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First I'll make a definition: $$\operatorname{Loc-int}(g):=\left\lbrace x\in[0,1] : \exists \epsilon>0\text{ s.t. }\int_{(x-\epsilon,x+\epsilon)\cap[0,1]}|g|dm<\infty\right\rbrace,$$ where $m$ is the Lebesgue measure.

I've been searching for weeks now for an example that suits the next terms: $$\forall \lambda \in(0,1) \exists f:[0,1] \to \mathbb{R}\text{ measurable, differentiable s.t. } m\left([0,1]/\operatorname{Loc-int}(f')\right)>\lambda.$$

Until now I've only managed to find an example of differentiable function that is not locally integrable at $x=0$. Any help would be appreciated!

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"...s.t....". Should there be some equality sign between that integral and $\,|g|\lambda\;$ or what? And anyway: what does that mean? Is it $\,|g(\lambda)|\;$ ? –  DonAntonio Aug 29 '13 at 11:01
    
for some reason it shows the question with a typo and doesn't let me edit.. –  Snir Aug 29 '13 at 11:01
    
You don't need all those dollar sign within the first ones around the set definition... –  DonAntonio Aug 29 '13 at 11:02
    
Try to take the equation to a separate line with $$ signs. –  Jonathan Y. Aug 29 '13 at 11:03
    
worked, thanks! –  Snir Aug 29 '13 at 11:05

2 Answers 2

I'll make a suggestion on the idea which might help. Let's take a Cantor set of positive measure (the idea is to cut off the intevals of length ratio sligthly less than one third; the exact value of this "slightly less" can be deduced). We can have that measure in $[0,1)$.

Now, on each inteval that we cut off we introduce a function that behaves like $(x-x_i)^2\sin\left(\frac{1}{(x-x_i)^2}\right)$ near the ends of the interval with $x_i$ being those ends. Thanks Samprity for this function.

THe hypothesis is that we have a function on the whole interval with a derivative being non-integrable on a set of positive measure (still need a formal proof of this fact, as JonathanY. mentioned).

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A Cantor set includes many more points than the ends of the removed segments (in fact, none of its measure--when it has any--is concentrated at such points), since as we know there are only countably many such endpoints; more importantly, perhaps, is that those endpoints--along with the rest of the set--are nowhere dense. –  Jonathan Y. Aug 29 '13 at 11:34
    
If I'll try to practice your suggestion, you're saying that in every step of the cantor set building we have a differentiable function whose derivative is not integrable in a larger number of points, but how do we know that the limit of those functions has the same properties? –  Snir Aug 29 '13 at 11:39
    
@Snir If on each step we scale down the added part of the function, we can arrive to differentiability by some argument of uniform convergence. On the other hand, once a point has a non-integrability, we won't change the function on some neighbourhood of the function, so we only increase the set of point of non-integrability. –  TZakrevskiy Aug 29 '13 at 12:11
    
@JonathanY. Yes, indeed. I think that there's some workaround, because basing on a Cantor set of positive measure back in th uni we built a function with a derivative being non Riemann-integrable. –  TZakrevskiy Aug 29 '13 at 12:13

This is an example of such type of functions.

$$f(x) = \begin{cases}x^2 \sin{(\frac{1}{x^2})} & \text{when } x \neq 0 \\ 0 &\text{when } x = 0\end{cases}$$

$f$ is differentiable in $\mathbb{R}$. Its derivative will be

$$f'(x) = \begin{cases}2x \sin{(\frac{1}{x^2})} - \frac{1}{x} \cos{(\frac{1}{x^2})} &\text{when } x\neq 0 \\ 0 &\text{when } x = 0\end{cases}$$

Now see $f'(x)$ is unbounded in any closed interval around $0$, like $[-1,1]$ as it contains an unbounded term $\frac{1}{x}$, so not Riemann integrable. So $f(x)$ is a differentiable function whose derivative is not integrable. I hope you can fined explanation in Lebesgue measure as I do not know it properly.

Thank you for a little Latex correction.

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Samprity, you can use the \begin{cases} f_1 & [case-1]\\ f_2 & [case-2]\end{cases} tags. –  Jonathan Y. Aug 29 '13 at 11:23
    
Than you Mr. Jonathan Y. –  Dutta Aug 29 '13 at 11:28
    
This is the exact example I've found already, but that is not good enough. I need a set of every positive measure less than 1, not just one point. –  Snir Aug 29 '13 at 11:35
    
Why downvote? Anybody? –  Dutta Aug 29 '13 at 18:10

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