Uniform distribution with respect to Hausdorff measure Suppose I have some nicely defined "fractal" subset of (to make life simpler) Euclidean space $\mathbb{E}^n,$ of some arbitrary Hausdorff dimension $s,$ such that the corresponding Hausdorff measure $H_s$ has positive mass. The first question is whether it makes sense to talk of a "uniform sample" (or Poisson point process) with respect to $H_s$ (since it is only an outer measure), and second question is: if it makes sense, have people figured out how to generate uniform $H_s$ variates (in simple cases, like the von Koch snowflake, or the more interesting cases, like limit sets of Kleinian groups)?
 A: There is a standard map from the interval $[0,1]$ onto the
snowflake curve, that maps Lebesgue measure into (constant multiple of) Hausdorff measure.  So take a sample in $[0,1]$ and map it to the snowflake.  
This works for all of the standard "self-similar" IFS constructions of fractals, when the IFS satisfies the open set condition.  There is a standard "parameter space" or "code space" with a well-understood measure on it that maps onto the fractal, and the image measure is (constant multiple of) Hausdorff measure, and also (constant multiple of) packing measure.
A: Given $A$ a loc. compact space and $\mu$ a loc. finite measure on $A$, there exists a unique Poisson point process $\eta$ with control measure $\mu$. (see for instance th.3.2.1 in "stochastic geometry", by Schneider & Weil, if there are no atoms, but it holds in full generality) .
If $A$ is your fractal and $\mu$ Hausdorff measure, then $\eta$ is what you are looking for, isn't it?
For the construction, everything depends on how you define your fractal set. The difficulty is exactly the same as drawing a uniform random point in a subset that has finite Hausdorff measure.
A: If one interprets "nicely defined" correctly, the answer is "yes, duh!".
One possible interpretation is to be the attractor of a Horseshoe map. Then the dynamics on this attractor can be conjugated to symbolic dynamics and it is clear what "drawing a random point" means, and how to generate them numerically. Of course, one also needs to check that the invariant measure has maximal Hausdorff dimension.
In order to illustrate this, let me give the simplest example. Consider the map $f(x) = x^2 - c$ for $c > 4$. It is well known, that there exists an invariant set $\Sigma = f^{-1}(\Sigma)$ with Hausdorff dimension $\in (0,1)$. $\Sigma$ can be described as
$$
 \Sigma = \{x:\quad \exists C>0 \forall n\geq 1, |f^n(x)|\leq C\}
$$
so the set of points with bounded orbit. If you define for $x\in \Sigma$ the sequence
$$
 x_n = \begin{cases} 1 & f^n(x) > 0, \\\ - 1,& f^n(x) < 0 \end{cases}
$$
Then standard results imply that the Bernouilli measure on $\{-1,1\}^{\mathbb{N}}$ is conjugate to the measure of maximal Hausdorff dimension, which is ergodic with respect to $(\Sigma, f)$. Hence, picking a random point just amounts to choosing a random string of $\pm 1$.
