Hausdorff measure of a curve I know this is a simple question for mathematicians, but as an engineer I need to understand with a simple example the idea behind the concept of Hausdorff measure.
Say you have a smooth curve $\Gamma$ in $\mathbb{R^2}$ embedded in a square $\Omega$. What is the Hausdorff measure of $\Gamma$? How do you compute it?

Just to see if I understood things correctly, consider the following picture, where I have a square $T$ and a piecewise linear curve $c$, where one of the pieces $f$ is overlapped with one portion of the boundary face of the square, call it $e$.

Is it correct to say that $\mathcal{H}^1(f \cap e) >0$?
 A: A smooth curve has Hausdorff dimension $1$.  If the curve is "simple" (does not intersect itself), then the Hausdorff measure of the curve is the same as the arc length of the curve.
Plug: Theorem 6.3.8
Edgar, Gerald, Measure, topology, and fractal geometry, Undergraduate Texts in Mathematics. New York, NY: Springer (ISBN 978-0-387-74748-4/hbk). xv, 268 p. (2008). ZBL1152.28008.
A: You can think of the Hausdorff measure very geometrically: cover your curve with a bunch of small circles $C_1, ..., C_n$. If these circles are small enough the intersection of each one with the curve will have approximately length $2r_i$ where $r_i$ is the radius of the ball $C_i$. If the balls don't have too many intersections, we can thus say that the curve has approximately length $\sum_i 2r_i$. Informally the (1-dimensional) Hausdorff measure of your curve is now the quantity you get by considering the limit as the radius of the largest of all balls tends to $0$. Consider figures 2.3.1 and 2.3.2 from "Geometric measure theory: a beginner's guide" (available in the internet archive) to get a good visual picture of this process (the book also contains a good explanation of the actual definition). Note how this process can be generalized quite a bit which will result in the definition you'd usually find in a textbook; we for example don't necessarily use circles (or balls in higher dimensions) but rather arbitrary sets for which we define the diameter to be the maximal ("supremal") distance between two points in the set. And the radius we've used will be replaced by multiples of the lebesgue measures of unit-balls.
Actually computing this measure from the definition can be tricky. For a lot of cases that you'd probably be interested in (submanifolds of $\mathbb{R}^n$) however you can get around this by instead computing the volume of your submanifold (see Hausdorff measure, volume form, reference).
Let's consider the example of finding the 1D Hausdorff-measure of the graph of $f \colon x \mapsto x^2$ as a subset of $[0,1]^2$ more or less from definition. In this case we already note that "chaining up balls" on the curve is hard so we switch to other sets: we use rectangular boxes (note that formally we'd have to proceed a bit differently and consider all possible coverings and so on but lets not worry about that). Let $n$ be some natural number and define $I_i := [\frac{i}{n}, \frac{i+1}{n}]$ for all $i=0,...,n-1$, then our curve is covered by $I_i \times [(\frac{i}{n})^2, (\frac{i+1}{n})^2]$. The maximal distance between two points in that set (so its diameter) is $\sqrt{\frac{1}{n^2} + (\frac{2i+1}{n^2})^2}$ which we can work out to be $\frac{\sqrt{(2i+1)^2 + n^2}}{n^2}$. This means for all $n$ we can approximate the length of our curve by $$\sum_{i=0}^{n-1} \operatorname{diam}(I_i \times f(I_i)) = \sum_{i=0}^{n-1} \frac{\sqrt{(2i+1)^2 + n^2}}{n^2}$$
and as $n \to \infty$ this should converge to the 1D Hausdorff-measure of the curve which is $\frac{1}{4}(2\sqrt{5} + \operatorname{arsinh}(2)) \approx 1.4789...$.
