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Union of countable many sets

  1. Apr 26, 2006 #1
    Please help me with this.

    Let be two functions of x, B(x) and A(x).

    Boas ("A primer or real functions", p. 137) says:

    If A(x) < B(x), there are rational numbers r and R such that A(x) < r < R < B(x) (OK). Since there are only countable many pairs of rational numbers (OK), the set where B(x) > A(x) is contained in the union of countably many sets of measure zero (why?) ...

    I know that a set composed by two numbers has measure zero, and I assume that "there are only countable many pairs of rational numbers", but I fail to see why this implies the next statement ("the set where B(x) > A(x) is contained in the union of countably many sets of measure zero").

    Thanks for your help.
     
  2. jcsd
  3. Apr 26, 2006 #2

    AKG

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    What you've stated is false, and I suspect you just haven't said it in English properly. Let B(x) = 5, A(x) = 3. Then the set where B(x) > A(x) is all of R, which is not in the union of countably many sets of measure zero. Note, a countable union of measure-zero sets is itself a measure-zero set.
     
  4. Apr 27, 2006 #3
    Well, the only thing where I tried to simplify Boas paragraph was this:

    Instead of B(x) and A(x) Boas puts f+(x) and f-(x), where f+(x) and f-(x) are defined as follows:

    f+(x) =

    lim sup {f(x+h) - f(x)}/ h
    h->0+


    f-(x) =

    lim inf {f(x+h) - f(x)}/ h
    h->0-
     
  5. Apr 27, 2006 #4

    matt grime

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    Or to put it another way, the only thing you did was fail to tell people that A and B are of a very special type.

    Are there any other hypotheses you're missing from f? Like is it piecewise continuouos, continuous, measurable, differentiable....? (If it is differentiable then the set of x where A(x)<B(x) is empty, for instance.)
     
  6. Apr 27, 2006 #5
    Sorry.

    Function f is monotonic (non decreasing).
     
  7. Apr 27, 2006 #6
    Matt, the complete theorem is:

    If the function f is monotonic, then it is differentiable almost everyewhere. From the proof of this theorem, which covers two pages of Boas little book, I took the paragraphs of my initial post.
     
  8. Apr 28, 2006 #7

    HallsofIvy

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    That certainly doesn't clarify your original post!

    Your original post said
    Do you mean that A and B are functions such that A(x)< B(x) for all x? In that case, it doesn't make sense to talk about "the set where B(x)> A(x)". It that was not what you meant, then what did you mean by "If A(x)< B(x)"? It might help if we knew what A(x) and B(x) are!
     
  9. Apr 28, 2006 #8
    Boas says:

    Let I be an interval. Let f be a non decreasing function.

    The set { x / x belongs to I and f-(x) < f+(x) } has measure zero. "For, if f+(x) > f-(x), there are rational numbers r and R such that f-(x) < r < R < f+(x). Since there are only countable many pairs or rational numbers (I assume this), the set where f+(x) > f-(x) is contained in the union of countably many sets of measure zero (why??) and so is itself of measure zero" (ok).
     
  10. Apr 28, 2006 #9

    HallsofIvy

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    I guess I now have to ask what you mean by f-(x) and f+(x)!
     
  11. Apr 28, 2006 #10
    Please see third post.

    Thanks anyway.
     
  12. Apr 28, 2006 #11

    AKG

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    Okay, there are two things to see here:

    1)

    [tex]\{x \in I\ |\ f_+(x) > f_-(x)\} = \bigcup _{(r,R) \in \mathbb{Q}^2}\{x \in I\ |\ f_+(x) > R > r > f_-(x)\}[/tex]

    2)

    For each pair (r,R) in Q², the set {x in I | f+(x) > R > r > f-(x)} is a zero set (a set of measure zero).

    If you're getting stuck with number 1), think about it some more, and keep in mind that if a > b, then there are rationals R and r such that a > R > r > b. For number 2), it's certainly not a trivial fact. My book uses something called the Chebyshev Lemma, as well as the Vitali Covering Lemma to prove it. Does your textbook just assert without any justification? If so, I recommend picking up another textbook.
     
  13. Apr 28, 2006 #12
    Those are Boas's words in his book.

    Boas assumes that there are only countable many pairs of rational numbers and (it seems so) that statement supports the next one: " (then)the set where f+(x) > f-(x) is contained in the union of countable many sets of measure zero", ... but I fail to see how.
     
  14. Apr 28, 2006 #13

    AKG

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    I thought I was clear. Please actually read this post:

    1) The number of pairs of rational points is countable. This is just set theory, and is really something you should know if you're doing a real analysis course. The rationals are countable, and any countable Cartesian product of countable sets is itself countable.

    2) The set {x in I | f+(x) > f-(x)} is the union of all sets of the form {x in I | f+(x) > R > r > f-(x)} where r and R are rational numbers, i.e. (r,R) is a rational pair. This is an easy fact, and I've already asked you to think about it.

    3) Each set {x in I | f+(x) > R > r > f-(x)} is a zero set. This is not a trivial fact. This requires, at least in my book, the Chebyshev Lemma and the Vitali Covering Lemma. I can't believe Boas just stated it like that. My book gives about one full page to prove this fact. I'm not going to tell you why its true, because that would involve restating those lemmas I mentioned. You should be able to look this up yourself. If your textbook really doesn't prove it (are you sure he didn't justify this earlier, or maybe it's in the next paragraph and you just haven't read far enough?), get another book. I have Pugh's Real Mathematical Analysis which I find to be a great book. This theorem you're looking at is given as Lebesgue's Last Theorem.

    Anyways, most of what I said was just repeat and unnecessary elaboration of what I've already said. You aren't clear on what exactly you don't understand, so please try to clarify. Also, I doubt this is what was confusing you, but since there are only countably many rational pairs (r,R), there are only countably many sets of the form {x in I | f+(x) > R > r > f-(x)}, and if each of these is a zero set, then:

    [tex]\bigcup _{(r,R) \in \mathbb{Q}^2}\{x \in I\ |\ f_+(x) > R > r > f_-(x)\}[/tex]

    is clearly a countable union of zero sets. If you understand point 2) above, you'll understand why he says that {x in I | f+(x) > f-(x)} is a countable union of __________ sets. If you understand point 3) above, then you'll understand why that blank __________ was filled in with "measure-zero". According to you, he's given you no way to understand point 3). So is this where you're having problems understanding?
     
    Last edited: Apr 28, 2006
  15. Apr 28, 2006 #14
    I will review slowly the previous chapters of Boas's book. Thank you AKG for your time and insights.

    Castilla.
     
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