Continuous strictly increasing function with derivative infinity at a measure 0 set Let $E\subset [0,1]$ with $\mu(E)=0$. Does there exist a continuous, strictly increasing function $f$ on $[0,1]$ so that $f'(x)=\infty$ for all $x\in E$ (in Lebesgue sense)?
I think there exist such a function, but I don't know how to construct.

I couldn't figure how the general case works but if $E$ is countable consider $E=\{q_n\;:\;n\in\mathbb{N}\}$ and take $f$ a function which is $0$ on $(-\infty,-1/2]$, $1$ on $[1/2,\infty)$, monotonic and has a derivative $f':\mathbb{R} \to [0,\infty]$ with $f'(0)=\infty$, additionally take a sequence $(a_n)_{n\in\mathbb{N}}\in \ell^1$, $a_n>0$. Then the function
$$
 g(x) = \sum_{n\in\mathbb{N}} a_nf(x-q_n)
$$
 is continuous and monotone. Further since for all $n$ $a_nf'(x-q_n)$ is a non negative function
$$
 g(x) = \int_{-1}^x \sum_n a_n f'(s-q_n)\mathrm{d}s
$$
and hence at every $q_m$ 
$$
 g'(x) = f'(0) + \sum_{n\not= m} f'(q_m-q_n) =\infty.
$$
