Let $A$ be the set of points given by the graph of $f(x) = x^2$ on the interval $[0,1]$:
$$
A = \{(x,x^2) \in \mathbb{R}^2\bigg| \ x \in[0,2] \}.
$$
I want to calculate the length of this one dimensional curve in $\mathbb{R}^2$ using the Hausdorff measure. Using the standard method of calculating this length via an integral with an arc-length parameterization I (meaning WolframAlpha) get:
$$
I(A) = \int_0^2 \sqrt{1 + \frac{dy}{dx}^2}dx = \int_0^2 \sqrt{1 + (2x)^2} dx = \sqrt{17} + \frac{1}{4} \sinh^{-1}(4) \approx 4.64678.
$$
The Hausdorff measure is defined as follows. Let $\alpha_m$ be the Lebesgue measure of the closed unt ball in $\mathbb{R}^m$. For $B \subset \mathbb{R}^n$, the $m$-dimesnsional Hausdorff measure of $B$ is defined as:
$$
H^m(B) = \lim_{\delta \to 0 } \inf \bigg\{\sum_j \alpha_m \bigg(\frac{1}{2}\bigg)^m (\text{diam}(S_j))^m \bigg| B \subset \bigcup_j S_j, \ \text{diam}(S_j) < \delta \bigg\}
$$
So in my case I want the one-dimensional Hausdorff measure of the set $A$ I defined earlier and the definition becomes:
$$
H^1(A) = \lim_{\delta \to 0 } \inf \bigg\{\sum_j (\text{diam}(S_j)) \bigg| A \subset \bigcup_j S_j, \ \text{diam}(S_j) < \delta \bigg\}
$$
I believe this should agree with earlier result obtained with the arc-length integral and we should get:
$$
H^1(A) = I(A) \sqrt{17} + \frac{1}{4} \sinh^{-1}(4) \approx 4.64678.
$$
But I have no idea how we can evaluate this Hausdorff measure of $A$ from the definition. It seems extremely complicated, and this is about the simplest example I can think of!
How do we pick the sets $S_j$ whose union covers $A$ and that have diameter less than $\delta$ – there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit. How can we evaluate it?
Best Answer
Of course there is a general theorem saying that for a smooth arc $\gamma$ the one-dimensional Hausdorff measure of this arc is equal to its length computed via an integral.
In the case at hand the length $L(\gamma)$ of the arc $$\gamma:\quad x\mapsto(x,x^2)\qquad(0\leq x\leq 2)$$ is defined as $$L(\gamma):=\sup_{\cal P}\sum_{k=1}^{N_{\cal P}}\sqrt{(x_k-x_{k-1})^2+(x_k^2-x_{k-1}^2)^2}\ ,\tag{1}$$ where the $\sup$ is taken over all partitions ${\cal P}$ of the $x$-interval $[0,2]$. It is proven in calculus 102 that this $L(\gamma)$ is equal to the integral you have calculated.
On the other hand arguments similar to those used in the proof of $(1)$, but "the other way around", show that $H^1(\gamma)$, as defined in the question, is equal to the same integral.