I want to justify why the Chaos game works to produce Sierpinski triangle. I use a theorem taken from Massopust Interpolation and Approximation with Splines and Fractals.

Suppose that $(X,d)$ is a compact metric space and $(X,F,P)$ is an IFS with probabilities. Futher assume that $m \in P(X)$ is the invariant fractal measure. Let $x_0 \in X$ be arbitrary and let $x_k = f_i(x_{k-1})$ for $k \in \mathbb{N}$ where $f_i \in F$ is chosen with probability $p_i \in P$.

Then, for almost all random sequences $\{x_k\}$, the following equality holds: $$m(A) = \lim_{k \to \infty} \frac{N(A \cap \{x_l:l = 0,1,\ldots,k)\})}{k+1}$$ for all $A \in B(H(X))$ with $m(fr(A)) = 0$ and where $N(B)$ denotes the number of points in set $B$.

The right hand side of the equation represents the fraction of points that lie on set $A$. So if I choose the IFS generating the Sierpinski triangle such as in this video I would need that $\mu(\mathcal{a}) \sim 1$ where $\mathcal{a}$ is the Sierpinski triangle.

I tried to compute by hand the invariant fractal measure for the IFS that produces the Sierpinski triangle but I was obtaining a wrong results. Is that the right way to go? How can I justify that the chaos game produces the Sierpinski triangle in the limit using this theorem?


Given a IFS with probabilities, $(X,\{f_i\})$ formed by a compact metric space $(X,d)$ and a finite number of contractive mappings $f_i:X \to X$ and a set of probabilities $p_i > 0$ with $\sum p_i = 1$, the measure $\mu$ such that $\mu = \sum p_i \cdot \mu \circ f_i ^{-1}$ is called $p$-balanced measure or invariant fractal measure.


This question gives a formal answer to Sierpinski Triangle and the Chaos Game.


The following is a small correction to Massopust Interpolation and Approximation with Splines and Fractals.

Relation between the fractal generated by the IFS $A$ and the invariant measure $m$

If the involved probabilities are strictly positive $A = supp \; m$.

Support of a Borel probability measure on a compact metric space has measure 1

  1. Since $X$ is a compact metric space, it is separable.

  2. Borel probability measures on separable metric spaces have full measure.

Recall $P(X)$ is the set of measures $\mu$ on $(X,B(X))$ such that $\mu(X) = 1$. This makes the elements of $P(X)$ Borel probability measures. So we can write:

$$1 = \lim_{k \to \infty} \frac{N(A \cap \{x_l:l = 0,1,\ldots,k)\})}{k+1}$$ for all $A \in B(H(X))$ with $m(fr(A)) = 0$. As we wanted. It would be interesting to comment on the conditions $A \in B(H(X))$ and $m(fr(A)) = 0$. To see if they are satisfied in the Sierpinsky triangle.


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