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Show that differentiable curves have measure zero in R^2

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data
    (a) Let [itex]\alpha:I=[a,b]→R^2[/itex] be a differentiable curve. Assume the parametrization is arc length. Show that for [itex]s_{1},s_{2}\in I, |\alpha(s_{1})-\alpha(s_{2})|≤|s_{1}-s_{2}|[/itex] holds.

    (b) Use the previous part to show that given [itex]\epsilon >0[/itex] there are finitely many two dimensional open discs [itex]B_{\epsilon}(x_{i}), i=1,..,n[/itex] such that [itex]\alpha (I)\subset \cup _{i=1..n}B_{\epsilon}(x_{i})[/itex] and [itex]\sum_{i}Area(B_{\epsilon}(x_{i}))<\epsilon[/itex].

    2. The attempt at a solution

    (a) For this I made an argument using the mean value theorem for equality. For all [itex]s\in I[/itex], we have [itex]|\alpha '(s)|=1[/itex] since the curve is parametrized by arc length. Then, given some [itex]s_{o}\in (s_{1},s_{2})[/itex] for some [itex]s_{1},s_{2}\in[a,b][/itex] we have by the mean value theorem

    [itex]\frac{\alpha(s_{1})-\alpha(s_{2})}{s_{1}-s_{2}}=\alpha'(s_{o})[/itex]

    whereby,

    [itex]\frac{|\alpha(s_{1})-\alpha(s_{2})|}{|s_{1}-s_{2}|}=1[/itex], so equality holds.

    I am unsure how I should show that it has to be less than. I will have to think of this more. What I am really stuck on is how to even attempt part (b). Any suggestions would be very appreciated.
     
  2. jcsd
  3. Apr 4, 2014 #2

    micromass

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    You seem to be applying the mean-value theorem like ##\alpha## was a function in one-variable. But in reality, it is a vector-valued function. Do you know the mean-value theorem for vector-valued functions?
     
  4. Apr 4, 2014 #3
    Actually, I hadn't caught that until you pointed it out. I am not familiar with that theorem.
     
  5. Apr 4, 2014 #4

    micromass

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    OK, doesn't matter, you can do it with integrals easily.

    LeTake ##v## a vector with ##\|v\|= 1##. Show that

    [tex]\|\alpha(b) - \alpha(a)\| = \int_a^b \alpha^\prime(t)\cdot v ~dt \leq \int_a^b\|\alpha^\prime(t)\|dt[/tex]

    then take a suitable ##v## to prove your conjecture.
     
  6. Apr 4, 2014 #5
    Ok, so I have taken care of that. Thanks for the suggestions, they were quite helpful. I have also shown the second part quite well enough. The only thing that remains is the following:

    (c) If h is a bounded continuous function on [itex]R^{2}[/itex], using the previous part, give a reasonable argument as to why

    [itex]∫_{\alpha}h dx dy=0[/itex]

    should hold.

    The thought I had was to use the change of variables theorem to integrate over a larger set containing [itex]\alpha(I)[/itex]. But I suppose you would have to define some local parametrization in [itex]R^2[/itex], unless you took the union of the disks as in part (b) to be the locally parametrized surface? That seems really odd though and doesn't make a lot of sense. . .
     
  7. Apr 4, 2014 #6

    micromass

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    So in general, if ##E## is any set of measure ##0## and if ##f## is some function (let's say continuous), then

    [tex]\int_E f = 0[/tex]

    Try something like

    [tex]\left|\int_E f\right| \leq \int_E |f|\leq \int_E \textrm{sup}(|f|)\leq \lambda(E)\textrm{sup}(|f|)[/tex]
     
  8. Apr 4, 2014 #7
    What is the significance of [itex]\lambda(E)[/itex] in this case? Could you define that for me? It seems rather arbitrarily chosen to me.
     
  9. Apr 4, 2014 #8

    micromass

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    The function ##\lambda## is the Lebesgue measure. Have you not seen this?
     
  10. Apr 5, 2014 #9
    Yes, now that makes sense. Our notation uses m rather than [itex]\lambda[/itex]. I appreciate your help.
     
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