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I have tried to read volume 3 of Shakarchi and Stein more than a few times now and I keep getting stuck in chapter one in the same place.

After going through a bunch of basic concepts from analysis we finally come to a lemma, which says that if a given rectangle is the almost disjoint union of other rectangles, then the area of the given rectangle is equal to the sum of the area of the inner rectangles. Ok duh.

So at this point I assume we are going to have a rigorous proof which is going to explain why drawing lines of area 0 in a rectangle doesn't decrease the area of the rectangle. Unfortunately the first sentence of the proof is "we consider the grid formed by extending the sides of all rectangles indefinitely."

Is that supposed to be rigorous? I can think of three different ways to extend the sides of all the inner rectangles indefinitely, and I still can't figure out which one they mean because the picture is pretty strange, and so now I figure I am wrong about why the authors feel the need to prove such an obvious statement.

The next sentence is "This construction yields finitely many rectangles...and a partition of the integers between 1 and M". I'm sorry, but is this a joke, or is this just a weak part of the book that I should skip over?

By the end of the proof I am left very confused as to what the point of this section is. Can someone please explain what I am missing here?

Thanks!

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Why do you think this is a joke? The approach looks perfectly sensible to me. –  Did Feb 21 '13 at 7:55
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To whoever downvoted: This is an eminently reasonable question. –  Potato Feb 21 '13 at 8:08
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@Potato: too much complaining. Too little mathematics. "I can think of three different ways to extend the sides..." why not list them? Maybe one of them is the correct one! What's wrong with "This construction yields finitely many rectangles?" Why does the OP think it is a joke? –  Willie Wong Feb 21 '13 at 8:11
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@WillieWong These are very understandable reactions for someone without a lot of mathematical maturity who is encountering Stein's handwaving for the first time. I think we should endeavor to read the question as charitably as possible, and there is definitely a legitimate mathematical concern here along with the frustration and confusion about rigor. –  Potato Feb 21 '13 at 8:15
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@Matt: you are to extend the sides of the rectangles. Not the rectangles themselves! If I say: we can draw a rectangular grid such that all the "corners" of the original collection of rectangles lie on grid points, do you have a better picture? –  Willie Wong Feb 21 '13 at 8:33

4 Answers 4

up vote 6 down vote accepted

The emphasis in the Lemma is on the word finite. The Lemma fails if you allow infinitely many rectangles. For example, by the Stein and Shakarchi definition, a degenerate rectangle is a rectangle of 0 volume. Now consider a given rectangle $R = [0,1]\times [0,1]$. It can be written as a union of infinitely many almost disjoint rectangles $$ R = \cup_{x\in [0,1]} [0,1]\times [x,x] $$

Each of the degenerate rectangles on the right hand side has 0 volume. The rectangle on the left hand side as positive volume.

The proof shows two steps that are useful in measure theory:

  1. A process of refining the rectangles while maintaining the finiteness of the total number. If you have taken a rigorous course of calculus dealing with Riemann or Darboux sums, you will recognize this process as similar to something you have seen before.
  2. An indication why finiteness is important: as it turns out that the importance of finiteness is manifest already at the level of the real line! (Zeno's paradox!) It shows how once we understand finiteness on the level of one dimension, going to higher dimensions is nothing different at all.

The proof does leave a few things to be desired: in particular it implicitly uses a two step argument which would be better spelled out.

  1. Implicitly the first step is to show that a decomposition of the rectangle into a finite rectangular grid preserves area.
  2. The second step is to show that by the refinement process, a decomposition of a rectangle into a finite almost-disjoint union of rectangles can be refined into a decomposition into a finite rectangular grid, such that each of the original rectangles are made up of finitely many of the "grid" rectangles.

Later on in the book you will read about a lot more of these types of ideas, the basic form of which goes like this:

We are given $U$ a set such that it is a union $\cup_\alpha V_\alpha$ of some sets. The $V_\alpha$ may not be the best sets, so we may not know much about the volume $|V_\alpha|$ and how it relates to the volume $|U|$. But it turns out that there is a deterministic procedure that replaces $V_\alpha$ by a collection $W_\beta$ such that $U\subset \cup_\beta W_\beta$, and that the $W_\beta$ are very well chosen, so that not only do we have control on how large $|W_\beta|$ is, we also know the relationship between $|U|$ and $|W_\beta|$. The deterministic procedure allows us to related $|W_\beta|$ to $|V_\alpha|$, and this in turn finally gives us what we really wanted: a relation between $|U|$ and $|V_\alpha|$.

In the proof, $U$ is the big rectangle $R$, the $V_\alpha$ are the almost-disjoint collection of smaller rectangles $R_i$, and $W_\beta$ are the small grid rectangles.


Let me re-write the proof in a way that is slightly less graphical and more precise, which may help the OP.

Step 1:

Suppose $R$ is a rectangle $[a,b]\times[c,d]$. Let $$a = s_0 < s_1 < s_2 \ldots < s_m = b $$ and $$c = t_0 < t_1 < t_2 \ldots < t_n = d $$ We write $R_{ij}$ for $i = 1\ldots m$ and $j = 1\ldots n$ for the small rectangles $$ R_{ij} = [s_{i-1}, s_i] \times [t_{j-1}, t_j] $$ A little bit of algebra shows you that $$ |R| = (s_m - s_0) \cdot (t_n - t_0) = \sum_{i = 1}^m\sum_{j = 1}^n (s_i - s_{i-1})\cdot(t_j - t_{j-1}) = \sum_{ij} |R_{ij}| $$

Step 2:

Let $R = [a,b]\times [c,d]$ and let $R_k$ be a finite collection of almost disjoint rectangles whose union is $R$. Write $R_k = [a_k, b_k]\times [c_k,d_k]$. Consider the set $$ S = \{ a_1, b_1, a_2, b_2, \ldots, a_K, b_K\} $$ and the set $$ T = \{ c_1, d_1, c_2, d_2, \ldots, c_K, d_K\} $$ (where by convention the sets $\{0,0\} = \{0\}$ [we can cancel repeated numbers]).

Order $S$ and $T$ increasingly and label the points $s_0 = a < s_1 \ldots s_m = b$ and $t_0 = c < t_1 \ldots t_n = d$. We now have a large rectangular grid. What is useful is that for each rectangle $R_k$, we can find $a_k,b_k\in S$ and $c_k,d_k\in T$, so by step 1, the volume $|R_k|$ can be written as the sum of the volumes of some numbers of the grid rectangles $[s_{i-1}, s_{i}] \times [t_{j-1}, t_{j}] = R_{ij}$. Similarly the big rectangle $R$ can also have its volume written as a sum of $|R_{ij}|$. It remains to conclude that each of the $R_{ij}$ is used exactly once in both of the counts. This last step follows from the "almost disjoint" property: that is, if $R_{ij} \subseteq R_k$ and $R_{i'j'} \subseteq R_{k'}$ where $k' \neq k$, then the ordered pairs $(i,j) \neq (i',j')$.

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Thank you for this, it was very helpful. –  Matt Calhoun Feb 23 '13 at 16:26

This is an example of a fact which is obvious. Now, there are lots of "obvious" statements in mathematics whose proofs are important to understand, at least at an introductory level, since the subtleties involved in the proofs often illustrate basic applications of a new technique.

In my opinion, this is not one of those proofs. In other words, this proof is simply a tedious verification of an easily understandable fact, and the methods are basic and have no relation to the techniques in the rest of the text. If you were to simply say "I understand that the sum of volumes of an almost disjoint union of rectangles should equal the volume of their union when their union is also a rectangle" and move on, you would be completely fine.

That the author(s) spend so much time on the proof is unfortunate, though I understand their desire not to bypass details in an introductory text.

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+1. This is often an unpopular opinion. Often when I express this, I'm met with something that amounts to "But real mathematicians prove everything rigorously". –  Katie Dobbs Feb 21 '13 at 10:52
    
I disagree with the implication that the author wasted time. The very next lemma begins "a modification of this argument then yields..." showing that the idea would become useful later (and probably even later in the book than that.) In this case, the proof's context justifies it. This would all have been lost if the idea had been dismissed as "obvious". –  rschwieb Feb 21 '13 at 17:55
    
@KatieDobbs I know the general sentiment you're talking about, but that is more of an issue for authors writing high school texts. Shakarchi and Stein are not making that mistake, rather they are introducing the idea of "refinements." –  rschwieb Feb 21 '13 at 18:04

A few comments before I answer the mathematical question. Yes, this is intuitively obvious. But mathematics is about finding rigorous ways to justify your intuitions, and given the strange beasts one encounters in analysis (nowhere differentiable functions, etc) we need a formal foundation. Stein's aim is to proceed purely formally and build a rigorous concept of measure that does not rely on intuitive arguments.

Now for the math. Let's work in $\mathbb R^2$ to keep things simple. We have a rectangle $[a_1, a_2]\times [b_1,b_2]$ that is the union of some smaller rectangles $[a_{i1}, a_{i2}]\times [b_{i1},b_{i2}]$ By extend all the boundaries, he means consider all lies of the form $y=a_{i1}$ and $x=b_{i2}$ and so forth. Order the vertical lines from left to right and the horizontal lines from bottom to top. For each consecutive pair of horizontal lines and each consecutive pair of vertical lines formed this way, we get a rectangle. These are precisely the rectangles that are drawn in the picture on the right in figure 2. Now, we can prove that these lie in the original rectangle. One such generic rectangle is $[a_{i1}, a_{i2}]\times [b_{j1},b_{j2}]$. Since $a_1 \le a_{i1}< a_{i2} \le a_2$ and likewise for the $b$ coordinates, $[a_{i1}, a_{i2}]\times [b_{j1},b_{j2}]$ is a subset of $[a_1, a_2]\times [b_1,b_2]$.

In a similar fashion, we can prove each of the new cut up rectangles lies in a unique original smaller rectangle. And going back to the definition of almost disjoint, we can prove the unions forming the original smaller rectangles are almost disjoint. You should be able to follow Stein at this point. Let me know if you need more detail.

The point is that while this argument seems handwavey, it can be made completely rigorous if you choose to do so. That is the important thing.

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Unfortunately the first sentence of the proof is "we consider the grid formed by extending the sides of all rectangles indefinitely."

Here is a construction of the grid in dimension $3$ (the general case being an obvious modification).

One starts from a rectangle $R=[x,y]\times[z,t]\times[u,v]$ which is the almost disjoint union of finitely many rectangles $R_k=[x_k,y_k]\times[z_k,t_k]\times[u_k,v_k]$.

Consider all the coordinates $x_k$ and $y_k$ involved and order them as $x=a_0\lt a_1\lt\cdots\lt a_N=y$. Likewise, consider all the coordinates $z_k$ and $t_k$ involved and order them as $z=b_0\lt b_1\lt\cdots\lt b_P=t$ and consider all the coordinates $u_k$ and $v_k$ involved and order them as $u=c_0\lt c_1\lt\cdots\lt c_Q=v$.

Now, fix a bijection from $\mathfrak J=[NPQ]$ to $[N]\times[P]\times[Q]$, say $j\mapsto(n(j),p(j),q(j))$. For every $j$ in $\mathfrak J$, define $$ \tilde R_j=[a_{n(j)-1},a_{n(j)}]\times[b_{p(j)-1},b_{p(j)}]\times[c_{q(j)-1},c_{q(j)}]. $$ Then the rectangle $R$ is the almost disjoint union of the $M=NPQ$ rectangles $\tilde R_j$ and each original rectangle $R_k$ is the union of some $\tilde R_j$, that is, $R_k=\bigcup\limits_{j\in J(k)}\tilde R_j$, where the sets $J(k)$ are disjoint and their union is $\mathfrak J$.

Is all this preferable to Shakarchi and Stein's presentation? I am not sure.

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Which is close to your option 1.). –  Did Feb 21 '13 at 8:52
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"is this preferable..." When done on a blackboard in front of students, S+S's presentation gets the point across quickly. It is one of those things where even an animated GIF would get the point through faster than describing it in words. (Somehow this reminds me of reading some of Thurston's geometric intuitions.) –  Willie Wong Feb 21 '13 at 8:52
    
@WillieWong Funny, Thurston's way of showing mathematics was in my mind when I wrote this answer. –  Did Feb 21 '13 at 8:54

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