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 $\sum \phi_n = \phi$.


  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)?

  • $\begingroup$ Do you want help with your proof of would be satisfied with an alternative argument? There are even homeomorphic functions whose derivative is zero at every point where it exists. Moreover, they appear naturally in theory of hyperbolic surfaces. $\endgroup$ – Moishe Kohan Apr 19 '14 at 3:02
  • $\begingroup$ Your proof of (3) is wrong. The inequality $u_N < v_N$ for large enough $N$ does not imply $\lim{u_n} < \lim{v_n}$, but only $\lim{u_n} \leq \lim{v_n}$. But that can easily be corrected : in fact you have $\sum_{n=1}^N \phi_n(y)-\phi_n(x) \geq \phi_k(q_k)-\phi_k(p_k)$ whence $\phi(y)-\phi(x) \geq \phi_k(q_k)-\phi_k(p_k)$. $\endgroup$ – Ewan Delanoy Apr 19 '14 at 7:35
  • 2
    $\begingroup$ www.math.sc.edu/~howard/Notes/fubini.ps.gz $\endgroup$ – Ewan Delanoy Apr 19 '14 at 12:54
  • $\begingroup$ @studiosus: help with my current proof would be best. $\endgroup$ – user1770201 Apr 19 '14 at 13:05

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 $\mathbb{R}$.

  • $\begingroup$ increasing instead of non-decreasing would be better. $\endgroup$ – vesszabo Apr 22 '14 at 17:31

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