Hi, I need some hints for the proof of the sard theorem in 1 dimension:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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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?

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# Homework Help: Sard theorem in dimension one.

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