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Measure zero

  1. Nov 1, 2011 #1
    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[itex]^{n}[/itex]. We say A has measure zero in R[itex]^{n}[/itex] if for every ε>0, there is a covering Q[itex]_{1}[/itex],Q[itex]_{2}[/itex],... of A by countably many rectangles such that [itex]\sum[/itex][itex]_{i=1}[/itex][itex]^{\infty}[/itex]v(Q[itex]_{i}[/itex])<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. jcsd
  3. Nov 1, 2011 #2

    CompuChip

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    A countable set is a set from which you can find an injection to the naturals.
     
  4. Nov 1, 2011 #3

    HallsofIvy

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    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.)
     
  5. Nov 1, 2011 #4
    Ah,
    I really thank you for your reply.
    But, I wanna check myself if I understand your reply well.

    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 [itex]\sum[/itex][itex]_{i=1}[/itex][itex]^{\infty}[/itex]v(Q[itex]_{i}[/itex]) mean in this case??
     
    Last edited: Nov 1, 2011
  6. Nov 1, 2011 #5

    HallsofIvy

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    I presume that the [itex]Q_i[/itex] 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 [itex]Q_i[/itex]. Notice that they are indexed by i going from 1 to infinity. That is, the sets are labeled [itex]Q_1[/itex], [itex]Q_2[/itex], ... 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).
     
  7. Nov 1, 2011 #6
    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[itex]_{i}[/itex] 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[itex]_{i}[/itex] 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?
     
  8. Nov 1, 2011 #7

    CompuChip

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    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.
     
  9. Nov 1, 2011 #8
    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?
     
  10. Nov 1, 2011 #9

    CompuChip

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    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
    [tex](a, b) \times (c, d) = \{ (x, y) \mid x \in (a, b), y \in (c, d)[/tex]
    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:
    [tex]\sum_{n = -\infty}^{\infty} \nu(Q_n) = \epsilon' + 2 \sum_{n = 1}^\infty \frac{\epsilon'}{n^2}[/tex]
    (this is straightforward, although it takes some sum manipulations to get there).

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

    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 [itex]\epsilon'[/itex] in the preceding by a suitable constant (e.g. [itex]\epsilon = \epsilon' / (1 + \pi^2 / 3)[/itex] should do) you can get the result smaller than [itex]\epsilon[/itex].
    (*) 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.
     
  11. Nov 1, 2011 #10
    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
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