For the Sierpinski Triangle, $S$, the $d$ dimensional Hausdorff measure is given by, $H^{d}(S)$. If a linear transformation, $W$ is applied to $S$, with

$$W(x,y)=\begin{bmatrix} 1/2 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}$$

We get a new set $S*$. Denote the Hausdorff measure by $H^{d}(S*)$. Are there any results for the value,


  • 4
    $\begingroup$ The easy estimate is that it is between $(1/2)^d$ and $1^d=1$. $\endgroup$ – Gerald Edgar Jun 18 '16 at 17:33
  • $\begingroup$ Related: math.stackexchange.com/questions/107443/… $\endgroup$ – Pedro Lauridsen Ribeiro Jun 18 '16 at 17:43
  • $\begingroup$ @ChristianRemling ... If we are in $2$-space, and $d>1$, can't we say more? Something like ${} \ge (1/2)\cdot 1^{d-1}$ perhaps. $\endgroup$ – Gerald Edgar Jun 18 '16 at 21:45
  • $\begingroup$ @GeraldEdgar: Yes, I think you're right. $\endgroup$ – Christian Remling Jun 18 '16 at 21:51
  • 1
    $\begingroup$ @zachary-w-robertson: The fraction $\rho$ is not fixed and depends on the location of the set $S$. For example, both horizontal and vertical intervals of length 1 have 1-dimensional Hausdorff measure 1, but after the linear transformation the horizontal interval will have 1-dimensional measure 1/2 and vertical will still keep 1. The only thing we can do is to give upper and lower bound for $\rho$ and these bounds depend on the maximal and minimal eigenvalues the linear transformation. $\endgroup$ – Taras Banakh Aug 13 '17 at 5:07

(Just an attempt to answer my question, I'm probably breaking all the rules, but hey I'm just a naïve mathematician at present)

Insert the Ansatz,

$$d\mu(w(S))=r^d d\mu(S)$$

Assume, that since $w_*S=\cup_i w_i(S)$ holds, where $w_*=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$, that

$$d\mu(w_*S)=r_*^dd\mu(S)=\sum_i d\mu(w_i(S))=3r^d d\mu(S)$$

$$\Rightarrow r_*^d=3r^d$$

Given that the Haussdorf measure for $r$ term shrinks the x axis, and the one for $r_*$ term stretches the y axis, which is by symmetry the same as stretching the x axis, I think its safe to assume the two constants are related by,


$$\Rightarrow 3r^d=r^{-d}$$ $$\Rightarrow 3r^{2d}=1$$ $$\Rightarrow r^{2d}=\frac{1}{3}$$ $$\Rightarrow r=3^{-\cfrac{1}{2d}}$$ $$\Rightarrow d\mu(w(S))=r^d\mu(S)=\cfrac{\sqrt{3}}{3} \cdot d\mu(S)$$ $$\Rightarrow \int_{w(S)} dH^d=\cfrac{\sqrt{3}}{3} \cdot \int_S dH^d$$ $$\Rightarrow H^d(w(S))=\cfrac{\sqrt{3}}{3}H^d(S)$$ $$\Rightarrow \cfrac{H^d(S_*)}{H^d(S)}=\cfrac{\sqrt{3}}{3}$$

  • $\begingroup$ I think this is just wrong. $\endgroup$ – Gerald Edgar Jul 19 '16 at 0:24
  • $\begingroup$ @GeraldEdgar No objection from me. I was just trying the obvious solution out...still haven't made any progress. I doubt a formula is easy to get. $\endgroup$ – Zachary W. Robertson Jul 19 '16 at 3:33
  • 7
    $\begingroup$ It's good to try things like this, but it's best not to post as an answer unless you have checked that it is correct. I would suggest deleting this answer and perhaps noting it as a possible approach within the question. Having "answers" that don't answer the question just confuses people; in particular, in some cases it can make it look like the question is resolved and cause people to ignore it. $\endgroup$ – Nate Eldredge Oct 16 '16 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.