I know how to use a ruler to approximate a length of an object (like a wire or a stick) in meters. I could also use the ruler to approximate a two dimensional area (like a table top or a parking lot) in $\text{meters}^2$ by dividing it into a grid and counting squares.

I read that we can estimate the length of the coast of Britain to have fractal dimension 1.25. Is there a value in $\text{meters}^{1.25}$ giving the 1.25-dimensional fractal measure of that coast? Can I calculate it using my ruler or rulers of different sizes/precisions? Or if I drew a Koch snowflake whose largest triangle had side length $1\text{ meter}$, could I find it's $\ln(4)/\ln(3)$-dimensional measure in some analogous way?

If there is such a thing as fractional-dimensional measure for dimension $d$, can we give a fractional unit like $\text{meters}^d$ physical meaning?

  • $\begingroup$ Don't have time for a full answer, but the idea of Hausdorff Measure plays a role. $\endgroup$
    – icurays1
    Aug 28, 2017 at 21:32
  • $\begingroup$ If a full answer is too long to fit nicely here at least describe some ideas and scenic points along the route. $\endgroup$ Jan 9, 2020 at 5:56
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    $\begingroup$ Zhu, Zhi Wei; Zhou, Zuo Ling; Jia, Bao Guo (October 2003). "On the Lower Bound of the Hausdorff Measure of the Koch Curve". Acta Mathematica Sinica. 19 (4): 715–728. doi:10.1007/s10114-003-0310-2 claims the $\log_3(4)$-dimensional Hausdorff measure of the Koch curve is between $0.5$ and $0.589$ (the upper bound is a result from a different paper) $\endgroup$
    – Claude
    Jan 9, 2020 at 14:59
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    $\begingroup$ @Claude See the edited the bounty notice. An explanation of the difficulties in handling the Koch snowflake will help the community. Or a study of some other fractal :-) $\endgroup$ Jan 10, 2020 at 14:53
  • $\begingroup$ For what it's worth I've learned a simple, non-rigorous way to understand Hausdorff dimension. Looking closely at any "edge" of the Koch snowflake, you can see it is self similar to 4 copies of itself, and it is 3 times bigger than any of these copies. Hence $\log(4)/\log(3) = log_3(4)$ is the Hausdorff dimension. This agrees with the dimension of Euclidean space in that if you look at a cube of space in $\mathbb{R}^n$ you can see it's broken up into $m^n$ cubes of side-length $1/m$ the original. That gives Hausdorff dimension of $\log(m^n)/\log(m) = n$. Can Hausdorff meausure be related? $\endgroup$ Jan 10, 2020 at 20:39

1 Answer 1


Definition: A metric space is a set $X$ together with a distance function $d(x,y)$ that assigns a nonnegative number to any pair of points $x$ and $y$ from $X$, and $d$ satisfies:

  • $d(x,y) = 0 \iff x=y $.
  • $d(x,y)=d(y,x)$, and
  • $d(x,y) \leq d(x,z) + d(z,y)$ ("the triangle inequality").

Definition: for a subset $A \subset X$, $$ \operatorname{diam} A := \sup \{d(x,y): x \in A, y \in A \} .$$

Definition: for a point $ x \in X $, and $r>0$, the ball of radius $r$ centered at is $$ B(x,r): = \{ y\in X: d(x,y) < r \}.$$

The story of the Hausdorff measure/dimension begins by extending our Euclidean lengths, areas, and volumes to arbitrary dimensions, in a rather crude way: in Euclidean $\mathbb{R}^n$, the $n$-dimensional volume of a ball is $$ \omega_n r^n,$$ where $\omega_n$ is a constant independent of radius. Notice that if radius is scaled by $c>0$ the volume scales by $c^n$. The numbers $\omega_n$ are the all famous ones in dimensions 1, 2, and 3 and for the higher dimensions $\omega_n$ is obtained from $\omega_{n-1}$, recursively, by applying Fubini's theorem: a slice of $n$-sphere is an $(n-1)$-sphere with radius depending on the height at which we slice. Think about cutting a potato, a 3-ball, and getting (2D) disks!

It follows also that $\omega_n$ has an explicit formula via the Gamma function. Therefore, one can plug in noninteger values in place of $n$'s and (formally) define $\omega_\alpha$.

(In what follows $\alpha \in [0,\infty)$.)

Assume you have a metric space and you suspect that it is $\alpha$-dimensional. Then you should assign an $\alpha$-dimensional volume of $$ \omega_\alpha r^\alpha ,$$ to the balls of radius $r$, i.e $B(x,r)$.

And a subset $A$ compared to a ball of same diameter has volume "comparable" (no precise argument here) to $$ \omega_\alpha (\frac{\text{diam} \ A}{2})^\alpha.$$

Now, given a subset $E\subset X$, to define its $\alpha$-dimensional Hausdorff measure, $H^\alpha(E)$, we first find for a $\delta >0$ the quantity $$ H^\alpha_\delta (E) = \inf \sum_i \omega_\alpha (\frac{\text{diam} \ A_i}{2})^\alpha,$$ where infimum is over all countable coverings $\cup_{i=1}^\infty A_i \supset E$ with no restrictions on $A_i$'s other than $\operatorname{diam} A_i < \delta$ for each $i$. That is, we look for the tightest coverings with sets of diameter less than $\delta$.

(In the definitions of $H^\alpha_\delta$ and $H^\alpha$ we allow infinities.)

Now, $$ H^\alpha(E) = \lim_{\delta \to 0} H^\alpha_\delta(E).$$ Notice that smaller $\delta$ makes $H^\alpha_\delta$ larger, since we do not allow some of the coverings that previously worked, hence the infimum goes up. So, this limit exists since $H^\alpha_\delta$ increases as $\delta \to 0$.

With all this buildup we are ready to harvest the fruits!

Lemma: Given any metric space $(X,d)$, any subset $A \subset X$, there exists an $\alpha_0 \in [0,\infty)$ such that \begin{align} H^\alpha(A) &= 0 &\forall \alpha > \alpha_0, \\ H^\alpha(A) &= \infty &\forall \alpha < \alpha_0. \end{align}

We call this $\alpha_0$ the Hausdorff dimension of the (sub)set $A$. The theorem tells us that too large measuring cups will not detect $A$ and smaller measuring cups won't contain $A$, which is quite intuitive and geometric: the area of a line is zero, and length of a (solid/filled) square is infinite.

This is how and why of a dimension of $1.25$ for the coast of Britain—this is the right $\alpha_0$ for it. Now, you ask then how much is this measure? This is the same as asking what happens right at $\alpha = \alpha_0$, which the lemma above leaves out?

There are three possibilities: $H^\alpha(E) = 0$, or $=\infty$, or it is a finite positive number. In each case the value deserves to be called the $\alpha$-dimensional measure/volume of $E$. Easy examples can be given for all cases. For instance, although a plane, extended indefinitely, is 2D, its 2D area is infinity nevertheless.

With specific and explicit metric spaces, specially those constructed through self-similar processes, it is usually straightforward to find the dimension—using the steps of the construction to find coverings that apply to the definition of $H^\alpha_\delta$. But finding the actual value $H^\alpha(E)$, may prove more difficult.

I do not know if this counts as "physical meaning" to non-integer Hausdorff dimensions and measures, but to me it is so natural and geometric!

  • $\begingroup$ This is a helpful explanation of the Hausdorff dimension. Can you give an example of a set of a fractal Hausdorff dimension AND calculate its Hausdorff measure (of the prescribed dimension), non-zero and finite. $\endgroup$ Jan 12, 2020 at 5:52
  • $\begingroup$ I know for the standard 1/3-Cantor set its appropriate dimensional measure is nonzero and positive. I am looking for a reference and if I find i will share it here... $\endgroup$ Jan 12, 2020 at 14:25
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    $\begingroup$ The book "Geometric Measure Theory, a beginner's guide" by Frank Morgan, 4th edition, in exercise 2.6 of chapter 2 we are asked to show that the usual Cantor set, $C$, its H-dimension is $\alpha=\frac{\ln 2}{\ln 3}$. The exercise asks us to "try to prove that $H^{\alpha}(C)=\frac{\omega_\alpha}{2^\alpha} \ ,$ or at least that $H^\alpha(C) > 0.$" $\endgroup$ Jan 12, 2020 at 14:41
  • $\begingroup$ That book, BTW, has many rather concrete examples of fractals and exotic spaces. $\endgroup$ Jan 12, 2020 at 14:43
  • $\begingroup$ It is a worthy exercise! $\endgroup$ Jan 12, 2020 at 15:02

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