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.


  • $\begingroup$ Did you get anywhere with this? I'm working on something similar. $\endgroup$ – user28877 Jan 29 '14 at 7:11
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    $\begingroup$ Regarding constructing some absolutely continuous function, whose poles of derivative is that measure zero set, this is not possible. Since the set for which a function has an infinite derivative is $G_{\delta}$ (proved by William H. Young in 1903) and there exists measure zero sets that are not $G_{\delta}$ (indeed, there exist measure zero sets that are not even Borel), the best you can hope for is to find a function that has an infinite derivative at every point in the given measure zero set and possibly also an infinite derivative at some points not in the given measure zero set. $\endgroup$ – Dave L. Renfro Jun 20 '14 at 18:50
  • $\begingroup$ Interesting comment @DaveL.Renfro. I was wondering if $A$ was the only place where the derivative was infinity. I found a solution to this one too: math.stackexchange.com/questions/567840/… Do you have any idea on this one: math.stackexchange.com/questions/594725/… $\endgroup$ – Tomás Jun 20 '14 at 19:12
  • $\begingroup$ @Tomás: If $A$ is not $G_{\delta},$ there will have to be points not in $A$ where the function you constructed has an infinite derivative. I looked at Strictly increasing, absolutely continuous function with vanishing derivative, and I can see why $F$ is strictly increasing (immediate from $m(A\cap I) > 0$ for all intervals $I)$ and absolutely continuous (easy to see $F$ is Lipschitz continuous), but I don't know what to do with the actual question right now. $\endgroup$ – Dave L. Renfro Jun 20 '14 at 20:42
  • $\begingroup$ Take a look there again please @DaveL.Renfro. I think I have solved the problem. Thank you. $\endgroup$ – Tomás Jun 20 '14 at 20:54

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.


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