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I'm learning measure theory, I'll make some statements I'm not very sure about. Please correct me if I'm wrong.

This problem from a job ad states:

Let $(X_1,X_2)$ be uniform over the unit Sierpinski triangle (represented in Cartesian coordinates). What is its covariance matrix?

But the proportion of area subtracted from an equilateral to form a Sierpinsky triangle (i.e. its Lebesgue measure, I think?) is:

$$ \sum_{n=1}^\infty \frac{3^{n-1}}{4^n} = \frac{1}{4} \frac{1 - 0}{1 - 3/4} = \frac{1}{4}\frac{1}{1/4} = 1 $$

So there should be no "measure" left in the Sierpinsky triangle itself (I think the Lebesgue measure of the Sierpinsky triangle is simply undefined?).

So my question is: is it possible to define such a uniform distribution over the Sierpinsky triangle? Thus, does the covariance of such a distribution actually exist?

Mark McClure
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rhaps0dy
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    I'm not sure if this is directly analogous, but it at least looks similar: if you take $[0,1]\times[0,1]$ and subtract $[0,1]\times(0,1]$, you're left with a set that has zero measure of area but it's nevertheless possible to define a uniform distribution on that leftover set. – David Z Nov 08 '18 at 22:06
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    Sure, the Lebesgue measure of the Sierspinski triangle is 0. But, on a set level, the Sierspinski triangle is roughly equivalent to the interval $[0,1]$ by taking a point in the triangle, $(0.x_1 x_2 \ldots, 0.y_1 y_2 \ldots)$ in binary, to the real number with representation $0.(2x_1+y_1) (2x_2+y_2) \ldots$ in base 3. (Though there are countably many points with multiple representations...) You should then be able to transfer Lebesgue measure on $[0,1]$ to the Sierpinski triangle via this map. Then... – Daniel Schepler Nov 08 '18 at 22:13
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    I would guess that transferred measure has nice properties such as that each of the component triangles has measure $\frac{1}{3}$, the contraction of a triangle to one of the component triangles respects measures aside from a factor of $\frac{1}{3}$, etc. – Daniel Schepler Nov 08 '18 at 22:14
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    You might also be able to define the same measure by starting with the algebra of "finite unions of subtriangles", assign the obvious measure to each such that each subtriangle at level $n$ gets measure $3^{-n}$, then go through the outer measure to Caratheodory measurability condition construction. – Daniel Schepler Nov 08 '18 at 22:16
  • David Z, I see. This seems to be a singular distribution, which is what is talked about in [this related question][https://math.stackexchange.com/questions/2796448/uniform-distribution-on-the-cantor-set-durretts-probability-theory-and-exam/2796609] that I just found. Though it's still not clear to me that an analogous distribution would necessarily have a covariance? – rhaps0dy Nov 08 '18 at 22:18
  • Once you have a measure on the Sierpinski triangle, then you can define Lebesgue integration with respect to that measure, presumably prove $x^2,xy,y^2$ are measurable functions, and calculate the Lebesgue integrals of those functions. – Daniel Schepler Nov 08 '18 at 22:19
  • Daniel Schepler, doesn't the outer measure of a Sierpinsky triangle have simply the same Lebesgue measure as a full triangle? My reasoning: the infimum countable union of triangles that covers the whole Sierpinsky triangle is a full triangle of the same size. – rhaps0dy Nov 08 '18 at 22:20
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    I'm not sure what you're asking in that last comment - but in the construction of the Sierpinski triangle, the set of boxes left at step $n$ has Lebesgue measure $(3/4)^n$ and contains the eventual set, so the Sierpinski triangle is Lebesgue measurable with measure 0. However, the new measure I'm defining on $S \subseteq \mathbb{R}^2$ is not really related to the Lesesgue measure on $\mathbb{R}^2$. – Daniel Schepler Nov 08 '18 at 22:36
  • Ah I see, off course the construction at step n covers the whole set, I was totally wrong there. I now also understand that you were talking about relating the measure on the triangle to the (unidimensional) interval $[0, 1]$, and not $\mathbb{R}^2$, I hadn't read that correctly. I think this qualifies as an answer too, thank you very much! – rhaps0dy Nov 08 '18 at 22:45

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Nobody's mentioned Hausdorff measure yet, which is a common and natural choice of measure on a fractal. The Hausdorff dimension of the Sierpinski triangle is $\ln 3 / \ln 2$, so we would be talking about Hausdorff measure of that dimension. Apparently it is known that this is a finite measure (though the actual Hausdorff measure of the triangle is not known exactly, see http://math.bme.hu/~morap/sierpinski.pdf) so it can be renormalized to be a probability measure.

Another construction is to let $\mu_n$ be normalized Lebesgue measure on the region remaining after removing the $n$th level of triangles (so there is still positive area remaining). It should be easy to prove that this sequence of probability measures converges weakly to some probability measure $\mu$, which seems to deserve the name of "uniform" measure on the gasket. I would guess it is the same as normalized Hausdorff measure, though I haven't sat down to check it carefully.

The latter is probably how you are actually supposed to solve the problem. Let $(X_{1,n}, X_{2,n})$ be uniformly distributed on the region remaining after removing the $n$th level, and compute $\lim_{n \to \infty} \operatorname{Var}(X_{1,n})$ and $\lim_{n \to \infty} \operatorname{Cov}(X_{1,n}, X_{2,n})$. (The definition of weak convergence in terms of integrating continuous functions will show you that this yields the moments of the limiting measure.)

In any case the measure is supported in a bounded set, so it is trivial that it has moments of all orders; in particular the covariance exists.

Nate Eldredge
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It's true that in many cases, for a (measurable) subset $S \subseteq \mathbb{R}^n$, you will be able to inherit a probability measure on $S$ from Lebesgue measure on $\mathbb{R}^n$, by defining $\mu(A) := \frac{m(A)}{m(S)}$ for $A \subseteq S$. It's also true that in this case, the Lebesgue measure of the Sierpinski triangle $S \subseteq \mathbb{R}^2$ is 0, meaning that definition won't work. So, we will have to go another route to construct a measure on $S$.

An analogous situation is that we can define a "uniform" measure on the unit sphere $S^n \subseteq \mathbb{R}^{n+1}$, even though $S^n$ has Lebesgue measure zero. (In this particular case, one way to construct the measure is to restrict the Riemannian manifold structure on $\mathbb{R}^{n+1}$ to $S^n$, then construct the volume form on $S^n$ corresponding to this Riemannian manifold structure on $S^n$. However, the Sierpinski triangle is certainly not a submanifold of $\mathbb{R}^2$, so that construction won't work here.)

One way to construct such a probability measure on $S$, which will be uniform, would be: start with the algebra of finite unions of subtriangles, and to each such object associate a measure in the natural way such that a subtriangle of level $n$ gets measure $3^{-n}$. Then, the construction of taking the corresponding outer measure, then forming the $\sigma$-algebra of sets measurable by the Caratheodory criterion will form a measure on $S$. (Though verifying that the initial measure satisfies the countable additivity condition will probably be fairly tricky, comparable to the trickiness of the check involved in constructing Lebesgue measure. This will be necessary to conclude that each subtriangle is measurable with respect to the final measure and has the expected measure $3^{-n}$.)

This measure will then, as expected, satisfy uniformity conditions such as translation invariance (when you take a translation from one subtriangle to another); and respecting contraction of $S$ to a subtriangle, with the contraction of $S$ to a subtriangle of level $n$ multiplying measures by $3^{-n}$. This measure will also certainly be singular with respect to Lebesgue measure (if you extend it to a measure on $\mathbb{R}^2$).

Daniel Schepler
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Apparently, you can define a distribution over the Cantor set, which is similar. The distribution has a CDF but a PDF that is "infinite" in all the points. See this question and its answers.

icurays1 [1] answers that this is an example of a singular distribution, another example of which is a Gaussian distribution with singular covariance. Thus, it appears that the distribution exists, and its moments probably also exist, using the general definition of the moment-generating function.

[1] icurays1 (https://math.stackexchange.com/users/49070/icurays1), Durrett: what is a "uniform" distribution on the Cantor set?, URL (version: 2016-11-13): https://math.stackexchange.com/q/2011492

rhaps0dy
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