Suppose we have $P:A\to[0,1]$, where $A$ is the fat cantor set denoted as $C$.
We produce $C$ by removing $1/4$ of $[0,1]$ around mid-point $1/2$
$$C_{1}=[0,3/8]\cup [5/8,1]$$
$$C_{1,1}=[0,3/8] \ \ \ \ \ \ C_{1,2}=[5/8,1]$$
We repeat, removing $1/16$ of $[0,1]$ around each midpoint of remaining intervals $C_{1,1}$ and $C_{1,2}$.
$$C_{2}=[0,5/32]\cup[7/32,12/32]\cup[5/8,25/32]\cup[27/32,1]$$
$$C_{2,1}=[0,5/32] \ \ \ \ \ \ \ C_{2,2}=[7/32,12/32] \ \ \ \ C_{2,3}=[5/8,25/32] \ \ \ C_{2,4}=[27/32,1]$$
Repeat the process until $\lim\limits_{n\to\infty}C_{n}=C$, where $\lim\limits_{n\to\infty} 1/4^n$ of $[0,1]$ around each midpoint of remaining intervals are removed.
I want to find the average, in terms of integration, such that it is equivelant to
$$\lim_{n\to\infty}\sum_{i=1}^{2^{n}} P(t_{n,i})(1/2^n)$$
Where $\bigcup\limits_{i=1}^{2^{n}}C_{n,i}=C_n$ and $t_{n,i}\in C_{n,i}$
If $P$ is continuous, the average should be $\dfrac{1}{m(C)} \int_C P(x)\; dx$ where $m$ is Lebesgue measure.
EDIT: It looks to me like if $C(i)$ is the $i$'th iteration starting at $C(0)=[0,1]$, $a_i = \int_{C(i)} x^2\; dx$ has generating function $$ g(x) = \sum_{i=0}^\infty a_i x^i = {\frac {35\,{x}^{3}-1066\,{x}^{2}+7456\,x-8192}{ 3 \left( x-1 \right) \left( x-2 \right) \left( x-16 \right) \left( x-8 \right) \left( x-32 \right) }}$$ and $\lim_{i \to \infty} a_i$ is $-$ the residue of this at $x=1$, namely $19/105$. Dividing by the measure of $C$, namely $1/2$, I get the average for $P(x)=x^2$ to be $38/105$.