# 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

• $D^+f(x)$ does not converge, as $x \in E$ belongs to infinite $(c_k, d_k)$, – Zhuanghua Liu Jul 5 '17 at 6:38

## 1 Answer

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 $$f(x+\epsilon_N)-f(x)\ge N\epsilon_N$$. Finally, we see that

$$\begin{equation} \begin{split} D^+f(x)&=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]\ge\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h} \dfrac{f(x+\epsilon_N)-f(x)}{t}\right]\\&=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h} \dfrac{N\epsilon_N}{t}\right]\ge N. \end{split} \end{equation}$$

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 :)