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](https://www.youtube.com/watch?v=IGlGvSXkRGI) 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?




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**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](https://mathoverflow.net/questions/128281/sierpinski-triangle-and-the-chaos-game?rq=1).



Formal justification of the Chaos game in the Sierpinski triangle
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](https://math.stackexchange.com/questions/974233/prove-if-x-is-a-compact-metric-space-then-x-is-separable/974252).
2. Borel probability measures on separable metric spaces [have full measure](https://mathoverflow.net/questions/62747/support-of-probability-measures-on-separable-metric-spaces?rq=1).


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.


