# Sard theorem in dimension one.

1. Dec 28, 2009

### 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?

Last edited: Dec 28, 2009
2. Dec 29, 2009

### 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))?

3. Dec 29, 2009

### 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 $$I_i$$,

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

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