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Given a set $A$ with Lebesgue measure 0, how does one construct an increasing function which has a finite derivative at all points except those at $A$. Namely, for all $x \in A$, $$\left | \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \right | = 0$$?

I can do this in the case where $A = \{a_j\}_1^\infty$ is countable: $$\sum_{j\geq 1} 2^{-j} \chi_{(a_j,\infty)}$$, or something to this effect.

Any tips to get started?

  • I might be able to help, but you'll need to rewrite your question more precisely and correctly. For example, at present your second sentence says something quite different than your first sentence. Also, is this a class or textbook problem, or is this something that came up in your research? If the latter, it will probably be enough for me to point out that Edward Marczewski, Jan S. Lipinski, and Solomon Marcus wrote several papers on this topic in the late 1950s through the 1960s, and others have followed up on their work. – Dave L. Renfro Jan 29 '14 at 16:23
  • Also, does your function have to be continuous at these points where the derivative fails to exists? I'm supposing it does, as monotone functions can only have countably many discontinuities, and this would leave only uninteresting measure zero sets to consider. But just for clarification. –  Jan 29 '14 at 18:35

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You can't get an arbitrary Lebesgue null set $A$ in this way. For one thing, the set of differentiability is Borel (and it's somewhere low in the Borel hierarchy, I don't remember exactly where).

If $A$ is compact, there is a simple construction: define $$w(x) = \sum_{n=1}^\infty (1-r_n^{-1}\operatorname{dist}(x,A))^+ \tag{1}$$ where as usual, $a^+=\max(a,0)$. Here $r_n>0$ is chosen so that the measure of the $r_n$-neighborhood of $A$ is less than $2^{-n}$. This ensures that the series (1) converges in $L^1(\mathbb R)$.

Every point of the complement of $A$ has a neighborhood in which the series is a finite sum. Thus, $w$ is continuous on the complement of $A$. On the other hand, $\lim_{x\to a}w(x)=\infty$ for every $a\in A$, because when $x$ gets close to $A$, more of the terms in (1) turn to $1$.

Therefore, an antiderivative of $w$ (call it $f$) has the desired property.

The construction can be easily extended to the case of a closed unbounded set $A$.

you can call me Al
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  • Where the differentiability set falls on the Borel hierarchy is tackled here: http://math.stackexchange.com/questions/105810/continuous-functions-are-differentiable-on-a-measurable-set –  Jan 30 '14 at 07:45