Vrftjkry

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 $D^+f(x)=lim_{hrightarrow 0}left[sup_{0<|t|le h}dfrac{f(x+t)-f(x)}{t} right]$ and $D^-f(x)=lim_{hrightarrow 0}left[inf_{0<|t|le h}dfrac{f(x+t)-f(x)}{t} right]$

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 $xin E$ 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

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 $xin I_k:=(c_k,d_k)$. Let $Nin mathbb{N}$ be such that $k_{N}$ is in the latter enumeration. Then, $xin I_{k_1}cap I_{k_2}cap...cap I_{k_N}$. 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=Nepsilon_N.
end{equation}

Hence, we observe that $f(x+epsilon_N)-f(x)ge Nepsilon_N$. Finally, we see that

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

But recall that $N$ can be arbitrarily large based on the fact that $xin I_k$ for infinitely many $k$. This tells us that $D^+f(x)$ is unbounded. Zhuanghua Liu gave you the correct hint :)