Existence of a Strictly Increasing, Continuous Function whose Derivative is 0 a.e. on $\mathbb{R}$

Question

This proof is almost done except for the step of showing that the function's derivative is $0$ a.e.

Let $I = \{[p_n, q_n]\}$ denote the set of all closed intervals in $\mathbb{R}$ with rational endpoints $p_i, q_i \in \mathbb{Q}$.

Let $\phi_n$ denote a variant of the Cantor Ternary Function (i.e., Devil's Staircase Function) extended to all $\mathbb{R}$ which satisfies $\phi_n(x) = 0$ for all $x < p_i$, $\phi_n(x) = 1/2^n$ for all $x > q_i$ and is non-decreasing, continuous, and of zero derivative a.e. on $[p_i,q_i]$ so that $\phi_n$ satisfies

1. $\phi_n'(x) = 0$ a.e. on $\mathbb{R}$
2. $\phi_n$ non-decreasing, continuous on $\mathbb{R}$
3. $\phi_n$ is bounded above by $1/2^n$
4. $\phi_n (p_n) < \phi_n (q_n)$ (trivial but important later)

Now let $\sum \phi_n = \phi$.

$\fbox{Claim}$

1. $\sum \phi_n \rightarrow \phi$ uniformly
2. $\phi$ is strictly increasing and continuous on $\mathbb{R}$ yet also satisfies $\phi'(x) = 0$ a.e. on $\mathbb{R}$.

$\fbox{Attempted Proof}$

1. First since $\forall n \in \mathbb{N}$, we have that $|\phi_n| \le 1/2^n$ with $\sum 1/2^n < \infty$, it follows that $\sum \phi_n \rightarrow \phi$ uniformly (a fact from Real Analysis).

2. Consider that the finite sum of $k$ continuous functions is itself a continuous function. Since all of the $\phi_n$ are continuous functions, we therefore have that $f_k = \sum_{n=1}^k \phi_n$ is also a continuous function. Moreover, since (1) implies that $f_k \rightarrow \phi$ unformly, we now have that $\phi$ is also continuous (via another Real Analysis fact that a uniformly convergent sequence of continuous functions converges to another continuous function).

3. To show $\phi$ is strictly increasing, consider $x,y \in \mathbb{R}$ s.t. $x < y$. Since $\exists k \in \mathbb{N}$ s.t. $x < p_k < q_k < y$, we have that $\phi_k(x) \le \phi_k(p_k) < \phi_k(q_k) \le \phi_k(y)$ implies that $\phi_k(x) < \phi_k(y)$ so that for $N \ge k$ we have $\sum_{n = 1}^N \phi_n(x) < \sum_{n = 1}^N \phi_n(y)$ and more generally $\phi(x) < \phi(y)$. This shows that indeed $\phi$ is strictly increasing on $\mathbb{R}$.

4. To show that $\phi'(x) = 0$ a.e. on $\mathbb{R}$, we will show that $\phi' = \sum \phi_n' = \sum 0_n$, where each $0_n$ is the derivative function of $\phi_n$ which is $0$ almost everywhere.

But I'm not sure how to complete (4)?

Answer 1

From your answer I see that it is ok to use standard facts from real analysis, so let's use them!

Fubini's theorem on differentiation. Assume $(f_n)_{n\in\mathbb{N}}$ is a sequence of non-decreasing functions on $[a,b]$, and the series $\sum_{n=1}^\infty f_n(x)$ converges for all $x\in [a,b]$, then $$ \left(\sum_{n=1}^\infty f_n(x)\right)'=\sum_{n=1}^\infty f_n'(x) $$ a.e. on $[a,b]$.

Now we turn to the original problem. Consider arbitrary interbal $[a,b]$. By assumption $\phi_n'=0$ a.e. on $\mathbb{R}$ and a fortiori on $[a,b]$. By Fubini's theorem on differentiation we get $$ \phi'(x)=\left(\sum_{n=1}^\infty \phi_n(x)\right)'=\sum_{n=1}^\infty \phi_n'(x)=\sum_{n=1}^\infty 0=0 $$ a.e. on $[a,b]$. Since $[a,b]$ is an arbitrary interval, then $\phi'=0$ a.e. on $\mathbb{R}$.