We know the dimension of the Koch snowflake's perimeter, but does it have a measure? I start with an equilateral triangle with side perimeter three meters. I can define a Koch snowflake by the following sequence of figures. Starting with that triangle, produce the next figure by replacing the middle third of each line segment with the other two sides of an equilateral triangle whose third vertex is exterior to the previous figure. We know that the Hausdorff dimension of the limit is (log 4)/(log 3).
Can we say that the measure of the snowflake's perimeter is three of a unit that we may express as one meter to the power ((log 4)/(log 3))? Something like that? Why, or why not?
I thought of this while fishing for a weekly brain teaser puzzle to pose to my office mates. I think it's too hard!
 A: The 2007 paper you linked in a comment, Bounds of the Hausdorff measure of the Koch curve, contains information which is the most recent I could find. In it, and the related Bounds of Hausdorff measure of the Sierpinski gasket, both by Baoguo Jia, an approach to estimating Hausdorff measures of self-similar sets satisfying the open set condition (citing some books by Falconer on fractal geometry for some facts about the Hausdorff measure for the proof that this method works) is outlined. However, the method is not very effective (it is not easy to calculate good approximations with certainty). 
In "Bounds of the Hausdorff measure of the Koch curve", the author proves that the $s$-dimensional (where $s=\log4/\log3$) Hausdorff measure of the Koch curve (base length 1) is bounded below by $$\left(2 \left(\frac{2 \sqrt{3}}{9}\right)^s\right)
   \exp\left(-\frac{12 s\sqrt{3}}{9}\right)=2\text{^}\left(-2-\frac{8}{\sqrt{3} \log (3)}+s\right)\approx0.0325239$$and bounded above by $$2 \left(\frac{2 \sqrt{3}}{9}\right)^s=2^{s-2}\approx0.599512\text.$$
At the end of the paper, they conjecture tighter bounds. Assuming their $6/81$ was meant to be ${\sqrt{876}}/{81}$ (the context makes this a reasonable typo), they conjecture a lower bound of $$\left(\frac{1}{122} 4^4
   \left(\frac{\sqrt{876}}{81}\right)^s\right)
   \exp\left(-\frac{12 s\sqrt{3} }{3^5}\right)=\frac{1}{61}73^{s/2}*2\text{^}\left(s-\frac{8}{27 \sqrt{3} \log
   (3)}\right)\approx0.528786$$ and an upper bound of $$\frac{1}{122} 4^4
   \left(\frac{\sqrt{876}}{81}\right)^s=\frac{1}{61}73^{s/2}*2^s\approx0.589052\text.$$
In short, if the horizontal base for the curve is $1\mathrm{m}$, and this paper is correct, then the $s$-dimensional Hausdorff measure (which is the appropriate way to measure the size of something a self-similar fractal like this) is definitely less than $0.6\mathrm{m}^{s}$, and probably more than $0.5\mathrm{m}^{s}$.
