Sard theorem in dimension one.

[boboYO]

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

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

For any C_k, evenly split [a,b] into intervals I_1,\dots,I_n such that \frac{b-a}{n}<\frac{1}{k}.For each I_i, if there is a point of C_k in it, say x, then

|fx-fy|<\varepsilon|x-y| for all y in I_i.

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

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

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

\bigcup_{k\in\mathbb{N}} f(C_k) = f\left(\bigcup_{k\in\mathbb{N}} C_k \right) is also bounded above by K\varepsilon. Since \varepsilon 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?

 

[Dick]

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))?

 

[boboYO]

(-\epsilon,\epsilon) is open and f'(x) is continuous so C_\epsilon is open too. An open set is a (at most) countable union of disjoint open intervals:

C_\epsilon=\bigcup_{i=1}^{\infty} I_i

For every I_i, if x and y are 2 points in I_i,

|fx-fy|&lt;\epsilon|x-y|&lt; \epsilon\mu(I_i), due to MVT. So \mu(f(I_i))&lt;\epsilon\mu(I_i)

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

Thanks for your help :)

 

[Dick]

Now that I understand. And it even looks right.