In Falconer's Fractal Geometry book, when discussing Graphs of Functions (chapter 11), he says that:
"If $\text{graph}(f)$ has continuous derivative, it is not difficult to see that $\text{graph}(f)$ has dimension $1$ and, indeed, is a regular $1$-set."
Definition. A $1$-set is a set $F$ with $0<\mathcal{H}^{1}(F)<\infty$, where $\mathcal{H}^{1}$ denotes the $1$-dimensional Hausdorff measure.
Could somebody please explain this to me? Also, why is it not difficult to see?
One of first things to know about Hausdorff measure is how it behaves under Lipschitz maps: $$ \mathcal H^s(f(E))\le L^s \mathcal H^s(E) $$ where $L$ is the Lipschitz constant of $f$. Suppose $f\in C^1[a,b]$, so there is a constant $M$ such that $|f'|\le M$ on $[a,b]$. Then the map $x\mapsto (x,f(x))$ is Lipschitz: $$ |(x_1,f(x_1)) - (x_2,f(x_2))| \le \sqrt{|x_1-x_2|^2 + M^2|x_1-x_2|^2} = \sqrt{1+M^2}|x_1-x_2| $$ Hence $$\mathcal H^1(\operatorname{graph}(f))\le \sqrt{1+M^2}\mathcal H^1([a,b]) = \sqrt{1+M^2}(b-a)$$ On the other hand, the inverse map $(x,f(x))\mapsto x$ is also Lipschitz, with $L=1$. Hence $$\mathcal H^1(\operatorname{graph}(f))\ge b-a$$ The result follows.
For functions defined on the real line, the Hausdorff measure of the graph will be infinite. But the Hausdorff dimension is still $1$, because it is stable under countable unions, and the union can be taken over the restrictions of $f$ to $[-n,n]$, $n\in\mathbb{N}$.