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Let $f\colon[a,b]\to\mathbb{R}$ be a real valued function and put $$A=\{x\in(a,b)\colon f\text{ is differentiable at }x,\; f'(x)=0\}.$$

Let $\lambda$ denote the Lebesgue measure on $[a,b]$, then the following holds:

  1. Let $\varepsilon>0$, $\alpha>0$ and $y\in f[A]$, then there exists an $x\in A$ and $\delta>0$ such that$$\forall a\in(x-\delta,x+\delta)\colon f(a)\in[y-\delta\varepsilon, y+\delta\varepsilon]$$ and $$\lambda\big([y-\delta\varepsilon, y+\delta\varepsilon]\big)<\alpha.$$
  2. Vitali's covering lemma implies that $\lambda^*(f[A])=0$, where $\lambda^*$ denote the Lebesgue outer measure.

Vitali's Covering lemma is stated as

Let $E$ be a set of finite Lebesgue outer measure and $\mathcal{F}$ a collection of intervals the that $E$ in the sense of Vitali. Then, for all $\varepsilon>0$ there exists a finite disjoint collection $\{I_1,\ldots, I_n\}$ such that $$\lambda^*\left(E\setminus\bigcup_{k=1}^nI_k\right)<\varepsilon.$$

I was wondering if anyone knows how to prove these two statements. Thank you in advance.

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What is the question? – Henrik Mar 12 '12 at 20:12
Thank you for giving us this information. However, this site is for asking questions, not for stating theorems. – GEdgar Mar 12 '12 at 20:13
Maybe it's sleep deprivation, but that comment made me laugh aloud for an entire minute (@GEdgar)! – The Chaz 2.0 Mar 12 '12 at 20:31
It might be useful to state which version of the Vitali covering lemma you are referencing. – user1736 Mar 13 '12 at 0:42

This is a partial solution!

For the first part:

Let $y\in A$, $\varepsilon>0$ and $\alpha>0$, then there is a $x\in A$ such that $y=f(x)$. Since $f$ is differentiable at $x$, there is a $\delta>0$ such that $$|u-x|<\delta\implies\left|\frac{f(u)-f(x)}{u-x}-f'(x)\right|\le\varepsilon.$$ Using that $f'(x)=0$, $$f(x)-\varepsilon|u-x|\le f(u)\le f(x)+\varepsilon|u-x|.$$ So $f(u)\in[y-\delta\varepsilon, y+\delta\epsilon]$. Because $\varepsilon$ was given arbitriraly and any smaller $\delta$ suffices, the measure of the interval $[y-\delta\varepsilon, y+\delta\varepsilon]$ made less than $\alpha$.

This is the appropriate Vitali covering lemma:

Let $E$ be a set of finite outer measure and $\mathcal{F}$ a collection of closed intervals that cover $E$ in the Vitali sense, that is for any $\alpha>0$ there and $x\in E$ there is an interval $I\in\mathcal{F}$ such that length$(I)<\alpha$.

Note that the collection $$\mathcal{F}=\{I(y,\alpha)=[y-\delta\varepsilon, y+\delta\varepsilon]\colon \mathrm{length}(I(y,\alpha))<\alpha, y\in f[A]\}$$ is clearly a Vitali cover for $f[A]$ by part 1. Unfortunately, $f[A]$ is a subset of $\mathbb{R}$, which doesn't have finite outer measure.

I don't know how to continu onwards. I guess that you need to consider for $R>0$ sufficiently large, the set $$E=f[A]\cap[-R,R]$$ and assume that $\lambda^*(E)>0$. This might lead to a contradiction.

Is there someone with a helpful suggestion?

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