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A set $R\subset\mathbb R^n$ with the form $[a_1,b_1]\times\cdots\times[a_n,b_n]$ where $a_i<b_i\ \forall i=1,\dots,n$ is called a $n$-rectangle, with volume $|R|=(b_1-a_1)\cdots(b_n-a_n)$.

A set $D\subset\mathbb R^n$ is of $\textbf{content zero}$ if $\ \forall\varepsilon>0,\ \exists\ n$-rectangles $R_1,\dots,R_m$ such that

$$D\subseteq\displaystyle\bigcup_{i=1}^mR_i\ \text{ and } \ \displaystyle\sum_{i=1}^m|R_i|<\varepsilon.$$

Show that any $n$-rectangle is not a set of content zero.

My attempt is to consider $\varepsilon=|R|$, then there exists $n$-rectangles $R_1,\dots,R_m$ such that $|R_1|+\dots+|R_m|<|R|$ and $R\subseteq R_1\cup\cdots\cup R_m$. This implies $|R|\leq\big|R_1\cup\cdots\cup R_m\big|$.

Now the problem is to show that $\big|R_1\cup\cdots\cup R_m\big|\leq|R_1|+\cdots+|R_m|$ which is a contradiction.

Any idea ?

PermQi
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  • Show that if $R\subset \bigcup_{i=1}^mR_i$ then $|R|\leq \sum_{i=1}^m|R_i|$ (each $R,R_i$ being an $n$-rectangle). This is intuitive, but can be tricky to prove… – peek-a-boo Jul 26 '23 at 16:40
  • can you give me a hint on this tricky fact ? – PermQi Jul 26 '23 at 16:48
  • the $n$-dimensional case is just extra notation, so focus on $n=1$ first. In this case, each $R_i$ is an interval (and you may without loss of generality assume $R_i\subset R$ by replacing $R_i$ with $R_i\cap R$). Now, collect all the endpoints of the $R_i$’s; this forms a partition of $R$. For example, if $R=[0,10]$ and $R_1=[0,7]$ and $R_2=[3,10]$, then we have the partition $P={0,3,7,10}$ of $R$. Now, a little bit of thought will give you the desired inequality (think back to this example, or a few other examples to see how to convert intuition into a formal proof). – peek-a-boo Jul 26 '23 at 16:51
  • So here is my thought. Consider the 1d-case. Suppose $R=[a,b]\subseteq R_1\cup\cdots\cup R_m,\ R_i=[a_i,b_i]$. Assume without loss of generality that $R_1$ is the first rectangle of the cover from the left side. Let \begin{align} S_1&=R_1=[a_1,b_1] \ S_2&=[b_1,b_1+b_2-a_2] \ S_i&=\left[\sum_1^{i-1}b_i-\sum_2^{i-1}a_i\ ,\ \sum_1^ib_i-\sum_2^ia_i\right] \end{align} Then $|S_i|=|R_i|=b_i-a_i$, and $R_1\cup\cdots\cup R_m=[\min a_i,\max b_i]\subseteq S_1\cup\cdots\cup S_m$ Hence, $\big|R_1\cup\cdots\cup R_m\big|\leq\big|S_1\cup\cdots\cup S_m\big|=|S_1|+\cdots+|S_m|=|R_1|+\cdots+|R_m|$ – PermQi Jul 26 '23 at 17:52
  • But doing this way, I feel something seems not rigorous. That is $R_1\cup\cdots\cup R_m\subseteq S_1\cup\cdots\cup S_m$, to have this we need to show that $\max b_i \leq\sum\limits_1^mb_i-\sum\limits_2^ma_i$, which is not simple. Then, I need to generalize into n-dimensional case, which I feel not quite clear. – PermQi Jul 26 '23 at 18:06
  • Whatever the $a_i,b_i$ are, rename all of them and arrange them in increasing order, $a=t_0<t_1<\dots<t_p=b$ (where $R=[a,b]$). Then, $|R|=b-a=\sum_{j=1}^p(t_j-t_{j-1})\leq\cdots$. If you’re still in doubt, look up the corresponding integration chapters of Spivak/Munkres. – peek-a-boo Jul 27 '23 at 02:30

1 Answers1

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A straightforward way to show this is to show that for any covering of a rectangle $D$ by rectangles $R_k$, we have $|D| \le \sum_k |R_k|$.

The proof is tedious & long winded, but straightforward. The idea is to first trim the $R_k$ so that they are all contained in $D$, then split $D$ into many sub rectangles, each of which is contained in at least one of the $R_k$.

Call two rectangles $R,R'$ overlapping iff they an interior point. I will call a set of the form $\prod_k [a_k,b_k]$ where at least one of the pairs $a-k,b_k$ satisfies $a_k=b_k$ a flat rectangle.

Note that if $ D \subset \cup_k R_k$ then $ D \subset \cup_k (R_k \cap D)$. We can discard any $R_i$ if $R_i \cap D = \emptyset$. However, it is possible that $R_i \cap D$ is a flat rectangle. It is straightforward to show that if $R_i$ is a flat rectangle, then $D \subset \cup_{k\neq i} (R_k \cap D)$. Repeating shows that with $R_k' = D \cap R_k$ and $K$ being the set of non empty non flat $R_k$ we have $D \subset \cup_{k \in K} R_k' \subset \cup_k R_k$. We also note that $|R_k'| \le |R_k|$ for each $k \in K$, so it is sufficient to show that $|D| \le \sum_{k \in K} |R_k'|$.

So, we can assume that $R_k \subset D$ to simplify subsequent notation.

If we have a rectangle $R= \prod_k [a_k,b_k]$ and we 'split' one of the sides (with $m_k \in (a_k, b_k)$) into $[a_k,b_k] = [a_k, m_k] \cup [m_k, b_k]$ to create two non overlapping rectangles $R_1,R_2$, and we see that $|R| = |R_1|+|R_2|$. Hence we can split a rectangle into many parts one side at a time and the sum of the volumes remains constant.

Now collect all the $a_1,b_1$s of each rectangle into a set $P_1$, and similarly for the other sides to get $P_2,..,P_n$. The $P_i$ form a partition of each side of $D$, and using the technique of the preceding paragraph, we can refine $D$ into a collection of rectangles $D_k$ such that $D = \cup_k D_k$ where $|D| = \sum_k |D_k|$ and each $D_k$ is completely contained in at least one of the $R_j$.

Let $I$ be a subset of indices such that $D_k \in \cup_{i \in I} R_i$ for each $k$, then we have $|D| = \sum_k |D_k| \le \sum_{i \in I} |R_k| \le \sum_k |R_k|$. In particular, $D$ is not of content zero.

copper.hat
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