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Measure zero and differentiability

  1. Nov 24, 2007 #1

    quasar987

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    [SOLVED] measure zero and differentiability

    1. The problem statement, all variables and given/known data
    I proved in the preceding exercise the following characterization of measure zero:

    "A subset E of R is of measure zero if and only if it has the following property:

    (***) There exists a sequence [tex]J_k=]a_k,b_k[[/tex] such that every x in E belongs to an infinity of J_k and

    [tex]\sum_{k=1}^{+\infty}(b_k-a_k)<+\infty[/tex]"

    Now the question is the following:

    Let E be of null measure and {J_k} be as above. Let also [tex]f_E:\mathbb{R}\rightarrow\mathbb{R}[/tex] be the increasing function defined by

    [tex]f_E(x) = \sum_{k=1}^{+\infty}\lambda(]-\infty,x]\cap J_k)[/tex]

    Show f is not differentiable at any point of E.


    2. Relevant equations

    Differentiable iff the limit of the differential quotient exists and is bounded iff the left and right derivative are equal

    3. The attempt at a solution

    Let h>0 and x0 be in E. I can write the right derivative and use the additivity of measure to simplify to

    [tex]D_rf_E(x_0)=\lim_{h\rightarrow 0}\frac{\sum_{k=1}^{+\infty}\lambda([x,x+h]\cap J_k)}{h}[/tex]

    then what? What difference does it make than x is in E? I mean, how does the fact that is belongs to an infinity of J_k comes in?
     
    Last edited: Nov 24, 2007
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  3. Nov 25, 2007 #2

    EnumaElish

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    Is "x + h" in J_k for some h > 0?
     
  4. Nov 25, 2007 #3

    quasar987

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    Well, each J_k is open and contains x, so for every k, there is an h, call it h_k such that x + h is in J_k, yes.
     
  5. Nov 27, 2007 #4

    quasar987

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