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Sard theorem in dimension one.

  1. Dec 28, 2009 #1
    Hi, I need some hints for the proof of the sard theorem in 1 dimension:

    Prove that the set of critical values (f(x) where f'(x)=0) of continuously differentiable f:[a,b]->R has measure 0.

    My attempt:

    Fix [tex]\varepsilon[/tex], Let Crit(f) be the set of critical points of f. We want to show that f(Crit(f)) has measure 0.
    Let [tex]C_k[/tex] be the set of all [tex]x[/tex] in Crit(f) such that [tex]|x-y|<\frac{1}{k} \implies |fx-fy|<\varepsilon|x-y|[/tex].

    Clearly [tex]\bigcup_{k\in\mathbb{N}} C_k \supset \mathrm{Crit}(f) [/tex], so it suffices to show that [tex]f\left(\bigcup_{k\in\mathbb{N}} C_k \right) [/tex] has measure 0.

    For any [tex]C_k[/tex], evenly split [tex][a,b][/tex] into intervals [tex]I_1,\dots,I_n[/tex] such that [tex]\frac{b-a}{n}<\frac{1}{k}[/tex].

    For each [tex]I_i[/tex], if there is a point of [tex]C_k[/tex] in it, say [tex]x[/tex], then

    [tex]|fx-fy|<\varepsilon|x-y|[/tex] for all y in [tex]I_i[/tex].

    Thus if [tex]I_i[/tex] contains a point of [tex]C_k[/tex], then [tex]f(I_i)[/tex] is contained in a open interval of at most [tex]2w \varepsilon[/tex] where w is the width of the interval.

    Thus [tex]f(C_k)[/tex] is contained in an open set of length at most [tex]2w\varepsilon n =2(b-a) \varepsilon=K\varepsilon[/tex].

    Since each [tex]f(C_k)[/tex] has measure bounded above by [tex]K \varepsilon[/tex], and [tex]C_{k}\subset C_{k+1}[/tex] for all k, then

    [tex]\bigcup_{k\in\mathbb{N}} f(C_k) = f\left(\bigcup_{k\in\mathbb{N}} C_k \right) [/tex] is also bounded above by [tex]K\varepsilon[/tex]. Since [tex]\varepsilon[/tex] was arbitrary we are done.


    I am not completely convinced because I did not use the hypothesis that the derivative is continuous. Did I make a mistake? Is there a simpler way do this?
    Last edited: Dec 28, 2009
  2. jcsd
  3. Dec 29, 2009 #2


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    I'm really having a hard time following that. It's not even clear to me that you even used that f is differentiable. Why don't you define C(epsilon) to be the set of all x such that f'(x)<epsilon. Then Crit(f) is contained in C(epsilon). Can you show the measure of f(C(epsilon)) is less than or equal to epsilon*measure(C(epsilon))?
  4. Dec 29, 2009 #3
    [tex](-\epsilon,\epsilon)[/tex] is open and f'(x) is continuous so [tex]C_\epsilon[/tex] is open too. An open set is a (at most) countable union of disjoint open intervals:

    [tex]C_\epsilon=\bigcup_{i=1}^{\infty} I_i [/tex]

    For every [tex]I_i[/tex], if x and y are 2 points in [tex]I_i[/tex],

    [tex]|fx-fy|<\epsilon|x-y|< \epsilon\mu(I_i) [/tex], due to MVT. So [tex]\mu(f(I_i))<\epsilon\mu(I_i)[/tex]

    Taking the union over i yields measure(f(Cε)) <ε*measure(Cε)<=ε(b-a) and we are done.

    Thanks for your help :)
    Last edited: Dec 29, 2009
  5. Dec 29, 2009 #4


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    Now that I understand. And it even looks right.
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