The 1-dimensional Hausdorff measure of a curve in the plane

Question

For a set $X\subseteq\mathbb{R}^2$, let $H^1(X)$ be its 1-dimensional Hausdorff measure. Suppose $X$ is a regular curve (say, a graph of a continuous function $f:\mathbb{R}\to\mathbb{R}$). In that case, does it hold that $H^1(X)$ is simply the length of the curve, as can be calculated by standard integration?

And in case the curve is non-regular - for example, if it is the range of a Brownian motion or a space-filling curve - in that case will $H^1(X)$ be $\infty$?

And last one: if I am not mistaken, both the range of a Brownian motion and a space-filling curve are of Hausdorff dimension 2; but do they differ in that that the 2-dimensional Hausdorff measure of the former is 0 and of the latter is the area of the filled space?

Answer 1

First question, yes:

"In any metric space$(X,d)$, if $\gamma:[0,1]\longrightarrow X$ is an injective Lipschitz function, then $H_1(\gamma([0,1])$ is the length of the curve."

See https://www.encyclopediaofmath.org/index.php/Hausdorff_measure.

For the third question, see "Relation with Hausdorff dimension" in the same site.

Answer 2

In order for the 1-dimensional Hausdorff measure of a curve in $\mathbb R^n$ to be equal to its length, the curve has to be BETTER than continuous. Lipschitz curves have these property, but the curves in the set of positive Wiener measure (related to the Brownian motion) are not that nice.

Note that almost all these curves are $C^{0,a}$, for $a \le 1/2$.