A question regarding the existence of a differentiable function with certain simple measure-theoretic properties

Question

Does there exist a set $[a, b]\subseteq \mathbb{R}\cup\{-\infty, \infty\}$ and a continuous function $F:[a, b] \rightarrow \mathbb{R}$ such that $F^\prime(x) := \lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}$ exists in $\mathbb{R}\cup\{-\infty, \infty\}$ for every $x \in (a, b)$ and $\{F(x): x \in (a, b) \land |F^\prime(x)|= \infty\}$ has positive Lebesgue outer measure?

Answer 1

Let $E\subset [0,1]$ be compact, nowhere dense, and have positive measure. We don't need the intricacies of Cantor sets for this: Simply choose an open subset $U$ of $(0,1)$ containing $\mathbb Q \cap (0,1),$ of measure less than $1,$ and let $E=[0,1]\setminus U.$

For $x\in \mathbb R,$ let $g(x)= d(x,E).$ Then $g$ is continuous everywhere, $g=0$ on $E,$ and $g>0$ everywhere else. Now define

$$G(x) = \int_0^x g(t)\,dt,\,\, x\in \mathbb R.$$

We have $G\in C^1(\mathbb R)$ with $G'(x)=g(x)$ everywhere.

Claim: $G$ is a strictly increasing bijection of $\mathbb R$ onto $\mathbb R.$ Hence $G$ is a homeomorphism of $\mathbb R$ onto $\mathbb R.$

Proof: Because $G'=g\ge 0,$ $G$ is at least nondecreasing. If $x<y,$ then $(x,y)\setminus E$ is open and nonempty. This follows from the nowhere density of $E.$ Therefore $G'=g>0$ in an open subset of $(x,y),$ showing $G(x)<G(y)$ as desired. It's easy to verify $G(x)\to \infty$ as $x\to \infty,$ and $G(x)\to -\infty$ as $x\to -\infty.$ By the IVT, $G$ is surjective. As is well known, this implies $G$ is a homeomorphism.

Now let $F= G^{-1}$ on $\mathbb R.$ As we know, $F:\mathbb R\to \mathbb R$ is a homeomorpism. By the usual results in calculus, $F'(y)$ exists for $y\in \mathbb R \setminus G(E)$ with the formula

$$F'(y)= \frac{1}{G'(G^{-1}(y))}.$$

Now let $y_0\in G(E).$ We wish to show

$$\tag 1 \lim_{y\to y_0} \frac{F(y)-F(y_0)}{y-y_0}=\infty.$$

There is a unique $x_0$ such that $G(x_0)=y_0.$ Because $G$ is a continuous bijection, $(1)$ holds iff

$$\lim_{x\to x_0} \frac{F(G(x))-F(G(x_0))}{G(x)-G(x_0)}=\infty.$$

But the last expression is the same as

$$\frac{x-x_0}{G(x)-G(x_0)}= \frac{1}{\dfrac{G(x)-G(x_0)}{x-x_0}}.$$

Because $G'(x_0)=0,$ the last expression $\to \infty$ as $x\to x_0.$ This gives $(1)$ as desired.

So we have shown $F$ is differentiable everywhere (in the extended sense), with $0<F'<\infty$ on $\mathbb R\setminus G(E)$ and $F'=\infty$ for $y\in G(E).$ Since $\{F(y): F'(y)=\infty\} =E,$ and $E$ has positive measure, we have our example.