Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
What is the question? –  user22705 Mar 12 '12 at 20:12
4  
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
add comment

1 Answer

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?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.