I am self-learning probability theory from the text Measure, Integral and Probability, by Capinski and Kopp.
Exercise problem 2.3 asks to investigate the Cantor-Lebesgue function, show that it is continuous and sketch its graph. I would like someone to verify, if my proof attempt is technically correct and sufficiently rigorous.
If it's not correct, a hint/clue, without giving away the entire proof would be nice.
[Capinski-2.3] If $x\in[0,1]$ has a ternary expansion $(a_n)$ that is $x = 0.a_1 a_2 \ldots$ with $a_n = 0,1,2$ define $N$ as the first index $n$ for which $a_n = 1$ and set $N = \infty$ if none of the $a_n$'s are $1$ (when $x \in C$). Now, set $b_n = \frac{a_n}{2}$ for $n < N$ and $b_N =1$, and let $F(x) = \sum_{n=1}^{\infty}\frac{b_n}{2^n}$ for each $x \in [0,1]$. Clearly, this function is monotone increasing and has $F(0) = 0$, $F(1) = 1$. Yet, it is constant on middle thirds.
Prove that the Lebesgue function is continuous and sketch its graph.
Proof.
Define $F_{N}(x)=\sum_{n=1}^{N}\frac{b_{n}}{2^{n}}$. We hand-draw the plots for $F_1$, $F_2$ and $F_3$ to better understand what's going on.
The plot of $F_{1}(x)$ is:
The plot of $F_2(x)$ is:
The plot of $F_3(x)$ is:
The Lebesgue function F is the pointwise limit of $F_{n}(x)$ and it has constant dyadic values of the form $\frac{p}{2^{k}}$, where $p\in\mathbf{Z}_{+}$, $0\leq p\leq2^{k}$, $k\in\mathbf{N}$ for all $x$ belonging to any of the open middle one-thirds.
It maps all points in the Cantor set $C$ to infinite binary strings of $0$s and $1$s. And F maps all terminating/recurring strings on the x-axis to dyadic points on the y-axis. Since every real in $[0,1]$ can be represented in binary or ternary number system, with $0=(0.000\ldots)_2$ and $1=(0.1111\ldots)_2$, it follows that the domain of $F$ is $[0,1]$ and the range of F is also $[0,1]$. Moreover, $F(0)=0$ and $F(1)=1$. The Lebesgue function jumps from $0$ to $1$ on a set of length zero.
Pick an arbitrary $\epsilon>0$ and let $c\in[0,1]$ and consider the open interval $V_{\epsilon}(f(c))=(F(c)-\epsilon,F(c)+\epsilon)$.
There exists $K\in\mathbf{N}$,such that $\frac{1}{2^{K}}<\epsilon$. We also know that there exists a closed interval $I_{K}=[\frac{p}{2^{K}},\frac{p+1}{2^{K}}]$ containing $F(c)$.
Consider the set $\{x:F(x) = \frac{m}{2^K}\}$. We pick the points $\sup \{x:F(x) = \frac{m}{2^K}\}$, for $m = p$, $p+1$. Thus, we have mapped the end-points of the interval $I_K$ on the $y$-axis to $\frac{q}{3^K},\frac{q}{3^{K+1}}$ on the $x$-axis. Since $F$ is increasing and bounded, it follows that for all $\frac{q}{3^K} \leq x \leq \frac{q+1}{3^K}$, we have $F(\frac{q}{3^K}) \leq F(x) \leq F(\frac{q+1}{3^K})$.
In fact, this map sends every closed interval of length $\frac{1}{2^{K}}$ on the $y$-axis whose end-points are dyadic numbers, to an interval of the length $\frac{1}{3^{K}}$ on the x-axis containing $c$.
Pick $\delta=\frac{1}{2\cdot3^{K}}$. Then, $\forall x\in(c-\delta,c+\delta)$ it follows that $F(x)\in V_{\epsilon}(F(c))\cap I_{K}\subset V_{\epsilon}(F(c))$. Thus, $F$ is continuous at $c$.
Since $c$ was arbitrary, this must be true for all $c\in[0,1]$. Consequently, $F$ is continuous on $[0,1]$. Q.E.D.