# Homework Help: Measure zero

1. Nov 1, 2011

### gotjrgkr

1. The problem statement, all variables and given/known data

While studying a book "analysis on manifolds" by munkres, I see a definition of measure zero. That is,
Let A be a subset of R$^{n}$. We say A has measure zero in R$^{n}$ if for every ε>0, there is a covering Q$_{1}$,Q$_{2}$,... of A by countably many rectangles such that $\sum$$_{i=1}$$^{\infty}$v(Q$_{i}$)<ε.

But, in this text, there's no remark about the countable set. I mean, it seems to me that countable set is not defined.

Could you tell me what the "countably many" mean in the above definition??
2. Relevant equations

3. The attempt at a solution

2. Nov 1, 2011

### CompuChip

A countable set is a set from which you can find an injection to the naturals.

3. Nov 1, 2011

### HallsofIvy

Munkres assumes that you already know what "countable" means. As CompuChip said, a set is countable if there is a one to one mapping to a subset of the positive integers. (Some text books use the term "countable" to mean an infinite set that can be mapped to a subset of the positive integers, some use the term to include finite sets.)

4. Nov 1, 2011

### gotjrgkr

Ah,

If I use the term "countable_#" for a set A to indicate that there's a bijection from the set A to the set of all natural numbers, the word "countable" in the text book would
mean either "finite" or "countable_#", right?? Am I right?

But, I still have a difficulty to understand the meaning of the definition of measure zero.
I don't have any idea how to see the case when for a given ε>0, there's a finite set of rectangles covering a given set. what does the $\sum$$_{i=1}$$^{\infty}$v(Q$_{i}$) mean in this case??

Last edited: Nov 1, 2011
5. Nov 1, 2011

### HallsofIvy

I presume that the $Q_i$ are non-overlapping sets whose union is some larger set, A, say. That says that the measure of the set A is the sum of the measures of each individual $Q_i$. Notice that they are indexed by i going from 1 to infinity. That is, the sets are labeled $Q_1$, $Q_2$, ... That labeling itself maps the sets to the positive integers so there must be a "countable" number of such sets (which are not necessarily countable themselves).

6. Nov 1, 2011

### gotjrgkr

I'm a little confused...
If a set A has measure zero, then
do you mean that the meaning of "countable" in the text is such that for any given ε>0, there is a set of rectangles Q$_{i}$ which is mapped to the set of all natural numbers by a bijection(I mean, this bijection is a one to one mapping of the set of rectagles Q$_{i}$ onto the set of all natural numbers) and whose sum of the volumes of rectangles is less than ε and union of the rectangles contains A?

7. Nov 1, 2011

### CompuChip

Maybe an explicit example will help here.

Suppose you take the line segment from (0, 0) to (1, 0) in R2 and pick a value for ε.
If you take a rectangle with corners (0, -ε/2), (0, ε/2), (1, ε/2) and (1, -ε/2) you have a set with one rectangle that covers the line segment and has total measure ε. This is a very easy way to demonstrate that the line segment has measure zero. If you insist that "countable" excludes "finite" you can of course get the same result but through a longer route :-)

If you consider the x-axis in R2, you can no longer do with finitely many rectangles even if you wanted.
What you can do, for example, is let Qn be a rectangle around each integer, of height ε / n2 (I think - you might want to check this). There will be infinitely many, but still countably many.

8. Nov 1, 2011

### gotjrgkr

So, in the first case, there's only one rectangle for the given ε, so that it's finite and it implies there is a finite set(that is, countable set) of rectangles(actually only one) which covers the line segment, right?
I can imagine what you say about the first case, but could you explain more specifically about the second case?

9. Nov 1, 2011

### CompuChip

OK.
What I meant is, let Q0 be the rectangle (-1/2, 1/2) x (-ε'/2, ε'/2). I use the normal Cartesian product notation, so
$$(a, b) \times (c, d) = \{ (x, y) \mid x \in (a, b), y \in (c, d)$$
and (a, b) is the open interval a < x < b.

For all other integers n, define
Qn = (n - 1/2, n + 1/2) x (-ε' / 2n2, ε' / 2n2).

Now you can compute the total surface:
$$\sum_{n = -\infty}^{\infty} \nu(Q_n) = \epsilon' + 2 \sum_{n = 1}^\infty \frac{\epsilon'}{n^2}$$
(this is straightforward, although it takes some sum manipulations to get there).

Using a standard result that
$$\sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
you can show that you get $\epsilon'(1 + \pi^2 / 3)$.

Some things I'll leave for you to think about:
(*) I have defined Qn for 0 and negative integers as well. However, the set of all these rectangles is still countable (hint: 0, 1, -1, 2, -2, ...).
(*) If you scale $\epsilon'$ in the preceding by a suitable constant (e.g. $\epsilon = \epsilon' / (1 + \pi^2 / 3)$ should do) you can get the result smaller than $\epsilon$.
(*) You might want to run through my calculations - I have (intentionally) skipped some of the more technical steps. Let me know if you can't follow.

10. Nov 1, 2011

### gotjrgkr

I got it!! I see what you mean..
Anyway, in the munkres's text, he also treats a finite set as countable set as well as an infinite set mapping to the set of positive integers by one-to-one and onto fashion.
I was confused because in some other books, for example, rudin's principle of mathematical analysis, a set A is said to be countable if there is a bijection between A and the set of natural numbers.
I wanted to know which one is used in the munkres's text.
I really appretiate all of you. It helps me a lot!!
(please, let me know if there is a wrong part in the statement above that I wrote.)

Last edited: Nov 1, 2011