Construct a set $s\subseteq[0,1]$ by sampling points in $[0,1]$ with uniform probability density $x\leq1$ so that $|s|=x$. Interpret this as a sampling frame during which data is captured. Now, consider a decreasing function $f(t)$ with domain $[0,1]$. One can calculate the expected integral sum of the observed values of $f(t)$ (where the expectation is over every possible permutation of $s$) as $$\int_0^{1}\!x f(t)\,dt.$$
Now, fix a truncation point $T\in\mathbb{N}$ and construct a new, periodic, function $g(t)$ as follows: $$g(t)=f(t)$$ if $t<1/T$, and $$g(t)=g[t-(1/T)]$$ if $t\geq 1/T$. How can one calculate the expected integral sum of the unique observed values of $g(t)$ given $s\subseteq[0,1]$?
I had thought that one could approach this problem by saying that repeating the first $1/T$ seconds of signal $f$ $T$ times was equivalent to supposing that $s\subseteq[0,1/T]$ with the density of each point in $s$ being $T\min\{x,1/T\}$. The integral would then be $$\int_0^{1/T}\!T\min\{x,1/T\} f(t)\,dt.$$
Suppose $x=1/T$. The above integral then becomes $$\int_0^{1/T}\!f(t)\,dt,$$ which corresponds to observing $f(t)$ at every point in $[0,1/T]$ with probability 1. This seems counterintuitive because there are many permutations of $s$ (such as $s=[0,1/2T]\cup[1/T,3/2T]$) in which some positive measure of $f(t)$ are observed more than once.