Decomposing a rectangle in $\mathbb{R}^d$

Question

Let $K=[a_1,b_1]\times\dots \times [a_d,b_d]$ be a closed rectangle in $\mathbb{R}^d$ and suppose that $K\subset\cup_{i=1}^n R_i$, where each $R_i$ is an open rectangle. In his measure theory book Cohn writes:

$K$ can be decomposed into a finite collection $\{K_j\}$ of rectangles that overlap only on their boundaries and are such that for each $j$ the interior of $K_j$ is included in some $R_i$.

If $\overline{R}_i$ denote the closure of $R_i$, then $K\cap\overline{R}_i$ is a closed rectangle. For each $1\leq m \leq d$, let $P_m$ denotes all the left and right $m$th coordinates appearing among the $K\cap\overline{R}_i$ $ (1\leq i\leq n)$. Then $P_m\subset [a_m,b_m]$, and if we order the elements of $P_m$ from left to right we obtain a partition of $[a_m,b_m]$ into a finite collection if closed intervals that overlap only at the boundary. Let $\{K_j\}$ denote the finite collection of all closed rectangles in $\mathbb{R}^d$ which are formed by taking $d$-products of those closed intervals (by choosing one closed interval in the partition of $[a_m,b_m]$ for each $m$). These rectangles intersect only at their boundaries and

$$K=\cup K_j$$

Indeed if $x\in K$, then $x_m \in [a_m,b_m]$ for each $m$, and so $x\in K_j$ for some $j$. Conversely if $x\in K_j$ for some $j$, then $x_m \in [a_m,b_m]$ for each $m$ and so $x\in K$.

Am stuck at proving that the interior of each $K_j$ is included in some $R_i$. It would suffice to show that $K_j\subset K\cap\overline{R}_i$ for some $i$. But I don't see how to show it.

Any help is greatly appreciated.

Answer 1

Use the $\sup$ norm for convenience.

For $k \in K$, let $\delta(k) = \sup \{ r | B(k,r) \subset R_i \text{ for some } i\}$. Since the $R_i$ are open we see that $\delta(k) > 0$ for all $k$.

Let $\delta= \inf_{k \in K} \delta(k)$, it is straightforward to show that $\delta>0$ and for any $k$ there is some $R_i$ such that $B(k,\delta) \subset R_i$.

Let $l_i = b_i-a_i$ and take the rectangles $$\Delta_{n,k_1,...,k_d} = [a_1+{k_1 \over n} l_1, a_1+{k_1+1 \over n} l_1] \times \cdots \times [a_d+{k_d \over n} l_d, a_d+{k_d+1 \over n} l_d] $$ with ${1 \over n} < \delta$. Note that for each $n$ we have $K = \cup_{k_i=0,...,n-1} \Delta_{n,k_1,...,k_d}$.

Since the center of each rectangle lies in some $R_i$, we see that each rectangle lies in some $R_i$.

Addendum:

To see why $\delta>0$: Suppose $\delta=0$ then there is some sequence $k_i$ such that $\delta(k_i) \to 0$. Without loss of generality we can assume $k_k \to k$. We know that $\delta(k) >0$ and hence for $k_i$ sufficiently close to $k$ we have $\delta(k_i) \ge {1 \over 2} \delta(k)$ which is a contradiction.

Answer 2

For completeness I will finish the approach I proposed originaly. This approach might be closer to what Cohn had in mind.

Assume $K^\circ\neq \emptyset$ and $K\cap R_i \neq \emptyset$ for each $i$. Write $K\cap \bar{R}_i=[a^i_1,b^i_1]\times\dots\times[a^i_d,b^i_d]$. Note that $K=\cup_{i=1}^n K\cap \bar{R}_i$. For each $j=1,\dots,d$, let

$$a_j=c_j^1<c^2_j<\dots<c_j^{m_j}=b_j$$

be the list of all distinct numbers among $a_j^i$ and $b_{j}^i$ for $1\leq i\leq n$. Denote by $\mathcal{L}$ any $n$-tuple of the form $\mathcal{L}=(l_1,\dots,l_d)$ with $1\leq l_j\leq m_j$. Then denote

$$K^{\mathcal{L}}=[c_1^{l_1},c_1^{l_1+1}]\times\dots\times[c_d^{l_d},c_d^{l_d+1}]$$

Clearly $K=\cup\{K^{\mathcal{L}}\}$ and the $K^{\mathcal{L}}$ overlap only on their boundaries. Suppose $(K^{\mathcal{L}})^\circ \cap (K\cap \bar{R}_i) \neq \emptyset$ for some $\mathcal{L}$ and $i$. Then $c_j^{l_j}<x_j<c_j^{l_j+1}$ and $a^i_j\leq x_j\leq b^i_j$ for $1\leq j\leq d$. By the choice of the $c_j$'s we must have $a^i_j\leq c_j^{l_j}<x_j<c_j^{l_j+1}\leq b^i_j$. Thus $K^{\mathcal{L}}\subset K\cap \bar{R}_i\subset \bar{R}_i$ and hence $(K^{\mathcal{L}})^\circ\subset\bar{R}_i^\circ=R_i$. This shows that the interior of $K^{\mathcal{L}}$ is included in some $R_i$.

Answer 3

I can offer a ¨bootstrap¨ approach, which should really be a comment, but is too long. I think if you do the proof in $\mathbb R^2$ by first drawing a picture, the idea will be much clearer and extends easily to $\mathbb R^n.$ We have $K=[a,b]\times [c,d]$ and $R_i=[a_i,b_i]\times [c_i,d_i].$ In the first place, because $K$ is compact and $(\cup_{i=1}^n R_i)^c$ is closed, and disjoint from $K,\ d(K,(\cup_{i=1}^n R_i)^c)=r>0,$ so you may assume that $\cup_{i=1}^n R_i$ is itself an open rectangle $R$ containing $K$ and concentric with $K$. Then, the picture tells the story. We will take $\overline R_i\cap \overline R_j$ for $1\le i,j\le n$ to produce the required rectangles. Explicitly: the $R_i$ are a collection of rectangles, at least two of which must overlap, whose union is $R$. Now, project each rectangle $\overline R_i\cap K$ on the $x$ axis and then on the $y$ axis. We get a partition $P_x$ of $[a,b]$ and $P_y$ of $[c,d]$. Using these partitions, we get a collection of almost disjoint rectangles whose union is $K$, with the desired property that the interior of each $[x_k,x_{k+1}]\times [y_l,y_{l+1}]$ lies strictly in some $R_i$: indeed, suppose $(z_1,w_1)$ and $(z_2,w_2)$ are both interior to $[x_k,x_{k+1}]\times [y_l,y_{l+1}]$ but lie in different $R_i.$ Without loss of generality $z_1<z_2.$ Then, by construction there is an $x\in P_x$ and such that $z_1<x<z_2.$ This is a contradiction because by construction, there are no partition points between $x_k$ and $x_{k+1}.$