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We construct the Cantor Function as follows.

Define $f_{0}(x)=x$ for any $x\in [0,1]$. And let

$f_{1}(x) = \begin{cases} (3/2)x \:, & 0\leq x\leq 1/3\\ 1/2 \:, & 1/3< x <2/3\\ (3/2)x-1/2 \:, & 2/3\leq x\leq 1. \end{cases}$

$f_{2}(x) = \begin{cases} (1/2)f_{1}(3x) \:, & 0\leq x\leq 1/3\\ f_{1}(x) \:, & 1/3< x <2/3\\ (1/2)f_{1}(3x-2)+1/2 \:, & 2/3\leq x\leq 1. \end{cases}$

Define $f_n$ for each $n\geq 2$ similarly.

I have been able to prove that each $f_n$ is continuous, but I want to show now that $\lim{(f_n)}$ is increasing. To do this, I think I should show that each $f_n$ is increasing.

I've tried to do this by testing each different possible domain for $a,b \in [0,1]$ with $a

I'm having trouble working out the induction proof. I can do the base case, but I'm not sure how to work out the rest. Specifically, how do you show that $f_k$ is increasing from the assumption that $f_i$ is increasing, with $1\leq i

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For present purposes, the following suffices:

Lemma: Suppose $[c, d] \subset [a, b]$ are closed intervals, that $f$ is a continuous, non-decreasing function on $[a, b]$, and that $g$ is a continuous function on $[a, b]$ satisfying:

  1. $g(c) = f(c)$ and $g(d) = f(d)$;

  2. $g$ is non-decreasing on $[c, d]$.

Then $g$ is non-decreasing on $[a, b]$.

To check that $x \leq y$ implies $g(x) \leq g(y)$, split into three cases: $x \leq c$; $d \leq y$; and $c \leq x \leq y \leq d$.

Each function $f_{n+1}$ in your sequence is obtained from $f_{n}$ by two operations of this type (given by your second set of formulas, with $1$ replaced by $n$ and $2$ replaced by $n + 1$), so $f_{n+1}$ is non-decreasing if $f_{n}$ is.