# 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.

• Is E given and you want to construct f or do you just need an example of such an E and f? Do you expect something from f? Differentiable everywhere away feom E ...? Commented May 10, 2016 at 15:10
• @user3808066 E is given and I need to construct an f for it. f is differentiable at every points since it is a strictly increasing function. Commented May 10, 2016 at 15:12
• Strictly increasing only implies allmost everywhere differentiability. You can have edges. Commented May 10, 2016 at 15:14
• And what does the in "Lebesque sense" mean? It is allready a condition on a zero set. Commented May 10, 2016 at 15:16
• See my 18 April 2007 sci.math post (follow-up comment here) and the stackexchange question Absolute continuous function with given set of discontinuities of derivatives.. Commented May 12, 2016 at 14:39

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 $q_m$ $$g'(x) = f'(0) + \sum_{n\not= m} f'(q_m-q_n) =\infty.$$