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!
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.