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Differentiability of Monotone Function's: Lebesgue's Theorem

  1. Mar 13, 2013 #1
    a)Does convergence imply being properly defined? So would it not be properly defined if it was divergent?

    b)I am having trouble why the last part (in the attachment) says, "Then, by (1), [itex]f(x_0) - f(x) \geq dfrac{2^k} for all [itex]x < x_0[/itex]." But does (1) tell us that it's "equal" instead of "greater than or equal to"?

    Thanks in advance

    Attached Files:

  2. jcsd
  3. Mar 13, 2013 #2
    a) We define the function

    [tex]f(x) = \sum_{q_n\leq x} \frac{1}{2^n}[/tex]

    But if the series in the RHS diverges for a certain ##x##. Then we would have ##f(x)=+\infty##. But ##+\infty## is not a real number. We want ##f(x)## to be a real number for any ##x##. This is only satisfied if the series converge.

    So yes, if we say that ##f(x)## is well-defined, then that actually means that the series converge.

    b) I'm not quite understanding why you expect the equality to hold. You have

    [tex]f(q_k) - f(x) = \sum_{x<q_n\leq q_k} \frac{1}{2^n} \geq \frac{1}{2^k}[/tex]

    The inequality in the end of course holds because ##1/2^k## is a term in the sum on the LHS. So the inequality means that you drop every term in the sum except ##1/2^k##. I don't get why you think that this should be an equality.
  4. Mar 13, 2013 #3
    Thanks, but I have one more question...

    [tex]f(q_k) - f(x) = \sum_{x<q_n\leq q_k} \frac{1}{2^n} \geq \frac{1}{2^n} [/tex], right? But then how do we have [itex]|f(x)-f(x_0)| = \frac{1}{2^n} < \frac{1}{2^n}[/itex]?

    I don't know, I think I'm confused about what n represents.
  5. Mar 13, 2013 #4


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    Basically, for any finite n, you can construct a ball about xo that excludes all the

    elements Sn:={q1,...,qn}, since min{ |xo-qi| i=1,2,..n} :=d >0

    meaning, for finitely-many elements qi , the distance from xo to any of them is

    non-zero, and the since the set Sn is finite, the set of distances from xo to any member in Sn has a minimum ( no need to worry about infs.) ,

    say that minimum is called d.

    Then, within the ball, say, B(x,d/2) , the only possible elements of C in the ball

    are {q(n+1),q(n+2),....} . Now, by definition of f(x), can

    you see why the difference can be made as small as possible (think of the elements of C between x and xo in the ball)?
    Last edited: Mar 13, 2013
  6. Mar 13, 2013 #5
    Thanks a lot, but I'm not really familiar with "balls"...
  7. Mar 13, 2013 #6


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    An open ball B(x,r) in the real line , i.e., x is any real number and d is a positive real is the set:

    B(x,r):={ y in Reals : |x-y|<d }
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