# Formal justification of the Chaos game in the Sierpinski triangle

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?

Definitions

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.

Notes

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.