Exercise 3.40 from Folland, Real Analysis Let $F$ denote the Cantor function on $[0, 1]$ (see $§1.5$), and set $F(x)= 0$ for $x<0$
and $F(x)=1$ for $x>1$. Let ${[a_n, b_n]}$ be an enumeration of the closed subintervals
of $[0,1]$ with rational endpoints, and let $F_n{(x)}=F((x-a_n)/(b_n-a_n))$. Then
let $G$ be the sum of ${2^{-n}F_n}$ as $n$ goes from $1$ to $\infty$, then $G$ is continuous and strictly increasing on $[0,1]$, and $G'=0$ a.e.
(Use Exercise 39.)
Source: Folland, Real Analysis, exercise $3.40$. 
Can anybody help please? I have no idea how to do this problem. Thanks a lot.
 A: The point of this exercise is to demonstrate the counterintuitive fact that there exist strictly increasing singular-continuous functions, i.e. strictly increasing continuous functions whose derivative vanishes almost everywhere.
The Cantor function $F$ is continuous, increasing (but not strictly) and $F^\prime=0$ a.e. (see an earlier chapter). Let $$G(x)=\sum_{n=1}^\infty 2^{-n} F_n(x)$$
The sum clearly converges uniformly on the real line since $0\le F_n(x)\le 1$ (Weierstrass $M$-test). Thus, since the $F_n$ are continuous, also $G$ is continuous. Because the $F_n$ are increasing, it is clear that $G$ must be increasing.
Exercise 39 allows us to conclude $G^\prime=0$ a.e. (essentially because $F_n^\prime(x)=0$ a.e. plus monotonicity).
So the only thing left to show is that $F$ is strictly increasing on the unit interval, i.e. if $0\le x<y\le 1$ then $F(x)<F(y)$. We may assume that $x$ and $y$ are rational (if not, then we can find rational $\tilde{x},\tilde{y}$ such that $0\le x<\tilde{x}<\tilde{y}<y\le 1$ and use monotonicity of $F$).
Then there must be an $m$ such that $[a_m,b_m]=[x,y]\subset [0,1]$. Thus we have
$$F_m(x)=F\left(\frac{x-a_m}{b_m-a_m}\right)=F(0)=0<1=F(1)=F\left(\frac{y-a_m}{b_m-a_m}\right)=F_m(y)$$
and therefore $G(x)<G(y)$ (the difference between the two is at least $2^{-m}$).
A: By construction $F_n$ is continuous and $F_n(\mathbb{R}) = [0, 1]$ for all $n \in \mathbb{N}.$
Therefore, if $n \in \mathbb{N}$ then
$$\left|G - \sum_{k = 1}^n 2^{-k} F_k\right| = \sum_{k = n + 1}^\infty 2^{-k} F_k \le \sum_{k = n + 1}^\infty 2^{-k} = 2^{-n}$$
and hence $G$ is the uniform limit of a sequence of continuous functions.
This implies that $G$ is continuous.
If $x, y \in \mathbb{R}$ and $x < y$ there exists $n \in \mathbb{N}$ such that $[a_n, b_n] \subseteq (x, y),$ so that $F_n(x) = 0$ while $F_n(y) = 1$ and hence
$$G(x) = F_n(x) + \sum_{\substack{k \in \mathbb{N} \\ k \neq n}} 2^{-k} F_k(x) < F_n(y) + \sum_{\substack{k \in \mathbb{N} \\ k \neq n}} 2^{-k} F_k(x) \le F_n(y) + \sum_{\substack{k \in \mathbb{N} \\ k \neq n}} 2^{-k} F_k(y) = G(y).$$
The complement of the Cantor set $C$ is open, so if $x \in (0, 1) \setminus C$ then $F$ is constant on a neighbourhood of $x,$ implying that $F'(x) = 0.$
Since $m(C) = 0$ and $F$ is constant on $(-\infty, 0)$ and $(1, \infty),$ this shows that $F' = 0$ almost everywhere.
If $n \in \mathbb{N},$ the preimages of $(0, 1)$ and $(0, 1) \setminus C$ under the map $x \mapsto \frac{x - a_n}{b_n - a_n}$ have measures $b_n - a_n$ and $\sum_{k = 1}^\infty 2^{k - 1} 3^{-k} (b_n - a_n) = b_n - a_n,$ which implies that $F_n' = 0$ almost everywhere by the chain rule.
Therefore, by the previous exercise $G' = \sum_{k = 1}^\infty 2^{-k} F_k' = 0$ almost everywhere.
