Let us consider more closely the question about space-filling curves. The Peano curve and the Hilbert curve, and several other variations of them, have parametrizations $[0,1]\to[0,1]^2$ that actually take the 1-dimensional Lebesgue measure on $[0,1]\, ,$ $\mathcal{L} _1\, ,$ to the 2-dimensional Lebesgue measure on $[0,1]^2\, ,$ $\mathcal{L} _2\, ,$ by push-forward. Notably, the Peano curve $\gamma: [0,1]\to[0,1]^2$ is originally defined in terms of ternary expansions of real numbers, which makes it particularly simple to check the independence of the coordinates. The Hilbert curve also admits a (slightly less) simple description in terms of binary expansions. Here are the details; I'll try to give you an abstract-nonsense-friendly description.
Let's start by some well known facts. The map that takes a sequence $a:=(a_1,a_2,\dots)\in\ 3^\mathbb{N}$ (here $3:=\{0,1,2\}$ and $\mathbb{N}$ is the set of positive integers) into its value as ternary expansion, namely $\mathrm{v}(a):= \sum_ {n\ge1} 3^{-n}a_ n\in[0,1]$, is a continuous surjective map (and bijective, up to removing a certain countable, null subset $D$ of the domain) :
$$\operatorname{v}:3^\mathbb{N}\to[0,1]\, .$$
Also, this map takes the measure $\mathbf{m}$, product of countably many copies of the uniform probability measure on $3$ to the measure $\mathcal{L} _1$ ("the base 3 digits of a real number are independent and uniformly distributed". Any other base of course works as well).
Now, at the level of the ternary sequences we have the nice and simple Cantor bijection $3^\mathbb{N}\to 3^\mathbb{N}\times 3^\mathbb{N}$
"split the sequence of digits into the sequence of odd-position digits and the sequence of even-position digits"
$$C:3^\mathbb{N}\ni (a_n)\mapsto \big(\, (a_{2n-1}), (a_{2n})\, \big) \in 3^\mathbb{N}\times 3^\mathbb{N} $$
which is easily seen to be a compact metric space homeomorphism that takes the measure $\mathbf{m}$ into the product measure $\mathbf{m}\otimes\mathbf{m}$. Note that this homeomorphism does not pass through the quotient map $\mathrm{v}$, for in general sequences with the same value do not produce sequences with the same value by extraction of a subsequence. However, removing the above mentioned countable set $D$ the map $\mathbf{v}$ becomes bijective and you do have correspondingly a bi-measurable, a.e. defined (or everywhere but non-continuous) map of the unit interval to the unit square that takes the measure $\mathcal{L} _1$ to the measure $\mathcal{L} _2$. All that is quite standard. Now your question cames quite naturally, as a request for a commutative square:
Can we find another measure preserving homeomorphism
$\Gamma:3^\mathbb{N}\to
> 3^\mathbb{N}\times 3^\mathbb{N}$ that
induces a map $\gamma:
> [0,1]\to[0,1]\times[0,1]$ through the map $\mathbf{v}$, that is, such that $\gamma\circ\mathbf{v}=(\mathbf{v}\times\mathbf{v})\circ\Gamma$ ?
Since $\mathbf{v}$ is a quotient map, this map $\gamma$ will be automatically a continuous surjection, that also takes $\mathcal{L} _1$ to $\mathcal{L} _2$. The answer is yes, and this is Peano's construction (he was not interested in the measure-theoretic property, but this also follows immediately from the definition). It's the way he constructed his example in the celebrated paper dated 1890 on Mathematische Annalen, " Sur une courbe, qui remplit toute une aire plane ".
Here's Peano definition of the map $\Gamma$: extract as before the sequence of the odd-position digits and the even-position digits, but first invert every odd-position digit whenever there are an odd number of odd even-position digits before it, and invert every even-position digit
whenever there are an odd number of odd odd-position digits before it. Here "invert" just means taking $x\in3$ to $2-x$, that is $0$ to $2$, $1$ to $1$, $2$ to $0$. Translating this definition in a formula, is not difficult to check it defines a homeomorphism of the form $\Gamma=C \circ \phi$, compatible with the map $\mathbf{v}$. The measure property is quite obvious, since the "inverting digit map" $ \phi:3^\mathbb{N}\to 3^\mathbb{N}$ is clearly an involutory preserving measure homeomorphism.
For the Hilbert curve, the digit description has to be done in terms of binary representation, and it is slightly less simple (I have it written somewhere and will look for it and quote here at request) but everything works as well.
Let me finish with an historical note. Of course, what is not easy in the short Peano's paper is to understand what's going on geometrically. He made no picture in this paper, although the graphical iterative construction was perfectly clear to him, and was with all probability his starting point —he made an ornamental tiling showing a picture of the curve in his home in Turin. His choice to avoid any appeal to graphical visualization was no doubt motivated by a desire for a well-founded, completely rigorous proof owing nothing to pictures, in the spirit of the program of arithmetization of analysis. In the conclusion of his paper he observed incidentally that the same construction may be made with all odd basis, and even basis too, although in the latter case, by slightly more complicated formulae (hence less elegant from his viewpoint). In order to make Peano's example more accessible to the mathematical community, a couple of years later Hilbert wrote on the same journal the very clear geometric construction that we know. He chose the Peano construction in base 2 because it is simpler from the graphical point of view.
$\alpha$ need not be a random variable.
The most natural choice for $(\Omega, \mathcal{F}, P)$ is product space: let $\Omega = [0,1]^{[0,1] \cup \{2\}}$ and for $A \subset [0,1] \cup 2$, let $\pi_A : \Omega \to [0,1]^A$ be the projection map. (For $a \in [0,1] \cup 2$, we let $\pi_a$ denote $\pi_{\{a\}} : \Omega \to [0,1]^{\{a\}} = [0,1]$.) Let $\mathcal{F}$ be the product $\sigma$-field on $\Omega$, which is by definition the smallest $\sigma$-field that makes all $\pi_a : \Omega \to [0,1]$ measurable. It is then not hard to show that every $B \in F$ is of the form $B = \pi_A^{-1}(C)$ for some countable $A \subset [0,1] \cup \{2\}$ and some $C \subset [0,1]^A$ which is measurable with respect to the product $\sigma$-field on $[0,1]^A$. That is, a measurable subset of $\Omega$ can only look at countably many coordinates.
Now for each countable $A \subset [0,1] \cup \{2\}$, let $\mu_A$ be the measure on $[0,1]^A$ which is the infinite product of Lebesgue measure, and for $B = \pi_A^{-1}(C) \in \mathcal{F}$, set $\mu(B) = \mu_A(C)$. It's easy to verify that $\mu$ is well defined and countably additive (note that $\bigcup_n \pi_{A_n}^{-1}(C_n)$ is of the form $\pi_A^{-1}(C)$ where $A = \bigcup_n A_n$ is countable). Moreover, $\mu$ is a probability measure and, under $\mu$, the $\pi_a$ are iid $U(0,1)$ random variables. So $(\Omega, \mathcal{F}, \mu)$ satisfies the hypotheses, taking $\xi_a = \pi_a$ and $u = \pi_2$.
We then define $\alpha$ as you say, via $\alpha(\omega) = \pi_{\pi_2(\omega)}(\omega)$. Then $\alpha$ is certainly not a random variable. If it were, then $\alpha^{-1}([0,1/2])$ would be of the form $\pi_A^{-1}(C)$ for $A$ countable and $C \subset [0,1]^A$. Let $b \in [0,1] \setminus A$. Define $\omega$ via $\omega(2)=b$, $\omega(b)=1$, and $\omega(a) = 0$ for all $a \in [0,1] \setminus \{b\}$. Define $\omega'$ similarly but with $\omega'(b)=0$. Then $\alpha^{-1}([0,1/2])$ contains $\omega'$ but not $\omega$, whereas $\pi_A^{-1}(C)$ must contain both of $\omega,\omega'$ or neither.
This doesn't rule out the possibility of being able to choose some more exotic $(\Omega, \mathcal{F}, P)$ (which should perhaps be left to the set theorists). But even if you could, I agree with fedja that no good can come of it. For example, working formally, you might observe that for each finite $A \subset [0,1]$, $\alpha$ is independent of $\{\xi_a : a \in A\}$ (since almost surely $u \notin A$), whence $\alpha$ is independent of $\{\xi_a : a \in [0,1]\}$, which appears to be absurd.
Best Answer
One of your concerns is (let me quote from your question)
Often I read that there is the possibility of having a family X1,…,Xn of random variables on the same space. I know no example—and would be happy to discover—of a problem truly modelled by this, whereas in most examples that I read there is either a single random variable
Here is what I do on the first day of my probability class.
The statistical experiment I describe is: Go to the road outside the college building and consider the first car that goes left to right after your arrival. As we do not know/cannot predict which car in the city might be there it is a statistical experiment. The sample space is the set of all cars in your city (or in your country).
Questions:
How many people are in that car?
What is the amount of petrol in the fuel tank at that time?
How many kilometers the car has travelled that day before you noticed?
What is the wavelength of the color of the car? (admittedly artificial)
All these are random variables on the same sample space.
Answer to question 1 might be useful to a person who sells eatables on the roadside? (more passengers means more business)
Answer to question 2 might help decide if it would be profitable to open a petrol-selling shop here.
I ask students to come up with examples of such statistical experiments instead of coin-tossing and dice-throwing ones.
I got this from a bright student:
Go to the library. Observe the first book that is borrowed by a user that day. Sample space is all books of the library.
Random variables are: Number of pages of that book, Price of that book, How many times it has been borrowed earlier.