We define the integral of a function $f(x,y)$ over a fractal $F$ to be,
$$(1) \quad \int_F f(x,y) \ d\mu(x,y)$$
Where $\mu$ is the normalized Hausdorff measure. Expressed another way, we have,
$$\int_F \ d\mu=\mu(F)=1$$
So, assuming $d\mu(W_n(s,t))=S_n(s,t) \cdot d\mu(s,t)$,
What is $S_n(s,t)$?
The Fractal we are considering for the integral is the attractor of the Dynamical System defined by,
$$F_{n+1}=W_1(F_n)\cup W_2(F_n)$$ $$W_1(x,y)=\begin{bmatrix} 1/2 & 0 \\ 1/2 & d_1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}$$ $$W_2(x,y)=\begin{bmatrix} 1/2 & 0 \\ -1/2 & d_2 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \end{bmatrix}+\begin{bmatrix} 1/2 \\ 1/2 \end{bmatrix}$$
Where $F_0$ is the condensation set, and the transformations $W_1$ and $W_2$ are applied to each element of the set. The attractor $F$ satisfies,
$$(2) \quad F=W_1(F)\cup W_2(F)$$
The above is an IFS for anyone familiar with Michael Barnsley's work. So I'd like to integrate over this particular fractal in a specific fashion. I want the integral of $f(x,y)=y$ over $F$. Measures are additive so I can rewrite $(1)$ using $(2)$.
$$\int_F y \ d\mu(x,y)=\int_{W_1(F)} y \ d\mu(x,y)+\int_{W_2(F)} y \ d\mu(x,y)$$
I'd like to apply a change of variables to both integrals here. For the first one, I want
$$\begin{bmatrix} x \\ y \end{bmatrix}=W_1(s,t)$$
For the second, I want,
$$\begin{bmatrix} x \\ y \end{bmatrix}=W_2(s,t)$$
Doing this isn't the hardest thing, I literally substitute. Its the simplification that's getting me. So after substitution, I get that we have,
$$=\int_F \frac{1}{2} s+d_1 t \ d\mu \left(\frac{1}{2}s,\frac{1}{2}s+d_1 t\right)+\int_F -\frac{1}{2} s+d_2 t+\frac{1}{2} \ d\mu \left( \frac{1}{2}s+\frac{1}{2},-\frac{1}{2}s+d_2 t+\frac{1}{2} \right)$$
So, assuming $d\mu(W_n(s,t))=S_n(s,t) \cdot d\mu(s,t)$,
What is $S_n(s,t)$?
Can this be generalized to linear transformations?
I know that $d\mu(w \cdot s+a, w \cdot t+b)=w^d \cdot d\mu(s,t)$ where $d$ is the exponent in the Hausdorff measure, which is the fractal dimension.
For this particular case, I know that if $d_1=d_2=0$ then the integral is equal to $1/4$. This seems to lead to the conclusion that,
$$d\mu \left(\frac{1}{2}s,\frac{1}{2}s \right)=d\mu(s,s)=\frac{1}{2} d\mu(s,t)$$