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

  • #1
psie
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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 -dimensional, closed rectangle with sides oriented parallel to the coordinate axes, or rectangle for short, is a subset of the form where for . The volume of is

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

Lemma 2.5. Suppose that is an -dimensional rectangle where each closed, bounded interval is an almost disjoint union of closed, bounded intervals , Define the rectangles Then

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

Proposition 2.6. If a rectangle is an almost disjoint, finite union of rectangles , then If is covered by rectangles , which need not be disjoint, then

Proof. Suppose that is an almost disjoint union of the rectangles . Then by 'extending the sides' of the , we may decompose into an almost disjoint collection of rectangles that is obtained by taking products of subintervals of partitions of the coordinate intervals into unions of almost disjoint, closed subintervals. Explicitly, we partition into where the are obtained by ordering the left and right th coordinates of all faces of rectangles in the collection , and define rectangles as in (1).

Each rectangle in the collection is an almost disjoint union of rectangles , and their union contains all such products exactly once, so by applying Lemma 2.5 to each and summing the results we see that Similarly, is an almost disjoint union of all the rectangles , so Lemma 2.5 implies that and (2) follows.

If a finite collection of rectangles covers , then there is an almost disjoint, finite collection of rectangles such that To obtain the , we replace by the rectangle , 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 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 ?
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.
 
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  • #2
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.
 
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  • #3
Upon closer thought, I think I have figured out the answers to my questions:

1. is an almost disjoint union of rectangles , each which is in turn an almost disjoint union of rectangles from the collection . So the union of the rectangles contains all such 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 , 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.
 
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