Two basic results from measure theory -- on volumes of rectangles

[psie]

TL;DR Summary

I have some questions regarding two basic results (but not so trivial to prove) in measure theory. It concerns the volumes of rectangles that cover another rectangle.

The notes I'm reading are from here. But I have summarized all the necessary details in this post. My question concerns the proposition, but it uses the definition below and the lemma.

Definition 2.1. An $n$-dimensional, closed rectangle with sides oriented parallel to the coordinate axes, or rectangle for short, is a subset $R\subset \mathbb R^n$ of the form $$R=[a_1,b_1]\times[a_2,b_2]\times\cdots\times[a_n,b_n]$$ where $-\infty<a_i\leq b_i<\infty$ for $i=1,\ldots,n$. The volume $\mu(R)$ of $R$ is $$\mu(R)=(b_1-a_1)(b_2-a_2)\ldots (b_n-a_n).$$

We say two rectangles are almost disjoint if they intersect at most along their boundaries.

Lemma 2.5. Suppose that $$R=I_{1} \times I_{2} \times \cdots \times I_{n}$$ is an $n$-dimensional rectangle where each closed, bounded interval $I_{i} \subset \mathbb{R}$ is an almost disjoint union of closed, bounded intervals $\left\{I_{i, j} \subset \mathbb{R}: j=1, \ldots, N_{i}\right\}$, $$I_{i}=\bigcup_{j=1}^{N_{i}} I_{i, j}$$ Define the rectangles \begin{equation*} S_{j_{1} j_{2} \cdots j_{n}}=I_{1, j_{1}} \times I_{2, j_{2}} \times \cdots \times I_{n, j_{n}} . \tag{1} \end{equation*} Then $$ \mu(R)=\sum_{j_{1}=1}^{N_{1}} \cdots \sum_{j_{n}=1}^{N_{n}} \mu\left(S_{j_{1} j_{2} \ldots j_{n}}\right)$$

I omit the proof for sake of brevity, but see the link above if you are interested.

Proposition 2.6. If a rectangle $R$ is an almost disjoint, finite union of rectangles $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$, then \begin{equation*} \mu(R)=\sum_{i=1}^{N} \mu\left(R_{i}\right) \tag{2} \end{equation*} If $R$ is covered by rectangles $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$, which need not be disjoint, then \begin{equation*} \mu(R) \leq \sum_{i=1}^{N}\mu\left(R_{i}\right) \tag{3} \end{equation*}

Proof. Suppose that $$R=\left[a_{1}, b_{1}\right]\times\left[a_{2}, b_{2}\right] \times \cdots\times\left[a_{n}, b_{n}\right]$$ is an almost disjoint union of the rectangles $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$. Then by 'extending the sides' of the $R_{i}$, we may decompose $R$ into an almost disjoint collection of rectangles $$\left\{S_{j_{1} j_{2} \ldots j_{n}}: 1 \leq j_{i} \leq N_{i} \text { for } 1 \leq i \leq n\right\}$$ that is obtained by taking products of subintervals of partitions of the coordinate intervals $\left[a_{i}, b_{i}\right]$ into unions of almost disjoint, closed subintervals. Explicitly, we partition $\left[a_{i}, b_{i}\right]$ into $$a_{i}=c_{i, 0} \leq c_{i, 1} \leq \cdots \leq c_{i, N_{i}}=b_{i}, \quad I_{i, j}=\left[c_{i, j-1}, c_{i, j}\right]$$ where the $c_{i, j}$ are obtained by ordering the left and right $i$th coordinates of all faces of rectangles in the collection $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$, and define rectangles $S_{j_{1} j_{2} \ldots j_{n}}$ as in (1).

Each rectangle $R_{i}$ in the collection is an almost disjoint union of rectangles $S_{j_{1} j_{2} \ldots j_{n}}$, and their union contains all such products exactly once, so by applying Lemma 2.5 to each $R_{i}$ and summing the results we see that $$\sum_{i=1}^{N} \mu\left(R_{i}\right)=\sum_{j_{1}=1}^{N_{1}} \cdots \sum_{j_{n}=1}^{N_{n}} \mu\left(S_{j_{1} j_{2} \ldots j_{n}}\right)$$ Similarly, $R$ is an almost disjoint union of all the rectangles $S_{j_{1} j_{2} \ldots j_{n}}$, so Lemma 2.5 implies that $$\mu(R)=\sum_{j_{1}=1}^{N_{1}} \cdots \sum_{j_{n}=1}^{N_{n}} \mu\left(S_{j_{1} j_{2} \ldots j_{n}}\right)$$ and (2) follows.

If a finite collection of rectangles $\left\{R_{1}, R_{2}, \ldots, R_{N}\right\}$ covers $R$, then there is an almost disjoint, finite collection of rectangles $\left\{S_{1}, S_{2}, \ldots, S_{M}\right\}$ such that $$R=\bigcup_{i=1}^{M} S_{i}, \quad \sum_{i=1}^{M} \mu\left(S_{i}\right) \leq \sum_{i=1}^{N} \mu\left(R_{i}\right)$$ To obtain the $S_{i}$, we replace $R_{i}$ by the rectangle $R \cap R_{i}$, and then decompose these possibly non-disjoint rectangles into an almost disjoint, finite collection of sub-rectangles with the same union; we discard 'overlaps' which can only reduce the sum of the volumes. Then, using (2), we get $$\mu(R)=\sum_{i=1}^{M} \mu\left(S_{i}\right) \leq \sum_{i=1}^{N} \mu\left(R_{i}\right)$$ which proves (3).

Questions:
1. Maybe this is silly, but I can not for the life of me make sense of the sentence "...and their union contains all such products exactly once...". Which union does the author mean and which products are meant?
2. How do we know $\sum_{i=1}^{M} \mu\left(S_{i}\right) \leq \sum_{i=1}^{N} \mu\left(R_{i}\right)$?
3. I don't understand why overlaps reduce the volume. Shouldn't the volume of the overlap be counted twice? I don't have clear picture in my head of what is going on.

 

[mathwonk]

if I were you, I would take n=2, N=3, and just make up my own proof, as this kind of thing is more than a bit tedious to read.

 

[psie]

Upon closer thought, I think I have figured out the answers to my questions:

1. $R$ is an almost disjoint union of rectangles $\{R_1,\ldots,R_N\}$, each which is in turn an almost disjoint union of rectangles from the collection $\left\{S_{j_{1} j_{2} \ldots j_{n}}: 1 \leq j_{i} \leq N_{i} \text { for } 1 \leq i \leq n\right\}$. So the union of the rectangles $\{R_1,\ldots,R_N\}$ contains all such $S_{j_{1} j_{2} \ldots j_{n}}$ exactly once.
2. This follows from the remark that discarding overlaps can only reduce the sum of the volumes. Here it's best to draw a picture, e.g. if $(R\cap R_1)\cap(R\cap R_2)\neq\emptyset$, then discard this part.
3. The author means the overlaps only add extra volume to the sum of volumes, hence discarding them reduces the sum of the volumes. It helps drawing a picture.