This Hausdorff dimension of the graph of an increasing function shows that:

Let $f$ be a continuous, strictly increasing function from $[0,1]$ to itself with $f(0)=0, f(1)=1$. Then $dim_H \; G = 1$ where $G$ is the graph of $f$.

I have at hand the Casino function, described as follows in Massopoust's *Interpolation and Approximation with Splines and Fractals*:

Let $X = [0,1] \times \mathbb{R}$, $N = 4$ and $Y = \{(x_v,y_v):0 = x_0 < \ldots x_N = 1, 0 = y_0 < \ldots < y_N = 1\}$. Define an IFS by $f_i(x,y) = \begin{pmatrix} x_i-x_{i-1} & 0 \\ 0 & y_i - y_{i-1} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} x_{i-1} \\ y_{i-1} \end{pmatrix} $ for $i = 1, \ldots, N$.

The associated RB operator $T$ is contractive and its unique fixed point is called a Casino function $c:[0,1] \to [0,1]$. These functions are monotone increasing and therfore $dim_H \; graph(c) = \dim_B \; graph(c) = 1$.

I was wondering how can I show that $dim_B \; graph(c) = 1$ and whether there is a general argument establishing:

Let $f$ be a continuous, strictly increasing function from $[0,1]$ to itself with $f(0)=0, f(1)=1$. Then $dim_B \; G = 1$ where $G$ is the graph of $f$.

I don't find an argument stablishing $dim_B \; G \le 1$.

Fractal Geometry. You might also want to look at p. 120 (beginning of Section 10.4) and Section 12.4 (pp. 148-150) of Tricot's 1995 bookCurves and Fractal Dimension, and this paper. $\endgroup$ – Dave L Renfro Mar 17 '19 at 19:10