \times \left[ a^{(d)}, b^{(d)} \right]$ be a box. We partition $B$ into dyadic cubes along each dimension, i.e., we find integers between $i_1,\ldots,i_d$ and $j_1,\ldots,j_d$ such that \begin{equation*} \begin{array}{ccc} i_1 \leq a_1 2^{n} < i_1 + 1 &\ldots &j_1 \leq b_1 2^{n} < j_1 + 1\ \vdots & \vdots & \vdots\ i_d \leq a_d 2^{n} < i_d + 1 &\ldots &j_d \leq b_d 2^{n} < j_d + 1 \end{array} \end{equation*} Notices that the following collection of dyadic cubes naturally covers $B$: $$\overline{B} := \bigcup_{k\in\mathbb{Z}^d, i \leq k \leq j + 1} \left[ \frac{k_1}{2^n}, \frac{k_1+1}{2^n}\right] \times \ldots \times \left[ \frac{k_d}{2^n}, \frac{k_d+1}{2^n}\right]$$ (Pick some $x:a_i \leq x_i \leq b_i$ from $B$. Then there exist integers $k_i \leq x_i 2^n < k_i + 1$ and by $i_i \leq a_i \leq x_i \leq b_i < j_i +1$ follows $i_i \leq k_i < k_i+1 \leq j_i + 1$.) The measure of this dyadic cube cover is by additivity: \begin{align*} m\left(\overline{B} \right) &= \sum_{k\in\mathbb{Z}^d, i \leq k \leq j + 1} \frac{1}{2^{nd}} = \left[\prod_{l=1}^{d} (j_l + 1 - i_l)\right] \frac{1}{2^{nd}}\[5pt] &\leq \left[\prod_{l=1}^{d} (b_l2^{n} + 1 - a_l2^{n} +1 )\right] \frac{1}{2^{nd}} =\prod_{l=1}^{d} b_l - a_l + \frac{2}{2^{n}} \end{align*} which can be verified to approximate $\prod_{l=1}^{d} b_l - a_l$ arbitrarily close for sufficiently large $n_\epsilon \in \mathbb{N}$. Also notice that the approximation error goes down with increasing $n$ (thus, the boundary holds for all $n> n_\epsilon$). Similarly, taking tighter borders we obtain a dyadic inner cover of $B$: $$\underline{B} := \bigcup_{k\in\mathbb{Z}^d, i+1 \leq k \leq j} \left[ \frac{k_1}{2^n}, \frac{k_1+1}{2^n}\right] \times \ldots \times \left[ \frac{k_d}{2^n}, \frac{k_d+1}{2^n}\right]$$ Which also approximates $B$ arbitrarily good: \begin{align*} m\left(\underline{B} \right) &= \sum_{k\in\mathbb{Z}^d, i+1 \leq k \leq j} \frac{1}{2^{nd}} = \left[\prod_{l=1}^{d} (j_l - i_l-1)\right] \frac{1}{2^{nd}}\[5pt] &\geq \left[\prod_{l=1}^{d} (b_l2^{n}-1 - a_2^{n} -1 )\right] \frac{1}{2^{nd}} =\prod_{l=1}^{d} b_l - a_l - \frac{2}{2^{n}} \end{align*} This naturally extends to elementary sets, since we can partition $F$ into $N$ disjoint boxes $B_l$ and then take $\epsilon/N$ precision for each box.\newline Now consider the original Jordan measurable set $E$. Fix an arbitrary $\epsilon >0$ for our limit. By definition of Jordan measurability, we can find an elementary outer cover $\overline{E}$ and an elementary inner cover $\underline{E}$ such that both do not differ in measure from $E$ by $\epsilon$. Now we cover both elementary covers by dyadic covers $\overline{\overline{E}}$ and $\underline{\underline{E}}$ of sidelength $2^{-n_\epsilon}$, which do not differ from original covers by more that $\epsilon/2$. From this we conclude \begin{align*} m\left(\overline{\overline{E}}\right) - m(E) &\leq m\left(\overline{E}\right) + \frac{\epsilon}{2} - m(E) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\ m(E) - m\left(\underline{\underline{E}}\right) &\leq \epsilon \end{align*} In other words, we have found $n_{\epsilon}$ large enough so that $$2^{-dn_\epsilon}\cdot \mathcal{E}^{*}(E,2^{-n_\epsilon})-m(E) \leq \epsilon$$ $$m(E)-2^{-dn_\epsilon}\cdot \mathcal{E}_{*}(E,2^{-n_\epsilon}) \leq \epsilon$$ More generally, $$\exists n_\epsilon \in \mathbb{N}: \quad \forall n \geq n_\epsilon: \quad \mathcal{E}^{*}(E,2^{-n_\epsilon})-m(E) < \epsilon$$ $$\exists n_\epsilon \in \mathbb{N}: \quad \forall n \geq n_\epsilon: \quad m(E) - \mathcal{E}_{*}(E,2^{-n_\epsilon}) < \epsilon$$ as desired.\end{itemize}

Prakash Anand · Answer

If $\lim _{n ightarrow \infty} 2^{-d n}\left(\mathcal{E}^{*}(A, n)-\mathcal{E}_{*}(A, n)ight)=0$ : Let $S_{*}(A, n)$ be the set of dyadic cubes of scale $2^{-n}$ that are contained in $A$, and let $S^{*}(A, n)$ be the set of dyadic cubes of scale $2^{-n}$ that intersect $A$. Then,
$\bigcup S_{*}(A, n) \subset A \subset \bigcup S^{*}(A, n)$, and these are elementary sets. Since dyadic cubes of the same scale are always disjoint, we get that
$$
2^{-d n} \mathcal{E}_{*}(A, n)=m\left(\bigcup S_{*}(A, n)ight) \leq J_{*}(A) \leq J^{*}(A) \leq m\left(\bigcup S^{*}(A, n)ight)=2^{-d n_{n}} \mathcal{E}^{*}(A, n) .
$$
Thus, taking $n ightarrow \infty$ we get that $J^{*}(A)=J_{*}(A)$. Moreover, we have that
$$
m(A)=\lim _{n ightarrow \infty} 2^{-d n} \mathcal{E}^{*}(A, n)=\lim _{n ightarrow \infty} 2^{-d n} \mathcal{E}_{*}(A, n) .
$$
If $A$ is Jordan measurable: Let $B=I_{1} 	imes \cdots 	imes I_{d}$ be a box where $I_{j}$ is the interval with endpoints $a_{j}<b_{j}$, define
- $(B)^{n}:=\left[a_{1}-2^{-n}, b_{1}+2^{-n}ight] 	imes \cdots 	imes\left[a_{d}-2^{-n}, b_{d}+2^{-n}ight]$, the enlargement of $B$ by $2^{-n}$ in each coordinate direction.
- $(B)_{n}:=\left(a_{1}+2^{-n}, b_{1}-2^{-n}ight) 	imes \cdots 	imes\left(a_{d}+2^{-n}, b_{d}-2^{-n}ight)$, shrinking $B$ by $2^{-n}$ in each coordinate direction.\par

Under these definitions, if a dyadic cube $D_{n}(z)$ of scale $2^{-n}$ intersects $B$ then it is contained in $(B)^{n}$; if $D_{n}(z)$ is not contained in $B$ then $D_{n}(z)$ does not intersect $(B)_{n}$.
Finally note that
$$
\begin{aligned}
m\left((B)^{n}ight) &-m\left((B)_{n}ight)=\prod_{j=1}^{d}\left(b_{j}-a_{j}+2 \cdot 2^{-n}ight)-\prod_{j=1}^{d}\left(b_{j}-a_{j}-2 \cdot 2^{-n}ight) \
&=\sum_{S \subset\{1, \ldots, d\}} \prod_{j \in S}\left(b_{j}-a_{j}ight) \cdot\left(\left(2 \cdot 2^{-n}ight)^{d-|S|}-\left(-2 \cdot 2^{-n}ight)^{d-|S|}ight) \
& \leq 2^{d} \cdot 4 \cdot 2^{-n} \cdot m(B) .
\end{aligned}
$$
Fix $\varepsilon>0$ and let $E \subset A \subset F$ be elementary sets such that $m(F) \leq m(A)+\varepsilon$ and $m(E) \geq m(A)-\varepsilon .$ Suppose that $E=\biguplus_{k=1}^{n} E_{k}$ and $F=\biguplus_{j=1}^{m} F_{j}$ where $E_{n}, F_{j}$ are boxes.\par

Let $D_{n}(z)$ be a dyadic cube of scale $2^{-n}$ such that $D_{n}(z)$ intersects $A$ but is not contained in $A$. Then, there exists $j$ such that $D_{n}(z)$ intersects some $F_{j}$, so $D_{n}(z)$ is contained in $\left(F_{j}ight)^{n} .$ Also, $D_{n}(z)$ is not contained in $E_{k}$ for all $k$, so for all $k$ we get that $D_{n}(z)$ does not intersect $\left(E_{k}ight)_{n}$. Thus, if we define
$$
(E)_{n}:=\biguplus_{k=1}^{n}\left(E_{k}ight)_{n} \quad 	ext { and } \quad(F)^{n}:=\bigcup_{j=1}^{m}\left(F_{j}ight)^{n}
$$
then
$$
\begin{aligned}
m\left((E)_{n}ight) &=\sum_{k=1}^{n} m\left(\left(E_{k}ight)_{n}ight) \geq \sum_{k=1}^{n} m\left(E_{k}ight) \cdot\left(1-2^{d} \cdot 2^{2-n}ight)=m(E) \cdot\left(1-2^{d} \cdot 2^{2-n}ight) \
m\left((F)^{n}ight) & \leq \sum_{j=1}^{m} m\left(\left(F_{j}ight)^{n}ight) \leq \sum_{j=1}^{m} m\left(F_{j}ight) \cdot\left(1+2^{d} \cdot 2^{2-n}ight)=m(F) \cdot\left(1+2^{d} \cdot 2^{2-n}ight)
\end{aligned}
$$
and since for any Jordan measurable set $J$ we have that $2^{-d n} \mathcal{E}_{+}(J) \leq m(J) \leq 2^{-d n} \mathcal{E}^{*}(J)$,
$$
\begin{aligned}
2^{-d n}\left(\mathcal{E}^{*}(A, n)-\mathcal{E}_{*}(A, n)ight) & \leq 2^{-d n}\left(\mathcal{E}_{*}\left((F)^{n}, night)-\mathcal{E}^{*}\left((E)_{n}, night)ight) \leq m\left((F)^{n}ight)-m\left((E)_{n}ight) \
& \leq m(F)-m(E)+2^{d} \cdot 2^{2-n} \cdot(m(F)+m(E)) \
& \leq \varepsilon+2^{d} \cdot 2^{2-n} \cdot(2 m(A)+\varepsilon)
\end{aligned}
$$
Taking $n ightarrow \infty$, we have that for all $\varepsilon>0$,
$$
0 \leq \lim _{n ightarrow \infty} 2^{-d n}\left(\mathcal{E}^{*}(A, n)-\mathcal{E}_{*}(A, n)ight) \leq \varepsilon
$$
so this limit must be 0 .

Exercise 1.1.14 (Metric entropy formulation of Jordan measurability)

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Exercise 1.1.14 (Metric entropy formulation of Jordan measurability)

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