Find the average of a function defined on a Fat Cantor Set

Question

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 such that

$$\lim_{n\to\infty}\sum_{i=1}^{2^{n}} P(t_{n,i})(1/2^n)$$

Where $\bigcup\limits_{i=1}^{2^{n}}C_{n,i}=C_n$ and $t_{n,i}\in C_{n,i}$

How would we use mesh and list manipulation to solve this? Would we get $\frac{1}{2}\int_{0}^{1} P dx$?

EDIT: I found when $P(x)=x^2$, the average I want should be $4/11$ but according to this answer, the average should be $38/105$? Which answer is correct?

Answer 1

For each $n\ge 1$ let $l_{n,i}$ and $r_{n,i}$ be the left and the right endpoints of the segment $C_{n,i}$. Since the function $x^2$ increases on $[0,1]$ and $ l_{n,i}\le t_{n,i}\le r_{n,i}$ for each $n$ and $j$, we have the bounds

$$S_{l,n}=\frac 1{2^n}\sum_{i=1}^{2^{n}} P(l_{n,i})\le \frac 1{2^n}\sum_{i=1}^{2^{n}} P(t_{n,i})\le \frac 1{2^n}\sum_{i=1}^{2^{n}} P(r_{n,i})=S_{r,n}.$$

Put $l_{0,1}=0$, $r_{0,1}=1$. Then for each $n\ge 0$ we have a recurrence

$l_{n+1,2j-1}=l_{n,j}$,

$r_{n+1,2j-1}=(l_{n,j}+r_{n,j})/2-0.5\cdot 4^{-n-1}$,

$l_{n+1,2j}=(l_{n,j}+r_{n,j})/2+0.5\cdot 4^{-n-1}$, and

$r_{n+1,2j}=r_{n,j}$.

This recurrence is a base for the following Pascal program calculating the values of $S_{l,n}$ and $S_{r,n}$ up to $n=9$.

program p3608100;
const
 NN=9;
var
 OFi:Text;
 n:Byte;
 j:Word;
 l,r:array[0..NN,1..1 shl NN]of Real;
 Sl,Sr:array[0..NN]of Real;
 pow4:Real;
begin
assign(OFi,'3608100.txt');
rewrite(OFi);
l[0,1]:=0;
r[0,1]:=1;
for n:=0 to NN-1 do begin
 pow4:=1 shl (n+1);
 pow4:=1/sqr(pow4);
 for j:=1 to 1 shl n do begin
  l[n+1,2*j-1]:=l[n,j];
  r[n+1,2*j-1]:=(l[n,j]+r[n,j])/2-0.5*pow4;
  l[n+1,2*j]:=(l[n,j]+r[n,j])/2+0.5*pow4;
  r[n+1,2*j]:=r[n,j];
 end;
 Sl[n+1]:=0;Sr[n+1]:=0;
 for j:=1 to 1 shl (n+1) do begin
  Sl[n+1]:=Sl[n+1]+sqr(l[n+1,j]);
  Sr[n+1]:=Sr[n+1]+sqr(r[n+1,j])
 end;
 Sl[n+1]:=Sl[n+1]/(1 shl (n+1));
 Sr[n+1]:=Sr[n+1]/(1 shl (n+1));
 writeln(OFi,Sl[n+1]:10:8,' ',Sr[n+1]:10:8);
end;
close(OFi);
end.

A Delphi version of the program calculated the following values $S_{l,n}$ and $S_{r,n}$ up to $n=12$ and with double precision:

0.195312500000000 0.570312500000000
0.287597656250000 0.443847656250000
0.327545166015625 0.397857666015625
0.345483779907227 0.378686904907227
0.353891015052795 0.370004296302795
0.357947923243046 0.365882493555546
0.359938955400139 0.363875722978264
0.360925024753669 0.362885779148201
0.361415686458713 0.362394156307346
0.361660422664158 0.362149180751317
0.361782641930041 0.362026901764331
0.361843714331917 0.361965814446739

Since $38/105= 0.36 (190476)$, these calculations confirm the answer by Robert Israel;

Answer 2

According @RobertIsrael the average should be $38/105$. This is what he states.

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 $38/105$.