Context: This is part of an absurd thought experiment in which an immortal, telekinetic rat tries to push a blinking button from a variable distance while piloting a spacecraft blindfolded at relativistic speeds. I need to calculate what the rat most likely believes is the probability that the button is lit. This is my attempt to do that.
Fix $\omega\in\Bbb R$, and let $L:\Bbb R\to\Bbb R$ be given by
$$L(t)=\lfloor\sin(\omega t)+1\rfloor$$
$\color{lightgray}{\text{[This is the signal controlling the light]}}$
For each $n\in\Bbb N$, let $[\Bbb R]^n=\{S\subseteq\Bbb R:|S|=n\}$. Let $[\Bbb R]^{<\Bbb N}=\bigcup_{n\in\Bbb N}[\Bbb R]^n$, and define $A:[\Bbb R]^{<\Bbb N}\to\Bbb R$ according to
$$\begin{align} A(T)&=\frac1{|T|}\sum_{t\in T}\lfloor\sin(\omega t)+1\rfloor\\ &=\frac1{|T|}\sum_{t\in T}L(t) \end{align}$$
That is, $A(T)$ is the average value of the image of $T$ under $L$.
$\color{lightgray}{\text{[Each finite subset of $\Bbb R$ corresponds to a possible set of times (as measured from the button)}\\ {\text{when information about the state of the button reaches the rat. Due to time dilation and}\\ \text{unpredictable movement (a result of the blindfold), we can assume nothing about the likelihood}\\ \text{of a given set.]}}}$
Questions:
How is the average value of $A$ defined?
What is the average value of $A$?
It seems extremely obvious that the answer should be $1/2$. Realistically, a large enough uniform random sampling of a sufficiently large interval $[a,b]$ will almost always land you with a value of $1/2$; but I don't know how to show that the average average is actually $1/2$, and that the apparent tendency towards this value isn't a miraculous coincidence.
Intuitively, I'd think that we'd want to take some kind limit-integral like
$$\overline A=\lim_{R\to[\Bbb R]^{<\Bbb N}}\mu(R)^{-1}\int_R A(T)\ dT$$
Given $|[\Bbb R]^{<\Bbb N}|=|\Bbb R|$ (obviously true), there must be some bijection $C:\Bbb R\to[\Bbb R]^{<\Bbb N}$. Assuming that we can define $C$ in a nice way, we could reasonably assert
$$\lim_{R\to[\Bbb R]^{<\Bbb N}}\mu(R)^{-1}\int_R A(T)\ dT=\lim_{a\to-\infty}\lim_{b\to\infty}\frac1{b-a}\int_a^b A(C(t))\ dt$$
but 1) this requires $A\circ C:\Bbb R\to\Bbb R$ to be continuous, which I'm not sure is even possible (and I don't know how to prove it either way), 2) there may not be a nice way to define $C$ even if such a continuous function does exist.
In order for this question to make sense, you need to give a distribution over the sets $T \in [\mathbb R]^{< \mathbb N}$. One way to do this would be to pick a distribution $p$ over $\mathbb N$, a distribution $\mu$ over $\mathbb R$, and propose that we first generate $N \sim p$, then each of the $N$ elements of $T$ independently from $\mu$.
The average value of $A$ will then depend on your choice of $p$ and $\mu$, and will not in general equal $\tfrac12.$ For example, if $\mu \sim U[0, \pi/\omega],$ the average will be 1. I'm assuming you're interested in the case where $\mu$ approximates a uniform distribution on all of $\mathbb R$; no such distribution exists, but we can take a limit, e.g. letting $\mu \sim U[-R, R]$ for $R \to \infty$.
In this case, we obtain
$$\begin{align*} \mathbb E[A] &= \sum_n p_n\ \mathbb E[A \mid N = n]\\ &= \sum_n p_n \int L\ d\mu\\ &= \frac{1}{2R}\int_{-R}^R \lfloor \sin(\omega t) + 1 \rfloor\ dt, \end{align*}$$
which equals $\tfrac12$ for any $R$, by symmetry. You can expect similar results for other limits, but the details will depend on what you want to assume about $p$ and $\mu$.