(I want to investigate the validity of this approach, as I already know this is the correct result)
I present a proof that $$\int_{K} (x+y) \ \mu(x,y)={{9+\sqrt 3} \over 18}$$ Where the region of integration is the Koch Curve, and $\mu(x,y)$ is measure such that $\mu(K)=1$ and $\mu(K/3)=1/4$. In terms of iteration, the Koch curve at an iteration $i$ can be constructed from $4^i$ lines of lenth $3^{-i}$. The measure of a particular iterate line is thus given by $4^{-i}$ since $\mu(K)$=1. The measure is approximated by taking the limit as $i$ approaches infinity. However, I'm not sure that the argument is mathematically sound.
The Koch Curve maps to itself under these four transformations.
$$ w_1=\begin{bmatrix} 1/3 & 0 \\ 0 & 1/3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}$$
$$w_2={1 \over 3} \cdot \begin{bmatrix} 1/2 & -\sqrt3/2 \\ \sqrt3/2 & 1/2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}+\begin{bmatrix} 1/3 \\ 0 \end{bmatrix} $$
$$w_3={1 \over 3} \cdot \begin{bmatrix} 1/2 & \sqrt3/2 \\ -\sqrt3/2 & 1/2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}+\begin{bmatrix} 1/2 \\ \sqrt3/6 \end{bmatrix} $$
$$w_4=\begin{bmatrix} 1/3 & 0 \\ 0 & 1/3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}+\begin{bmatrix} 2/3 \\ 0 \end{bmatrix} $$
where x and y are the coordinate points of the Koch Curve. Note that since the transformations share the $1/3$ scaling coefficient, we may rewrite the integral as...
$$\int_K (x+y) \mu(x,y)=\int_{K/3} (x+y) \mu(x,y)+\int_{K/3} (1/2 \cdot x-\sqrt3/2 \cdot y +1/3)+(\sqrt3/2 \cdot x+1/2 \cdot y) \mu(x,y)+\int_{K/3} (1/2 \cdot x+\sqrt3/2 \cdot y +1/2)+(1/2 \cdot y-\sqrt3/2 \cdot x+\sqrt3/6) \mu(x,y)+\int_{K/3} (x+2/3+y) \mu(x,y)$$
Since the measure density is uniform we can avoid rewriting the measure function. Adding like terms and noting that the integral and measure functions for each integral are the same, we arrive at...
$$\int_K (x+y) \mu(x,y)=3 \cdot \int_{K/3} (x+y) \mu(x,y)+\left({{\sqrt3 +9} \over 2} \right) \cdot \int_K \mu(x,y)$$
This result means (I'd presumably do this by induction correct?) that we may iteratively write the integral as... $$\int_K (x+y) \mu(x,y)= 3^n \cdot \int_{K/{3^n}} (x+y) \mu(x,y)+\sum_{k=0}^n 3^{k-1} \cdot \left({{\sqrt3 +9} \over 2} \right) \cdot 3^k \cdot \int_{K/{3^k}} \mu(x,y)$$
By the definition of the measure, the integral on the right evaluates to $4^{-k}$...
$$\int_K (x+y) \mu(x,y)= 3^n \cdot \int_{K/{3^n}} (x+y) \mu(x,y)+\sum_{k=0}^n 1/3 \cdot \left({{\sqrt3 +9} \over 2} \right) \cdot 4^{-k}$$
As n approaches infinity, I can argue that the measure goes down proportional with $4^n$, however I don't know how to put that in a more rigorous format. The final result is...
$$\int_{K} (x+y) \ \mu(x,y)={{9+\sqrt 3} \over 18}$$
Questions: Is my argument rigorous enough? Is my argument for the vanishing of the right hand integral to hand wavy? I'm interested in making my argument more professional. Identifying possible flaws or risks in generalizing this approach would also be appreciated. A better method of evaluating this integral would also be interesting to consider...
