On the connection between classes of functions of bounded variation and classes of functions with fractal graph

For a real-valued function $f$ continuous on a closed interval, the modulus of fractality $\nu(f, \varepsilon)$ is defined for every $\varepsilon > 0$ as the minimum number of squares with sides of length $\varepsilon$ parallel to the coordinate axes that can cover the graph of $f$. For a nonincreasing function $\mu: (0, +\infty) \to (0, +\infty)$, we consider the class $F^{\mu}$ of functions continuous on a closed interval and such that $\nu(f, \varepsilon) = O(\mu(\varepsilon))$. The relationship between the classes $F^{\mu_1}$ and $F^{\mu_2}$ is described for various $\mu_1$ and $\mu_2$. A connection is established between the classes $F^{\mu}$ and the classes of continuous functions of bounded variation $BV_{\Phi}[a, b] \cap C[a, b]$ for arbitrary convex functions $\Phi$. Namely, there is an inclusion
$$ BV_{\Phi}[a,b] \cap C[a,b] \subset F^{\frac{\Phi^{\,-1}(\varepsilon)}{\varepsilon^2}}. $$
A counterexample is constructed showing that this inclusion cannot be improved. It is further shown that the equality of the classes $F^{\mu}$ and $BV_{\Phi}[a,b] \cap C[a,b]$ occurs only in the case
$$ BV[a, b] \cap C[a,b] = F^{1/\varepsilon}, $$
where $BV[a,b]$ are functions of classical bounded variation. For other cases, a counterexample is constructed showing that if $\mu(\varepsilon)$ grows faster than $\dfrac{1}{\varepsilon}$ as $\varepsilon \to +0$, then the class $F^{\mu} $ is not a subclass of any of the classes $BV_{\Phi}[a, b]$.