Show that $f$ defined on the interval $(a,b)$ is not differentiable for every point in $E$ with $m(E)=0$ 
Let $E$ have measure zero contained in the open interval $(a,b)$. In a previous problem I showed that there is a countable collection of open intervals, $\{(c_k,d_k)\}_k$, contained in $(a,b)$ for which each point in $E$ is contained in infinitely many intervals of the collection and $\sum_k |(c_k,d_k)|=\sum_kd_k-c_k<\infty.$ Define $f(x)=\sum_k|(c_k,d_k)\cap(-\infty,x)|$ for $x \in (a,b)$. Show $f$ is increasing and fails to be differentiable for every point in $E$.

The absolute value bars mean measure of the interval. I have already shown $f$ was increasing by showing that if $a \le u <v \le b$ then $$\begin{align} f(v)-f(u) &=\sum_k |(c_k,d_k)\cap [(-\infty,u)\cup[u,v)] |-\sum_k|(c_k,d_k)\cap (-\infty,u)|\\&=\sum_k|(c_k,d_k)\cap [u,v)] |\ge 0 \end{align}$$
Let $D^+f(x)=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]$ and $D^-f(x)=\lim_{h\rightarrow 0}\left[\inf_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]$
To show f is not differentiable at any point in $E$ then I need to show that $D^+f(x) \not= D^-f(x)$ for every $x \in E$. This is where I am having trouble. First from using what I did above to show that $f$ is increasing I found that for $t>0$
$$ A_t(f(x))= \dfrac{f(x+t)-f(x)}{t}=\dfrac{1}{t}\sum_{k=1}^{\infty}|(c_k,d_k)\cap[x,x+t)| $$
So at this point what we know is that if $x\in E$ then $x$ belongs to $(c_k,d_k)$ for infinitely many $k$ and $\sum_{k=1}^{\infty}|(c_k,d_k)\cap[x,x+t)|$ converges since $\sum_{k=1}^{\infty}|(c_k,d_k)|<\infty$. But Im not sure how to find the $\sup_{0<|t|\le h}$ and $\inf_{0<|t|\le h}$ of $A_t(f(x))$ I think that $\inf_{0<|t|\le h}A_t(f(x))=0$ and $\sup_{0<|t|\le h}A_t(f(x))>0$ but I dont know how to show it.
Any help is appreciated, thanks 
 A: I don't know if you are still seeking for the answer (probably not...), but here's my try.
Indeed, I cannot figure out the supremum and the infimum from what you got. Let's think about it differently.
Let $\{k_1,k_2,...,k_n,...\}$ be the collection of natural numbers for which $x\in I_k:=(c_k,d_k)$. Let $N\in \mathbb{N}$ be such that $k_{N}$ is in the latter enumeration. Then, $x\in I_{k_1}\cap I_{k_2}\cap...\cap I_{k_N}$. We know that finite intersections of open sets is open. Therefore, the latter finite intersection is open. This means we can pick an $\epsilon_N>0$ small enough for $x+\epsilon_N$ to remain in the intersection. Consequently, $[x,x+\epsilon_N)\subset I_{k_1}\cap I_{k_2}\cap...\cap I_{k_N}$. In particular, this means that the $N^{th}$-partial sum of $f(x+\epsilon_N)-f(x)$ is
\begin{equation}
    [f(x+\epsilon_N)-f(x)]_N=\sum_{k=1}^{N}\ell((c_k,d_k)\cap [x,x+\epsilon_N))=\sum_{k=1}^{N}\epsilon_N=N\epsilon_N.
\end{equation}
Hence, we observe that $f(x+\epsilon_N)-f(x)\ge N\epsilon_N$. This also holds with $\epsilon_N$ replaced for $t$, for all $0<t<\epsilon_N$. Finally, we see that
\begin{equation}
\begin{split}
     D^+f(x)&=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]\ge\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h} \dfrac{Nt}{t}\right]=N.
\end{split}
\end{equation}
But recall that $N$ can be arbitrarily large based on the fact that $x\in I_k$ for infinitely many $k$. This tells us that $D^+f(x)$ is unbounded.  Zhuanghua Liu gave you the correct hint :)
