Suppose $A \subset [0,1]$ has Lebesgue measure zero. How can I construct a strictly increasing absolutely continuous function $f : [0,1] \to \mathbf{R}$ with $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\infty$$ for each $x \in A$?
We know that for any absolutely continuous function, its derivative do not blow except a measure zero set. This problem is doing converse, that is, give a specific measure zero set, and constructing some absolutely continuous function, whose poles of derivative is that measure zero set.
Thanks.
Best Answer
You can construct $f$ as follows: let $m$ denote Lebesgue measure. Once $m(A)=0$, we can choose for each $i=1,2,\cdots$, a sequence of intervals $I_{ij}$ such that $$A\subset \cup_{j=1}^\infty I_{ij},\ \sum_{j=1}^\infty m(I_{ij})\le \frac{1}{2^i}. \tag{1}$$
Note that each $x$ belongs to infinitely many intervals $I_{ij}$. Define $f:[0,1]\to\mathbb{R}$ by $$f(x)=\sum _{i,j=1}^\infty m(I_{ij}\cap [0,x]).$$
Note that $f$ is increasing. Also $f$ satisfies $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\infty,\ \forall x\in A.$$
Indeed, as $x$ belongs to infinitely many intervals $I_{ij}$, we can assume without of generality that $x$ belongs to $J_k=(x-a_k,x+a_k)$ for all $k=1,2,\cdots$, that each $J_k=I_{ij}$ for some $i,j$ and $a_k\to 0$. Note that
\begin{eqnarray} \frac{f(x-h)-f(x)}{h} &=& \frac{1}{h}\sum _{i,j=1}^\infty m(I_{ij}\cap [x-h,x]) \nonumber \\ &\ge& \frac{1}{h}\sum _{k=1}^\infty m(J_k\cap [x-h,x]) \nonumber \\ &\ge& \frac{1}{h}(Nh+\sum _{k=N+1}^\infty a_k), \end{eqnarray}
where $N$ depending on $h$ is choosen in such a way that $h\le a_N$. As $h\to 0^+$, we see from the previous inequality that the left derivative is infinity. With an similar reasoning, you can conclude that the right derivative is also infinity.
To prove that $f$ is absolutely continuous, note that $$\sum _{k=1}^N |f(x_{k+1})-f(x_k)|=\sum _{k=1}^n \sum_{i,j=1}^\infty m(I_{ij}\cap[x_k,x_{k+1}]). \tag{2}$$
Once $(1)$ is satisfied, you conclude from $(2)$ that $f$ is absolutely continuous. To get a strictly increasing functions, just consider $g(x)=x+f(x)$.
Note: The original construction is due to professor Porter and it is contained here.